9918 lines
247 KiB
Markdown
9918 lines
247 KiB
Markdown
Page 194
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**Exercise Set 4.1**
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In 1-4 justify your answers by using the definitions of even, odd, prime, and
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composite numbers.
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1. Assume that $k$ is a particular integer.
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a. Is $-17$ an odd integer?
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$$ -17 = 2(-9) + 1 $$
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Let $k = -9$, so our expression becomes by substitution:
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$$ -17 = 2k + 1 $$
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Since $-17$ can be represented by the form $2k + 1$ where $k = -9$ and $k$ is an
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integer, by the definition of an odd number, $-17$ is an odd integer.
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b. Is $0$ neither even nor odd?
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No, $0$ can be represented as $0 = 2(0)$, and let $k = 0$, so $0 = 2k$, where
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$k$ is an integer, and by definition of an even number, $0$ is even.
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c. Is $2k - 1$ odd?
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Yes, $2k - 1 = 2(k - 1) + 1$ where $k - 1$ is an integer by the difference of
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integers. Let $m = k - 1$, so our expression becomes $2k - 1 = 2m + 1$, and
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since $m$ is an integer, we can conclude that $2k - 1$ is an odd integer by
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definition of odd integers.
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2. Assume that $c$ is a particular integer.
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a. Is $-6c$ an even integer?
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Yes $-6c = 2(-3c)$, where $-3c$ is an integer by the product of integers, and
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since $-6c$ can be expressed as $2 \cdot \text{ some integer}$, it is even by
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the definition of even integers.
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b. Is $8c + 5$ an odd integer?
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Yes, $8c + 5 = 8c + 4 + 1 = 2(4c + 2) + 1$ where $4c + 2$ is an integer by the
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sum of products of integers. Since $8c + 5$ can be expressed as
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$2(\text{some integer}) + 1$, we can conclude that $8c + 5$ is an odd integer by
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the definition of odd integers.
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c. Is $(c^2 + 1) - (c^2 - 1) - 2$ an even integer?
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Yes, if evaluate the statement by laws of algebra, we get:
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$$ (c^2 + 1) - (c^2 - 1) - 2 = c^2 + 1 - c^2 + 1 - 2 = 1 + 1 - 2 = 0 $$
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And as established in 1b, $0$ is an even integer, so $(c^2 + 1) - (c^2 - 1) - 2$
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can be expressed in the form of $2 \cdot \text{ some integer}$, so by the
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definition of integers, $(c^2 + 1) - (c^2 - 1) - 2$ is an even integer.
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3. Assume that $m$ and $n$ are particular integers?
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a. Is $6m + 8n$ even?
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Yes, $6m + 8n = 2(3m + 4n)$. $3m + 4n$ is an integer by the sum of products of
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integers. Since $6m + 8n$ can be expressed as $2 \cdot \text{ some integer}$, by
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the definition of even integers, $6m + 8n$ is even.
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b. Is $10mn + 7$ odd?
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$10mn + 7 = 10mn + 6 + 1 = 2(5mn + 3) + 1$. $5mn + 3$ is an integer by the
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product and sum of integers. Since $10mn + 7$ can be expressed as
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$2(\text{some integer}) + 1$, $10mn + 7$ is an odd integer.
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c. If $m > n > 0$, is $m^2 - n^2$ composite?
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Not necessarily. Consider $m = 3$ and $n = 2$, then $m^2 - n^2 = 9 - 4 = 5$,
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which is a prime number.
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4. Assume that $r$ and $s$ are particular integers.
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a. Is $4rs$ even?
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Yes, $4rs = 2(2rs)$, where $2rs$ is an integer by the product of integers. Since
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$4rs = 2(\text{ some integer})$, by the definition of an even integer, $4rs$ is
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an even integer.
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b. Is $6r + 4s^2 + 3$ odd?
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$6r + 4s^2 + 3 = 6r + 4s^2 + 2 + 1 = 2(3r + 2s^2 + 1) + 1$. $3r + 2s^2 + 1$ is
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an integer by product and sum of integers. Let $k = 3r + 2s^2 + 1$, and so
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$6r + 4s^2 + 3 = 2k + 1$. By definition of an odd integer, $6r + 4s^2 + 3$ is an
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odd integer.
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c. If $r$ and $s$ are both positive, is $r^2 + 2rs + s^2$ composite?
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Since $r^2 + 2rs + s^2 = (r + s)^2 = (r + s)(r + s)$ and $r + s \geq 2$. And
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since $r + s > 1$, the product of $(r + s)(r + s)$ is composite.
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Prove the statements in 5-11.
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5. There are integers $m$ and $n$ such that $m > 1$ and $n > 1$ and
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$\dfrac{1}{m} + \dfrac{1}{n}$ is an integer.
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For example, let $m = 2$ and $n = 2$, then $\dfrac{1}{2} + \dfrac{1}{2} = 1$,
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and $1$ is an integer.
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6. There are distinct integers $m$ and $n$ such that
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$\dfrac{1}{m} + \dfrac{1}{n}$ is an integer.
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For example, let $m = -2$, and $n = 2$, then $\dfrac{1}{-2} + \dfrac{1}{2} = 0$,
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and $0$ is an integer.
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7. There are real numbers $a$ and $b$ such that
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$$ \sqrt{a + b} = \sqrt{a} + \sqrt{b} $$
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For example, let $a = 0$ and $b = 9$, then
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$\sqrt{0 + 9} = \sqrt{9} = 3 = 0 + 3 = \sqrt{0} + \sqrt{9}$.
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8. There is an integer $n > 5$ such that $2^n - 1$ is prime.
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For example, let $n = 7$, then $2^7 - 1 = 127$, and $127$ is prime.
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9. There is a real number $x$ such that $x > 1$ and $2^x > x^{10}$.
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For example, let $x = \dfrac{1}{2}$, then $2^{\frac{1}{2}} \approx 1.414213562 >
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0.0009765625 = \left(\frac{1}{2}\right)^{10}$
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**Definition:** An integer $n$ is called a **perfect square** if, and only if,
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$n = k^2$ for some integer $k$.
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10. There is a perfect square that can be written as a sum of two other perfect
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squares.
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Let $n = 4$ and $m = 3$, and let $l = k^2$ be the sum of their squares:
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$$ l = k^2 = n^2 + m^2 = (4)^2 + (3)^2 = 16 + 9 = 25 $$
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$$ l = k^2 = 25 $$
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So $l = 25$ can be written as $n^2 + m^2$ where $n = 4$ and $m = 3$, and since
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both $n$ and $m$ are integers, we can say that $l$ is a perfect square by
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definition of a perfect square and $l$ can be written as the sum of two other
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perfect squares.
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11. There is an integer $n$ such that $2n^2 - 5n + 2$ is prime.
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For example let $n = 3$, then
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$2n^2 - 5n + 2 = 2(3)^2 - 5(3) + 2 = 2(9) - 15 + 2 = 18 - 15 + 2 = 3 + 2 = 5$,
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and $5$ is prime.
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In 12-13, (a) write a negation for the given statement, and (b) use a
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counterexample to disprove the given statement. Explain how the counterexample
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actually shows that the given statement is false.
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12. For all real numbers $a$ and $b$, if $a < b$ the $a^2 < b^2$.
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(a)
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Original:
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$$ \forall a \in \mathbb{R} \forall b \in \mathbb{R}((a < b) \to (a^2 < b^2)) $$
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Negation:
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$$ \exists a \in \mathbb{R} \exists b \in \mathbb{R}((a < b) \wedge (a^2 \geq b^2)) $$
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There exist real numbers $a$ and $b$ such that $a < b$ and $a^2 \geq b^2$.
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(b)
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Counterexample:
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Let $a = -2$ and let $b = -1$. The hypothesis $a < b$ holds as $-2 < -1$ is
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true, but the conclusion of the original statement $a^2 < b^2$ is false as
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$(-2)^2 = 4 \cancel{<} 1 = (-1)^2$.
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Since the original statement claims that the implication holds true for all real
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numbers $a$ and $b$, a single counterexample is sufficient to show that the
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statement is false.
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13. For every integer $n$, if $n$ is odd then $\dfrac{n - 1}{2}$ is odd.
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(a)
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Original:
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Let $P(n) = n \text{ is odd}$
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Let $Q(m) = m \text{ is odd}$
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$$ \forall n \in \mathbb{Z}(P(n) \to Q\left(\frac{n - 1}{2}\right)) $$
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Negation:
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$$ \exists n \in \mathbb{Z}(P(n) \wedge \neg Q\left(\frac{n - 1}{2}\right)) $$
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There exists some integer $n$ such that $n$ is odd and $\dfrac{n - 1}{2}$ is not
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odd.
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(b)
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Counterexample:
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Let $n = 1$. $n$ is odd as $1$ can be expressed as $n = 1 = 2(k) + 1$, where
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$k = 0$. This means that $1$ is odd by the definition of an odd integer, and the
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hypothesis of the original statement is true. The conclusion of the original
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statement, however, is false, as
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$\dfrac{n - 1}{2} = \dfrac{1 - 1}{2} = \dfrac{0}{2} = 0$, and $0$ is not odd.
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Since the original statement claims that the implication holds true for all
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integers $n$, a single counterexample is sufficient to show that the statement
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is false.
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Disprove each of the statements in 14-16 by giving a counterexample. In each
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case explain how the counterexample actually disproves the statement.
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14. For all integers $m$ and $n$, if $2m + n$ is odd then $m$ and $n$ are both
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odd.
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Let $m = 2$ and let $n = 1$, the hypothesis $2m + n$ is odd is true as
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$2(2) + 1 = 5$, and $5$ is odd, but the conclusion that both $m$ and $n$ are odd
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is false, as $m$ is even.
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15. For every integer $p$, if $p$ is prime then $p^2 - 1$ is even.
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Let $p = 2$. The hypothesis holds true as $2$ is prime, but the conclusion
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"$p^2 - 1$ is even" is false for this $p$ as $(2)^2 - 1 = 4 - 1 = 3$, and $3$ is
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not even.
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16. For every integer $n$, if $n$ is even then $n^2 + 1$ is prime.
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Let $n = 0$, the hypothesis "$n$ is even" holds true for this $n$ as $0$ is
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even. The conclusion "$n^2 + 1$ is prime" fails for this $n$ as
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$0^2 + 1 = 0 + 1 = 1$, and $1$ is not prime.
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In 17-20, determine whether the property is true for all integers, true for no
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integers, or true for some integers and false for other integers. Justify your
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answers.
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17. $(a + b)^2 = a^2 + b^2$
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This property is true for some integers and not others.
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For example where it is true, consider $a = 0$ and $b = 1$, then
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$(0 + 1)^2 = 1^2 = 1 = 0 + 1 = (0)^2 + (1)^2$ holds true for at least two
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integers.
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For example where it is false, consider $a = 1$ and $b = 1$, then
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$(1 + 1)^2 = 2^2 = 4 \neq 2 = 1 + 2 = (1)^2 + (1)^2$. Since this provides a
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counterexample, this property cannot hold true for all integers.
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Therefore, this property holds true for some integers and not others.
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18. $\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{a + c}{b + d}$
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This is true for $a = c = 0$ and $b = d = 1$as:
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$$ \frac{0}{1} + \frac{0}{1} = 0 + 0 = 0 = \frac{0}{2} = \frac{0 + 0}{1 + 1} $$
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This is false for $a = b = c = d = 1$, as:
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$$ \frac{1}{1} + \frac{1}{1} = 1 + 1 = 2 \neq 1 = \frac{2}{2} = \frac{1 + 1}{1 + 1} $$
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Therefore, this property holds true for some integers and not others.
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19. $-a^n = (-a)^n$
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This is true for $a = -1$ and $n = 1$.
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$$ -(-1)^1 = (-(-1))^1 $$
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$$ -(-1) = (-(-1)) $$
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$$ 1 = 1 $$
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This is false for $a = -1$ and $n = 2$.
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$$ -(-1)^2 = (-(-1))^2 $$
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$$ -(1) = (1)^2 $$
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$$ -1 \neq 1 $$
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Therefore, this property holds true for some integers and not others.
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20. The average of any two odd integers is odd.
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Let $m$ and $n$ be odd integers. Let $m = 2k + 1$ and $n = 2p + 1$ where $k$ and
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$p$ are any integers.
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We are asserting that $\dfrac{m + n}{2}$ is odd. By substitution, we can express
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this as:
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$$ \frac{m + n}{2} = \frac{(2k + 1) + (2p + 1)}{2} = \frac{2k + 2p + 2}{2} = \frac{2(k + p + 1)}{2} = k + p + 1 $$
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In order to prove that $k + p + 1$ is odd, we need to be able to express it in
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the form of $2(\text{some integer}) + 1$ by the definition of an odd integer.
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An example where this is true is if $k = 2$ and $p = 4$, then
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$k + p + 1 = 2 + 4 + 1 = 7$, and $7$ is an odd integer.
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A counterexample where this is false is if $k = 3$ and $p = 4$, then
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$k + p + 1 = 3 + 4 + 1 = 8$, and $8$ is not an odd integer.
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Therefore, this property holds true for some integers and not others.
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Prove the statement in 21 and 22 by the method of exhaustion.
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21. Every positive even integer less than 26 can be expressed as a sum of three
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or fewer perfect squares. (For instance, $10 = 1^2 + 3^2$ and $16 = 4^2$.)
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Let's first establish all positive even integers less than 26:
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$$ \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} $$
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$$ 2 = 1^2 + 1^2 $$
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$$ 4 = 2^2 $$
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$$ 6 = 2^2 + 1^2 + 1^2 $$
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$$ 8 = 2^2 + 2^2 $$
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$$ 10 = 3^2 + 1^2 $$
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$$ 12 = 2^2 + 2^2 + 2^2 $$
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$$ 14 = 3^2 + 2^2 + 1^2 $$
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$$ 16 = 4^2 $$
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$$ 18 = 4^2 + 1^2 + 1^2 $$
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$$ 20 = 4^2 + 2^2 $$
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$$ 22 = 3^2 + 3^2 + 2^2 $$
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$$ 24 = 4^2 + 2^2 + 2^2 $$
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22. For each integer $n$ with $1 \leq n \leq 10$, $n^2 - n + 11$ is a prime
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number.
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Let's establish all possible values for $n$:
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$$ \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} $$
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$n = 1$:
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$$ (1)^2 - (1) + 11 = 1 - 1 + 11 = 11 \text{ is prime} $$
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$n = 2$:
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$$ (2)^2 - (2) + 11 = 4 - 2 + 11 = 13 \text{ is prime} $$
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$n = 3$:
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$$ (3)^2 - (3) + 11 = 9 - 3 + 11 = 17 \text{ is prime} $$
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$n = 4$:
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$$ (4)^2 - (4) + 11 = 16 - 4 + 11 = 23 \text{ is prime} $$
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$n = 5$:
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$$ (5)^2 - (5) + 11 = 25 - 5 + 11 = 31 \text{ is prime} $$
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$n = 6$:
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$$ (6)^2 - (6) + 11 = 36 - 6 + 11 = 41 \text{ is prime} $$
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$n = 7$:
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$$ (7)^2 - (7) + 11 = 49 - 7 + 11 = 53 \text{ is prime} $$
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$n = 8$:
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$$ (8)^2 - (8) + 11 = 64 - 8 + 11 = 67 \text{ is prime} $$
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$n = 9$:
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$$ (9)^2 - (9) + 11 = 81 - 9 + 11 = 83 \text{ is prime} $$
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$n = 10$:
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$$ (10)^2 - (10) + 11 = 100 - 10 + 11 = 101 \text{ is prime} $$
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Each of the statements in 23-26 is true. For each, (a) rewrite the statement
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with the quantification implicit as If _____, then _____, and (b) write the
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first sentence of a proof (the "starting point") and the last sentence of a
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proof (the "conclusion to be shown"). (Note that you do not need to understand
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the statements in order to be able to do these exercises.)
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23. For every integer $m$, if $m > 1$ then $0 < \dfrac{1}{m} < 1$.
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(a) If an integer is greater than 1, then its reciprocal is between 0 and 1.
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(b)
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Starting Point: Suppose $m$ is any integer such that $m > 1$.
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To Show: $0 < \dfrac{1}{m} < 1$
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24. For every real number $x$, if $x > 1$ then $x^2 > x$.
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(a) If a real number is greater than 1, then it's square is greater than itself.
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(b)
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Starting Point: Suppose $x$ is any real number such that $x > 1$.
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To Show: $x^2 > x$.
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25. For all integers $m$ and $n$, if $mn = 1$ then $m = n = 1$ or $m = n = -1$.
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(a) If the product of any two integers is equal to 1, then both integers either
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equal 1 or -1.
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(b)
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Starting Point: Suppose $m$ and $n$ are any integers such that $mn = 1$.
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To Show: $m = n = 1$ or $m = n = -1$.
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26. For every real number $x$, if $0 < x < 1$ then $x^2 < x$.
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(a) If a real number is between 0 and 1, then its square is less than itself.
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(b)
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Starting Point: Suppose $x$ is any real number such that $0 < x < 1$.
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To Show: $x^2 < x$.
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27. Fill in the blanks in the following proof.
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**Theorem:** For every odd integer $n$, $n^2$ is odd.
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**Proof:** Suppose $n$ is any ___ (a) ___. By definition of odd, $n = 2k + 1$
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for some integer $k$. Then
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$$ n^2 = \left(___(b)____\right)^2 \quad \text{ by substitution} $$
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$$ \quad = 4k^2 + 4k + 1 \quad \text{ by multiplying out} $$
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$$ \quad = 2(2k^2 + 2k) + 1 \quad \text{ by factoring out a 2} $$
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Now $2k^2 + 2k$ is an integer because it is a sum of products of integers.
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Therefore $n^2$ equals $2 \cdot (\text{an integer}) + 1$, and so ___ (c) ___ is
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odd by definition of odd.
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Because we have not assumed anything about $n$ except that it is an odd integer,
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it follows from the principle of ___ (d) ___ that for _every_ odd integer $n$,
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$n^2$ is odd.
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a. odd integer.
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b. $2k + 1$
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c. $n^2$
|
|
|
|
d. universal generalization
|
|
|
|
In each of 28-31:
|
|
|
|
a. Rewrite the theorem in three different ways:
|
|
|
|
as $\forall$ _____, if _____ then _____, as $\forall$ _____, _____ (without
|
|
using the words _if_ or _then_),
|
|
|
|
and as If _____, then _____ (without using an explicit universal quantifier).
|
|
|
|
b. Fill in the blanks in the proof of the theorem.
|
|
|
|
28.
|
|
|
|
**Theorem:** the sum of any two odd integers is even.
|
|
|
|
**Proof:** Suppose $m$ and $n$ are any _[particular but arbitrarily chosen]_ odd
|
|
integers.
|
|
|
|
_[We must show that $m + n$ is even.]_
|
|
|
|
By __ (a) __, $m = 2r + 1$ and $n = 2s + 1$ for some integers $r$ and $s$.
|
|
|
|
Then
|
|
|
|
$$ m + n = (2r + 1) + (2s + 1) \quad \text{k by \_\_ (b) \_\_} $$
|
|
|
|
$$ \quad = 2r + 2s + 2 $$
|
|
|
|
$$ \quad = 2(r + s + 1) \quad \text{ by algebra} $$
|
|
|
|
Let $u = r + s + 1$. Then $u$ is an integer because $r$, $s$, and $1$ are
|
|
integers and because __ \(c\) __.
|
|
|
|
Hence $m + n = 2u$, where $u$ is an integer, and so, by __ (d) __, $m + n$ is
|
|
even _[as was to be shown]._
|
|
|
|
a.
|
|
|
|
**Theorem:** the sum of any two odd integers is even.
|
|
|
|
$\forall$ integers $m$ and $n$, if $m$ and $n$ are odd, then $m + n$ is even.
|
|
|
|
$\forall$ odd integers $m$ and $n$, $m + n$ is even.
|
|
|
|
If any two integers are odd, then their sum is even.
|
|
|
|
b.
|
|
|
|
(a) the definition of an odd integer
|
|
|
|
(b) substitution
|
|
|
|
(\c\) any sum of integers is an integer
|
|
|
|
(d)$ the definition of an even integer
|
|
|
|
29.
|
|
|
|
**Theorem:** The negative of any integer is even.
|
|
|
|
**Proof:** Suppose $n$ is any _[particular but arbitrarily chosen]_ even
|
|
integer.
|
|
|
|
_[We must show that $-n$ is even.]_
|
|
|
|
By __ (a) __, $n = 2k$ for some integer $k$.
|
|
|
|
Then
|
|
|
|
$$ -n = -(2k) \quad \text{ by \_\_ (b) \_\_} $$
|
|
|
|
$$ \quad = 2(-k) \quad \text{ by algebra} $$
|
|
|
|
Let $r = -k$. Then $r$ is an integer because $(-1)$ and $k$ are integers and __
|
|
\(c\) __.
|
|
|
|
Hence $-n = 2r$, where $r$ is an integer, and so $-n$ is even by __ (d) __ _[as
|
|
was to be shown]._
|
|
|
|
a.
|
|
|
|
**Theorem:** The negative of any integer is even.
|
|
|
|
$\forall$ integers $n$, if $n$ is negative, then $n$ is even.
|
|
|
|
$\forall$ negative integers $n$, $n$ is even.
|
|
|
|
If an integer is negative, then it is even.
|
|
|
|
b.
|
|
|
|
(a) the definition of an even integer
|
|
|
|
(b) substitution
|
|
|
|
\(c\) the product of any two integers is an integer
|
|
|
|
(d) the definition of an even integer
|
|
|
|
30.
|
|
|
|
**Theorem 4.1.2:** The sum of any even integer and any odd integer is odd.
|
|
|
|
**Proof:** Suppose $m$ 8s any even integer and $n$ is __ (a) __. By definition
|
|
of even, $m = 2$ for some __ (b) __, and by definition of odd, $n = 2s + 1$ for
|
|
some integer $s$. By substitution and algebra,
|
|
|
|
$$ m + n = \text{\_\_ (c) \_\_} = 2(r + s) + 1 $$
|
|
|
|
Since $r$ and $s$ are both integers, so is their sum $r + s$. Hence $m + n$ has
|
|
the form twice some integer plus one, and so __ (d) __ by definition of odd.
|
|
|
|
a.
|
|
|
|
**Theorem 4.1.2:** The sum of any even integer and any odd integer is odd.
|
|
|
|
$\forall$ integers $m$ and $n$, if $m$ is an even integer and $n$ is an odd
|
|
integer, then $m + n$ is odd.
|
|
|
|
$\forall$ even integers $m$ and odd integers $n$, $m + n$ is odd.
|
|
|
|
If $m$ is an even integer and $n$ is any odd integer, then $m + n$ is odd.
|
|
|
|
b.
|
|
|
|
(a) any odd integer
|
|
|
|
(b) integer $r$
|
|
|
|
\(c\) $2r + (2s + 1)$
|
|
|
|
(d) $m + n$ is odd
|
|
|
|
31.
|
|
|
|
**Theorem:** Whenever $n$ is an odd integer, $5n^2 + 7$ is even.
|
|
|
|
**Proof:** Suppose $n$ is any _[particular but arbitrarily chosen]_ odd integer.
|
|
|
|
_[We must show that $5n^2 + 7$ is even.]_
|
|
|
|
By definition of odd, $n$ = __ (a) __ for some integer $k$.
|
|
|
|
Then
|
|
|
|
$$ 5n^2 + 7 = \text{\_\_ (b) \_\_} \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 5(4k^2 + 4k + 1) + 7 $$
|
|
|
|
$$ \quad = 20k^2 + 20k + 12 $$
|
|
|
|
$$ \quad = 2(10k^2 + 10k + 6) \quad \text{ by algebra} $$
|
|
|
|
Let $t =$ __ \(c\) __. Then $t$ is an integer because products and sums of
|
|
integers are integers.
|
|
|
|
Hence $5n^2 + 7 = 2t$, where $t$ is an integer, and thus __ (d) __ by definition
|
|
of even _[as was to be shown]._
|
|
|
|
a.
|
|
|
|
**Theorem:** Whenever $n$ is an odd integer, $5n^2 + 7$ is even.
|
|
|
|
$\forall$ integers $n$, if $n$ is an odd integer, then $5n^2 + 7$ is even.
|
|
|
|
$\forall$ odd integers $n$, $5n^2 + 7$ is even.
|
|
|
|
If $n$ is an odd integer, then $5n^2 + 7$ is even.
|
|
|
|
b.
|
|
|
|
(a) $2k + 1$
|
|
|
|
(b) $5(2k + 1)^2 + 7$
|
|
|
|
\(c\) $10k^2 + 10k + 6$
|
|
|
|
(d) $5n^2 + 7$ is even
|
|
|
|
---
|
|
|
|
**Exercise Set 4.2**
|
|
|
|
Page 204
|
|
|
|
Prove the statements in 1-11. In each case use only the definitions of the terms
|
|
and the Assumptions listed on page 161, not any previously established
|
|
properties of odd and even integers. Follow the directions given in this section
|
|
for writing proofs of universal statements.
|
|
|
|
1. For every integer $n$, if $n$ is odd then $3n + 5$ is even.
|
|
|
|
**Theorem:** Suppose $n$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $n$ is odd, $n = 2k + 1$ for some integer $k$.
|
|
|
|
Then
|
|
|
|
$$ 3n + 5 = 3(2k + 1) + 5 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 6k + 3 + 5 $$
|
|
|
|
$$ \quad = 6k + 8 $$
|
|
|
|
$$ \quad = 2(3k + 4) \quad \text{ by algebra} $$
|
|
|
|
Let $t = 3k + 4$.
|
|
|
|
Then $3n + 5 = 2(3k + 4) = 2t$, where $t$ is an integer because products and
|
|
sums of integers are integers.
|
|
|
|
Therefore $3n + 5$ is even by the definition of even integers.
|
|
|
|
Q.E.D.
|
|
|
|
2. For ever integer $m$, if $m$ is even then $3m + 5$ is odd.
|
|
|
|
**Theorem:** Suppose $m$ is any even integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $m$ is even, $m = 2k$ for some integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ 3m + 5 = 3(2k) + 5 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 6k + 5 $$
|
|
|
|
$$ \quad = 6k + 4 + 1 $$
|
|
|
|
$$ \quad = 2(3k + 2) + 1 \quad \text{ by algebra} $$
|
|
|
|
Let $t = 3k + 2$.
|
|
|
|
Then $3m + 5 = 2(3k + 2) + 1 = 2t + 1$ where $t$ is an integer because the
|
|
product and sum of integers are integers.
|
|
|
|
Therefore $3m + 5$ is odd by the definition of odd integers.
|
|
|
|
Q.E.D.
|
|
|
|
3. For every integer $n$, $2n - 1$ is odd.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
**Proof:**
|
|
|
|
Then:
|
|
|
|
$$ 2n - 1 = 2n - 2 + 1 \quad \text{ by algebra} $$
|
|
|
|
$$ \quad = 2(n - 1) + 1 \quad \text{ by factoring} $$
|
|
|
|
Let $t = n - 1$.
|
|
|
|
Then $2n - 1 = 2(n - 1) + 1 = 2t + 1$ where $t$ is an integer because the
|
|
difference of integers is an integer.
|
|
|
|
Therefore $2n - 1$ is odd by the definition of an odd integer.
|
|
|
|
Q.E.D.
|
|
|
|
4. **Theorem 4.2.2:** The difference of any even integer minus any odd integer
|
|
is odd.
|
|
|
|
**Theorem:** Suppose $m$ is any even integer and $n$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $m$ is even and $n$ is odd, $m = 2k$ and $n = 2s + 1$ where $k$ is some
|
|
integer and $s$ is some integer.
|
|
|
|
Then
|
|
|
|
$$ m - n = 2k - (2s + 1) \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 2k - 2s - 1 $$
|
|
|
|
$$ \quad = 2k - 2s - 2 + 1 $$
|
|
|
|
$$ \quad = 2(k - s - 1) + 1 $$
|
|
|
|
Let $t = k - s - 1$.
|
|
|
|
Then $m - n = 2(k - s - 1) + 1 = 2t + 1$ where $t$ is an integer because the
|
|
difference of integers is an integer.
|
|
|
|
Therefore $m - n$ is odd by the definition of odd integers.
|
|
|
|
Q.E.D.
|
|
|
|
5. If $a$ and $b$ are any odd integers, then $a^2 + b^2$ is even.
|
|
|
|
**Theorem:** Suppose $a$ is any odd integer and $b$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $a$ is an odd integer and $b$ is an odd integer, $a = 2k + 1$ and
|
|
$b = 2s + 1$ where $k$ is some integer and $s$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ a^2 + b^2 = (2k + 1)^2 + (2s + 1)^2 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = (2k + 1)(2k + 1) + (2s + 1)(2s + 1) \quad \text{ by exponentiation} $$
|
|
|
|
$$ \quad = (4k^2 + 4k + 1) + (4s^2 + 4s + 1) $$
|
|
|
|
$$ \quad = 4k^2 + 4k + 1 + 4s^2 + 4s + 1 $$
|
|
|
|
$$ \quad = 4k^2 + 4k + 4s^2 + 4s + 2 $$
|
|
|
|
$$ \quad = 2(2k^2 + 2k + 2s^2 + 2s + 1) $$
|
|
|
|
Let $t = 2k^2 + 2k + 2s^2 + 2s + 1$.
|
|
|
|
Then $a^2 + b^2 = 2(2k^2 + 2k + 2s^2 + 2s + 1) = 2t$ where $t$ is an integer
|
|
because the product and sum of integers is an integer.
|
|
|
|
Therefore $a^2 + b^2$ is even by the definition of even integers.
|
|
|
|
Q.E.D.
|
|
|
|
6. If $k$ is any odd integer and $m$ is any even integer, then $k^2 + m^2$ is
|
|
odd.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $k$ is any odd integer and $m$ is any even integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $k$ is an odd integer and $m$ is an even integer, $k = 2a + 1$ and
|
|
$m = 2b$ where $a$ is some integer and $b$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ k^2 + m^2 = (2a + 1)^2 + (2b)^2 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = (2a + 1)(2a + 1) + (2b)(2b) \quad \text{ by exponentiation} $$
|
|
|
|
$$ \quad = (4a^2 + 4a + 1) + (4b^2) $$
|
|
|
|
$$ \quad = 4a^2 + 4a + 4b^2 + 1 $$
|
|
|
|
$$ \quad = 2(2a^2 + 2a + 2b^2) + 1 $$
|
|
|
|
Let $t = 2a^2 + 2a + 2b^2$.
|
|
|
|
Then $k^2 + m^2 = 2(2a^2 + 2a + 2b^2) + 1 = 2t + 1$ where $t$ is an integer
|
|
because the product and sum of integers is an integer.
|
|
|
|
Therefore $k^2 + m^2$ is odd by the definition of an odd integer.
|
|
|
|
Q.E.D.
|
|
|
|
7. The difference between the squares of any two consecutive integers is odd.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $n$ is an integer, $n + 1$ is a consecutive integer of $n$.
|
|
|
|
Then:
|
|
|
|
$$ n^2 - (n + 1)^2 = n^2 - (n + 1)(n + 1) \quad \text{ by exponentiation} $$
|
|
|
|
$$ \quad = n^2 - (n^2 + 2n + 1) $$
|
|
|
|
$$ \quad = n^2 - n^2 - 2n - 1 \quad \text{ by distribution} $$
|
|
|
|
$$ \quad = -2n - 1 \quad \text{ by distribution} $$
|
|
|
|
$$ \quad = -2n - 2 + 1 $$
|
|
|
|
$$ \quad = 2(-n - 1) + 1 $$
|
|
|
|
Let $t = -n - 1$.
|
|
|
|
Then $n^2 - (n + 1)^2 = 2(-n - 1) + 1 = 2t + 1$ where $t$ is an integer because
|
|
the product and difference of integers is an integer.
|
|
|
|
Therefore $n^2 - (n + 1)^2$ is odd by the definition of an odd integer.
|
|
|
|
Q.E.D.
|
|
|
|
8. For any integers $m$ and $n$, if $m$ is even and $n$ is odd then $5m + 3n$ is
|
|
odd.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $m$ is any even integer and $n$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $m$ is an even integer and $n$ is an odd integer, $m = 2k$ and
|
|
$n = 2s + 1$ where $k$ is some integer and $s$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ 5m + 3n = 5(2k) + 3(2s + 1) \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 10k + 6s + 3 $$
|
|
|
|
$$ \quad = 10k + 6s + 2 + 1 $$
|
|
|
|
$$ \quad = 2(5k + 3s + 1) + 1 $$
|
|
|
|
Let $t = 5k + 3s + 1$.
|
|
|
|
Then $5m + 3n = 2(5k + 3s + 1) + 1 = 2t + 1$ where $t$ is an integer because the
|
|
products and sums of integers is an integer.
|
|
|
|
Therefore $5m + 3n$ is odd by the definition of an odd integer.
|
|
|
|
Q.E.D.
|
|
|
|
9. If an integer greater than $4$ is a perfect square, then the immediately
|
|
preceding integer is not prime.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any integer where $n > 4$ and $n$ is a perfect square.
|
|
|
|
**Proof:**
|
|
|
|
Since $n$ is a perfect square and $n > 4$, then $n = k^2$ for some integer $k$
|
|
where $k > 2$ or $k < -2$.
|
|
|
|
Then:
|
|
|
|
$$ n - 1 = k^2 - 1 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = (k + 1)(k - 1) \quad \text{ by algebra} $$
|
|
|
|
In order for $n - 1$ to be prime, either $k + 1$ or $k - 1$ must be equal to
|
|
$1$.
|
|
|
|
If $k > 2$, then both $k + 1 > 1$ and $k - 1 > 1$ are true.
|
|
|
|
If $k < -2$, then both $k + 1 < 1$ and $k - 1 < 1$ are true.
|
|
|
|
Therefore neither $k + 1$ nor $k - 1$ can ever be equal to $1$.
|
|
|
|
Therefore $n - 1$ is not prime by the definition of a prime number.
|
|
|
|
Q.E.D.
|
|
|
|
10. If $n$ is any even integer, then $(-1)^n = 1$.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any even integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $n$ is an even integer, then $n = 2k$ where $k$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ (-1)^n = (-1)^{2k} \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = (-1)^{2 \cdot k} $$
|
|
|
|
$$ \quad = ((-1)^2)^k $$
|
|
|
|
$$ \quad = 1^k $$
|
|
|
|
$$ \quad = 1 \quad \text{ by the laws of exponents} $$
|
|
|
|
Therefore $(-1)^n = 1$.
|
|
|
|
Q.E.D.
|
|
|
|
11. If $n$ is any odd integer, then $(-1)^n = -1$.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $n$ is an odd integer, then $n = 2k + 1$ where $k$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ (-1)^n = (-1)^{2k + 1} \quad \text{ by substitution} $$
|
|
|
|
$$ (-1)^n = (-1)^{2 \cdot k} \cdot (-1)^1 $$
|
|
|
|
$$ (-1)^n = ((-1)^2)^k \cdot (-1)^1 $$
|
|
|
|
$$ (-1)^n = 1^k \cdot -1 $$
|
|
|
|
$$ (-1)^n = 1 \cdot -1 $$
|
|
|
|
$$ (-1)^n = -1 \quad \text{ by the laws of exponents} $$
|
|
|
|
Therefore $(-1)^n = -1$.
|
|
|
|
Q.E.D.
|
|
|
|
Prove that the statements in 12-14 are false.
|
|
|
|
12. There exists an integer $m \geq 3$ such that $m^2 - 1$ is prime.
|
|
|
|
Take the negation first:
|
|
|
|
For all integers $m \geq 3$, $m^2 - 1$ is not prime.
|
|
|
|
**Theorem:**
|
|
|
|
There is no integer $m \geq 3$ such that $m^2 - 1$ is prime.
|
|
|
|
**Proof:**
|
|
|
|
By algebra, we know that:
|
|
|
|
$$ m^2 - 1 = (m + 1)(m - 1) $$
|
|
|
|
We also know that for $m^2 - 1$ to be prime, either $m + 1$ or $m - 1$ must be
|
|
equal to $1$.
|
|
|
|
Since $m \geq 3$, we know that both $m + 1 \geq 4$ and $m - 1 \geq 2$ are both
|
|
true. Thus both factors are greater than 1.
|
|
|
|
Therefore $m^2 - 1$ is a product of two integers greater than $1$, so it is not
|
|
prime.
|
|
|
|
Therefore $m^2 - 1$ is not prime by the definition of prime numbers.
|
|
|
|
Q.E.D.
|
|
|
|
13. There exists an integer $n$ such that $6n^2 + 27$ is prime.
|
|
|
|
Take the negation first:
|
|
|
|
For all integers $n$, $6n^2 + 27$ is not prime.
|
|
|
|
**Theorem:**
|
|
|
|
There is no integer $n$ such that $6n^2 + 27$ is prime.
|
|
|
|
**Proof:**
|
|
|
|
By algebra we know that:
|
|
|
|
$$ 6n^2 + 27 = 3(2n^2 + 9) $$
|
|
|
|
Since $n^2$ is always positive or $0$, by the laws of exponentiation and by
|
|
algebra, we can conclude that $2n^2 + 9 \geq 9$ is true.
|
|
|
|
Since $3 > 1$ and $2n^2 + 9 > 1$, we then know that $6n^2 + 27$ is a product of
|
|
two integers greater than $1$, so it is not prime.
|
|
|
|
$6n^2 + 27$ is not prime by the definition of prime numbers.
|
|
|
|
Q.E.D.
|
|
|
|
14. There exists an integer $k \geq 4$ such that $2k^2 - 5k + 2$ is prime.
|
|
|
|
Take the negation first:
|
|
|
|
For all integers $k \geq 4$, $2k^2 - 5k + 2$ is not prime.
|
|
|
|
**Theorem:**
|
|
|
|
There is no integer $k \geq 4$ such that $2k^2 - 5k + 2$ is prime.
|
|
|
|
**Proof:**
|
|
|
|
By algebra we know:
|
|
|
|
$$ 2k^2 - 5k + 2 = (k - 2)(2k - 1) $$
|
|
|
|
Since we know that $k \geq 4$, we know that $k - 2 \geq 2$ and $2k - 1 \geq 7$.
|
|
|
|
Since $k - 2 > 1$ and $2k - 1 > 1$, we then know that $2k^2 - 5k + 2$ is a
|
|
product of two integers greater than $1$, so it is not prime.
|
|
|
|
Therefore $2k^2 - 5k + 2$ is not prime by definition of prime numbers.
|
|
|
|
Q.E.D.
|
|
|
|
Find the mistakes in the "proofs" shown in 15-19.
|
|
|
|
15.
|
|
|
|
**Theorem:** For every integer $k$, if $k > 0$ then $k^2 + 2k + 1$ is composite.
|
|
|
|
**"Proof:** For $k = 2$, $k > 0$ and $k^2 + 2k + 1 = 2^2 + 2 \cdot 2 + 1 = 9$.
|
|
And since $9 = 3 \cdot 3$, then $9$ is composite. Hence the theorem is true."
|
|
|
|
Answer: This proof just shows that the theorem is true for a single case,
|
|
$k = 2$, in order to prove a universal claim as the theorem presents, the proof
|
|
must prove the conclusion true for every integer $k$ where $k > 0$.
|
|
|
|
16.
|
|
|
|
**Theorem:** The difference between any odd integer and any even integer is odd.
|
|
|
|
**"Proof:** Suppose $n$ is any odd integer, and $m$ is any even integer. By
|
|
definition of odd, $n = 2k + 1$ where $k$ is an integer, and by definition of
|
|
even, $m = 2k$ where $k$ is an integer. Then
|
|
|
|
$$ n - m = (2k + 1) - 2k = 1 $$
|
|
|
|
Answer: This proof makes the mistake of using $k$ to represent two different
|
|
quantities. By setting $n = 2k + 1$ and $m = 2k$, the proof implies that
|
|
$n = m + 1$, and thus deduces the conclusion for only this situation. This proof
|
|
falsely then "proves" that the difference between _any_ even and odd integer
|
|
will always equal $1$, but taking most examples of even and odd integers as
|
|
cases for this would show that this is false. In essence, this proof makes the
|
|
mistake of assigning the same variable name to represent two different integers,
|
|
and then by algebra comes to a false conclusion.
|
|
|
|
17.
|
|
|
|
**Theorem:** For every integer $k$, if $k > 0$, then $k^2 + 2k + 1$ is
|
|
composite.
|
|
|
|
**"Proof:** Suppose $k$ is any integer such that $k > 0$. If $k^2 + 2k + 1$ is
|
|
composite, then $k^2 + 2k + 1 = rs$ for some integers $r$ and $s$ such that
|
|
|
|
$$ 1 < r < k^2 + 2k + 1 $$
|
|
|
|
and
|
|
|
|
$$ 1 < s < k^2 + 2k + 1 $$
|
|
|
|
Since
|
|
|
|
$$ k^2 + 2k + 1 = rs $$
|
|
|
|
and both $r$ and $s$ are strictly between $1$ and $k^2 + 2k + 1$, then
|
|
$k^2 + 2k + 1$ is not prime. Hence $k^2 + 2k + 1$ is composite as was to be
|
|
shown."
|
|
|
|
Answer: This proof makes the mistake of assuming what is to be proved. Instead
|
|
of proving that $k^2 + 2k + 1$ is composite, it assumes the definition of
|
|
composite numbers applies to the expression and then extrapolates logic about
|
|
$r$ and $s$ that cannot be known because it has not yet been proven that
|
|
$k^2 + 2k +1$ is composite. This starts at the line starting with "Since", which
|
|
cannot be asserted as that is an assertion of the conclusion, not the
|
|
hypothesis.
|
|
|
|
Teacher's answer: This incorrect proof assumes what is to be proved. The word
|
|
_since_ in the third sentence is completely unjustified. The second sentence
|
|
tells only what happens _if_ $k^2 + 2k + 1$ is composite. But at that point in
|
|
the proof, it has not been established that $k^2 + 2k + 1$ _is_ composite. In
|
|
fact, that is exactly what is to be proved.
|
|
|
|
18.
|
|
|
|
**Theorem:** The product of any even integer and any odd integer is even.
|
|
|
|
**"Proof:** Suppose $m$ is any even integer and $n$ is any odd integer. If
|
|
$m \cdot n$ is even, then by definition of even there exists an integer $r$ such
|
|
that $m \cdot n = 2r$. Also since $m$ is even, there exists an integer $p$ such
|
|
that $m = 2p$, and since $n$ is odd there exists an integer $q$ such that
|
|
$n = 2q + 1$. Thus
|
|
|
|
$$ mn = (2p)(2q + 1) = 2r $$
|
|
|
|
where $r$ is an integer. By definition of even, then, $m \cdot n$ is even, as
|
|
was to be shown."
|
|
|
|
Answer: This incorrect proof exhibits confusion between what is known and what
|
|
is still to be shown. The writer correctly uses the definitions of even and odd
|
|
integers to express $m$ and $n$ as $2p$ and $2q + 1$, but assumes the conclusion
|
|
that $mn$ must be an expression of $2r$, which is exactly what is to be shown,
|
|
but has not yet been proven. In essence, they have jumped to the conclusion.
|
|
|
|
19.
|
|
|
|
**Theorem:** The sum of any two even integers equals $4k$ for some integer $k$.
|
|
|
|
**"Proof:** Suppose $m$ and $n$ are any two even integers. By definition of
|
|
even, $m = 2k$ for some integer $k$ and $n = 2k$ for some integer $k$. By
|
|
substitution,
|
|
|
|
$$ m + n = 2k + 2k = 4k $$
|
|
|
|
That is what was to be shown."
|
|
|
|
Answer: This incorrect proof suffers from multiple problems. One is that it uses
|
|
the same variable name $k$ to represent two potentially different integers when
|
|
expressing both $m$ and $n$ as even integers. The writer then incorrectly sums
|
|
them to $4k$ and concludes they have proven the conclusion, but the form of $4k$
|
|
does not explicitly show that $m + n$ is even by the definition of even
|
|
integers.
|
|
|
|
In 20-38 determine whether the statement is true or false. Justify your answer
|
|
with a proof or a counterexample, as appropriate. In each case use only the
|
|
definitions of the terms and the Assumptions listed on page 161, not any
|
|
previously established properties.
|
|
|
|
20. The product of any two odd integers is odd.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any odd integer and $m$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $n$ and $m$ are odd integers, $n = 2k + 1$ and $m = 2s + 1$ where $k$ is
|
|
some integer and $s$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ n \cdot m = (2k + 1)(2s + 1) \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 4ks + 2s + 2k + 1 $$
|
|
|
|
$$ \quad = 2(2ks + s + k) + 1 \quad \text{ by algebra} $$
|
|
|
|
Let $t = 2ks + s + k$.
|
|
|
|
Then $n \cdot m = 2(2ks + s + k) + 1 = 2t + 1$ where $t$ is an integer because
|
|
the products and sums of integers is an integer.
|
|
|
|
Therefore $n \cdot m$ is odd by the definition of odd integers.
|
|
|
|
Q.E.D.
|
|
|
|
21. The negative of any odd integer is odd.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $n$ is odd, $n = 2k + 1$ where $k$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ -n = -(2k + 1) \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = -2k - 1 $$
|
|
|
|
$$ \quad = -2k - 2 + 1 $$
|
|
|
|
$$ \quad = 2(-k - 1) + 1 $$
|
|
|
|
Let $t = -k - 1$.
|
|
|
|
Then $-n = 2(-k - 1) + 1 = 2t + 1$ where $t$ is an integer because the products
|
|
and differences of integers is an integer.
|
|
|
|
Therefore $-n$ is odd by definition of an odd integer.
|
|
|
|
Q.E.D.
|
|
|
|
22. For all integers $a$ and $b$, $4a + 5b + 3$ is even.
|
|
|
|
False. Intuition says if $a = b = 0$ then $4a + 5b + 3 = 3$ which is not even.
|
|
Let's prove this more formally.
|
|
|
|
Take the negation:
|
|
|
|
There exists some integer $a$ and some integer $b$ such that $4a + 5b + 3$ is
|
|
not even.
|
|
|
|
**Theorem:** There is some integer $a$ and some integer $b$ such that
|
|
$4a + 5b + 3$ is not even.
|
|
|
|
Let $a = 0$ and let $b = 0$
|
|
|
|
Then:
|
|
|
|
$$ 4a + 5b + 3 = 4(0) + 5(0) + 3 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 0 + 0 + 3 $$
|
|
|
|
$$ \quad = 3 $$
|
|
|
|
Since $3$ is not even, $4a + 5b + 3$ is not even for the given $a$ and $b$.
|
|
|
|
Therefore there exists integers $a$ and $b$ such that $4a + 5b + 3$ is not even
|
|
and the original given statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
23. The product of any even integer and any integer is even.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any even integer and $m$ is any integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $n$ is even, $n = 2k$ for some integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ n \cdot m = (2k)(m) \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 2km $$
|
|
|
|
$$ \quad = 2(km) $$
|
|
|
|
Let $t = km$.
|
|
|
|
Then $n \cdot m = 2(km) = 2t$ where $t$ is an integer because the product of
|
|
integers is an integer.
|
|
|
|
Therefore $n \cdot m$ is even by definition of even integers.
|
|
|
|
Q.E.D.
|
|
|
|
24. If a sum of two integers is even, then one of the summands is even. (In the
|
|
expression $a + b$, $a$ and $b$ are called **summands**.)
|
|
|
|
This is false, quickly consider $1 + 3 = 4$ where both the summands are odd, but
|
|
the sum is even.
|
|
|
|
Take the negation first:
|
|
|
|
There exists two integers whose sum is even but neither integer is even.
|
|
|
|
**Claim:**
|
|
|
|
There is some integer $a$ and there is some integer $b$ such that $a + b$ is
|
|
even and neither $a$ nor $b$ is even.
|
|
|
|
**Proof:**
|
|
|
|
Let $a = 1$ and $b = 3$.
|
|
|
|
Then:
|
|
|
|
$$ a + b = 4 $$
|
|
|
|
$$ \quad = 2(2) $$
|
|
|
|
Then $a + b$ is even by the definition of even integers.
|
|
|
|
Then:
|
|
|
|
$$ a = 1 $$
|
|
|
|
$$ \quad = 2(0) + 1 $$
|
|
|
|
And:
|
|
|
|
$$ b = 3 $$
|
|
|
|
$$ \quad = 2(1) + 1 $$
|
|
|
|
Then both $a$ and $b$ are odd by the definition of odd integers.
|
|
|
|
Therefore $a + b$ is even, but $a$ and $b$ are not even for the given $a$ and
|
|
$b$, therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
25. The difference of any two even integers is even.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $m$ is an even integer and $n$ is an even integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $m$ and $n$ are even integers, $m = 2k$ and $n = 2s$ where $k$ is some
|
|
integer and $s$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ n - m = (2s) - (2k) \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 2s - 2k $$
|
|
|
|
$$ \quad = 2(s - k) \quad \text{ by algebra} $$
|
|
|
|
Let $t = s - k$.
|
|
|
|
Then $n -m = 2(s - k) = 2t$ where $t$ is an integer because the difference of
|
|
integers is an integer.
|
|
|
|
Therefore $n - m$ is even by the definition of even integers.
|
|
|
|
Q.E.D.
|
|
|
|
26. For all integers $a$, $b$, and $c$, if $a$, $b$, and $c$ are consecutive,
|
|
then $a + b + c$ is even.
|
|
|
|
This is false. Take the negation for the claim.
|
|
|
|
**Claim:**
|
|
|
|
There exists some integer $a$, some integer $b$, and some integer $c$ such that
|
|
$a$, $b$, and $c$ are consecutive and $a + b + c$ is not even.
|
|
|
|
**Proof:**
|
|
|
|
Let $a = 2$, $b = 3$, $c = 4$.
|
|
|
|
Then:
|
|
|
|
$$ a + b + c = 2 + 3 + 4 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 9 $$
|
|
|
|
$$ \quad = 8 + 1 $$
|
|
|
|
$$ \quad = 2(4) + 1 $$
|
|
|
|
Therefore for the given $a$, $b$, and $c$, $a + b + c$ is not even, by the
|
|
definition of an odd number.
|
|
|
|
Therefore the given $a$, $b$, and $c$ are consecutive numbers, but their sum is
|
|
not even. The statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
27. The difference of any two odd integers is even.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any odd integer and $m$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
Since $n$ and $m$ are odd integers, $n = 2k + 1$ and $m = 2s + 1$ where $k$ is
|
|
some integer and $s$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ n - m = (2k + 1) - (2s + 1) \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 2k + 1 - 2s - 1 $$
|
|
|
|
$$ \quad = 2k - 2s $$
|
|
|
|
$$ \quad = 2(k - s) $$
|
|
|
|
Let $t = k - s$.
|
|
|
|
Then $n - m = 2(k - s) = 2t$ where $t$ is an integer because the difference of
|
|
integers is an integer.
|
|
|
|
Therefore $n - m$ is even by definition of an even integer.
|
|
|
|
Q.E.D.
|
|
|
|
28. For all integers $n$ and $m$, if $n - m$ is even then $n^3 - m^3$ is even.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any integer and $m$ is any integer and $n - m$ is even.
|
|
|
|
**Proof:**
|
|
|
|
Since we know that $n - m$ is even, $n - m = 2k$ where $k$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ n^3 - m^3 = (n - m)(n^2 + nm + m^2) \quad \text{ by factoring} $$
|
|
|
|
$$ \quad = 2k(n^2 + nm + m^2) \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 2[k(n^2 + nm + m^2)] $$
|
|
|
|
Let $t = k(n^2 + nm + m^2)$.
|
|
|
|
Then $n^3 - m^3 = 2t$ where $t$ is an integer because products and sums of
|
|
integers is an integer.
|
|
|
|
Therefore $n^3 - m^3$ is even by the definition of even integers.
|
|
|
|
Q.E.D.
|
|
|
|
29. For every integer $n$, if $n$ is prime then $(-1)^n = -1$.
|
|
|
|
This is false when $n = 2$. Let's prove our claim.
|
|
|
|
**Claim:**
|
|
|
|
There exists some integer $n$ such that $n$ is prime and $(-1)^n \neq -1$.
|
|
|
|
**Proof:**
|
|
|
|
Let $n = 2$.
|
|
|
|
Then:
|
|
|
|
$$ (-1)^n = (-1)^2 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 1 $$
|
|
|
|
$$ 1 \neq -1 $$
|
|
|
|
Therefore since there is a prime number for $n$ such that $(-1)^n \neq -1$, the
|
|
given statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
30. For every integer $m$, if $m > 2$ then $m^2 - 4$ is composite.
|
|
|
|
This is false. If $m = 3$, then $m^2 - 4 = 9 - 4 = 5$ which is not composite.
|
|
Let's prove our claim.
|
|
|
|
**Claim:**
|
|
|
|
There exists some integer $m$ such that $m > 2$ and $m^2 - 4$ is not composite.
|
|
|
|
**Proof:**
|
|
|
|
Let $m = 3$.
|
|
|
|
Then:
|
|
|
|
$$ m^2 - 4 = (3)^2 - 4 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 9 - 4 $$
|
|
|
|
$$ \quad = 5 $$
|
|
|
|
$$ \quad = (1)(5) $$
|
|
|
|
Since $m^2 - 4$ cannot be written as the product of 2 factors where both factors
|
|
are greater than $1$, $m^2 - 4$ is not composite.
|
|
|
|
Therefore since there is some integer $m$ such that $m > 2$ and $m^2 - 4$ is not
|
|
composite, this statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
31. For every integer $n$, $n^2 - n + 11$ is a prime number.
|
|
|
|
This is false for when $n = 11$, let's formalize our claim.
|
|
|
|
**Claim:**
|
|
|
|
There exists some integer $n$ such that $n^2 - n + 11$ is not a prime number.
|
|
|
|
**Proof:**
|
|
|
|
Let $n = 11$.
|
|
|
|
Then:
|
|
|
|
$$ n^2 - n + 11 = (11)^2 - (11) + 11 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 121 - 11 + 11 $$
|
|
|
|
$$ \quad = 121 $$
|
|
|
|
$$ \quad = (11)(11) $$
|
|
|
|
Therefore $n^2 - n + 11$ is not a prime number since it is divisible by a number
|
|
other than $1$ and itself for this given $n$.
|
|
|
|
Thus there exists some integer $n$ such that $n^2 - n + 11$ is not a prime
|
|
number, and therefore the given statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
32. For every integer $n$, $4(n^2 + n + 1) - 3n^2$ is a perfect square.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
**Proof:**
|
|
|
|
Then:
|
|
|
|
$$ 4(n^2 + n + 1) - 3n^2 = 4n^2 + 4n + 4 - 3n^2 \quad \text{ by distribution} $$
|
|
|
|
$$ \quad = n^2 + 4n + 4 $$
|
|
|
|
$$ \quad = (n + 2)(n + 2) $$
|
|
|
|
$$ \quad = (n + 2)^2 $$
|
|
|
|
Let $t = n + 2$.
|
|
|
|
Then $4(n^2 + n + 1) - 3n^2 = t^2$ where $t$ is an integer because the sum of
|
|
integers is an integer.
|
|
|
|
Therefore $4(n^2 + n + 1) - 3n^2$ is a perfect square by the definition of
|
|
perfect squares.
|
|
|
|
Q.E.D.
|
|
|
|
33. Every positive integer can be expressed as a sum of three or fewer perfect
|
|
squares.
|
|
|
|
This is false.
|
|
|
|
**Claim:**
|
|
|
|
There exists some positive integer $x$ such that $x$ cannot be expressed as the
|
|
sum of three or fewer perfect squares.
|
|
|
|
**Proof:**
|
|
|
|
Let $x = 7$. We check all sums of three nonnegative perfect squares
|
|
$a^2 + b^2 + c^2$, where $a, b, c \in \{0, 1, 2\}$ because $3^2 = 9 > 7$.
|
|
|
|
Possible squares: $0^2 = 0$, $1^2 = 1$, $2^2 = 4$.
|
|
|
|
Now we check all sums
|
|
|
|
1. Using only $0$ and $1$:
|
|
|
|
- $0 + 0 + 1 = 1$, $0 + 1 + 1 = 2$, $1 + 1 + 1 = 3$
|
|
|
|
All of these are too small and do not add up to $7$.
|
|
|
|
2. Using a $4$ ($2^2$) with $0$ and $1$:
|
|
|
|
- $4 + 0 + 0 = 4$, $4 + 0 + 1 = 5$, $4 + 1 + 1 = 6$, $4 + 4 + 0 = 8$
|
|
|
|
All of these do not equal $7$.
|
|
|
|
No combination sums to $7$.
|
|
|
|
Therefore, since all possible combinations from the given set of numbers that
|
|
could potentially sum to $7$ when each individual number is squared have been
|
|
exhausted, it can be concluded that $x = 7$ cannot be expressed as the sum of
|
|
three or fewer perfect squares.
|
|
|
|
Therefore there exists at least one integer $x$ such that $x$ cannot be
|
|
expressed as a sum of three or fewer perfect squares, and this statement is
|
|
false.
|
|
|
|
Q.E.D.
|
|
|
|
34. (Two integers are **consecutive** if, and only if, one is one more than the
|
|
other.) Any product of four consecutive integers is one less than a perfect
|
|
square.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
**Proof:**
|
|
|
|
Then:
|
|
|
|
$$ (n)(n + 1)(n + 2)(n + 3) = (n(n + 3))(n + 1)(n + 2) $$
|
|
|
|
$$ \quad = (n^2 + 3n)(n^2 + 3n + 2)$$
|
|
|
|
$$ \quad = (n^2 + 3n)((n^2 + 3n) + 2)$$
|
|
|
|
Let $x = n^2 + 3n$.
|
|
|
|
Then:
|
|
|
|
$$ \quad = (x)((x) + 2) $$
|
|
|
|
$$ \quad = x^2 + 2x $$
|
|
|
|
$$ \quad = x^2 + 2x + 1 - 1 $$
|
|
|
|
$$ \quad = (x^2 + 2x + 1) - 1 $$
|
|
|
|
$$ \quad = (x + 1)(x + 1) - 1 $$
|
|
|
|
$$ \quad = (x + 1)^2 - 1 $$
|
|
|
|
Then remove the substitution:
|
|
|
|
$$ \quad = ((n^2 + 3n) + 1)^2 - 1 $$
|
|
|
|
$$ \quad = (n^2 + 3n + 1)^2 - 1 $$
|
|
|
|
Since $n^2 + 3n + 1$ is an integer because the products and sum of integers is
|
|
an integer, this means that $(n^2 + 3n + 1)^2$ is a perfect square and
|
|
$(n^2 + 3n + 1)^2 - 1$ is one less than a perfect square.
|
|
|
|
Therefore the product of any four consecutive integers is one less than a
|
|
perfect square.
|
|
|
|
Q.E.D.
|
|
|
|
35. If $m$ and $n$ are any positive integers and $mn$ is a perfect square, then
|
|
$m$ and $n$ are perfect squares.
|
|
|
|
This is false.
|
|
|
|
**Claim:**
|
|
|
|
There is a positive integer $m$ and there is a positive integer $n$ such that
|
|
$mn$ is a perfect square and $m$ and $n$ are not perfect squares.
|
|
|
|
**Proof:**
|
|
|
|
Let $m = 2$ and $n = 8$.
|
|
|
|
Then:
|
|
|
|
$$ mn = (2)(8) \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 16 $$
|
|
|
|
$$ \quad = 4^2 $$
|
|
|
|
Then $mn$ is a perfect square, but $m$ and $n$ are not perfect squares.
|
|
|
|
Therefore there exists some $m$ and there exists some $n$ such that $mn$ is a
|
|
perfect square and $m$ and $n$ are not perfect squares, proving the statement
|
|
false.
|
|
|
|
Q.E.D.
|
|
|
|
36. The difference of the squares of any two consecutive integers is odd.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
**Proof:**
|
|
|
|
Then:
|
|
|
|
$$ (n + 1)^2 - n^2 = (n + 1)(n + 1) - n^2 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = n^2 + 2n + 1 - n^2 $$
|
|
|
|
$$ \quad = 2n + 1 $$
|
|
|
|
Therefore $(n + 1)^2 - n^2$ is odd by the definition of an odd integer.
|
|
|
|
Q.E.D.
|
|
|
|
37. For all nonnegative real numbers $a$ and $b$,
|
|
$\sqrt{ab} = \sqrt{a}\sqrt{b}$. (Note that if $x$ is a nonnegative real
|
|
number, then there is a unique nonnegative real number $y$, denoted
|
|
$\sqrt{x}$, such that $y^2 = x$.)
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $a$ is any nonnegative real number and $b$ is any nonnegative real
|
|
number.
|
|
|
|
**Proof:**
|
|
|
|
Since $a \geq 0$ and $b \geq 0$, we know that $\sqrt{a}$ and $\sqrt{b}$ are
|
|
defined nonnegative real numbers such that:
|
|
|
|
$$ (\sqrt{a})^2 = a $$
|
|
|
|
and
|
|
|
|
$$ (\sqrt{b})^2 = b $$
|
|
|
|
Then:
|
|
|
|
$$ (\sqrt{a}\sqrt{b})^2 = (\sqrt{a})^2(\sqrt{b})^2 = ab \quad \text{ by the product of powers property} $$
|
|
|
|
We then know that $\sqrt{a}\sqrt{b} \geq 0$ because both factors are
|
|
nonnegative.
|
|
|
|
So $\sqrt{a}\sqrt{b}$ is a nonnegative real number whose square is $ab$.
|
|
|
|
Therefore $\sqrt{ab} = \sqrt{a}\sqrt{b}$ by the definition of square root
|
|
(uniqueness of the nonnegative number whose square is $ab$).
|
|
|
|
Q.E.D.
|
|
|
|
38. For all nonnegative real numbers $a$ and $b$,
|
|
|
|
$$ \sqrt{a + b} = \sqrt{a} + \sqrt{b} $$
|
|
|
|
This is false.
|
|
|
|
**Claim:**
|
|
|
|
There is some nonnegative real number $a$ and some nonnegative real number $b$
|
|
such that $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$.
|
|
|
|
**Proof:**
|
|
|
|
Let $a = 9$ and $b = 16$.
|
|
|
|
Then:
|
|
|
|
$$ \sqrt{a + b} = \sqrt{9 + 16} \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = \sqrt{25} $$
|
|
|
|
$$ \quad = 5 $$
|
|
|
|
Then:
|
|
|
|
$$ 5 \stackrel{?}{=} \sqrt{a} + \sqrt{b} $$
|
|
|
|
$$ 5 \stackrel{?}{=} \sqrt{9} + \sqrt{16} \quad \text{ by substitution} $$
|
|
|
|
$$ 5 \stackrel{?}{=} 3 + 4 $$
|
|
|
|
$$ 5 \neq 7 $$
|
|
|
|
Therefore for the given $a$ and $b$, we have shown that
|
|
$\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$, thus proving the statement false.
|
|
|
|
Q.E.D.
|
|
|
|
39. Suppose that integers $m$ and $n$ are perfect squares. Then
|
|
$m + n + 2\sqrt{mn}$ is also a perfect square. Why?
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $m$ is any perfect square and $n$ is any perfect square.
|
|
|
|
**Proof:**
|
|
|
|
Since $m$ and $n$ are perfect squares, $m = k^2$ and $n = s^2$ for some integer
|
|
$k$ and some integer $s$.
|
|
|
|
Then:
|
|
|
|
$$ m + n + 2\sqrt{mn} = (k^2) + (s^2) + 2\sqrt{(k^2)(s^2)} \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = k^2 + s^2 + 2\sqrt{k^2}\sqrt{s^2} \quad \text{ by the product of powers property} $$
|
|
|
|
$$ \quad = k^2 + s^2 + 2ks $$
|
|
|
|
$$ \quad = k^2 + 2ks + s^2 \quad \text{ by the commutative property} $$
|
|
|
|
$$ \quad = (k + s)(k + s) $$
|
|
|
|
$$ \quad = (k + s)^2 $$
|
|
|
|
Therefore $m + n + 2\sqrt{mn}$ is a perfect square by the definition of a
|
|
perfect square.
|
|
|
|
Q.E.D.
|
|
|
|
40. If $p$ is a prime number, must $2^p - 1$ also be prime? Prove or give a
|
|
counterexample.
|
|
|
|
False.
|
|
|
|
**Claim:**
|
|
|
|
There is some prime number $p$ such that $2^p - 1$ is not prime.
|
|
|
|
Let $p = 11$.
|
|
|
|
Then:
|
|
|
|
$$ 2^p - 1 = 2^{11} - 1 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 2048 - 1 $$
|
|
|
|
$$ \quad = 2047 $$
|
|
|
|
$$ \quad = (23)(89) $$
|
|
|
|
Thus there is a case where $p$ is a prime number and $2^p - 1$ is not prime, and
|
|
therefore the given statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
41. If $n$ is a nonnegative integer, must $2^{2n} + 1$ be prime? Prove or give a
|
|
counterexample.
|
|
|
|
False.
|
|
|
|
**Claim:**
|
|
|
|
There exists a nonnegative integer $n$ such that $2^{2n} + 1$ is not prime.
|
|
|
|
**Proof:**
|
|
|
|
Let $n = 5$.
|
|
|
|
Then:
|
|
|
|
$$ 2^{2n} + 1 = 2^{2 \cdot 5} + 1 \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = 2^{10} + 1 $$
|
|
|
|
$$ \quad = 1024 + 1 $$
|
|
|
|
$$ \quad = 1025 $$
|
|
|
|
$$ \quad = (25)(41) $$
|
|
|
|
Thus there exists an nonnegative integer $n$ such that $2^{2n} + 1$ is not
|
|
prime, and therefore the given statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
---
|
|
|
|
**Exercise Set 4.3**
|
|
|
|
Page 210
|
|
|
|
The numbers in 1-7 are all rational. Write each number as a ratio of two
|
|
integers.
|
|
|
|
1. $-\dfrac{35}{6}$
|
|
|
|
$$ -\frac{35}{6} = -\frac{35}{6} $$
|
|
|
|
2. $4.6037$
|
|
|
|
$$ 4.6037 = \frac{46037}{10000} $$
|
|
|
|
3. $\dfrac{4}{5} + \dfrac{2}{9}$
|
|
|
|
$$ \frac{4}{5} + \frac{2}{9} = \frac{9(4) + 5(2)}{45} = \frac{46}{45} $$
|
|
|
|
4. $0.37373737\dots$
|
|
|
|
Let $x = 0.37373737\dots$, then $100x = 37.373737\dots$, so
|
|
$100x - x = 99x = 37$. Therefore:
|
|
|
|
$$ x = 0.37373737\dots = \frac{37}{99} $$
|
|
|
|
5. $0.56565656\dots$
|
|
|
|
Let $x = 0.56565656\dots$, then $100x = 56.565656\dots$, so
|
|
$100x - x = 99x = 56$. Therefore:
|
|
|
|
$$ x = 0.56565656\dots = \frac{56}{99} $$
|
|
|
|
6. $320.5492492492\dots$
|
|
|
|
$$ x = 320.5492492492\dots $$
|
|
|
|
$$ 10000x = 3205492.492492492\dots $$
|
|
|
|
$$ 10x = 3205.492492492\dots $$
|
|
|
|
$$ 10000x - 10x = 9990x = 3202287 $$
|
|
|
|
$$ x = \frac{3202287}{9990} $$
|
|
|
|
7. $52.4672167216721\dots$
|
|
|
|
$$ x = 52.4672167216721\dots $$
|
|
|
|
$$ 100000x = 5246721.672167216721\dots $$
|
|
|
|
$$ 10x = 524.672167216721\dots$$
|
|
|
|
$$ 100000x - 10x = 99990x = 5246197 $$
|
|
|
|
$$ x = 52.4672167216721\dots = \frac{5246197}{99990} $$
|
|
|
|
8. The zero product property, says that if a product of two real numbers is $0$,
|
|
then one of the numbers must be $0$.
|
|
|
|
a. Write this property formally using quantifiers and variables.
|
|
|
|
Let $P(x)$ be "$x = 0$."
|
|
|
|
Let $Q(y)$ be "$y = 0$."
|
|
|
|
Let $R(x, y)$ be "$(x)(y) = 0$."
|
|
|
|
$$ \forall x \in \mathbb{R} \forall y \in \mathbb{R} (R(x, y) \to (P(x) \vee Q(y))) $$
|
|
|
|
b. Write the contrapositive of your answer to part (a).
|
|
|
|
$$ \forall x \in \mathbb{R} \forall y \in \mathbb{R} ((\neg P(x) \wedge \neg Q(y)) \to \neg R(x, y)) $$
|
|
|
|
c. Write an informal version (without quantifier symbols or variables) for your
|
|
part to part (b).
|
|
|
|
If any two real numbers do not equal zero, then their product does not equal
|
|
zero.
|
|
|
|
9. Assume that $a$ and $b$ are both integers and that $a \neq 0$ and $b \neq 0$.
|
|
Explain why $\dfrac{(b - a)}{(ab^2)}$ must be a rational number.
|
|
|
|
A rational number is a ratio of integers with a nonzero denominator. The given
|
|
fraction
|
|
|
|
$$ \frac{(b - q)}{(ab^2)} $$
|
|
|
|
is rational, the numerator is an integer as the difference of integers are
|
|
integers, and the denominator is an integer because the product of integers are
|
|
integers, also the assumption states that both $a$ and $b$ are not $0$, so the
|
|
denominator cannot be $0$ by the zero product property. Hence the given fraction
|
|
is a rational number.
|
|
|
|
10. Assume that $m$ and $n$ are both integers and that $n \neq 0$. Explain why
|
|
$\dfrac{(5m - 12n)}{(4n)}$ must be a rational number.
|
|
|
|
Given that $m$ and $n$ are both integers, in the given fraction
|
|
|
|
$$ \frac{(5m -12n)}{4n} $$
|
|
|
|
The numerator $5m - 12n$ is an integer because the difference of integers are
|
|
integers. The denominator $4n$ is an integer because the product of integers are
|
|
integers. Also, the since $n \neq 0$, $4n \neq 0$ by the zero product property.
|
|
Hence the given fraction is a rational number.
|
|
|
|
11. Prove that every integer is a rational number.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $x$ is any integer.
|
|
|
|
**Proof:**
|
|
|
|
Then:
|
|
|
|
$$ x = x \cdot 1 $$
|
|
|
|
$$ \dfrac{x}{1} = x $$
|
|
|
|
Then $x$ is an integer and $1$ is an integer where $1 \neq 0$. Hence $x$ can be
|
|
expressed as a quotient of integers with a nonzero denominator and therefore $x$
|
|
is a rational number by definition of a rational number.
|
|
|
|
Q.E.D.
|
|
|
|
12. Let $S$ be the statement "The square of any rational number is rational." A
|
|
formal version of $S$ is "For every rational number $r$, $r^2$ is rational."
|
|
Fill in the blanks in the proof for $S$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose that $r$ is __ (a) __. By definition of rational, $r = \dfrac{a}{b}$ for
|
|
some __ (b) __ with $b \neq 0$. By substitution,
|
|
|
|
$$ r^2 = \text{\_\_ (c) \_\_} = \frac{a^2}{b^2} $$
|
|
|
|
Since $a$ and $b$ are both integers, so are the products $a^2$ and __ (d) __.
|
|
Also $b^2 \neq 0$ by the __ (e) __. Hence $r^2$ is a ratio of two integers with
|
|
a non-zero denominator,n and so __ (f) __ by definition of rational.
|
|
|
|
a. a rational number
|
|
|
|
b. integers $a$ and $b$
|
|
|
|
c. $\left(\frac{a}{b}\right)^2$
|
|
|
|
d. $b^2$
|
|
|
|
e. zero product property
|
|
|
|
f. $r^2$ is a rational number
|
|
|
|
13. Consider the following statement: The negative of any rational number is
|
|
rational.
|
|
|
|
a. Write the statement formally using a quantifier and a variable.
|
|
|
|
$$ \forall q \in \mathbb{Q} (-q \in \mathbb{Q}) $$
|
|
|
|
Alternatively:
|
|
|
|
$$ \forall q ((q \in \mathbb{Q}) \in (-q \in \mathbb{Q})) $$
|
|
|
|
b. Determine whether the statement is true or false and justify your answer.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $q$ is any rational number.
|
|
|
|
**Proof:**
|
|
|
|
Since $q$ is a rational number, $q$ can be expressed as $\dfrac{a}{b}$ where $a$
|
|
and $b$ are integers and $b \neq 0$.
|
|
|
|
Then:
|
|
|
|
$$ -q = -\left(\frac{a}{b}\right) \quad \text{ by substitution} $$
|
|
|
|
$$ -q = \frac{-a}{b} $$
|
|
|
|
Then the numerator $-a$ is an integer because the product of integers are
|
|
integers. The denominator $b$ is an integer and $b \neq 0$ by assumption of $q$
|
|
as a rational number. Hence $-q$ can be expressed as the ratio of two integers
|
|
with a nonzero denominator, and therefore $-q$ is a rational number by
|
|
definition of rational numbers.
|
|
|
|
Q.E.D.
|
|
|
|
14. Consider the statement: The cube of any rational number is a rational
|
|
number.
|
|
|
|
a. Write the statement formally using a quantifier and a variable.
|
|
|
|
$$ \forall q ((q \in \mathbb{Q}) \to (q^3 \in \mathbb{Q})) $$
|
|
|
|
b. Determine whether the statement is true or false and justify your answer.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $q$ is any rational number.
|
|
|
|
**Proof:**
|
|
|
|
Since $q$ is a rational number, $q = \dfrac{a}{b}$ where $a$ and $b$ are
|
|
integers and $b \neq 0$.
|
|
|
|
Then:
|
|
|
|
$$ q^3 = \left(\frac{a}{b}\right)^3 \quad \text{ by substitution} $$
|
|
|
|
$$ q^3 = \frac{a^3}{b^3} \quad \text{ by power of a quotient} $$
|
|
|
|
Then the numerator $a^3$ is an integer because the products of integers are
|
|
integers. Also the denominator $b^3$ is an integer because the product of
|
|
integers are integers and $b^3 \neq 0$ by the zero product property.
|
|
|
|
Thus $q^3$ can be expressed as a ratio of two integers with a nonzero
|
|
denominator and therefore $q^3$ is a rational number by definition of a rational
|
|
number.
|
|
|
|
Q.E.D.
|
|
|
|
Determine which of the statements in 15-19 are true and which are false. Prove
|
|
each true statement directly from the definitions, and give a counterexample for
|
|
each false statement. For a statement that is false, determine whether a small
|
|
change would make it true. If so, make the change and prove the new statement.
|
|
Follow the directions for writing proofs on page 173.
|
|
|
|
15. The product of any two rational numbers is a rational number.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $q$ and $r$ are rational numbers.
|
|
|
|
**Proof:**
|
|
|
|
Since $q$ and $r$ are rational numbers, then $q = \dfrac{a}{b}$ and
|
|
$r = \dfrac{c}{d}$ where $a$, $b$, $c$, and $d$ are some integers and $b \neq 0$
|
|
and $d \neq 0$.
|
|
|
|
Then:
|
|
|
|
$$ qr = \left(\frac{a}{b}\right)\left(\frac{c}{d}\right) \quad \text{ by substitution} $$
|
|
|
|
$$ qr = \frac{ac}{bd} $$
|
|
|
|
Then the numerator $ac$ is an integer because the product of integers are
|
|
integers. The denominator $bd$ is an integer because the product of integers are
|
|
integers, and $bd \neq 0$ by the of the zero product property.
|
|
|
|
Thus $qr$ can be expressed as a ratio of two integers with a nonzero denominator
|
|
and therefore $qr$ is a rational number by the definition of a rational number.
|
|
|
|
Q.E.D.
|
|
|
|
16. The quotient of any two rational numbers is a rational number.
|
|
|
|
This is false.
|
|
|
|
**Claim:**
|
|
|
|
There exists some rational number $q$ and some rational number $r$ such that
|
|
$\dfrac{q}{r}$ is not a rational number.
|
|
|
|
**Proof:**
|
|
|
|
Let $q = \dfrac{1}{2}$ and $r = \dfrac{0}{1}$.
|
|
|
|
Then:
|
|
|
|
$$ \frac{q}{r} = \frac{\dfrac{1}{2}}{\dfrac{0}{1}} \quad \text{ by substitution} $$
|
|
|
|
$$ \quad = \frac{1}{2 \cdot 0} $$
|
|
|
|
$$ \quad = \text{ undefined} $$
|
|
|
|
So the numerator of the given $\dfrac{q}{r}$ is $1$ which is an integer, but the
|
|
denominator is $0$, which means $\dfrac{q}{r}$ is not any number, and therefore
|
|
not a rational number.
|
|
|
|
Thus there exists two rational numbers whose quotients are not a rational
|
|
number, therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
A small change that would make this true were if the statement were reworded as:
|
|
|
|
For any two rational numbers, the quotient of those two numbers is a rational
|
|
number as long as the rational number in the divisor doesn't equal $0$.
|
|
|
|
17. The difference of any two rational numbers is a rational number.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose that $q$ and $r$ are any rational numbers.
|
|
|
|
**Proof:**
|
|
|
|
Since $q$ and $r$ are rational numbers, $q = \dfrac{a}{b}$ and
|
|
$r = \dfrac{c}{d}$ where $a$, $b$, $c$, and $d$ are some integers and $b \neq 0$
|
|
and $d \neq 0$.
|
|
|
|
Then:
|
|
|
|
$$ q - r = \frac{a}{b} - \frac{c}{d} \text{ by substitution} $$
|
|
|
|
$$ \quad = \frac{ad - cb}{bd} $$
|
|
|
|
Then the numerator $ad - cb$ is an integer because the difference and products
|
|
of integers are integers. The denominator $bd$ is an nonzero integer because the
|
|
products of integers are integers and because of the zero product property.
|
|
|
|
Thus $q - r$ can be expressed as a ratio of two integers with a nonzero
|
|
denominator, and therefore $q - r$ is a rational number by the definition of a
|
|
rational number.
|
|
|
|
Q.E.D.
|
|
|
|
18. If $r$ and $s$ are any two rational numbers, then $\dfrac{r + s}{2}$ is
|
|
rational.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $r$ and $s$ are any two rational numbers.
|
|
|
|
**Proof:**
|
|
|
|
Since $r$ and $s$ are rational numbers, then $r = \dfrac{a}{b}$ and
|
|
$s = \dfrac{c}{d}$ where $a$, $b$, $c$, and $d$ are integers and $b \neq 0$ and
|
|
$d \neq 0$.
|
|
|
|
Then:
|
|
|
|
By substitution:
|
|
|
|
$$ \frac{r + s}{2} = \frac{\dfrac{a}{b} + \dfrac{c}{d}}{2} $$
|
|
|
|
$$ \quad = \frac{1}{2}\left(\frac{a}{b} + \frac{c}{d}\right) $$
|
|
|
|
$$ \quad = \frac{a}{2b} + \frac{c}{2d} $$
|
|
|
|
$$ \quad = \frac{ad + bc}{2bd} $$
|
|
|
|
Then the numerator $ad + bc$ is an integer because the products and sums of
|
|
integers are integers. The denominator $2bd$ is a nonzero integer because the
|
|
products of integers are integers and because of the zero product property.
|
|
|
|
Thus $\dfrac{r + s}{2}$ can be expressed as the ratio of two integers with a
|
|
nonzero denominator, and therefore $\dfrac{r + s}{2}$ is a rational number by
|
|
the definition of a rational number.
|
|
|
|
Q.E.D.
|
|
|
|
19. For all real numbers $a$ and $b$, if $a < b$ then
|
|
$a < \dfrac{a + b}{2} < b$.
|
|
|
|
(You may use the properties of inequalities in T17-T27 of Appendix A.)
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $a$ and $b$ are any real numbers and that $a < b$.
|
|
|
|
**Proof:**
|
|
|
|
Then:
|
|
|
|
By T19:
|
|
|
|
$$ a + a < a + b $$
|
|
|
|
$$ 2a < a + b $$
|
|
|
|
By T20:
|
|
|
|
$$ a < \frac{a + b}{2} $$
|
|
|
|
And:
|
|
|
|
By T19:
|
|
|
|
$$ a + b < b + b $$
|
|
|
|
$$ a + b < 2b $$
|
|
|
|
By T20:
|
|
|
|
$$ \frac{a + b}{2} < b $$
|
|
|
|
Therefore $a < \dfrac{a + b}{2} < b$.
|
|
|
|
Q.E.D.
|
|
|
|
20. Use the results of exercises 18 and 19 to prove that given any two rational
|
|
numbers $r$ and $s$ with $r < s$, there is another rational number between
|
|
$r$ and $s$. An important consequence is that there are infinitely many
|
|
rational numbers in between any two distinct rational numbers. See Section
|
|
7.4.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $r$ is any rational number and $s$ is any rational number where $r < s$.
|
|
|
|
**Proof:**
|
|
|
|
By 18, we know that $\dfrac{r + s}{2}$ is a rational number.
|
|
|
|
By 19, we know that if $r < s$, then $r < \dfrac{r + s}{2} < s$.
|
|
|
|
Therefore there exists some rational number $\dfrac{r + s}{2}$ that is between
|
|
$r$ and $s$, _[as was to be shown]_.
|
|
|
|
Q.E.D.
|
|
|
|
Use the properties of even and odd integers that are listed in Example 4.3.3 to
|
|
do exercises 21-23. Indicate which properties you use to justify your reasoning.
|
|
|
|
21. True or false? If $m$ is any even integer and $n$ is any odd integer, then
|
|
$m^2 + 3n$ is odd. Explain.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $m$ is any even integer and $n$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
By 3, the product of any two odd integers is odd, then:
|
|
|
|
$$ 3n \text{ is odd} $$
|
|
|
|
Since $m$ is even, $m = 2k$ for some integer $k$.
|
|
|
|
Then:
|
|
|
|
By substitution:
|
|
|
|
$$ m^2 = (2k)^2 $$
|
|
|
|
$$ \quad = 4k^2 $$
|
|
|
|
$$ \quad = 2(2k^2) $$
|
|
|
|
Then $m^2$ is even by the definition of an even integer.
|
|
|
|
$$ m^2 \text{ is even} $$
|
|
|
|
By 5, the sum of any odd integer and any even integer is odd.
|
|
|
|
Thus $m^2 + 3n$ is odd, therefore the statement is true.
|
|
|
|
Q.E.D.
|
|
|
|
22. True or false? If $a$ is any odd integer, then $a^2 + a$ is even. Explain.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $a$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
Then:
|
|
|
|
$$ a^2 = a \cdot a $$
|
|
|
|
By 3, the product of any two odd integers is odd, then:
|
|
|
|
$$ a^2 \text{ is odd} $$
|
|
|
|
By 2, the sum and difference of any two odd integers are even, then:
|
|
|
|
$$ a^2 + a \text{ is even} $$
|
|
|
|
Therefore the statement is true.
|
|
|
|
Q.E.D.
|
|
|
|
23. True or false? If $k$ is any even integer and $m$ is any odd integer, then
|
|
$(k + 2)^2 - (m - 1)^2$ is even. Explain.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $k$ is any even integer and $m$ is any odd integer.
|
|
|
|
**Proof:**
|
|
|
|
By 1, the sum of any two even integers is even, then:
|
|
|
|
$$ k + 2 \text{ is even} $$
|
|
|
|
By 1, the product of any two even integers is even, then:
|
|
|
|
$$ (k + 2)^2 = (k + 2)(k + 2) $$
|
|
|
|
$$ (k + 2)^2 \text{ is even} $$
|
|
|
|
By 2 the difference of any two odd integers is even, then:
|
|
|
|
$$ m - 1 \text{ is even} $$
|
|
|
|
By 1 the product of any two even integers is even, then:
|
|
|
|
$$ (m - 1)^2 = (m - 1)(m - 1) $$
|
|
|
|
$$ (m - 1)^2 \text{ is even} $$
|
|
|
|
By 1 the difference of any two even integers is even, then:
|
|
|
|
$$ (k + 2)^2 - (m - 1)^2 \text{ is even} $$
|
|
|
|
Therefore the statement is true.
|
|
|
|
Q.E.D.
|
|
|
|
Derive the statements in 24-26 as corollaries of Theorems 4.3.1, 4.3.2, and the
|
|
results of exercises 12, 13, 14, 15, and 17.
|
|
|
|
24. For any rational numbers $r$ and $s$, $2r + 3s$ is rational.
|
|
|
|
**Theorem:**
|
|
|
|
By 15, the product of any two rational numbers is a rational number, then:
|
|
|
|
$$ 2r \text{ is rational} $$
|
|
|
|
and
|
|
|
|
$$ 3s \text{ is rational} $$
|
|
|
|
By Theorem 4.3.2, the sum of any two rational numbers is rational, then:
|
|
|
|
$$ 2r + 3s \text{ is rational} $$
|
|
|
|
Therefore the statement is true.
|
|
|
|
Q.E.D.
|
|
|
|
**Proof:**
|
|
|
|
25. If $r$ is any rational number, then $3r^2 - 2r + 4$ is rational.
|
|
|
|
By 15, the product of any two rational numbers is a rational number, then:
|
|
|
|
$$ r^2 = r \cdot r $$
|
|
|
|
$$ r^2 \text{ is rational} $$
|
|
|
|
$$ 3r^2 = 3 \cdot r^2 $$
|
|
|
|
$$ 3r^2 \text{ is rational} $$
|
|
|
|
and
|
|
|
|
$$ 2r = 2 \cdot r $$
|
|
|
|
$$ 2r \text{ is rational} $$
|
|
|
|
By 17, the difference of any two rational numbers is a rational number, then:
|
|
|
|
$$ 3r^2 - 2r \text{ is rational} $$
|
|
|
|
By Theorem 4.3.2, the sum of any two rational numbers is rational, then:
|
|
|
|
$$ (3r^2 - 2r) + 4 \text{ is rational} $$
|
|
|
|
Therefore the statement is true.
|
|
|
|
Q.E.D.
|
|
|
|
26. For any rational number $s$, $5s^3 + 8s^2 - 7$ is rational.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $s$ is any rational number.
|
|
|
|
**Proof:**
|
|
|
|
By 15, the product of any two rational numbers is a rational number, then:
|
|
|
|
$$ s^2 = s \cdot s $$
|
|
|
|
$$ s^2 \text{ is rational} $$
|
|
|
|
$$ s^3 = s^2 \cdot s $$
|
|
|
|
$$ s^3 \text{ is rational} $$
|
|
|
|
$$ 5s^3 = 5 \cdot s^3 $$
|
|
|
|
$$ 5s^3 \text{ is rational} $$
|
|
|
|
and
|
|
|
|
$$ 8s^2 = 8 \cdot s^2 $$
|
|
|
|
$$ 8s^2 \text{ is rational} $$
|
|
|
|
By 17, the difference of any two rational numbers is a rational number, then:
|
|
|
|
$$ 8s^2 - 7 \text{ is rational} $$
|
|
|
|
By Theorem 4.3.2, the sum of any two rational numbers is rational, then:
|
|
|
|
$$ 5s^3 + (8s^2 - 7) \text{ is rational} $$
|
|
|
|
Therefore the statement is true.
|
|
|
|
Q.E.D.
|
|
|
|
27. It is a fact that if $n$ is any nonnegative integer, then
|
|
|
|
$$ 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots + \frac{1}{2^n} = \frac{1 - \left(\dfrac{1}{2^{n + 1}}\right)}{1 - \left(\dfrac{1}{2}\right)} $$
|
|
|
|
(A more general form of this statement is proved in Section 5.2.) Is the
|
|
right-hand side of this equation rational? If so, express it as a ratio of two
|
|
integers.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $n$ is any nonnegative integer.
|
|
|
|
**Proof:**
|
|
|
|
Consider:
|
|
|
|
$$ \frac{1 - \left(\dfrac{1}{2^{n + 1}}\right)}{1 - \left(\dfrac{1}{2}\right)} $$
|
|
|
|
The denominator can be simpilifed as:
|
|
|
|
$$ 1 - \frac{1}{2} = \frac{1}{2} $$
|
|
|
|
Then:
|
|
|
|
$$ \frac{1 - \left(\dfrac{1}{2^{n + 1}}\right)}{\frac{1}{2}} $$
|
|
|
|
$$ \quad = 2\left(1 - \frac{1}{2^{n + 1}}\right)$$
|
|
|
|
$$ \quad = 2 - \frac{2}{2^{n + 1}} $$
|
|
|
|
$$ \quad = 2 - \frac{1}{2^n} $$
|
|
|
|
$$ \quad = \frac{2 \cdot 2^n - 1}{2^n}$$
|
|
|
|
$$ \quad = \frac{2^{n + 1} - 1}{2^n}$$
|
|
|
|
Since $n$ is a nonnegative integer, the numerator $2^{n + 1} - 1$ is an integer
|
|
because the products and differences of integers are integers. And the
|
|
denominator $2^n$ is a positive integer because of the products of integers and
|
|
because $n$ is a nonnegative integer.
|
|
|
|
Therefore right-hand side of the given equation is a rational number by the
|
|
definition of rational numbers.
|
|
|
|
Q.E.D.
|
|
|
|
28. Suppose $a$, $b$, $c$, and $d$ are integers and $a \neq c$. Suppose also
|
|
that $x$ is a real number that satisfies the equation
|
|
|
|
$$ \frac{ax + b}{cx + d} = 1 $$
|
|
|
|
Must $x$ be rational? If so, express $x$ as a ratio of two integers.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $a$, $b$, $c$, and $d$ are any integers and suppose $x$ is a real number
|
|
that satisfies the equation:
|
|
|
|
$$ \frac{ax + b}{cx + d} = 1 $$
|
|
|
|
_Proof:_*
|
|
|
|
Consider:
|
|
|
|
$$ \frac{ax + b}{cx + d} = 1 $$
|
|
|
|
$$ ax + b = cx + d $$
|
|
|
|
$$ ax - cx = d - b $$
|
|
|
|
$$ x(a - c) = d - b $$
|
|
|
|
$$ x = \frac{d - b}{a - c} $$
|
|
|
|
Then the numerator $d - b$ is an integer because the difference of integers are
|
|
integers. The denominator $a - c$ must be a nonzero integer because the
|
|
difference of integers are integers and because $a \neq c$.
|
|
|
|
Therefore $x$ must be a rational number by the definition of rational numbers.
|
|
|
|
Q.E.D.
|
|
|
|
29. Suppose $a$, $b$, and $c$ are integers and $x$, $y$, and $z$ are nonzero
|
|
real numbers that satisfy the following equations:
|
|
|
|
$$ \frac{xy}{x + y} = a \quad \text{ and } \quad \frac{xz}{x + z} = b \quad \text{ and } \quad \frac{yz}{y + z} = c $$
|
|
|
|
Is $x$ rational? If so, express it as ratio of two integers.
|
|
|
|
Omitted.
|
|
|
|
30. Prove that if one solution for a quadratic equation of the form
|
|
$x^2 + bx + c = 0$ is rational (where $b$ and $c$ are rational), then the
|
|
other solution is also rational. (Use the fact that if the solutions of the
|
|
equation are $r$ and $s$, then $x^2 + bx + c = (x - r)(x - s)$.)
|
|
|
|
**Theorem:**
|
|
|
|
Suppose there is any rational number $r$ that is a solution to a quadratic
|
|
equation of the form:
|
|
|
|
$$ x^2 + bx + c = 0 $$
|
|
|
|
Where $b$ and $c$ are rational.
|
|
|
|
And suppose $s$ is the other solution to the given equation.
|
|
|
|
**Proof:**
|
|
|
|
Given that both $r$ and $s$ are solutions to the given equation, then:
|
|
|
|
$$ x^2 + bx + c = (x - r)(x - s) = x^2 - rx - sx + rs $$
|
|
|
|
This means that:
|
|
|
|
$$ bx = (-r - s)x $$
|
|
|
|
$$ b = -1(r + s) $$
|
|
|
|
And:
|
|
|
|
$$ c = rs $$
|
|
|
|
Let's analyze $b$ and isolate $s$.
|
|
|
|
$$ b = -1(r + s) $$
|
|
|
|
$$ -b = r + s $$
|
|
|
|
$$ -b - r = s $$
|
|
|
|
Since both $b$ and $r$ are rational numbers, then $b = \dfrac{g}{h}$ and
|
|
$r = \dfrac{i}{j}$ where $g$, $h$, $i$, and $j$ are some integers and $h \neq 0$
|
|
and $j \neq 0$.
|
|
|
|
Then:
|
|
|
|
$$ s = -b - r = -\left(\frac{g}{h}\right) - \left(\frac{i}{j}\right) \text{ by substitution} $$
|
|
|
|
$$ \quad = \frac{-1(gj + hi)}{hj} $$
|
|
|
|
The numerator $-1(gj + hi)$ is an integer because the sum and products of
|
|
integers are integers. The denominator is a nonzero integer because the products
|
|
of integers are integers and because of the zero product property.
|
|
|
|
Therefore $s$ can be expressed as a ratio of two integers where the denominator
|
|
is nonzero. Thus $s$ is a rational number by the definition of rational numbers,
|
|
therefore the other solution is rational.
|
|
|
|
Q.E.D.
|
|
|
|
31. Prove that if a real number $c$ satisfies a polynomial equation of the form
|
|
|
|
$$ r_3x^3 + r_2x^2 + r_1x + r_0 = 0 $$
|
|
|
|
where $r_0$, $r_1$, $r_2$, and $r_3$ are rational numbers, then $c$ satisfies an
|
|
equation of the form
|
|
|
|
$$ n_3x^3 + n_2x^2 + n_1x + n_0 = 0 $$
|
|
|
|
where $n_0$, $n_1$, $n_2$, and $n_3$ are integers.
|
|
|
|
**Definition:** A number $c$ is called a **root** of a polynomial $p(x)$ if, and
|
|
only if, $p(c) = 0$.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $c$ is any real number that satisfies a polynomial equation of the form
|
|
|
|
$$ r_3x^3 + r_2x^2 + r_1x + r_0 = 0 $$
|
|
|
|
where $r_0$, $r_1$, $r_2$, and $r_3$ are rational numbers.
|
|
|
|
**Proof:**
|
|
|
|
Since $c$ is a real number that satisfies the given equation, then:
|
|
|
|
$$ r_3c^3 + r_2c^2 + r_1c + r_0 = 0 $$
|
|
|
|
Since $r_3$, $r_2$, $r_1$, and $r_0$ are rational numbers, then
|
|
$r_3 = \dfrac{a_3}{b_3}$, $r_2 = \dfrac{a_2}{b_2}$, $r_1 = \dfrac{a_1}{b_1}$,
|
|
and $r_0 = \dfrac{a_0}{b_0}$ where $a_3$, $a_2$, $a_1$, $a_0$ are some integers
|
|
and $b_3$, $b_2$, $b_1$, $b_0$ are some nonzero integers.
|
|
|
|
Then, by substitution:
|
|
|
|
$$ r_3c^3 + r_2c^2 + r_1c + r_0 = \left(\frac{a_3}{b_3}\right)c^3 + \left(\frac{a_2}{b_2}\right)c^2 + \left(\frac{a_1}{b_1}\right)c + \frac{a_0}{b_0} = 0 $$
|
|
|
|
$$ \quad = \frac{a_3b_2b_1b_0c^3 + a_2b_3b_1b_0c^2 + a_1b_3b_2b_0c + a_0b_3b_2b_1}{b_3b_2b_1b_0} = 0 $$
|
|
|
|
$$ \quad = a_3b_2b_1b_0c^3 + a_2b_3b_1b_0c^2 + a_1b_3b_2b_0c + a_0b_3b_2b_1 = 0 $$
|
|
|
|
Let $n_3 = a_3b_2b_1b_0$, and $n_2 = a_2b_3b_1b_0$, and $n_1 = a_1b_3b_2b_0$,
|
|
and $n_0 = a_0b_3b_2b_1$.
|
|
|
|
Then $n_3$, $n_2$, $n_1$, and $n_0$ are integers because of the product of
|
|
integers.
|
|
|
|
Thus we can write the given equation as:
|
|
|
|
$$ n_3c^3 + n_2c^2 + n_1c + n_0 = 0 $$
|
|
|
|
Where $c$ is a real number that satisfies the equation:
|
|
|
|
$$ n_3x^3 + n_2x^2 + n_1x + n_0 = 0 $$
|
|
|
|
Q.E.D.
|
|
|
|
32. Prove that for every real number $c$, if $c$ is a root of a polynomial with
|
|
rational coefficients, then $c$ is a root of a polynomial with integer
|
|
coefficients.
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $c$ is a root of a polynomial with rational coefficients.
|
|
|
|
**Proof:**
|
|
|
|
Then
|
|
|
|
$$ r_nx^n + r_{n - 1}x^{n - 1} + \dots + r_1x + r_0 = 0 $$
|
|
|
|
where each $r_i$ is rational.
|
|
|
|
Then each $r_i$ can be written as a ratio of integers with nonzero denominators.
|
|
Let $D$ be a common multiple of all denominators of the $r_i$. Multiplying the
|
|
equation by $D$ gives
|
|
|
|
$$ s_nx^n + s_{n - 1}x^{n - 1} + \dots + s_1x + s_0 = 0 $$
|
|
|
|
where each $s_i$ is an integer.
|
|
|
|
Thus $c$ is a root of a polynomial with integer coefficients.
|
|
|
|
Q.E.D.
|
|
|
|
Use the properties of even and odd integers that are listed in Example 4.3.3 to
|
|
do exercises 33 and 34.
|
|
|
|
33. When expressions of the form $(x - r)(x - s)$ are multiplied out, a
|
|
quadratic polynomial is obtained. For instance,
|
|
$(x - 2)(x - (-7)) = (x - 2)(x + 7) = x^2 + 5x - 14$.
|
|
|
|
a. What can be said about the coefficients of the polynomial obtained by
|
|
multiplying out $(x - r)(x - s)$ when both $r$ and $s$ are odd integers? When
|
|
both $r$ and $s$ are even integers? When one of $r$ and $s$ is even and the
|
|
other odd?
|
|
|
|
_Case when both $r$ and $s$ are odd integers:_
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $r$ and $s$ are odd integers.
|
|
|
|
**Conclusion:**
|
|
|
|
Let $x$ be some real number.
|
|
|
|
Then:
|
|
|
|
$$ (x - r)(x - s) = x^2 - rx - sx - rs = x^2 + (-1)(r + s)x + (-1)rs $$
|
|
|
|
We know the coefficient of $x^2$ is $1$.
|
|
|
|
By 2, we know the sum of any two odd integers are even, then:
|
|
|
|
We know the coefficient of $-1(r + s)$ is even.
|
|
|
|
By 3, we know the product of any two odd integers is odd, then:
|
|
|
|
We know the coefficient of $rs$ is odd.
|
|
|
|
_Case when both $r$ and $s$ are even integers:_
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $r$ and $s$ are even integers.
|
|
|
|
**Conclusion:**
|
|
|
|
Let $x$ be some real number.
|
|
|
|
Then:
|
|
|
|
$$ (x - r)(x - s) = x^2 - rx - sx - rs = x^2 + (-1)(r + s)x + (-1)rs $$
|
|
|
|
We know the coefficient of $x^2$ is $1$.
|
|
|
|
By 1 we know the sum of any two even integers is even, then:
|
|
|
|
We know the coefficient of $(-1)(r + s)x$ is even.
|
|
|
|
By 1 we know the product of any two even integers is even, then:
|
|
|
|
We know the coefficient of $(-1)rs$ is even.
|
|
|
|
_Case where $r$ and $s$ is even and the other odd:_
|
|
|
|
**Theorem:**
|
|
|
|
Suppose $r$ and $s$ are any integers where one is even and the other is odd.
|
|
|
|
**Conclusion:**
|
|
|
|
Let $x$ be some real number.
|
|
|
|
Then:
|
|
|
|
$$ (x - r)(x - s) = x^2 - rx - sx - rs = x^2 + (-1)(r + s)x + (-1)rs $$
|
|
|
|
We know the coefficient of $x^2$ is $1$.
|
|
|
|
By 5, we know the sum of any odd integer and any even integer is odd, then:
|
|
|
|
We know the coefficient of $(-1)(r +s)x$ is odd.
|
|
|
|
By 4, we know the product of any even integer and any odd integer is even, then:
|
|
|
|
We know the coefficient of $(-1)rs$ is even.
|
|
|
|
b. It follows from part (a) that $x^2 - 1253x + 255$ cannot be written as a
|
|
product of two polynomials with integer coefficients. Explain why this is so.
|
|
|
|
Because in all cases from part (a), the middle coefficient and the third
|
|
coefficient were always either even and odd or odd and even. Since both 1253 and
|
|
255 are odd, this expression cannot be expressed as the product of two
|
|
polynomials with integer coefficients.
|
|
|
|
34. Observe that
|
|
|
|
$$ (x - r)(x -s)(x - t) = x^3 - (r + s + t)x^2 + (rs + rs + st)x - rst $$
|
|
|
|
a. Derive a result for cubic polynomials similar to the result in part (a) of
|
|
exercise 33 for quadratic polynomials.
|
|
|
|
_Case where $r$ and $s$ and $t$ are all even_:
|
|
|
|
The coefficient of $x^3$ is $1$.
|
|
|
|
By 1, the sum of any two even integers is even, then:
|
|
|
|
$$ r + s \text{ is even} $$
|
|
|
|
$$ (r + s) + t \text{ is even} $$
|
|
|
|
The coefficient of $(r + s + t)x^2$ is even.
|
|
|
|
By 1, the sum and product of any two even integers is even, then:
|
|
|
|
$$ rs \text{ is even} $$
|
|
|
|
$$ st \text{ is even} $$
|
|
|
|
$$ (rs + rs + rt) \text{ is even} $$
|
|
|
|
The coefficient of $(rs + rs + st)x$ is even.
|
|
|
|
By 1, the product of any two even integers is even, then:
|
|
|
|
$$ rs \text{ is even} $$
|
|
|
|
$$ rst \text{ is even} $$
|
|
|
|
The coefficient of $rst$ is even.
|
|
|
|
_Case where $r$ is odd and $s$ and $t$ are even_:
|
|
|
|
The coefficient of $x^3$ is $1$.
|
|
|
|
By 1 the sum of any two even integers is even, then:
|
|
|
|
$$ s + t \text{ is even} $$
|
|
|
|
By 5 the sum of any odd integer and any even integer is odd.
|
|
|
|
$$ r + (s + t) \text{ is odd} $$
|
|
|
|
The coefficient of $(r + s + t)x^2$ is odd.
|
|
|
|
By 4, the product of any even integer and any odd integer is even, then:
|
|
|
|
$$ rs \text{ is even} $$
|
|
|
|
By 1, the sum and product of any two even integers is even.
|
|
|
|
$$ st \text{ is even} $$
|
|
|
|
$$ rs + st \text{ is even} $$
|
|
|
|
$$ rs + (rs + st) \text{ is even} $$
|
|
|
|
The coefficient of $(rs + rs + st)x$ is even.
|
|
|
|
$$ (rs)t \text{ is even} $$
|
|
|
|
The coefficient of $rst$ is even.
|
|
|
|
_Case where $r$ and $s$ are odd and $t$ is even_:
|
|
|
|
The coefficient of $x^3$ is $1$.
|
|
|
|
By 2, the sum of any two odd integers is even.
|
|
|
|
$$ r + s \text{ is even} $$
|
|
|
|
By 1, the sum of any two even integers is even.
|
|
|
|
$$ (r + s) + t \text{ is even} $$
|
|
|
|
The coefficient of $(r + s + t)x^2$ is even.
|
|
|
|
By 3, the product of any two odd integers is odd.
|
|
|
|
$$ rs \text{ is odd} $$
|
|
|
|
By 4, the product of any even integer and any odd integer is even.
|
|
|
|
$$ st \text{ is even} $$
|
|
|
|
By 2, the sum of any two odd integers is even.
|
|
|
|
$$ rs + rs \text{ is even} $$
|
|
|
|
By 1, the sum of any two even integers is even.
|
|
|
|
$$ (rs + rs) + st \text{ is even} $$
|
|
|
|
The coefficient of $(rs + rs + st)x$ is even
|
|
|
|
By 4, the product of any even integer and any odd integer is even.
|
|
|
|
$$ (rs)t \text{ is even} $$
|
|
|
|
The coefficient of $rst$ is even.
|
|
|
|
_Case where $r$ and $s$ and $t$ are all odd_:
|
|
|
|
The coefficient of $x^3$ is $1$.
|
|
|
|
By 2, the sum of any two odd integers is even.
|
|
|
|
$$ r + s \text{ even} $$
|
|
|
|
By 5, the sum of any odd integer and any even integer is odd.
|
|
|
|
$$ (r + s) + t \text{ is odd} $$
|
|
|
|
The coefficient of $(r + s + t)x^2$ is odd.
|
|
|
|
By 3, The product of any two odd integers is odd.
|
|
|
|
$$ rs \text{ is odd} $$
|
|
|
|
$$ st \text{ is odd} $$
|
|
|
|
By 2, the sum of any two odd integers is even.
|
|
|
|
$$ rs + rs \text{ is even} $$
|
|
|
|
By 5, the sum of any odd integer and any even integer is odd.
|
|
|
|
$$ (rs + rs) + st \text{ is odd} $$
|
|
|
|
The coefficient of $(rs + rs + st)x$ is odd.
|
|
|
|
$$ (x - r)(x -s)(x - t) = x^3 - (r + s + t)x^2 + (rs + rs + st)x - rst $$
|
|
|
|
By 3, The product of any two odd integers is odd.
|
|
|
|
$$ (rs)t \text{ is odd.} $$
|
|
|
|
The coefficient of $rst$ is odd.
|
|
|
|
b. Can $x^3 + 7x^2 - 8x - 27$ be written as a product of three polynomials with
|
|
integer coefficients? Explain.
|
|
|
|
In all cases, the order of the second through fourth terms are never: "odd,
|
|
even, odd". Therefore the given polynomial $x^3 + 7x^2 - 8x - 27$ can be written
|
|
as a product of three polynomials with integer coefficients.
|
|
|
|
In 35-39 find the mistakes in the "proofs" that the sum of any two rational
|
|
numbers is a rational number.
|
|
|
|
35.
|
|
|
|
**"Proof:** Any two rational numbers produce a rational number when added
|
|
together. So if $r$ and $s$ are particular but arbitrarily chosen rational
|
|
numbers, then $r + s$ is rational."
|
|
|
|
This proof assumes what is to be proved.
|
|
|
|
36.
|
|
|
|
**"Proof:** Let rational numbers $r = \dfrac{1}{4}$ and $s = \dfrac{1}{2}$ be
|
|
given. Then $r + s = \dfrac{1}{4} + \dfrac{1}{2} = \dfrac{3}{4}$, which is a
|
|
rational number. This is what was to be shown."
|
|
|
|
This proof is arguing from examples.
|
|
|
|
37.
|
|
|
|
**"Proof:** Suppose $r$ and $s$ are rational numbers. By definition of rational,
|
|
$r = \dfrac{a}{b}$ for some integers $a$ and $b$ with $b \neq 0$, and
|
|
$s = \dfrac{a}{b}$ for some integers $a$ and $b$ with $b \neq 0$. Then
|
|
|
|
$$ r + s = \frac{a}{b} + \frac{a}{b} = \frac{2a}{b} $$
|
|
|
|
Let $p = 2a$. Then $p$ is an integer since it is a product of integers. Hence
|
|
$r + s = \dfrac{p}{b}$, where $p$ and $b$ are integers and $b \neq 0$. Thus
|
|
$r + s$ is a rational number by definition of rational. This is what was to be
|
|
shown."
|
|
|
|
This incorrect proof uses the same letter to mean two different things.
|
|
|
|
38.
|
|
|
|
**"Proof:** Suppose $r$ and $s$ are rational numbers. Then $r = \dfrac{a}{b}$
|
|
and $s = \dfrac{c}{d}$ for some integers $a$, $b$, $c$, and $d$ with $b \neq 0$
|
|
and $d \neq 0$ (by definition of rational.) Then
|
|
|
|
$$ r + s = \frac{a}{b} + \frac{c}{d} $$
|
|
|
|
But this is a sum of two fractions, which is a fraction. So $r - s$ is a
|
|
rational number since a rational number is a fraction."
|
|
|
|
This incorrect proof exhibits confusion between what is known and what is still
|
|
to be shown. Additionally, they simply abandon what is to be shown since what is
|
|
to be shown is $r + s$ is rational, not $r - s$ is rational.
|
|
|
|
39.
|
|
|
|
**"Proof:** Suppose $r$ and $s$ are rational numbers. If $r + s$ is rational,
|
|
then by definition of rational $r + s = \dfrac{a}{b}$ for some integers $a$ and
|
|
$b$ with $b \neq 0$. Also since $r$ and $s$ are rational, $r = \dfrac{i}{j}$ and
|
|
$s = \dfrac{m}{n}$ for some integers $i$, $j$, $m$, and $n$ with $j \neq 0$ and
|
|
$n \neq 0$. It follows that
|
|
|
|
$$ r + s = \frac{i}{j} + \frac{m}{n} = \frac{a}{b} $$
|
|
|
|
which is a quotient of two integers with a nonzero denominator. Hence it is a
|
|
rational number. This is what is to be shown.
|
|
|
|
This incorrect prove is assuming what is to be proved.
|
|
|
|
---
|
|
|
|
**Exercise Set 4.4**
|
|
|
|
Page 220
|
|
|
|
Give a reason for your answer in each of 1-13. Assume that all variables
|
|
represent integers.
|
|
|
|
1. Is $52$ divisible by $13$?
|
|
|
|
Yes $52 = 13 \cdot 4$
|
|
|
|
2. Does $7 \mid 56$?
|
|
|
|
Yes $56 = 7 \cdot 8$
|
|
|
|
3. Does $5 \mid 0$?
|
|
|
|
Yes, $0 = 5 \cdot 0$
|
|
|
|
4. Does $3$ divide $(3k + 1)(3k + 2)(3k + 3)$?
|
|
|
|
Yes $3 | 3(3k + 1)(3k + 2)(k + 1)$
|
|
|
|
5. Is $6m(2m + 10)$ divisible by $4$?
|
|
|
|
Yes $6m(2m + 10) = 12m^2 + 60m = 4 \cdot (3m^2 + 15m)$
|
|
|
|
6. Is $29$ a multiple of $3$?
|
|
|
|
No, $\dfrac{29}{3} \approx 9.666666\dots$ which is not an integer.
|
|
|
|
7. Is $-3$ a factor of $66$?
|
|
|
|
Yes, $66 = -3(-22)$
|
|
|
|
8. Is $6a(a + b)$ a multiple of $3a$?
|
|
|
|
Yes, $6a(a + b) = 3a(2)(a + b)$
|
|
|
|
9. Is $4$ a factor of $2a \cdot 34b$?
|
|
|
|
Yes. $2a \cdot 34b = 68ab = 4(17ab) $
|
|
|
|
10. Does $7 \mid 34$?
|
|
|
|
No $\dfrac{34}{7} = 4 + \dfrac{6}{7}$ which is not an integer.
|
|
|
|
11. Does $13 \mid 73$?
|
|
|
|
No $\dfrac{73}{13} = 5 + \dfrac{8}{13}$ which is not an integer.
|
|
|
|
12. If $n = 4k + 1$, does $8$ divide $n^2 - 1$?
|
|
|
|
Yes.
|
|
|
|
$$ n^2 - 1 = (4k + 1)^2 - 1 = 16k^2 + 8k + 1 - 1 = 16k^2 + 8k = 8(2k^2 + k) $$
|
|
|
|
13. If $n = 4k + 3$, does $8$ divide $n^2 - 1$?
|
|
|
|
Yes.
|
|
|
|
$$ n^2 - 1 = (4k + 3)^2 - 1 = 16k^2 + 24k + 9 - 1 = 16k^2 + 24k + 8 = 8(2k^2 + 3k + 1) $$
|
|
|
|
14. Fill in the blanks in the following proof that for all integers $a$ and $b$,
|
|
if $a \mid b$ then $a \mid (-b)$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$ and $b$ are integers such that __ (a) __. By definition of
|
|
divisibility, there exists an integer $r$ such that __ (b) __. By substitution,
|
|
|
|
$$ -b = -(ar) = a(-r) $$
|
|
|
|
Let $t = $ __ (c) __. Then $t$ is an integer because $t = (-1) \cdot r$, and
|
|
both $-1$ and $r$ are integers. Thus, by substitution, $-b = at$, where $t$ is
|
|
an integer, and by the definition of divisibility, __ (d) __, as was to be
|
|
shown.
|
|
|
|
a. $a \mid b$
|
|
|
|
b. $b = a \cdot r$
|
|
|
|
c. $-r$
|
|
|
|
d. $a | (-b)$
|
|
|
|
Prove statements 15 and 16 directly from the definition of divisibility.
|
|
|
|
15. For all integers $a$, $b$, and $c$, if $a \mid b$ and $a \mid c$ then
|
|
$a \mid (b + c)$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$, $b$, and $c$ are any integers such that $a \mid b$ and $a \mid c$
|
|
|
|
By definition of divisibility, $b = ar$ and $c = as$ for some integers $r$ and
|
|
$s$.
|
|
|
|
Then:
|
|
|
|
$$ b + c = ar + as = a(r + s) $$
|
|
|
|
Let $t = r + s$, where $t$ is an integer because the sum of integers are
|
|
integers. And thus $b + c = a \cdot t$. By the definition of divisibility then
|
|
$a \mid (b + c)$.
|
|
|
|
Q.E.D.
|
|
|
|
16. For all integers $a$, $b$, and $c$, if $a \mid b$ then $a \mid c$ then
|
|
$a \mid (b - c)$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$, $b$, and $c$ are any integers such that $a \mid b$ and $a \mid c$
|
|
|
|
By definition of divisibility, $b = ar$ and $c = as$ for some integers $r$ and
|
|
$s$.
|
|
|
|
Then:
|
|
|
|
$$ b - c = ar - as = a(r - s) $$
|
|
|
|
Let $t = r - s$, where $t$ is an integer because the difference of integers are
|
|
integers. And thus $b - c = a \cdot t$. By the definition of divisibility then
|
|
$a \mid (b - c)$.
|
|
|
|
Q.E.D.
|
|
|
|
17. For all integers $a$, $b$, $c$, and $d$, if $a \mid c$ and $b \mid d$ then
|
|
$ab \mid cd$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$, $b$, $c$, and $d$ are any integers such that $a \mid c$ and
|
|
$b \mid d$.
|
|
|
|
By definition of divisibility, $c = ar$ and $d = bs$ for some integers $r$ and
|
|
$s$.
|
|
|
|
Then:
|
|
|
|
$$ cd = (ar)(bs) = ab(rs) $$
|
|
|
|
Let $t = rs$, where $t$ is an integer because the product of integers are
|
|
integers. Then $cd = ab \cdot t$. By the definition of divisibility then
|
|
$ab \mid cd$.
|
|
|
|
Q.E.D.
|
|
|
|
18. Consider the following statement: The negative of any multiple of $3$ is a
|
|
multiple of $3$.
|
|
|
|
a. Write the statement formally using a quantifier and a variable.
|
|
|
|
$$ \forall x \in \mathbb{Z}((3 \mid x) \to (3 \mid -x)) $$
|
|
|
|
b. Determine whether the statement is true or false and justify your answer.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $x$ is any integer such that $3 \mid x$.
|
|
|
|
By definition of a multiple, $x = 3k$ for some integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ -x = -(3k) = 3(-k) $$
|
|
|
|
Then $-k$ is an integer because the product of integers are integers. Therefore,
|
|
by the definition of divisibility, $3 \mid -x$.
|
|
|
|
Q.E.D.
|
|
|
|
19. Show that the following statement is false: For all integers $a$ and $b$, if
|
|
$3 \mid (a + b)$ then $3 \mid (a - b)$.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $a = 1$ and $b = 2$.
|
|
|
|
Then:
|
|
|
|
$$ a + b = 1 + 2 = 3 $$
|
|
|
|
So $3 \mid 3$ is true. But, then:
|
|
|
|
$$ a - b = 1 - 2 = -1 $$
|
|
|
|
$$ 3 \nmid -1 $$
|
|
|
|
Thus, for the given $a$ and $b$, $3 \mid (a + b)$, but $3 \nmid (a - b)$.
|
|
Therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
For each statement in 20-32, determine whether the statement is true or false.
|
|
Prove the statement directly from the definitions if it is true, and give a
|
|
counterexample if it is false.
|
|
|
|
20. The sum of any three consecutive integers is divisible by $3$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
Then:
|
|
|
|
$$ n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1) $$
|
|
|
|
Let $t = n + 1$, where $t$ is an integer because the sum of integers are
|
|
integers. Then $n + (n + 1) + (n + 2) = 3t$. Therefore, by the definition of
|
|
divisibility, $3 \mid n + (n + 1) + (n + 2)$.
|
|
|
|
Q.E.D.
|
|
|
|
21. The product of any two even integers is a multiple of $4$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $x$ and $y$ are any two even integers.
|
|
|
|
Since $x$ and $y$ are even integers, $x = 2k$ and $y = 2m$ where $k$ and $m$ are
|
|
some integers.
|
|
|
|
Then:
|
|
|
|
$$ xy = (2k)(2m) = 4km $$
|
|
|
|
Let $t = km$, where $t$ is an integer because the product of integers are
|
|
integers. Then $xy = 4t$. Therefore, by the definition of divsibility,
|
|
$4 \mid xy$.
|
|
|
|
Q.E.D.
|
|
|
|
22. A necessary condition for an integer to be divisible by $6$ is that it be
|
|
divisible by $2$.
|
|
|
|
_Quick Note:_
|
|
|
|
Recall that "$R$ ... necessary condition for $S$" means $S \to R$. Thus this is
|
|
saying:
|
|
|
|
If an integer is divisible by $6$, then it is divisible by $2$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $x$ is any integer and $6 \mid x$.
|
|
|
|
Then $x = 6k$ where $k$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ x = 6k = 2(3k) $$
|
|
|
|
Let $t = 3k$ where $t$ is an integer because the product of integers is
|
|
integers. Then $x = 2t$. Therefore by the definition of divisibility,
|
|
$2 \mid x$.
|
|
|
|
Q.E.D.
|
|
|
|
23. A sufficient condition for an integer to be divisible by $8$ is that it be
|
|
divisible by $16$.
|
|
|
|
_Quick Note:_
|
|
|
|
"$R$ is a sufficient condition for $S$" means "if $R$ then $S$."
|
|
|
|
This reads then as:
|
|
|
|
If an integer is divisible by $16$, then it is divisible by $8$.
|
|
|
|
**Proof:**
|
|
|
|
Let $x$ be any integer and $16 \mid x$.
|
|
|
|
Since $16 \mid x$, $x = 16k$ where $k$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ x = 16k = 8(2k) $$
|
|
|
|
Let $t = 2k$ where $t$ is an integer by the product of integers. Then $x = 8t$,
|
|
and thus $8 \mid x$ by the definition of divisibility.
|
|
|
|
Q.E.D.
|
|
|
|
24. For all integers $a$, $b$, and $c$, if $a \mid b$ and $a \mid c$ then
|
|
$a \mid (2b - 3c)$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$, $b$, and $c$ are any integers where $a \mid b$ and $a \mid c$.
|
|
|
|
Since $a \mid b$ and $a \mid c$, $b = ak$ and $c = am$ where $k$ and $m$ are
|
|
some integers.
|
|
|
|
Then:
|
|
|
|
$$ 2b - 3c = 2(ak) - 3(am) $$
|
|
|
|
$$ \quad = a(2k - 3m) $$
|
|
|
|
Let $t = 2k - 3m$, where $t$ is an integer by the difference and product of
|
|
integers. Then $2b - 3c = at$. Thus $a \mid 2b - 3c$ by the definition of
|
|
divisibility.
|
|
|
|
Q.E.D.
|
|
|
|
25. For all integers $a$, $b$, and $c$, if $a$ is a factor of $c$ and $b$ is a
|
|
factor of $c$ then $ab$ is a factor of $c$.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $a = 4$, $b = 8$, and $c = 8$.
|
|
|
|
Then $a \mid c$ is $4 \mid 8$ and $b \mid c$ is $8 \mid 8$, but $ab \mid c$ is
|
|
$32 \mid 8$, which is false since $32 \nmid 8$. Therefore this statement is
|
|
false.
|
|
|
|
Q.E.D.
|
|
|
|
26. For all integers $a$, $b$, and $c$, if $ab \mid c$ then $a \mid c$ and
|
|
$b \mid c$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$, $b$, and $c$ are integers and $ab \mid c$.
|
|
|
|
Since $ab \mid c$, $c = abk$ where $k$ is some integer.
|
|
|
|
Then:
|
|
|
|
$$ c = abk = a(bk) $$
|
|
|
|
And:
|
|
|
|
$$ c = abk = b(ak) $$
|
|
|
|
Let $t = bk$ and $u = ak$ where $t$ and $u$ are integers by the product of
|
|
integers. Then $c = at$ and $c = bu$. Therefore, by the definition of
|
|
divisibility, $a \mid c$ and $b \mid c$.
|
|
|
|
Q.E.D.
|
|
|
|
27. For all integers $a$, $b$, and $c$, if $a \mid (b + c)$ then $a \mid b$ or
|
|
$a \mid c$.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $a = 3$, $b = 4$, and $c = 5$.
|
|
|
|
Then $a \mid (b + c)$ is $3 \mid 9$, which is true. Then, however, $a \mid b$ is
|
|
$3 \mid 4$, which is false since $3 \nmid 4$ and $a \mid c$ becomes $3 \mid 5$,
|
|
which is also false since $3 \nmid 5$.
|
|
|
|
Thus for the given $a$, $b$, and $c$, $a \mid (b + c)$, but $a \nmid b$ and
|
|
$a \nmid c$. Therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
28. For all integers $a$, $b$, and $c$, if $a \mid bc$ then $a \mid b$ or
|
|
$a \mid c$.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $a = 6$, $b = 2$, $c = 3$.
|
|
|
|
Then $a \mid bc$ is $6 \mid 6$, which is true. Then, however, $a \mid b$ is
|
|
$6 \mid 2$, which is false since $6 \nmid 2$ and $a \mid c$ is $6 \mid 3$ which
|
|
also false since $6 \nmid 3$.
|
|
|
|
Thus for the given $a$, $b$, and $c$, $a \mid bc$ is true, but then $a \nmid b$
|
|
and $a \nmid c$. Therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
29. For all integers $a$ and $b$, if $a \mid b$ then $a^2 \mid b^2$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$ and $b$ are any integers where $a \mid b$.
|
|
|
|
Since $a \mid b$, then $b = ak$ for some integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ b^2 = (ak)^2 $$
|
|
|
|
$$ b^2 = a^2k^2 $$
|
|
|
|
Let $t = k^2$ where $t$ is an integer by the product of integers. Then
|
|
$b^2 = a^2t$. Therefore, by the definition of divisibility, $a^2 \mid b^2$.
|
|
|
|
Q.E.D.
|
|
|
|
30. For all integers $a$ and $n$, if $a \mid n^2$ and $a \leq n$ then
|
|
$a \mid n$.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $a = -36$ and $n = 6$.
|
|
|
|
Then $a \mid n^2$ is $-36 \mid 36$, which is true and $a \leq n$ is
|
|
$-36 \leq 6$, which is also true. Then, however, $a \mid n$ is $-36 \mid 6$,
|
|
which is false since $-36 \nmid 6$.
|
|
|
|
Thus for the given $a$ and $n$, $a \mid n^2$ and $a \leq n$, but $a \nmid n$.
|
|
Therefore this statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
31. For all integers $a$ and $b$, if $a \mid 10b$ then $a \mid 10$ or
|
|
$a \mid b$.
|
|
|
|
**Proof:**
|
|
|
|
Let $a = 4$, $b = 2$.
|
|
|
|
Then $a \mid 10b$ is $4 \mid 20$ is true, but then $a \mid 10$ is $4 \mid 10$ is
|
|
false and $a \mid b$ is $4 \mid 2$ is false.
|
|
|
|
Thus for the given $a$ and $b$, $a \mid 10b$, but then $a \nmid 10$ and
|
|
$a \nmid b$. Therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
32. A fast-food chain has a contest in which a card with numbers on it is given
|
|
to each customer who makes a purchase. If some of the numbers on the card
|
|
add up to $100$, then the customer wins $100. A certain customer receives a
|
|
card containing the numbers
|
|
|
|
72, 21, 15, 36, 69, 81, 9, 27, 42, and 63.
|
|
|
|
Will the customer win $100? Why or why not?
|
|
|
|
No, each of the given numbers is divisible by $3$, but $3 \nmid 100$, therefore
|
|
the sum of any of the given numbers can never equal $100.
|
|
|
|
33. Is it possible to have a combination of nickels, dimes, and quarters that
|
|
add up to $4.72? Explain.
|
|
|
|
No, nickels, dimes and quarters represent 5¢, 10¢, and 25¢ respectively. They
|
|
have a common divisor of 5, but $4.72 or 472¢, is not divisible by 5, and so it
|
|
is not possible to have a combination of nickels, dimes and quarters that will
|
|
add up to $4.72.
|
|
|
|
34. Consider a string consisting of _a_'s, _b_'s, and _c_'s where the number of
|
|
_b_'s is three times the number of _a_'s and the number of _c_'s is five
|
|
times the number of _a_'s. Prove that the length of the string is divisible
|
|
by $3$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer that represents the number of _a_ characters in a
|
|
string.
|
|
|
|
Suppose also that $3n$ is the number of _b_ characters in the string and $5n$
|
|
represents the number of _c_ characters in the string.
|
|
|
|
Let $L$ be the length of the string.
|
|
|
|
Then the length of the string is:
|
|
|
|
$$ L = n + 3n + 5n = 9n = 3(3n) $$
|
|
|
|
Let $t = 3n$ where $t$ is an integer by the product of integers. Then $L = 3t$.
|
|
Thus, by the definition of divisibility, $3 \mid L$.
|
|
|
|
Q.E.D.
|
|
|
|
35. Two athletes run a circular track at a steady pace so that the first
|
|
completes one round in 8 minutes and the second in 10 minutes. If they both
|
|
start from the same spot at 4 P.M., when will be the first time they return
|
|
to the start?
|
|
|
|
We are looking at the LCM (least common multiple) of both $8$ and $10$ in this
|
|
case, which is $40$. Then the two athletes will return to the start for the
|
|
first time at 4:40 P.M.
|
|
|
|
36. It can be shown (see exercises 44-48) that an integer is divisible by 3 if,
|
|
and only if, the sum of its digits is divisible by 3; an integer is
|
|
divisible by 9 if, and only if, the sum of its digits is divisible by 9; an
|
|
integer is divisible by 5 if, and only if, its right-most digit is a 5 or a
|
|
0; and an integer is divisible by 4 if, and only if, the number formed by
|
|
its right-most two digits is divisible by 4. Check the following integers
|
|
for divisibility by 3, 4, 5, and 9.
|
|
|
|
a. 637,425,403,705,125
|
|
|
|
Divisible by 3?
|
|
|
|
$$ 3 \stackrel{?}{\mid} (6 + 3 + 7 + 4 + 2 + 5 + 4 + 0 + 3 + 7 + 0 + 5 + 1 + 2 + 5) $$
|
|
|
|
$$ 3 \stackrel{?}{\mid} 54 $$
|
|
|
|
Yes, $\dfrac{54}{3} = 18$
|
|
|
|
Divisible by 4?
|
|
|
|
$$ 4 \stackrel{?}{\mid} 25 $$
|
|
|
|
No, $\dfrac{25}{4} = 6 + \dfrac{1}{4}$
|
|
|
|
Divisible by 5?
|
|
|
|
Last digit a $5$ or $0$?
|
|
|
|
Yes, last digit is $5$.
|
|
|
|
Divisible by 9?
|
|
|
|
$$ 9 \stackrel{?}{\mid} (6 + 3 + 7 + 4 + 2 + 5 + 4 + 0 + 3 + 7 + 0 + 5 + 1 + 2 + 5) $$
|
|
|
|
$$ 9 \stackrel{?}{\mid} 54 $$
|
|
|
|
Yes, $\dfrac{54}{9} = 6$
|
|
|
|
b. 12,858,306,120,312
|
|
|
|
Divisible by 3?
|
|
|
|
$$ 3 \stackrel{?}{\mid} (1 + 2 + 8 + 5 + 8 + 3 + 0 + 6 + 1 + 2 + 0 + 3 + 1 + 2) $$
|
|
|
|
$$ 3 \stackrel{?}{\mid} 42 $$
|
|
|
|
Yes $\dfrac{42}{3} = 14$
|
|
|
|
Divisible by 4?
|
|
|
|
$$ 4 \stackrel{?}{\mid} 12 $$
|
|
|
|
Yes $\dfrac{12}{4} = 3$
|
|
|
|
Divisible by 5?
|
|
|
|
Last digit a $5$ or $0$?
|
|
|
|
No.
|
|
|
|
Divisible by 9?
|
|
|
|
$$ 9 \stackrel{?}{\mid} (1 + 2 + 8 + 5 + 8 + 3 + 0 + 6 + 1 + 2 + 0 + 3 + 1 + 2) $$
|
|
|
|
$$ 9 \stackrel{?}{\mid} 42 $$
|
|
|
|
No, $\dfrac{42}{9} = 4 + \dfrac{2}{3}$.
|
|
|
|
c. 517,924,440,926,512
|
|
|
|
Divisible by 3?
|
|
|
|
$$ 3 \stackrel{?}{\mid} (5 + 1 + 7 + 9 + 2 + 4 + 4 + 4 + 0 + 9 + 2 + 6 + 5 + 1 + 2) $$
|
|
|
|
$$ 3 \stackrel{?}{\mid} 61 $$
|
|
|
|
No, $\dfrac{61}{3} = 20 + \dfrac{1}{3}$
|
|
|
|
Divisible by 4?
|
|
|
|
$$ 4 \stackrel{?}{\mid} 12 $$
|
|
|
|
Yes $\dfrac{12}{4} = 3$
|
|
|
|
Divisible by 5?
|
|
|
|
Last digit a $5$ or $0$?
|
|
|
|
No.
|
|
|
|
Divisible by 9?
|
|
|
|
$$ 9 \stackrel{?}{\mid} (5 + 1 + 7 + 9 + 2 + 4 + 4 + 4 + 0 + 9 + 2 + 6 + 5 + 1 + 2) $$
|
|
|
|
$$ 9 \stackrel{?}{\mid} 61 $$
|
|
|
|
No, $\dfrac{61}{9} = 6 + \dfrac{7}{9}$
|
|
|
|
d. 14,328,083,360,232
|
|
|
|
Divisible by 3?
|
|
|
|
$$ 3 \stackrel{?}{\mid} (1 + 4 + 3 + 2 + 8 + 0 + 8 + 3 + 3 + 6 + 0 + 2 + 3 + 2) $$
|
|
|
|
$$ 3 \stackrel{?}{\mid} 45 $$
|
|
|
|
Yes $\dfrac{45}{3} = 15$
|
|
|
|
Divisible by 4?
|
|
|
|
$$ 4 \stackrel{?}{\mid} 32 $$
|
|
|
|
Yes $\dfrac{32}{4} = 8$
|
|
|
|
Divisible by 5?
|
|
|
|
Last digit a $5$ or $0$?
|
|
|
|
No.
|
|
|
|
Divisible by 9?
|
|
|
|
$$ 9 \stackrel{?}{\mid} (1 + 4 + 3 + 2 + 8 + 0 + 8 + 3 + 3 + 6 + 0 + 2 + 3 + 2) $$
|
|
|
|
$$ 9 \stackrel{?}{\mid} 45 $$
|
|
|
|
Yes, $\dfrac{45}{9} = 5$
|
|
|
|
37. Use the unique factorization theorem to write the following integers in
|
|
standard factored form.
|
|
|
|
a. 1,176
|
|
|
|
$$ 1176 = 8 \cdot 147 = 8 \cdot 7 \cdot 21 = 2 \cdot 2 \cdot 2 \cdot 7 \cdot 7 \cdot 3 $$
|
|
|
|
$$ \quad = 2^3 \cdot 7^2 \cdot 3 $$
|
|
|
|
b. 5,733
|
|
|
|
$$ 5733 = 49 \cdot 117 = 7 \cdot 7 \cdot 9 \cdot 13 = 7^2 \cdot 3^2 \cdot 13 $$
|
|
|
|
$$ \quad = 7^2 \cdot 3^2 \cdot 13 $$
|
|
|
|
c. 3,675
|
|
|
|
$$ 3675 = 25 \cdot 147 = 5 \cdot 5 \cdot 7 \cdot 21 = 5^2 \cdot 7 \cdot 7 \cdot 3 $$
|
|
|
|
$$ \quad = 5^2 \cdot 7^2 \cdot 3 $$
|
|
|
|
38. Let $n = 8,424$.
|
|
|
|
a. Write the prime factorization for $n$.
|
|
|
|
$$ 8,424 = 24 \cdot 351 = 6 \cdot 4 \cdot 3 \cdot 117 $$
|
|
|
|
$$ \quad = 3 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 39 $$
|
|
|
|
$$ \quad = 3^3 \cdot 2^3 \cdot 39 $$
|
|
|
|
$$ \quad = 3^3 \cdot 2^3 \cdot 3 \cdot 13 $$
|
|
|
|
$$ \quad = 3^4 \cdot 2^3 \cdot 13 $$
|
|
|
|
b. Write the prime factorization for $n^5$.
|
|
|
|
$$ n = 3^4 \cdot 2^3 \cdot 13 $$
|
|
|
|
$$ n^5 = (3^4 \cdot 2^3 \cdot 13)^5 $$
|
|
|
|
0$ n^5 = 3^{20} \cdot 2^{15} \cdot 13^5 $$
|
|
|
|
c. Is $n^5$ divisible by 20? Explain.
|
|
|
|
$$ 20 = 2^2 \cdot 5 $$
|
|
|
|
So in order for $20 \mod n^5$, then $2^2 \mid n^5$ and $5 \mid n^5$,
|
|
$2^2 \mid n^5$ is possible since one of the prime factorizations of $n^5$ is
|
|
$2^{15}$ but none of the prime factorizations of $n^5$ is $5$. Therefore
|
|
$20 \nmid n^5$.
|
|
|
|
d. What is the least positive integer $m$ so that $8,424 \cdot m$ is a perfect
|
|
square?
|
|
|
|
To make $8424 \cdot m$ square, all prime exponents must be even. Let's examine
|
|
our prime factorization form of $n$:
|
|
|
|
$$ n = 3^4 \cdot 2^3 \cdot 13 $$
|
|
|
|
Only our last two terms need an additional factor each to make their exponents
|
|
even, so to make $8424m$ a perfect square, $m = 2 \cdot 13 = 26$, this will
|
|
make:
|
|
|
|
$$ 8424m = 3^4 \cdot 2^4 \cdot 13^2 $$
|
|
|
|
39. Suppose that in standard factored form
|
|
$a = p_1^{e_1}p_2^{e_2} \dots p_k^{e_k}$, where $k$ is a positive integer;
|
|
$p_1, p_2, \dots, p_k$ are prime numbers; and $e_1, e_2, \dots, e_k$ are
|
|
positive integers.
|
|
|
|
a. What is the standard factored form for $a^3$?
|
|
|
|
$$ a^3 = p_1^{3e_1}p_2^{3e_2} \dots p_k^{3e_k} $$
|
|
|
|
b. Find the least positive integer $k$ such that
|
|
$2^4 \cdot 3^5 \cdot 7 \cdot 11^2 \cdot k$ is a perfect cube (that is, it equals
|
|
an integer to the third power). Write the resulting product as a perfect cube.
|
|
|
|
$$ k = 2^2 \cdot 3 \cdot 7^2 \cdot 11 $$
|
|
|
|
40.
|
|
|
|
a. If $a$ and $b$ are integers and $12a = 25b$, does $12 \mid b$? does
|
|
$25 \mid a$? Explain.
|
|
|
|
Because $12a = 25b$, the unique factorization theorem guarantees that the
|
|
standard factored forms of $12a$ and $25b$ must be the same. Thus $25b$ contains
|
|
the factors $2^2 \cdot 3 (= 12)$. But since neither $2$ nor 3$ divides $25$, the
|
|
factors $2^2 \cdot 3$ must all occur in $b$, and hence $12 \mid b$. Similarly,
|
|
$12a$ contains the factors $5^2 = 25$, and since $5$ is not a factor of $124,
|
|
the factors $5^2$ must occur in $a$. So $25 \mid a$.
|
|
|
|
b. If $x$ and $y$ are integers and $10x = 9y$, does $10 \mid y$? does
|
|
$9 \mid x$? Explain.
|
|
|
|
Because $10x = 9y$, the unique factorization theorem guarantees that the
|
|
standard factored forms of $10x$ and $9y$ must be the same. Thus $10x$ contains
|
|
the factors $3^2$. But since $3$ does not divide $10$, the factor $3^2$ must all
|
|
occur in $x$, and hence $9 \mid x$. Similary $10x$ contains factors $2 \cdot 5$,
|
|
and since neither $5$ nor $2$ are factors of $9$, the factors $2 \cdot 5$ must
|
|
occur in $y$. So $10 \mid y$.
|
|
|
|
41. How many zeros are at the end of $45^8 \cdot 88^5$? Explain how you can
|
|
answer this question without actually computing the number. (_Hint:_
|
|
$10 = 2 \cdot 5$.)
|
|
|
|
If we find the standard factored form of $45^8 \cdot 88^5$ and then find the
|
|
total amount of $10$'s, this should tell us how many zeros are at the end of
|
|
$45^8 \cdot 88^5$.
|
|
|
|
$$ 45^8 \cdot 88^5 = (9 \cdot 5)^8 \cdot (4 \cdot 22)^5 $$
|
|
|
|
$$ \quad = (3^2 \cdot 5)^8 \cdot (2^2 \cdot 2 \cdot 11)^5 $$
|
|
|
|
$$ \quad = (3^2 \cdot 5)^8 \cdot (2^3 \cdot 11)^5 $$
|
|
|
|
$$ \quad = 3^{16} \cdot 5^8 \cdot 2^{15} \cdot 11^5 $$
|
|
|
|
The total amount of $10$'s we can find is the maximum amount of $5 \cdot 2$'s we
|
|
can factor from this expression, which is $8$ (while we have 15 $2$'s, we only
|
|
have 8 $5$'s).
|
|
|
|
Therefore there are $8$ zeros in $45^8 \cdot 88^5$.
|
|
|
|
42. If $n$ is an integer and $n > 1$, then $n!$ is the product of $n$ and every
|
|
other positive integer that is less than $n$. For example,
|
|
$5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$.
|
|
|
|
a. Write $6!$ in standard factored form.
|
|
|
|
$$ 6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 $$
|
|
|
|
$$ 6! = 3 \cdot 2 \cdot 5 \cdot 2 \cdot 2 \cdot 3 \cdot 2 $$
|
|
|
|
$$ 6! = 2^4 \cdot 3^2 \cdot 5 $$
|
|
|
|
b. Write $20!$ in standard factored form.
|
|
|
|
Writing this out would be exhaustive, instead let's use
|
|
[Legendre's formula](https://en.wikipedia.org/wiki/Legendre%27s_formula).
|
|
|
|
$$ 20! = 2^a \cdot 3^b \cdot 5^c \cdot 7^d \cdot 11^e \cdot 13^f \cdot 17^g \cdot 19^h $$
|
|
|
|
Take the floor of every possible prime factor, and those are the exponents of
|
|
each.
|
|
|
|
$$ \left\lfloor \frac{20}{2} \right\rfloor + \left\lfloor \frac{20}{4} \right\rfloor + \left\lfloor \frac{20}{8} \right\rfloor + \left\lfloor \frac{20}{16} \right\rfloor $$
|
|
|
|
$$ \quad = 10 + 5 + 2 + 1 = 18 $$
|
|
|
|
$$ 2^{18} $$
|
|
|
|
$$ \left\lfloor \frac{20}{3} \right\rfloor + \left\lfloor \frac{20}{9} \right\rfloor $$
|
|
|
|
$$ 6 + 2 = 8 $$
|
|
|
|
$$ 3^8 $$
|
|
|
|
$$ \left\lfloor \frac{20}{5} \right\rfloor $$
|
|
|
|
$$ 5^4 $$
|
|
|
|
$$ \left\lfloor \frac{20}{7} \right\rfloor $$
|
|
|
|
$$ 7^2 $$
|
|
|
|
$$ \boxed{20! = 2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19} $$
|
|
|
|
c. Without computing the value of $(20!)^2$ determine how many zeros are at the
|
|
end of this number when it is written in decimal form. Justify your answer.
|
|
|
|
We can square the answer for $20!$ from part b:
|
|
|
|
$$ (20!)^2 = (2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19)^2 $$
|
|
|
|
$$ \quad = 2^{36} \cdot 3^{16} \cdot 5^8 \cdot 7^4 \cdot 11^2 \cdot 13^2 \cdot 17^2 \cdot 19^2 $$
|
|
|
|
Then take all combinations of 2 and 5, and take the minimum exponent between
|
|
them as this will tell us how many zeros are at the end of this number when it
|
|
is written in decimal form.
|
|
|
|
$$ \min(36, 8) = 8 $$
|
|
|
|
So there are 8 zeros at the end of $(20!)^2$.
|
|
|
|
43. At a certain university 2/3 of the mathematics students and 3/5 of the
|
|
computer science students have taken a discrete mathematics course. The
|
|
number of mathematics students who have taken the course equals the number
|
|
of computer science students who have taken the course. If there are at
|
|
least 100 mathematics students at the university, what are the least
|
|
possible number of mathematics students and the least possible number of
|
|
computer science students at the university?
|
|
|
|
Let $M$ be the number of mathematics students and $C$ be the number of computer
|
|
science students.
|
|
|
|
Given:
|
|
|
|
$$ \frac{2}{3}M = \frac{3}{5}C $$
|
|
|
|
Set them equal:
|
|
|
|
$$ \frac{2}{3}M = \frac{3}{5}C = x $$
|
|
|
|
So:
|
|
|
|
$$ M = \frac{3}{2}x $$
|
|
|
|
$$ C = \frac{5}{3}x $$
|
|
|
|
To make both integers, $x$ must be a multiple of $6$:
|
|
|
|
$$ x = 6k $$
|
|
|
|
Then:
|
|
|
|
$$ M = 9k $$
|
|
|
|
$$ C = 10k $$
|
|
|
|
Now use the given condition:
|
|
|
|
$$ M \geq 100 $$
|
|
|
|
$$ 9k \geq 100 \Rightarrow k \geq 12 $$
|
|
|
|
Smallest $k = 12$:
|
|
|
|
$$ M = 9 \cdot 12 = 108 $$
|
|
|
|
$$ C = 10 \cdot 12 = 120 $$
|
|
|
|
Math Students: 108
|
|
|
|
Computer Science Students: 120
|
|
|
|
**Definition:** Given any nonnegative integer $n$, the **decimal
|
|
representation** of $n$ is an expression of the form
|
|
|
|
$$ d_kd_{k + 1} \dots d_2d_1d_0 $$
|
|
|
|
where $kr is a nonnegative integer, $d_0, d_1, d_2, \dots, d_k$ (called the
|
|
**decimal digits** of $n$) are integers from $0$ to $9$ inclusive, $dk \neq 0$
|
|
unless $n = 0$ and $k = 0$, and
|
|
|
|
$$ n = d_k \cdot 10^k + d_{k + 1} \cdot 10^{k + 1} + \dots + d_2 \cdot 10^2 + d_1 \cdot 10 + d_0 $$
|
|
|
|
(For example, $2,503 = 2 \cdot 10^3 + 5 \cdot 10^2 + 0 \cdot 10 + 3$.)
|
|
|
|
44. Prove that if $n$ is any nonnegative integer whose decimal representation
|
|
ends in $0$, then $5 \mid n$. (_Hint:_ If the decimal representation of a
|
|
nonnegative integer $n$ ends in $d_0$, then $n = 10m + d_0$ for some integer
|
|
$m$.)
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any nonnegative integer whose decimal representation ends in $0$.
|
|
|
|
Because $n$ is a nonnegative integer, it can be shown in decimal representation
|
|
as:
|
|
|
|
$$ n = 10m + d_0 $$
|
|
|
|
Where $m$ is some integer.
|
|
|
|
Since we know that $n$'s decimal representation ends in $0$, that means that
|
|
$d_0 = 0$. So $n$ is:
|
|
|
|
$$ n = 10m $$
|
|
|
|
Then:
|
|
|
|
$$ n = 5(2m) $$
|
|
|
|
Then $2m$ is an integer by the product of integers. By the definition of
|
|
divisibility, $5 \mid n$.
|
|
|
|
Q.E.D.
|
|
|
|
45. Prove that if $n$ is any nonnegative integer whose decimal representation
|
|
ends in $5$, then $5 \mid n$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any nonnegative integer whose decimal representation ends in $5$.
|
|
|
|
Because $n$ is a nonnegative integer, it can be shown in decimal representation
|
|
as:
|
|
|
|
$$ n = 10m + d_0 $$
|
|
|
|
Where $m$ is some integer.
|
|
|
|
Since we know that $n$'s decimal representation ends in $5$, that means that
|
|
$d_0 = 5$. So $n$ is:
|
|
|
|
$$ n = 10m + 5 $$
|
|
|
|
Then:
|
|
|
|
$$ n = 5(2m + 1) $$
|
|
|
|
Then $2m + 1$ is an integer by the product and sum of integers. By the
|
|
definition of divisibility, $5 \mid n$.
|
|
|
|
Q.E.D.
|
|
|
|
46. Prove that if the decimal representation of a nonnegative integer $n$ ends
|
|
in $d_1d_0$ and if $4 \mid (10d_1 + d_0)$, then $4 \mid n$. (_Hint:_ If the
|
|
decimal representation of a nonnegative integer $n$ ends in $d_1d_0$, then
|
|
there is an integer $s$ such that $n = 100s + 10d_1 + d_0$.)
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any nonnegative integer whose decimal representation ends in
|
|
$d_1d_0$ and $4 \mid (10d_1 + d_0)$.
|
|
|
|
Since $4 \mid (10d_1 + d_0)$, then $10d_1 + d_0 = 4k$ for some integer $k$.
|
|
|
|
And since the decimal representation of $n$ ends in $d_1d_0$, then there is an
|
|
integer $s$ such that $n = 100s + 10d_1 + d_0$.
|
|
|
|
$$ n = 100s + 10d_1 + d_0 $$
|
|
|
|
Then, by substitution:
|
|
|
|
$$ n = 100s + 4k $$
|
|
|
|
$$ n = 4(25s + k) $$
|
|
|
|
Where $25s + k$ is an integer by the product and sum of integers. Thus
|
|
$4 \mid n$.
|
|
|
|
Q.E.D.
|
|
|
|
47. Observe that
|
|
|
|
$$
|
|
7,524 = 7 \cdot 1,000 + 5 \cdot 100 + 2 \cdot 10 + 4 \\
|
|
\quad = 7(999 + 1) + 5(99 + 1) + 2(9 + 1) + 4 \\
|
|
\quad = (7 \cdot 99 + 7) + (5 \cdot 99 + 5) + (2 \cdot 9 + 2) + 4 \\
|
|
\quad = (7 \cdot 999 + 5 \cdot 99 2 \cdot 9) + (7 + 5 + 2 + 4) \\
|
|
\quad = (7 \cdot 111 \cdot 9 + 5 \cdot 11 \cdot 9 + 2 \cdot 9) + (7 + 5 + 2 + 4) \\
|
|
\quad = (7 \cdot 111 + 5 \cdot 11 + 2) \cdot 9 + (7 + 5 + 2 + 4) \\
|
|
\quad = (\text{an integer divisible by 9})i + (\text{the sum of the digits of } 7,524)
|
|
$$
|
|
|
|
Since the sum of the digits of $7,524$ is divisible by $9$, $7,524$ can be
|
|
written as a sum of two integers each of which is divisible by $9$. It follows
|
|
from exercise 15 that $7,524$ is divisible by $9$.
|
|
|
|
Generalize the argument given in this example to any nonnegative integer $n$. In
|
|
other words, prove that for any nonnegative integer $n$, if the sum of the
|
|
digits of $n$ is divisible by $9r, then $n$ is divisible by $9$.
|
|
|
|
Omitted.
|
|
|
|
48. Prove that for any nonnegative integer $n$, if the sum of the digits of $n$
|
|
is divisible by $3$, then $n$ is divisible by $3$.
|
|
|
|
Omitted.
|
|
|
|
49. Given a positive integer $n$ written in decimal form, the alternating sum of
|
|
the digits of $n$ is obtained by starting with the right-most digit,
|
|
subtracting the digit immediately to its left, adding the next digit to the
|
|
left, subtracting the next digit, and so forth. For example, the alternating
|
|
sum of the digits of 180,928 is $8 - 2 + 9 - 0 + 8 - 1 = 22$. Justify the
|
|
fact that for any nonnegative integer $n$, if the alternating sum of the
|
|
digits of $n$ is divisible by 11, then $n$ is divisible by 11.
|
|
|
|
Omitted.
|
|
|
|
50. The integer 123,123 has the form _abc,abc_, where _a_, _b_, and _c_ are
|
|
integers from $0$ through $9$. Consider all six-digit integers of this form.
|
|
Which prime numbers divide every one of these integers? Prove your answer.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
**Exercise Set 4.5**
|
|
|
|
Page 232
|
|
|
|
For each of the values of $n$ and $d$ given in 1-6, find integers $q$ and $r$
|
|
such that $n = dq + r$ and $0 \leq r < d$.
|
|
|
|
1. $n = 70$, $d = 9$
|
|
|
|
$$ d \mid n $$
|
|
|
|
$$ d \mod r $$
|
|
|
|
$$ d \mid n = 7 $$
|
|
|
|
$$ n = dq + r $$
|
|
|
|
$$ n = 9(7) + 7 $$
|
|
|
|
$$ q = 7, r = 7 $$
|
|
|
|
2. $n = 62$, $d = 7$
|
|
|
|
$$ n = dq + r $$
|
|
|
|
$$ 62 = (7)(8) + (6) $$
|
|
|
|
$$ q = 8, r = 6 $$
|
|
|
|
3. $n =36$, $d = 40$
|
|
|
|
$$ 36 = (40)(0) + (36) $$
|
|
|
|
$$ q = 0, r = 36 $$
|
|
|
|
4. $n = 3$, $d = 11$
|
|
|
|
$$ 3 = (11)(0) + (3) $$
|
|
|
|
$$ q = 0, r = 3 $$
|
|
|
|
5. $n = -45$, $d = 11$
|
|
|
|
$$ -45 = (11)(-5) + 10 $$
|
|
|
|
Note that $r$ must be nonnegative, hence why negatives look different here.
|
|
|
|
$$ q = -5, r = 10 $$
|
|
|
|
6. $n = -27$, $d = 8$
|
|
|
|
$$ -27 = (8)(-4) + 5 $$
|
|
|
|
$$ q = -4, r = 5 $$
|
|
|
|
**Evaluate the expressions in 7-10.**
|
|
|
|
7.
|
|
|
|
a. $43\ div\ 9$
|
|
|
|
$$ 43\ div\ 9 = 4 $$
|
|
|
|
b. $43 \mod 9$
|
|
|
|
$$ 43 \mod 9 = 7 $$
|
|
|
|
8.
|
|
|
|
a. $50\ div\ 7$
|
|
|
|
$$ 50\ \div\ 7 = 7 $$
|
|
|
|
b. $50 \mod 7$
|
|
|
|
$$ 50 \mod 7 = 1 $$
|
|
|
|
9.
|
|
|
|
a. $28\ div\ 5$
|
|
|
|
$$ 28\ div\ 5 = 5 $$
|
|
|
|
b. $28 \mod 5$
|
|
|
|
$$ 28 \mod 5 = 3 $$
|
|
|
|
10.
|
|
|
|
a. $30\ div\ 2$
|
|
|
|
$$ 30\ div\ 2 = 15 $$
|
|
|
|
b. $30 \mod 2$
|
|
|
|
$$ 30 \mod 2 = 0 $$
|
|
|
|
11. Check the correctness of formula (4.5.1) given in Example 4.5.3 for the
|
|
following values of $\text{Day}T$ and $N$.
|
|
|
|
4.5.1
|
|
|
|
$$ n = dq + r \text{ and } 0 \leq r < d $$
|
|
|
|
a. $\text{Day}T = 6(\text{Saturday}) \text{ and } N = 15$
|
|
|
|
$$ \text{Day}N = (\text{Day}T + N) \mod 7 $$
|
|
|
|
$$ \quad = (6 + 15) \mod 7 $$
|
|
|
|
$$ \quad = 21 \mod 7 = 0 $$
|
|
|
|
b. $\text{Day}T = 0(\text{Sunday}) \text{ and } N = 7$
|
|
|
|
$$ \text{Day}N = (\text{Day}T + N) \mod 7 $$
|
|
|
|
$$ \quad = (0 + 7) \mod 7 $$
|
|
|
|
$$ \quad = 7 \mod 7 = 0 $$
|
|
|
|
c. $\text{Day}T = 4(\text{Thursday}) \text{ and } N = 12$
|
|
|
|
$$ \text{Day}N = (\text{Day}T + N) \mod 7 $$
|
|
|
|
$$ \quad = (4 + 12) \mod 7 $$
|
|
|
|
$$ \quad = 16 \mod 7 = 2 $$
|
|
|
|
12. Justify formula (4.5.1) for general values of $\text{Day}T$ and $N$.
|
|
|
|
Omitted
|
|
|
|
13. On a Monday a friend says he will meet you again in 30 days. What day of the
|
|
week will that be?
|
|
|
|
$$ \text{Day}N = (\text{Day}T + N) \mod 7 $$
|
|
|
|
$$ \quad = (1 + 30) \mod 7 $$
|
|
|
|
$$ \quad = 31 \mod 7 = 3 $$
|
|
|
|
0 + 3 days = Wednesday
|
|
|
|
14. If today is Tuesday, what day of the week will it be 1,000 days from today?
|
|
|
|
$$ \text{Day}N = (\text{Day}T + N) \mod 7 $$
|
|
|
|
$$ \text{Day}N = (2 + 1000) \mod 7 $$
|
|
|
|
$$ \text{Day}N = 1002 \mod 7 = 1 $$
|
|
|
|
Monday.
|
|
|
|
15. January 1, 2000, was a Saturday, and 2000 was a leap year. What day of the
|
|
week will January 1, 2050, be?
|
|
|
|
Leap years occur roughly every 4 years. In the 50 year time span that means
|
|
$50 \div 4 = 12.5$. Taking the ceiling of this we have 13 additional days.
|
|
|
|
The amount of days for 50 years is $50 \cdot 365 = 18250$, so
|
|
$N = 18250 + 13 = 18263$.
|
|
|
|
Now we can plug this in:
|
|
|
|
$$ \text{Day}N = (\text{Day}T + N) \mod 7 $$
|
|
|
|
$$ \quad = (6 + 18263) \mod 7 $$
|
|
|
|
$$ \quad = 18269 \mod 7 = 6 $$
|
|
|
|
So January 1, 2050 will be a Saturday.
|
|
|
|
16. Suppose $d$ is a positive and $n$ is any integer. If $d \mid n$, what is the
|
|
remainder obtained when the quotient remainder theorem is applied to $n$
|
|
with divisor $d$?
|
|
|
|
$r = 0$
|
|
|
|
Because $d \mid n$, $n = dq + 0$ for some integer $q$. Thus the remainder $r$,
|
|
is $0$.
|
|
|
|
17. Prove directly from the definitions that for every integer $n$,
|
|
$n^2 - n + 3$ is odd. Use division into two cases: $n$ is even and $n$ is
|
|
odd.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
_Case $n$ is even:_
|
|
|
|
Since $n$ is even, $n = 2k$ for some integer $k$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ n^2 - n + 3 = (2k)^2 - (2k) + 3 $$
|
|
|
|
$$ \quad = 4k^2 - 2k + 3 $$
|
|
|
|
$$ \quad = 4k^2 - 2k + 2 + 1 $$
|
|
|
|
$$ \quad = 2(2k^2 - k + 1) + 1 $$
|
|
|
|
Let $m = 2k^2 - k + 1$, where $m$ is an integer by the product and sum of
|
|
integers. Therefore $n^2 - n + 3$ is odd by the definition of odd integers.
|
|
|
|
_Case $n$ is odd:_
|
|
|
|
Since $n$ is odd, $n = 2s + 1$ for some integer $s$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ n^2 - n + 3 = (2s + 1)^2 - (2s + 1) + 3 $$
|
|
|
|
$$ n^2 - n + 3 = (2s + 1)(2s + 1) - 2s - 1 + 3 $$
|
|
|
|
$$ n^2 - n + 3 = (4s^2 + 4s + 1) - 2s - 1 + 3 $$
|
|
|
|
$$ n^2 - n + 3 = 4s^2 + 2s + 3 $$
|
|
|
|
$$ n^2 - n + 3 = 4s^2 + 2s + 2 + 1 $$
|
|
|
|
$$ n^2 - n + 3 = 2(2s^2 + s + 1) + 1 $$
|
|
|
|
Let $p = 2s^2 + s + 1$ where $p$ is an integer by the product and sum of
|
|
integers. Therefore $n^2 - n + 3$ is odd by the definition of odd integers.
|
|
|
|
In both cases $n^2 - n + 3$ is odd.
|
|
|
|
Q.E.D.
|
|
|
|
18.
|
|
|
|
a. Prove that the product of any two consecutive integers is even.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
Then the product of $n$ and its consecutive integer is:
|
|
|
|
$$ n(n + 1) = n^2 + n $$
|
|
|
|
_Case where $n$ is even:_
|
|
|
|
Since $n$ is even, $n = 2k$ for some integer $k$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ n^2 + n = (2k)^2 + (2k) $$
|
|
|
|
$$ \quad = 4k^2 + 2k $$
|
|
|
|
$$ \quad = 2(2k^2 + k) $$
|
|
|
|
Let $m = 2k^2 + k$ where $m$ is an integer by the product and sum of integers.
|
|
Therefore $n^2 + n$ is even by the definition of even integers.
|
|
|
|
_Case where $n$ is odd:_
|
|
|
|
Since $n$ is odd, $n = 2s + 1$, for some integer $s$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ n^2 + n = (2s + 1)^2 + (2s + 1) $$
|
|
|
|
$$ n^2 + n = (2s + 1)(2s + 1) + 2s + 1 $$
|
|
|
|
$$ n^2 + n = 4s^2 + 4s + 1 + 2s + 1 $$
|
|
|
|
$$ n^2 + n = 4s^2 + 6s + 2 $$
|
|
|
|
$$ n^2 + n = 2(2s^2 + 3s + 1) $$
|
|
|
|
Let $p = 2s^2 + 3s + 1$, where $p$ is an integer by the product and sum of
|
|
integers. Therefore $n^2 + n$ is even by the definition of even integers.
|
|
|
|
In both cases $n^2 + n$ is even, therefore the product of any two consecutive
|
|
integers is even.
|
|
|
|
Q.E.D.
|
|
|
|
b. The result of part (a) suggests that the second approach in the discussion of
|
|
Example 4.5.7 might be possible after all. Write a new proof of Theorem 4.5.3
|
|
based on this observation.
|
|
|
|
4.5.3 Demonstrates this proof:
|
|
|
|
Prove: The square of any odd integer has the form $8m + 1$ for some integer $m$.
|
|
|
|
But suggests another approach might be possible.
|
|
|
|
"You could try another approach by arguing that since $n$ is odd, you can
|
|
represent it as $2q + 1$ for some integer $q$. Then
|
|
$n^2 = (2q + 1)^2 = 4q^2 + 4q + 1 = 4(q^2 + q) + 1$". It is clear from this
|
|
analysis that $n^2$ can be rewritten in the form $4m + 1$, but it may not be
|
|
clear that it can be written as $8m + 1$.
|
|
|
|
Given part a, we can now prove this. Let's do that now:
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any odd integer.
|
|
|
|
Since $n$ is odd, $n = 2q + 1$ for some integer $q$.
|
|
|
|
Then:
|
|
|
|
$$ n^2 = (2q + 1)^2 $$
|
|
|
|
$$ n^2 = (2q + 1)(2q + 1) $$
|
|
|
|
$$ n^2 = 4q^2 + 4q + 1 $$
|
|
|
|
$$ n^2 = 4(q^2 + q) + 1 $$
|
|
|
|
By part a, we know that $q^2 + q$ is even. So $q^2 + q = 2m$, for some integer
|
|
$m$.
|
|
|
|
$$ n^2 = 4(2m) + 1 $$
|
|
|
|
$$ n^2 = 8m + 1 $$
|
|
|
|
Therefore $n^2$ has the form $8m + 1$ for some integer $m$.
|
|
|
|
Q.E.D.
|
|
|
|
19. Prove directly from the definitions that for all integers $m$ and $n$, if
|
|
$m$ and $n$ have the same parity, then $5m + 7n$ is even. Divide into two
|
|
cases: $m$ and $n$ are both even and $m$ and $n$ are both odd.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $m$ and $n$ are any two integers with the same parity.
|
|
|
|
_Case $m$ and $n$ are even_:
|
|
|
|
Since $m$ and $n$ are even, $m = 2k$ and $n = 2p$ for some integers $k$ and $p$.
|
|
|
|
Then:
|
|
|
|
$$ 5m + 7n = 5(2k) + 7(2p) $$
|
|
|
|
$$ \quad = 10k + 14p $$
|
|
|
|
$$ \quad = 2(5k + 7p) $$
|
|
|
|
Let $s = 5k + 7p$ where $s$ is an integer by the product and sum of integers.
|
|
|
|
Therefore $5m + 7n$ is even by the definition of even integers.
|
|
|
|
_Case $m$ and $n$ are odd_:
|
|
|
|
Since $m$ and $n$ are odd, $m = 2t + 1$ and $n = 2v + 1$ for some integers $t$
|
|
and $v$.
|
|
|
|
Then:
|
|
|
|
$$ 5m + 7n = 5(2t + 1) + 7(2v + 1) $$
|
|
|
|
$$ 5m + 7n = 10t + 5 + 14v + 7 $$
|
|
|
|
$$ 5m + 7n = 10t + 14v + 12 $$
|
|
|
|
$$ 5m + 7n = 2(5t + 7v + 6) $$
|
|
|
|
Let $w = 5t + 7v + 6$ where $w$ is an integer by the product and sum of
|
|
integers.
|
|
|
|
Therefore $5m + 7n$ is even by the definition of even integers.
|
|
|
|
In both cases, $5m + 7n$ is even. Therefore for all integers $m$ and $n$, if $m$
|
|
and $n$ have the same parity, then $5m + 7n$ is even.
|
|
|
|
Q.E.D.
|
|
|
|
20. Suppose $a$ is any integer. If $a \mod 7 = 4$, what is $5a \mod 7$? In other
|
|
words, if division of $a$ by $7$ gives a remainder of $4$, what is the
|
|
remainder when $5a$ is divided by $7$? Your solution should show that you
|
|
obtain the same answer no matter what integer you start with.
|
|
|
|
$$ a \mod 7 = 4 $$
|
|
|
|
$$ 5a \mod 7 = ? $$
|
|
|
|
Since $a \mod 7 = 4$, this means that the remainder obtained when $a$ is divided
|
|
by $7$ is $4$. This means there is some integer $q$ so that
|
|
|
|
$$ a = 7q + 4 $$
|
|
|
|
Thus
|
|
|
|
$$ 5a = 5(7q + 4) = 35q + 20 $$
|
|
|
|
And when put into the form defined by the quotient-remainder theorem:
|
|
|
|
$n = dq + r$, recall our original divisor was $7$, so:
|
|
|
|
$$ \quad = 7(5q + 2) + 6 $$
|
|
|
|
So,
|
|
|
|
$$ 5a \mod 7 = 6 $$
|
|
|
|
21. Suppose $b$ is any integer. If $b \mod 12 = 5$, what is $8b \mod 12$? In
|
|
other words, if division of $b$ by $12$ gives a remainder of $5$, what is
|
|
the remainder when $8b$ is divided by $12$? Your solution should show that
|
|
you obtain the same answer no matter what integer you start with.
|
|
|
|
$$ b \mod 12 = 5 $$
|
|
|
|
$$ 8b \mod 12 = ? $$
|
|
|
|
$$ b = 12d + 5 $$
|
|
|
|
$$ 8b = 8(12d + 5) $$
|
|
|
|
$$ \quad = 96d + 40 $$
|
|
|
|
$$ \quad = 12(8d + 3) + 4 $$
|
|
|
|
$$ 8b \mod 12 = 4 $$
|
|
|
|
22. Suppose $c$ is any integer. If $c \mod 15 = 3$, what is $10c \mod 15$? In
|
|
other words, if division of $c$ by $15$ gives a remainder of $3$, what is
|
|
the remainder when $10c$ is divided by $15$? Your solution should show that
|
|
you obtain the same answer no matter what integer you start with.
|
|
|
|
$$ c \mod 15 = 3 $$
|
|
|
|
$$ 10c \mod 15 = ? $$
|
|
|
|
$$ c = 15d + 3 $$
|
|
|
|
$$ 10c = 10(15d + 3) $$
|
|
|
|
$$ 10c = 150d + 30 $$
|
|
|
|
$$ 10c = 15(10d + 2) + 0 $$
|
|
|
|
$$ 10c \mod 15 = 0 $$
|
|
|
|
23. Prove that for every integer $n$, if $n \mod 5 = 3$ then $n^2 \mod 5 = 4$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer where $n \mod 5 = 3$.
|
|
|
|
Since $n \mod 5 = 3$, $n = 5d + 3$ for some integer $d$.
|
|
|
|
Then:
|
|
|
|
$$ n^2 = (5d + 3)^2 $$
|
|
|
|
$$ n^2 = (5d + 3)(5d + 3) $$
|
|
|
|
$$ n^2 = 25d^2 + 30d + 9 $$
|
|
|
|
$$ n^2 = 5(5d^2 + 6d + 1) + 4 $$
|
|
|
|
Let $s = 5d^2 + 6d + 1$ where $s$ is an integer by the product and sum of
|
|
integers. Then the remainder is $4$ by the quotient remainder theorem.
|
|
|
|
Therefore $n^2 \mod 5 = 4$.
|
|
|
|
Q.E.D.
|
|
|
|
24. Prove that for all integers $m$ and $n$, if $m \mod 5 = 2$ and
|
|
$n \mod 5 = 1$ then $mn \mod 5 = 2$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $m$ and $n$ are any integers where $m \mod 5 = 2$ and $n \mod 5 = 1$.
|
|
|
|
Since $m \mod 5 = 2$ and $n \mod 5 = 1$, $m = 5d + 2$ and $n = 5q + 1$ for some
|
|
integers $d$ and $q$.
|
|
|
|
Then:
|
|
|
|
$$ mn = (5d + 2)(5q + 1) $$
|
|
|
|
$$ mn = 25dq + 5d + 10q + 2 $$
|
|
|
|
$$ mn = 5(5dq + d + 2q + 0) + 2 $$
|
|
|
|
$$ mn = 5(5dq + d + 2q) + 2 $$
|
|
|
|
Let $u = 5dq + d + 2q$ where $u$ is an integer by the product and sum of
|
|
integers. Then $mn$ has a remainder of $2$ when divided by $5$ by the
|
|
quotient-remainder theorem.
|
|
|
|
Therefore $mn \mod 5 = 2$.
|
|
|
|
Q.E.D.
|
|
|
|
25. Prove that for all integers $a$ and $b$, if $a \mod 7 = 5$ and
|
|
$b \mod 7 = 6$ then $ab \mod 7 = 2$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$ and $b$ are any integers where $a \mod 7 = 5$ and $b \mod 7 = 6$.
|
|
|
|
Since $a \mod 7 = 5$ and $b \mod 7 = 6$, $a = 7d + 5$ and $b = 7q + 6$ for some
|
|
integers $d$ and $q$.
|
|
|
|
Then:
|
|
|
|
$$ ab = (7d + 5)(7q + 6) $$
|
|
|
|
$$ \quad = 49dq + 35q + 42d + 30 $$
|
|
|
|
$$ \quad = 7(7dq + 5q + 6d + 4) + 2 $$
|
|
|
|
Let $u = 7dq + 5q + 6d + 4$ where $u$ is an integer by the product and sum of
|
|
integers. Then by the quotient-remainder theorem, $ab$ when divided by $7$ has a
|
|
remainder of $2$.
|
|
|
|
Therefore $ab \mod 7 = 2$.
|
|
|
|
Q.E.D.
|
|
|
|
26. Prove that a necessary and sufficient condition for an integer $n$ to be
|
|
divisible by a positive integer $d$ is that $n \mod d = 0$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer and $d$ is a positive integer where $d \mid n$.
|
|
|
|
Since $d \mid n$ and $d \neq 0$, $n = dq$ for some integer $q$ by the definition
|
|
of divisibility.
|
|
|
|
Then:
|
|
|
|
$$ n = dq $$
|
|
|
|
$$ \quad = dq + 0 $$
|
|
|
|
Then, $n$ when divided by $d$ has a remainder of $0$ by the quotient-remainder
|
|
theorem.
|
|
|
|
Thus $n \mod d = 0$.
|
|
|
|
Suppose then that $n \mod d = 0$ where $n$ is any integer and $d$ is a positive
|
|
integer.
|
|
|
|
Since $n \mod d = 0$ and $d \neq 0$, then $n = dq + 0$ for some integer $q$ by
|
|
the quotient-remainder theorem.
|
|
|
|
Then:
|
|
|
|
$$ n = dq + 0 $$
|
|
|
|
$$ n = dq $$
|
|
|
|
Thus $d \mid n$ by the definition of divisibility.
|
|
|
|
Therefore it has been shown that a necessary and sufficient condition for an
|
|
integer $n$ to be divisible by a positive integer $d$ is that $n \mod d = 0$.
|
|
|
|
Q.E.D.
|
|
|
|
27. Use the quotient-remainder theorem with divisor equal to $2$ to prove that
|
|
the square of any integer can be written in one of the two forms $4k$ or
|
|
$4k + 1$ for some integer $k$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
_Case 1: $n$ is even:_
|
|
|
|
Since $n$ is even, $n = 2q$ for some integer $q$.
|
|
|
|
Then:
|
|
|
|
$$ n^2 = (2q)^2 $$
|
|
|
|
$$ \quad = 4q^2 $$
|
|
|
|
Let $k = 2q^2$ where $k$ is an integer by the product of integers.
|
|
|
|
Then $n^2$ can be written in the form of $4k$.
|
|
|
|
_Case 2: $n$ is odd:_
|
|
|
|
Since $n$ is odd, $n = 2q + 1$ for some integer $q$.
|
|
|
|
Then:
|
|
|
|
$$ n^2 = (2q + 1)^2 $$
|
|
|
|
$$ \quad = (2q + 1)(2q + 1) $$
|
|
|
|
$$ \quad = 4q^2 + 4q + 1 $$
|
|
|
|
$$ \quad = 4(q^2 + q) + 1 $$
|
|
|
|
Let $k = q^2 + q$ where $k$ is an integer by the product and sum of integers.
|
|
|
|
Then $n^2$ can be written in the form of $4k + 1$.
|
|
|
|
Therefore by Case 1, $n^2$ can be written in the form $4k$, and by Case 2 $n^2$
|
|
can be written in the form $4k + 1$.
|
|
|
|
Q.E.D.
|
|
|
|
28.
|
|
|
|
a. Prove: Given any set of three consecutive integers, one of the integers is a
|
|
multiple of $3$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
_Case 1: $n$ is a multiple of $3$:_
|
|
|
|
Since $n$ is a multiple of $3$, $n = 3d$ where $d$ is some integer.
|
|
|
|
Then $n + 1 = 3d + 1$ and $(n + 1) \mod 3 = 1$ by the quotient-remainder
|
|
theorem.
|
|
|
|
Then $n + 2 = 3d + 2$ and $(n + 2) \mod 3 = 2$A by the quotient -remainder
|
|
theorem.
|
|
|
|
Therefore $n$ is a multiple of $3$ but $n + 1$ and $n + 2$ are not.
|
|
|
|
_Case 2: $n + 1$ is a multiple of $3$:_
|
|
|
|
Since $n + 1$ is a multiple of $3$, $n + 1 = 3d$ where $d$ is an integer such
|
|
that $d > 0$ by the definition of divisibility.
|
|
|
|
Then $n = 3d - 1 = 2d + 3 - 1 = 2d + 2$ and $n \mod 3 = 2$ by the
|
|
quotient-remainder theorem.
|
|
|
|
Then $n + 2 = 3d + 1 and $(n + 2) \mod 3 = 1$ by the quotient-remainder theorem.
|
|
|
|
Therefore $n + 1$ is a multiple of $3$ but $n$ and $n + 2$ are not.
|
|
|
|
_Case 3: $n + 2$ is a multiple of $3$:_
|
|
|
|
Since $n + 2$ is a multiple of $3$, $n + 2 = 3d$ where $d$ is an integer such
|
|
that $d > 0$ by the definition of divisibility.
|
|
|
|
Then $n + 1 = 3d - 1 = 2d + 3 - 1 = 2d + 2$ and $(n + 1) \mod 3 = 2$ by the
|
|
quotient-remainder theorem.
|
|
|
|
Then $n = 3d - 2 = 2d + 3 - 2 = 2d + 1$ and $n \mod 3 = 1$ by the
|
|
quotient-remainder theorem.
|
|
|
|
Therefore $n + 2$ is a multiple of $3$ but $n$ and $n + 1$ are not.
|
|
|
|
In all three cases, in any given set of three consecutive integers, one of the
|
|
integers is a multiple of 3.
|
|
|
|
Q.E.D.
|
|
|
|
b. Use the result of part (a) to prove that any product of three consecutive
|
|
integers is a multiple of 3.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
By a., either $n$ or $n + 1$ or $n + 2$ is a multiple of $3$.
|
|
|
|
_Case $n$ is a multiple of $3$_:
|
|
|
|
Since $n$ is a multiple of $3$, $n = 3d$ for some integer $d$.
|
|
|
|
Then:
|
|
|
|
$$ n(n + 1)(n + 2) = (3d)(n + 1) (n + 2) $$
|
|
|
|
$$ \quad = 3\left[(d)(n + 1) (n + 2)\right] $$
|
|
|
|
Let $m = \left[(d)(n + 1) (n + 2)\right]$ where $m$ is an integer by the product
|
|
and sum of integers. Then $3 \mid (n)(n + 1)(n + 2)$ by the definition of
|
|
divisibility.
|
|
|
|
Therefore $n(n + 1)(n + 2)$ is a multiple of $3$.
|
|
|
|
_Case $n + 1$ is a multiple of $3$_:
|
|
|
|
Since $n+ 1$ is a multiple of $3$, $n + 1 = 3d$ for some integer $d$.
|
|
|
|
Then:
|
|
|
|
$$ n(n + 1)(n + 2) = n(3d)(n + 2) $$
|
|
|
|
$$ \quad = 3\left[n(d)(n + 2)\right] $$
|
|
|
|
Let $m = \left[n(d)(n + 2)\right]$ where $m$ is an integer by the product and
|
|
sum of integers. Then $3 \mid (n)(n + 1)(n + 2)$ by the definition of
|
|
divisibility.
|
|
|
|
Therefore $n(n + 1)(n + 2)$ is a multiple of $3$.
|
|
|
|
_Case $n + 2$ is a multiple of $3$_:
|
|
|
|
Since $n + 2$ is a multiple of $3$, $n + 2 = 3d$ for some integer $d$.
|
|
|
|
Then:
|
|
|
|
$$ n(n + 1)(n + 2) = n(n + 1)(3d) $$
|
|
|
|
$$ \quad = 3\left[n(n + 1)(d)\right] $$
|
|
|
|
Let $m = \left[n(n + 1)(d)\right]$ where $m$ is an integer by the product and
|
|
sum of integers. Then $3 \mid (n)(n + 1)(n + 2)$ by the definition of
|
|
divisibility.
|
|
|
|
Therefore $n(n + 1)(n + 2)$ is a multiple of $3$.
|
|
|
|
In all three cases, any product of three consecutive integers is a multiple
|
|
of 3.
|
|
|
|
Q.E.D.
|
|
|
|
29.
|
|
|
|
a. Use the quotient-remainder theorem with divisor equal to $3$ to prove that
|
|
the square of any integer has the form $3k$ or $3k + 1$ for some integer $k$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
By the quotient-remainder theorem, $n$ can be represented as:
|
|
|
|
$n = 3q, \text{ or } n = 3q + 1 \text{ or } n = 3q + 2$ for some integer $q$.
|
|
|
|
_Case $n = 3q$:_
|
|
|
|
Then:
|
|
|
|
$$ n^2 = (3q)^2 $$
|
|
|
|
$$ \quad = 9q^2 $$
|
|
|
|
$$ \quad = 3(3q^2) $$
|
|
|
|
Let $k = 3q^2$ where $k$ is an integer by the product of integers.
|
|
|
|
Then $n^2$ has the form $3k$.
|
|
|
|
_Case $n = 3q + 1$:_
|
|
|
|
Then:
|
|
|
|
$$ n^2 = (3q + 1)^2 $$
|
|
|
|
$$ \quad = (3q + 1)(3q + 1) $$
|
|
|
|
$$ \quad = 9q^2 + 6q + 1 $$
|
|
|
|
$$ \quad = 3(3q^2 + 2q) + 1 $$
|
|
|
|
Let $k = 3q^2 + 2q$ where $k$ is an integer by the product and sum of integers.
|
|
|
|
Then $n^2$ has the form $3k + 1$.
|
|
|
|
_Case $n = 3q + 2$:_
|
|
|
|
Then:
|
|
|
|
$$ n^2 = (3q + 2)^2 $$
|
|
|
|
$$ n^2 = (3q + 2)(3q + 2) $$
|
|
|
|
$$ n^2 = 9q^2 + 12q + 4 $$
|
|
|
|
$$ n^2 = 9q^2 + 12q + 3 + 1 $$
|
|
|
|
$$ n^2 = 3(3q^2 + 4q + 1) + 1 $$
|
|
|
|
Let $k = 3q^2 + 4q + 1$ where $k$ is an integer by the product and sum of
|
|
integers.
|
|
|
|
Then $n^2$ has the form of $3k + 1$.
|
|
|
|
Therefore, in all cases, the square of any integer has the form $3k$ or $3k + 1$
|
|
for some integer $k$.
|
|
|
|
Q.E.D.
|
|
|
|
b. Use the $\mod$ notation to rewrite the result of part (a).
|
|
|
|
Therefore, in all cases, the square of any integer $n$ in mod notation can be
|
|
represented as $n^2 \mod 3 = 0$ or $n^2 \mod 3 = 1$.
|
|
|
|
30.
|
|
|
|
a. Use the quotient-remainder theorem with divisor equal to $3$ to prove that
|
|
the product of any two consecutive integers has the form $3k$ or $3k + 2$ for
|
|
some integer $k$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
By the quotient-remainder theorem, $n$ can be represented as:
|
|
|
|
$$ n = 3q \text{ or } n = 3q + 1 \text{ or } n = 3q + 2 $$
|
|
|
|
for some integer $q$.
|
|
|
|
Case $n = 3q$:
|
|
|
|
Then:
|
|
|
|
$$ n(n + 1) = (3q)(3q + 1) $$
|
|
|
|
$$ \quad = 9q^2 + 3q $$
|
|
|
|
$$ \quad = 3(q^2 + q) $$
|
|
|
|
Let $k = q^2 + q$ where $k$ is an integer by the product and sum of integers.
|
|
|
|
Then $n(n + 1)$ has the form $3k$.
|
|
|
|
Case $n = 3q + 1$:
|
|
|
|
Then:
|
|
|
|
$$ n(n + 1) = (3q + 1)((3q + 1) + 1) $$
|
|
|
|
$$ \quad = (3q + 1)(3q + 2) $$
|
|
|
|
$$ \quad = 9q^2 + 9q + 2 $$
|
|
|
|
$$ \quad = 3(3q^2 + 3q) + 2 $$
|
|
|
|
Let $k = 3q^2 + 3q$ where $k$ is an integer by the product of integers.
|
|
|
|
Then $n(n + 1)$ has the form $3k + 2$.
|
|
|
|
Case $n = 3q + 2$:
|
|
|
|
Then:
|
|
|
|
$$ n(n + 1) = (3q + 2)((3q + 2) + 1) $$
|
|
|
|
$$ \quad = (3q + 2)(3q + 3) $$
|
|
|
|
$$ \quad = 9q^2 + 9q + 6 $$
|
|
|
|
$$ \quad = 3(3q^2 + 3q + 2) $$
|
|
|
|
Let $k = 3q^2 + 3q + 2$ where $k$ is an integer by the product and sum of
|
|
integers.
|
|
|
|
Then $n(n + 1)$ has the form $3k$.
|
|
|
|
In all three cases, the product of any two consecutive integers has the form
|
|
$3k$ or $3k + 2$.
|
|
|
|
Q.E.D.
|
|
|
|
b. Use the $\mod$ notation to rewrite the result of part (a).
|
|
|
|
In all three cases, the product of any two consecutive integers, $n$ and $n + 1$
|
|
can be written in mod notation as $n(n + 1) \mod 3 = 0$ or
|
|
$n(n + 1) \mod 3 = 2$.
|
|
|
|
In 31-33, you may use the properties listed in Example 4.3.3.
|
|
|
|
31.
|
|
|
|
a. Prove that for all integers $m$ and $n$, $m + n$ and $m - n$ are either both
|
|
odd or both even.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $m$ and $n$ are any integers.
|
|
|
|
_Case both $m$ and $n$ are even:_
|
|
|
|
Property 1: The sum and difference of any two even integers are even.
|
|
|
|
By property 1 both $m + n$ and $m - n$ are even.
|
|
|
|
_Case both $m$ and $n$ are odd:_
|
|
|
|
Property 2: The sum and difference of any two odd integers are even.
|
|
|
|
By property 2 both $m + n$ and $m - n$ are even.
|
|
|
|
_Case where $m$ is odd and $n$ is even:_
|
|
|
|
Property 5: The sum of any odd integer minus any even integer is odd.
|
|
|
|
By property 5 both $m + n$ and $m - n$ are odd.
|
|
|
|
_Case where $m$ is even and $n$ is odd:_
|
|
|
|
Property 5: The sum of any odd integer minus any even integer is odd.
|
|
|
|
By property 5 both $m + n$ and $m - n$ are odd.
|
|
|
|
In all cases $m + n$ and $m - n$ are both odd or are both even.
|
|
|
|
Q.E.D.
|
|
|
|
b. Find all solutions to the equation $m^2 - n^2 = 56$ for which both $m$ and
|
|
$n$ are positive integers.
|
|
|
|
$$ m^2 - n^2 = (m + n)(m - n) = 56 $$
|
|
|
|
$$ 56 = 2 * 28 = 2 * 2 * 14 = 2 * 2 * 2 * 7 = 2^3 * 7 = 8 * 7 $$
|
|
|
|
Therefore $(m + n)(m - n) = (8)(7)$ or $(m - n)(m + n) = (8)(7)$. By part a, $m$
|
|
and $n$ must either both be odd or both be even.
|
|
|
|
$m + n = 14$ and $m - n = 4$ where $m = 9$ and $n = 5$.
|
|
|
|
Or also:
|
|
|
|
$m + n = 28$ and $m - n = 2$ where $m = 15$ and $n = 13$.
|
|
|
|
c. Find all solutions to the equation $m^2 - n^2 = 88$ for which both $m$ and
|
|
$n$ are positive integers.
|
|
|
|
$$ m^2 - n^2 = (m + n)(m - n) = 88 $$
|
|
|
|
$$ 88 = 2 * 44 = 2^2 * 22 = 2^3 * 11 $$
|
|
|
|
By part a, $m$ and $n$ must either both be odd or both be even.
|
|
|
|
$m + n = 22$ and $m - n = 4$ where $m = 13$ and $n = 9$
|
|
|
|
$m + n = 44$ and $m - n = 2$ where $m = 23$ and $n = 21$
|
|
|
|
32. Given any integers $a$, $b$, and $c$, if $a - b$ is even and $b - c$ is
|
|
even, what can you say about the parity of $2a - (b + c)$? Prove your
|
|
answer.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$, $b$, and $c$ are any integers where $a - b$ is even and $b - c$ is
|
|
even.
|
|
|
|
Note that $$
|
|
2a - (b + c) = a + a - (b + c) \\
|
|
= a + a - b - c \\
|
|
= a - b + a - c \\
|
|
= (a - b) + (a - c)
|
|
$$
|
|
|
|
Since we know that $(a - b)$ is even and $b - c$ is even.
|
|
|
|
Property 1: The difference of any two even integers are even.
|
|
|
|
By Property 1, we then know that $(a - b) + (a - c)$ is even.
|
|
|
|
Therefore the parity of $2a - (b + c)$ is even.
|
|
|
|
Q.E.D.
|
|
|
|
33. Given any integers $a$, $b$, and $c$, if $a - b$ is odd and $b - c$ is even,
|
|
what can you say about the parity of $a - c$? Prove your answer.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$, $b$ and $c$ are any integers where $a - b$ is odd and $b - c$ is
|
|
even.
|
|
|
|
Note that $a - c = (a - b) + (b - c)$
|
|
|
|
By property 5, we know that the sum of any odd integer and even integer is odd.
|
|
|
|
Therefore the parity of $a - c$ is odd.
|
|
|
|
Q.E.D.
|
|
|
|
34. Given any integer $n$, if $n > 3$, could $n$, $n + 2$, and $n + 4$ all be
|
|
prime? Prove or give a counterexample.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Case where $n = 4$:
|
|
|
|
Let $n = 4$.
|
|
|
|
$$
|
|
n = 4 \\
|
|
n + 2 = 6 \\
|
|
n + 4 = 8
|
|
$$
|
|
|
|
Thus for the given $n$ $n > 3$, but $n$ is not prime, $n + 2$ is not prime, and
|
|
$n + 4$ is not prime. Therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
Prove each of the statements in 35-43.
|
|
|
|
35. The fourth power of any integer has the form $8m$ or $8m + 1$ for some
|
|
integer $m$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
By the quotient remainder theorem, $n$ can be written as
|
|
|
|
$$
|
|
n = 4q \text{ or} \\
|
|
n = 4q + 1 \text{ or} \\
|
|
n = 4q + 2 \text{ or} \\
|
|
n = 4q + 3 \text{ or} \\
|
|
$$
|
|
|
|
for some integer $q$.
|
|
|
|
_Case $n = 4q$:_
|
|
|
|
Then:
|
|
|
|
$$ n^4 = (4q)^4 $$
|
|
|
|
$$ \quad = 256q^4 $$
|
|
|
|
$$ \quad = 8(32q^4) $$
|
|
|
|
Let $m = 32q^4$ where $m$ is an integer by the product of integers.
|
|
|
|
Thus $n^4$ is in the form $8m$.
|
|
|
|
_Case $n = 4q + 1$:_
|
|
|
|
Then:
|
|
|
|
$$ n^4 = (4q + 1)^4 $$
|
|
|
|
$$ \quad = 256q^4 + 256q^3 + 96q^2 + 16q + 1 $$
|
|
|
|
$$ \quad = 8(32q^4 + 32q^3 + 12q^2 + 2q) + 1 $$
|
|
|
|
Let $m = 32q^4 + 32q^3 + 12q^2 + 2q$ where $m$ is an integer by the product and
|
|
sum of integers.
|
|
|
|
Thus $n^4$ is in the form $8m + 1$.
|
|
|
|
_Case $n = 4q + 2$:_
|
|
|
|
Then:
|
|
|
|
$$ n^4 = (4q + 2)^4 $$
|
|
|
|
$$ n^4 = 256q^4 + 512q^3 + 384q^2 + 128q + 16 $$
|
|
|
|
$$ n^4 = 8(32q^4 + 64q^3 + 48q^2 + 16q + 2) $$
|
|
|
|
Let $m = 32q^4 + 64q^3 + 48q^2 + 16q + 2$ where $m$ is an integer by the product
|
|
and sum of integers.
|
|
|
|
Thus $n^4$ is in the form $8m$.
|
|
|
|
_Case $n = 4q + 3$:_
|
|
|
|
Then:
|
|
|
|
$$ n^4 = (4q + 3)^4 $$
|
|
|
|
$$ \quad = 256q^4 + 768q^3 + 864q^2 + 432q + 81 $$
|
|
|
|
$$ \quad = (256q^4 + 768q^3 + 864q^2 + 432q + 80) + 1 $$
|
|
|
|
$$ \quad = 8(32q^4 + 96q^3 + 108q^2 + 54q + 8) + 1 $$
|
|
|
|
Let $m = 32q^4 + 96q^3 + 108q^2 + 54q + 8$ where $m$ is an integer by the
|
|
product and sum of integers.
|
|
|
|
Thus $n^4$ is in the form $8m + 1$.
|
|
|
|
In all cases, the fourth power of any integer has the form $8m$ or $8m + 1$ for
|
|
some integer $m$.
|
|
|
|
Q.E.D.
|
|
|
|
36. The product of any four consecutive integers is divisible by $8$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
By the quotient remainder theorem, $n$ can be represented as:
|
|
|
|
$$
|
|
n = 4q \text{ or} \\
|
|
n = 4q + 1 \text{ or} \\
|
|
n = 4q + 2 \text{ or} \\
|
|
n = 4q + 3 \text{ or} \\
|
|
$$
|
|
|
|
_Case $n = 4q$:_
|
|
|
|
$$ n(n + 1)(n + 2)(n + 3) = (4q)(4q + 1)(4q + 2)(4q + 3) $$
|
|
|
|
$$ \quad = 256q^4 + 384q^3 + 176q^2 + 24q $$
|
|
|
|
$$ \quad = 8(32q^4 + 48q^3 + 22q^2 + 3q) $$
|
|
|
|
Let $m = 32q^4 + 48q^3 + 22q^2 + 3q$ where $m$ is an integer by the product and
|
|
sum of integers.
|
|
|
|
Then $8 \mid n(n + 1)(n + 2)(n + 3)$.
|
|
|
|
_Case $n = 4q + 1$:_
|
|
|
|
$$ n(n + 1)(n + 2)(n + 3) = ((4q + 1))((4q + 1) + 1)((4q + 1) + 2)((4q + 1) + 3) $$
|
|
|
|
$$ \quad = 256q^4 + 640q^3 + 560q^2 + 200q + 24 $$
|
|
|
|
$$ \quad = 8(32q^4 + 80q^3 + 70q^2 + 25q + 3) $$
|
|
|
|
Let $m = 32q^4 + 80q^3 + 70q^2 + 25q + 3$ where $m$ is an integer by the product
|
|
and sum of integers.
|
|
|
|
Then $8 \mid n(n + 1)(n + 2)(n + 3)$.
|
|
|
|
_Case $n = 4q + 2$:_
|
|
|
|
$$ n(n + 1)(n + 2)(n + 3) = ((4q + 2))((4q + 2) + 1)((4q + 2) + 2)((4q + 2) + 3) $$
|
|
|
|
$$ \quad = 256q^4 + 896q^3 + 1136q^2 + 616q + 120 $$
|
|
|
|
$$ \quad = 8(32q^4 + 112q^3 + 142q^2 + 77q + 15) $$
|
|
|
|
Let $m = 32q^4 + 112q^3 + 142q^2 + 77q + 15$ where $m$ is an integer by the
|
|
product and sum of integers.
|
|
|
|
Then $8 \mid n(n + 1)(n + 2)(n + 3)$.
|
|
|
|
_Case $n = 4q + 3$:_
|
|
|
|
$$ n(n + 1)(n + 2)(n + 3) = ((4q + 3))((4q + 3) + 1)((4q + 3) + 2)((4q + 3) + 3) $$
|
|
|
|
$$ \quad = 256q^4 + 1152q^3 + 1904q^2 + 1368q + 360 $$
|
|
|
|
$$ \quad = 8(32q^4 + 144q^3 + 238q^2 + 171q + 45) $$
|
|
|
|
Let $m = 32q^4 + 144q^3 + 238q^2 + 171q + 45$ where $m$ is an integer by the
|
|
product and sum of integers.
|
|
|
|
Then $8 \mid n(n + 1)(n + 2)(n + 3)$.
|
|
|
|
In all cases $8 \mid n(n + 1)(n + 2)(n + 3)$. Therefore the product of any four
|
|
consecutive integers is divisible by 8.
|
|
|
|
Q.E.D.
|
|
|
|
37. For any integer $n$, $n^2 + 5$ is not divisible by $4$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
_Case $n$ is even:
|
|
|
|
Since $n$ is even, $n = 2k$ for any integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ n^2 + 5 = (2k)^2 + 5 $$
|
|
|
|
$$ \quad = 4k^2 + 4 + 1 $$
|
|
|
|
$$ \quad = 4(k^2 + 1) + 1 $$
|
|
|
|
Then $(n^2 + 5) \mod 4 = 1$ and $n^2 + 5$ is not divisible by $4$.
|
|
|
|
_Case $n$ is odd:
|
|
|
|
Since $n$ is odd, $n = 2k + 1$ for any integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ n^2 + 5 = (2k + 1)^2 + 5 $$
|
|
|
|
$$ \quad = (2k + 1)(2k + 1) + 5 $$
|
|
|
|
$$ \quad = 4k^2 + 4k + 1 + 5 $$
|
|
|
|
$$ \quad = 4k^2 + 4k + 6 $$
|
|
|
|
$$ \quad = 4k^2 + 4k + 4 + 2 $$
|
|
|
|
$$ \quad = 4(k^2 + k + 1) + 2 $$
|
|
|
|
Let $m = k^2 + k + 1$ where $m$ is an integer by the product and sum of
|
|
integers. Then $(n^2 + 5) \mod 4 = 2$ and $n^2 + 5$ is not divisible by $4$.
|
|
|
|
In both cases, $n^2 + 5$ is not divisible by $4$.
|
|
|
|
Q.E.D.
|
|
|
|
38. For every integer $m$, $m^2 = 5k$, or $m^w = 5k + 1$, or $m^2 = 5k + 4$ for
|
|
some integer $k$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $m$ is any integer.
|
|
|
|
By the quotient-remainder theorem, we can say $m$ is:
|
|
|
|
$$
|
|
m = 5q \\
|
|
m = 5q + 1 \\
|
|
m = 5q + 2 \\
|
|
m = 5q + 3 \\
|
|
m = 5q + 4 \\
|
|
$$
|
|
|
|
_Case $m = 5q$:_
|
|
|
|
$$ m^2 = (5q)^2 $$
|
|
|
|
$$ m^2 = 25q^2 $$
|
|
|
|
$$ m^2 = 5(5q^2) $$
|
|
|
|
Let $k = 5q^2$, where $k$ is any integer by the product of integers.
|
|
|
|
Then $m^2 = 5k$.
|
|
|
|
_Case $m = 5q + 1$:_
|
|
|
|
$$ m^2 = (5q + 1)^2 $$
|
|
|
|
$$ m^2 = (5q + 1)(5q + 1) $$
|
|
|
|
$$ \quad = 25q^2 + 10q + 1 $$
|
|
|
|
$$ \quad = 5(5q^2 + 2q) + 1 $$
|
|
|
|
Let $k = 5q^2 + 2q$, where $k$ is any integer by the product of integers.
|
|
|
|
Then $m^2 = 5k + 1$.
|
|
|
|
_Case $m = 5q + 2$:_
|
|
|
|
$$ m^2 = (5q + 2)^2 $$
|
|
|
|
$$ \quad = (5q + 2)(5q + 2) $$
|
|
|
|
$$ \quad = 25q^2 + 20q + 4 $$
|
|
|
|
$$ \quad = 5(5q^2 + 4q) + 4 $$
|
|
|
|
Let $k = 5q^2 + 4q$, where $k$ is any integer by the product of integers.
|
|
|
|
Then $m^2 = 5k + 4$.
|
|
|
|
_Case $m = 5q + 3$:_
|
|
|
|
$$ m^2 = (5q + 3)^2 $$
|
|
|
|
$$ \quad = (5q + 3)(5q + 3) $$
|
|
|
|
$$ \quad = 25q^2 + 30q + 9 $$
|
|
|
|
$$ \quad = 25q^2 + 30q + 5 + 4 $$
|
|
|
|
$$ \quad = 5(5q^2 + 6q + 1) + 4 $$
|
|
|
|
Let $k = 5q^2 + 6q + 1$, where $k$ is any integer by the product of integers.
|
|
|
|
Then $m^2 = 5k + 4$.
|
|
|
|
_Case $m = 5q + 4$:_
|
|
|
|
$$ m^2 = (5q + 4)^2 $$
|
|
|
|
$$ \quad = (5q + 4)(5q + 4) $$
|
|
|
|
$$ \quad = 25q^2 + 40q + 16 $$
|
|
|
|
$$ \quad = 25q^2 + 40q + 15 + 1 $$
|
|
|
|
$$ \quad = 5(5q^2 + 8q + 3) + 1 $$
|
|
|
|
Let $k = 5q^2 + 8q + 3$, where $k$ is any integer by the product of integers.
|
|
|
|
Then $m^2 = 5k + 1$.
|
|
|
|
In all cases, $m^2 = 5k$ or $m^2 = 5k + 1$ or $m^2 = 5k + 4$. Therefore for
|
|
every integer $m$, $m^2 = 5k$ or $m^2 = 5k + 1$ or $m^2 = 5k + 4$ for some
|
|
integer $k$.
|
|
|
|
Q.E.D.
|
|
|
|
39. Every prime number except $2$ and $3$ has the form $6q + 1$ or $6q + 5$ for
|
|
some integer $q$.
|
|
|
|
Suppose $p$ is any prime number where $p \neq 2$ and $p \neq 3$.
|
|
|
|
By the quotient-remainder theorem, if $p$ were any integer, we could express $p$
|
|
as:
|
|
|
|
$$
|
|
p = 6q \\
|
|
p = 6q + 1 \\
|
|
p = 6q + 2 \\
|
|
p = 6q + 3 \\
|
|
p = 6q + 4 \\
|
|
p = 6q + 5 \\
|
|
$$
|
|
|
|
But since $p$ is a prime number that is not $2$ and not $3$, this narrows us
|
|
down to:
|
|
|
|
$$
|
|
p = \cancel{6q} \text{ even, not prime} \\
|
|
p = 6q + 1 \\
|
|
p = \cancel{6q + 2} \text{ even not prime} \\
|
|
p = \cancel{6q + 3} \text{ divisible by 3, not prime} \\
|
|
p = \cancel{6q + 4} \text{ even not prime} \\
|
|
p = 6q + 5 \\
|
|
$$
|
|
|
|
Therefore, every prime number $p$ except $2$ and $3$ has the form $6q + 1$ or
|
|
$6q + 5$.
|
|
|
|
Q.E.D.
|
|
|
|
40. If $n$ is any odd integer, then $n^4 \mod 16 = 1$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any odd integer.
|
|
|
|
Sine $n$ is odd, $n = 2k + 1$ for any integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ n^4 = (2k + 1)^4 $$
|
|
|
|
$$ \quad = (2k + 1)(2k + 1)(2k + 1)(2k + 1) $$
|
|
|
|
$$ \quad = 16k^4 + 32k^3 + 24k^2 + 8k + 1 $$
|
|
|
|
$$ \quad = 16k^4 + 32k^3 + 16k^2 + 8k^2 + 8k + 1 $$
|
|
|
|
$$ \quad = 16(k^4 + 2k^3 + k^2) + 8k^2 + 8k + 1 $$
|
|
|
|
$$ n^4 \mod 16 = 8k^2 + 8k + 1 (\mod 16) $$
|
|
|
|
$$ = 8(k^2 + k) + 1 (\mod 16) $$
|
|
|
|
$$ = 8(k(k + 1)) + 1 (\mod 16) $$
|
|
|
|
Note here that $k(k + 1)$ is even as the product of two consecutive products is
|
|
even. Therefore $k(k + 1) = 2m$ for some integer m$.
|
|
|
|
$$ = 8(2m) + 1 (\mod 16) $$
|
|
|
|
$$ = 16m + 1 (\mod 16) $$
|
|
|
|
$$ n^4 \mod 16 = 1 $$
|
|
|
|
Q.E.D.
|
|
|
|
41. For all real numbers $x$ and $y$, $|x| \cdot |y| = |xy|$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $x$ and $y$ are any real numbers.
|
|
|
|
_Case $x < 0$ and $y < 0$:_
|
|
|
|
$$ |x| = -x \text{ by the definition of absolute value} $$
|
|
|
|
$$ |y| = -y \text{ by the definition of absolute value} $$
|
|
|
|
$$ |x| \cdot |y| = -x \cdot -y = xy $$
|
|
|
|
Then:
|
|
|
|
$$ |xy| = xy $$
|
|
|
|
Since $x$ and $y$ are both negative, their product $xy$ is positive.
|
|
|
|
Therefore:
|
|
|
|
$$ |x| \cdot |y| = |xy| $$
|
|
|
|
_Case $x \geq 0$ and $y < 0$:_
|
|
|
|
$$ |x| = x $$
|
|
|
|
$$ |y| = -y $$
|
|
|
|
$$ |x| \cdot |y| = x \cdot -y = -xy $$
|
|
|
|
Since $x$ is nonnegative and $y$ is negative, $xy \leq 0$.
|
|
|
|
Therefore:
|
|
|
|
$$ |xy| = -xy $$
|
|
|
|
So,
|
|
|
|
$$ |x| \cdot |y| = |xy| $$
|
|
|
|
_Case $x < 0$ and $y \geq 0$:_
|
|
|
|
$$ |x| = -x $$
|
|
|
|
$$ |y| = y $$
|
|
|
|
$$ |x| \cdot |y| = -xy $$
|
|
|
|
Since $x$ is negative and $y$ is nonnegative, $xy \leq 0$.
|
|
|
|
Therefore:
|
|
|
|
$$ |xy| = -xy $$
|
|
|
|
So,
|
|
|
|
$$ |x| \cdot |y| = |xy| $$
|
|
|
|
_Case $x \geq 0$ and $y \geq 0$:_
|
|
|
|
$$ |x| = x $$
|
|
|
|
$$ |y| = y $$
|
|
|
|
$$ |x| \cdot |y| = xy $$
|
|
|
|
Since $x$ is nonnegative and $y$ is nonnegative, $xy \geq 0$
|
|
|
|
$$ |xy| = xy $$
|
|
|
|
So,
|
|
|
|
$$ |x| \cdot |y| = |xy| $$
|
|
|
|
In all cases, $|x| \cdot |y| = |xy|$. Therefore for all real numbers $x$ and
|
|
$y$, $|x| \cdot |y| = |xy|$.
|
|
|
|
Q.E.D.
|
|
|
|
42. For all real numbers $r$ and $c$ with $c \geq 0$, $-c \leq r \leq c$ if, and
|
|
only if, $|r| \leq c$. _(Hint: Proving $A$ if, and only if, $B$ requires
|
|
proving both if $A$ then $B$ and if $B$ then $A$.)_
|
|
|
|
Suppose $r$ and $c$ are real numbers where $c \geq 0$ and $-c \leq r \leq c$.
|
|
|
|
_Case where $r < 0$:_
|
|
|
|
$$ |r| = -r $$
|
|
|
|
Since we assumed that:
|
|
|
|
$$ -c \leq r \leq c $$
|
|
|
|
Then:
|
|
|
|
$$ c \geq -r $$
|
|
|
|
Or:
|
|
|
|
$$ -r \leq c $$
|
|
|
|
Since:
|
|
|
|
$$ |r| = -r $$
|
|
|
|
It follows that:
|
|
|
|
$$ |r| \leq c $$
|
|
|
|
_Case where $r \geq 0$:_
|
|
|
|
$$ |r| = r $$
|
|
|
|
Since we assumed that:
|
|
|
|
$$ -c \leq r \leq c $$
|
|
|
|
Then:
|
|
|
|
$$ r \leq c $$
|
|
|
|
Since:
|
|
|
|
$$ |r| = r $$
|
|
|
|
It follows that:
|
|
|
|
$$ |r| \leq c $$
|
|
|
|
Therefore $|r| \leq c$.
|
|
|
|
In both cases for all real numbers $r$ and $c$ with $c \geq 0$, it has been
|
|
shown that if $-c \leq r \leq c$, then $|r| \leq c$.
|
|
|
|
And then suppose $r$ and $c$ are real numbers where $c \geq 0$ and $|r| \leq c$.
|
|
|
|
_Case where $r < 0$:_
|
|
|
|
$$ |r| = -r $$
|
|
|
|
Since:
|
|
|
|
$$ |r| \leq c $$
|
|
|
|
Then:
|
|
|
|
$$ -r \leq c $$
|
|
|
|
$$ r \geq -c $$
|
|
|
|
Or:
|
|
|
|
$$ -c \leq r $$
|
|
|
|
_Case where $r \geq 0$:_
|
|
|
|
$$ |r| = r $$
|
|
|
|
Since:
|
|
|
|
$$ |r| \leq c $$
|
|
|
|
$$ r \leq c $$
|
|
|
|
It follows from both cases then that $-c \leq r \leq c$.
|
|
|
|
Therefore it has been shown that for all real numbers $r$ and $c$ with
|
|
$c \geq 0$, $-c \leq r \leq c$ if, and only if, $|r| \leq c$.
|
|
|
|
Q.E.D.
|
|
|
|
43. For all real numbers $a$ and $b$, $\lvert|a| - |b|\rvert \leq |a - b|$.
|
|
|
|
Omitted.
|
|
|
|
44. A matrix $\mathbb{M}$ has 3 rows and 4 columns.
|
|
|
|
$$
|
|
\left[\begin{array}{cccc}
|
|
a_{11} & a_{12} & a_{13} & a_{14} \\
|
|
a_{21} & a_{22} & a_{23} & a_{24} \\
|
|
a_{31} & a_{32} & a_{33} & a_{34} \\
|
|
\end{array}\right]
|
|
$$
|
|
|
|
The 12 entries in the matrix are to be stored in _row major_ form in locations
|
|
7,609 to 7,620 in a computer's memory. This means that the entries in the first
|
|
row (reading left to right) are stored first, then the entries in the second
|
|
row, and finally the entries in the third row.
|
|
|
|
a. Which location will $a_{22}$ be stored in?
|
|
|
|
$$
|
|
\left[\begin{array}{cccc}
|
|
7609 & 7610 & 7611 & 7612 \\
|
|
7613 & \boxed{7614} & a_{23} & a_{24} \\
|
|
a_{31} & a_{32} & a_{33} & a_{34} \\
|
|
\end{array}\right]
|
|
$$
|
|
|
|
b. Write a formula (in $i$ and $j$) that gives the integer $n$ so that $a_{ij}$
|
|
is stored in location 7,609 + $n$.
|
|
|
|
Row-major order means we count:
|
|
|
|
- 4 entries per row.
|
|
|
|
So before row $i$, there are $4(i - 1)$ entries.
|
|
|
|
Within row $i$, entry $j$ adds $j - 1$.
|
|
|
|
So:
|
|
|
|
$$ n = 4(i - 1) + (j - 1) $$
|
|
|
|
c. Find formulas (in $n$) for $r$ and $s$ so that $a_{rs}$ is stored in location
|
|
$7,609 + n$.
|
|
|
|
We start from:
|
|
|
|
$$ n = 4(i - 1) + (j - 1) $$
|
|
|
|
So:
|
|
|
|
$$ n + 1 = 4(i - 1) + j $$
|
|
|
|
Divide:
|
|
|
|
$$ i = \lfloor \frac{n}{4} \rfloor + 1 $$
|
|
|
|
Remainder gives:
|
|
|
|
$$ j = (n \mod 4) + 1 $$
|
|
|
|
So:
|
|
|
|
$$ r = \lfloor \frac{n}{4} \rfloor + 1, \quad s = (n \mod 4) + 1 $$
|
|
|
|
45. Let $\mathbb{M}$ be a matrix with $m$ rows and $n$ columns, and suppose that
|
|
the entries of $\mathbb{M}$ are stored in a computer's memory in row major
|
|
form (see exercise 44) in locations $N$, $N + 1$, $N + 2$, $\dots$,
|
|
$N + mn - 1$. Find formulas in $k$ for $r$ and $s$ so that $a_{rs}$ is
|
|
stored in location $N + k$.
|
|
|
|
Row-major order means: Each row has $n$ entries.
|
|
|
|
We are given location: $N + k$
|
|
|
|
So $k$ counts how far into the matrix we are (starting at $0$).
|
|
|
|
Row index $r$.
|
|
|
|
Each full row uses $n$ positions, so
|
|
|
|
$$ r = \lfloor \frac{k}{n} \rfloor + 1 $$
|
|
|
|
Column index $s$.
|
|
|
|
Position inside the row is the remainder:
|
|
|
|
$$ s = (k \mod n) + 1 $$
|
|
|
|
Final answer:
|
|
|
|
$$ r = \lfloor \frac{k}{n} \rfloor + 1, \quad s = (k \mod n) + 1 $$
|
|
|
|
46. If $m$, $n$, and $d$ are integers, $d > 0$ and $m \mod d = n \mod d$, does
|
|
it necessarily follow that $m = n$? That $m - n$ is divisible by $d$? Prove
|
|
your answers.
|
|
|
|
Omitted
|
|
|
|
47. If $m$, $n$, and $d$ are integers $d > 0$, and $d \mid (m - n)$, what is
|
|
the relation between $m 'mod d'$ and $n \mod d$? Prove your answer.
|
|
|
|
Omitted
|
|
|
|
48. If $m$, $n$, $a$, $b$, and $d$ are integers, $d > 0$ and $m \mod d = a$ and
|
|
$n \mod d = b$, is $(m + n) \mod d = a + b$? Is
|
|
$(m + n) \mod d = (a + b) \mod d$? Prove your answers.
|
|
|
|
Omitted
|
|
|
|
49. If $m$, $n$, $a$, $b$, and $d$ are integers, $d > 0$, and $m \mod d = a$ and
|
|
$n \mod d = b$, is $(mn) \mod d = ab$? Is $(mn) \mod d = ab \mod d$? Prove
|
|
your answers.
|
|
|
|
Omitted
|
|
|
|
50. Prove that if $m$, $d$, and $k$ are integers and $d > 0$, then
|
|
$(m + dk) \mod d = m \mod d$.
|
|
|
|
Omitted
|
|
|
|
---
|
|
|
|
**Exercise Set 4.6**
|
|
|
|
Page 240
|
|
|
|
Compute $\lfloor x \rfloor$ and $\lceil x \rceil$ for each of the values of $x$
|
|
in 1-4.
|
|
|
|
1. $37.999$
|
|
|
|
$\lfloor 37.999 \rfloor = 37$
|
|
|
|
$\lceil 37.999 \rceil = 38$
|
|
|
|
2. $\dfrac{17}{4}$
|
|
|
|
$\dfrac{17}{4} = 4 + \dfrac{1}{4} = 4.25$
|
|
|
|
$\lfloor \dfrac{17}{4} \rfloor = 4$
|
|
|
|
$\lceil \dfrac{17}{4} \rceil = 5$
|
|
|
|
3. $-14.00001$
|
|
|
|
$\lfloor -14.00001 \rfloor = -15$
|
|
|
|
$\lceil -14.00001 \rceil = -14$
|
|
|
|
4. $-\dfrac{32}{5}$
|
|
|
|
$-\dfrac{32}{5} = -6.4$
|
|
|
|
$\lfloor -\dfrac{32}{5} \rfloor = -7$
|
|
|
|
$\lceil -\dfrac{32}{5} \rceil = -6$
|
|
|
|
5. Use the floor notation to express $259\ div\ 11$ and $259 \mod 11$.
|
|
|
|
$$ 259\ div\ 11 = \lfloor \frac{259}{11} \rfloor = \lfloor 23.54545454\dots \rfloor = 23 $$
|
|
|
|
$$ 259 \mod 11 = 259 - 11 \cdot \lfloor \frac{259}{11} \rfloor $$
|
|
|
|
$$ = 259 - 11 \cdot 23 $$
|
|
|
|
$$ = 6 $$
|
|
|
|
6. If $k$ is an integer, what is $\lceil k \rceil$? Why?
|
|
|
|
$$ \lceil k \rceil = k $$
|
|
|
|
The ceiling of $k$ is $k$. Since $k$ is already an integer, then given the
|
|
definition for ceiling:
|
|
|
|
$$ \lceil k \rceil = n \Leftrightarrow n - 1 < k \leq n $$
|
|
|
|
We can substitute in $n = k$:
|
|
|
|
$$ k - 1 < k \leq k $$
|
|
|
|
And both parts are true, so:
|
|
|
|
$$ \lceil k \rceil = k $$
|
|
|
|
7. If $k$ is an integer, what is $\left\lceil k + \dfrac{1}{2} \right\rceil$?
|
|
Why?
|
|
|
|
$$ \left\lceil k + \frac{1}{2} \right\rceil = k + 1 $$
|
|
|
|
By the definition of ceiling:
|
|
|
|
$$ \ceil k + \dfrac{1}{2} \rceil = n \Leftrightarrow n - 1 < k + \frac{1}{2} \leq n $$
|
|
|
|
Now substitute in $n = k + 1$:
|
|
|
|
$$ (k + 1) - 1 < k + \frac{1}{2} \leq k + 1 $$
|
|
|
|
$$ k < k + \frac{1}{2} \leq k + 1 $$
|
|
|
|
And both parts are true, so:
|
|
|
|
$$ \lceil k + \frac{1}{2} \rceil = k + 1 $$
|
|
|
|
8. Seven pounds of raw material are needed to manufacture each unit of a certain
|
|
product. Express the number of units that can be produced from $n$ pounds of
|
|
raw material using either the floor or the ceiling notation. Which notation
|
|
is more appropriate?
|
|
|
|
$7$ pounds of raw material per product is main ratio.
|
|
|
|
This means that the number of units $q$ from $n$ pounds can be expressed using
|
|
the quotient-remainder theorem as:
|
|
|
|
$$ n = 7q $$
|
|
|
|
Where the remainder is removed as no more units can be created from leftover raw
|
|
material.
|
|
|
|
And when converted to floor notation, this is:
|
|
|
|
$$ q = \left\lfloor \frac{n}{7} \right\rfloor $$
|
|
|
|
The reason floor is more appropriate is that leftover material would not
|
|
realistically be able to be utilized to make another unit of said product.
|
|
|
|
9. Boxes, each capable of holding 36 units, are used to ship a product from the
|
|
manufacturer to a wholesaler. Express the number of boxes that would be
|
|
required to ship $n$ units of the product using either the floor or the
|
|
ceiling notation. Which notation is more appropriate?
|
|
|
|
Each box can hold up to $36$ units of product. Shipping $n$ units of product can
|
|
be expressed using the quotient remainder theorem as:
|
|
|
|
$$ n = 36q + r $$
|
|
|
|
Where $r$ is the remaining products that did not fill another box.
|
|
|
|
Then we convert to ceiling notation:
|
|
|
|
$$ q = \left\lceil \frac{n}{36} \right\rceil $$
|
|
|
|
The ceiling notation is more appropriate as even after packing all boxes full,
|
|
you cannot simply not ship the remaining product, so another box filled with the
|
|
remaining products is added to the shipment.
|
|
|
|
10. If 0 = Sunday, 1 = Monday, 2 = Tuesday, ..., 6 = Saturday, then January 1 of
|
|
year $n$ occurs on the day of the week given by the following formula:
|
|
|
|
$$ \left(n + \left\lfloor \frac{n - 1}{4} \right\rfloor - \left\lfloor \frac{n - 1}{100} \right\rfloor + \left\lfloor \frac{n - 1}{400} \right\rfloor\right) \mod 7 $$
|
|
|
|
a. Use this formula to find January 1 of
|
|
|
|
i. 2050
|
|
|
|
6 = Saturday
|
|
|
|
ii. 2100
|
|
|
|
5 = Friday
|
|
|
|
iii. the year of your birth.
|
|
|
|
Omitted
|
|
|
|
b. Interpret the different components of this formula.
|
|
|
|
11. State a necessary and sufficient condition for the floor of a real number to
|
|
equal that number.
|
|
|
|
It is a necessary and sufficient condition for any real number, $x$ to to
|
|
satisfy $\lfloor x \rfloor = x$ that $x$ be an integer.
|
|
|
|
12. Let $S$ be the statement: For any odd integer $n$,
|
|
$\left\lfloor \dfrac{n}{2} \right\rfloor = \dfrac{(n - 1)}{2}$. Then $S$ is
|
|
true, but the following "proof" is incorrect. Find the mistake.
|
|
|
|
**"Proof:**
|
|
|
|
Suppose $n$ is any odd integer. Then $n = 2k + 1$ for some integer $k$.
|
|
Consequently,
|
|
|
|
$$ \left\lfloor \frac{2k + 1}{2} \right\rfloor = \frac{(2k + 1) - 1}{2} = \frac{2k}{2} = k $$
|
|
|
|
But $n = 2k + 1$. Solving for $k$ gives $k = \dfrac{(n - 1)}{2}$. Hence, by
|
|
substitution, $\left\lfloor \dfrac{n}{2} \right\rfloor = \dfrac{(n - 1)}{2}$."
|
|
|
|
The mistake is in the initial substitution, by setting:
|
|
|
|
$$ \left\lfloor \frac{2k + 1}{2} \right\rfloor = \frac{(2k + 1) - 1}{2} $$
|
|
|
|
The author of this proof assumes the what is to be proved.
|
|
|
|
13. Prove that if $n$ is any even integer, then
|
|
$\left\lfloor \dfrac{n}{2} \right\rfloor = \dfrac{n}{2}$.
|
|
|
|
**Proof:** Suppose $n$ is any even integer.
|
|
|
|
Since $n$ is an even integer, $n = 2k$ for some integer $k$.
|
|
|
|
Then, by substitution:
|
|
|
|
$$ \left\lfloor \frac{n}{2} \right\rfloor = \left\lfloor \frac{2k}{2} \right\rfloor $$
|
|
|
|
$$ = \left\lfloor k \right\rfloor $$
|
|
|
|
Because $k$ is an integer, and by the definition of floor, $k \leq k < k + 1$.
|
|
We then know that:
|
|
|
|
$$ \lfloor k \rfloor = k $$
|
|
|
|
It follows then that:
|
|
|
|
$$ \left \lfloor \frac{2k}{2} \right\rfloor = \frac{2k}{2} $$
|
|
|
|
$$ \left \lfloor \frac{n}{2} \right\rfloor = \frac{n}{2} $$
|
|
|
|
Q.E.D.
|
|
|
|
14. Show that the following statement is false.
|
|
|
|
For all real numbers $x$ and $y$,
|
|
$\lfloor x - y \rfloor = \lfloor x \rfloor - \lfloor y \rfloor$.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $x = \dfrac{1}{2}$ and $y = \dfrac{3}{4}$
|
|
|
|
Then:
|
|
|
|
$$ \lfloor x - y \rfloor = \left\lfloor \frac{1}{2} - \frac{3}{4} \right\rfloor $$
|
|
|
|
$$ = -1 $$
|
|
|
|
Then consider:
|
|
|
|
$$ \lfloor x \rfloor - \lfloor y - \rfloor = \left\lfloor \frac{1}{2} \right\rfloor - \left\lfloor \frac{3}{4} \right\rfloor $$
|
|
|
|
$$ = 0 - 0 $$
|
|
|
|
$$ = 0 $$
|
|
|
|
Thus:
|
|
|
|
$$ -1 \neq 0 $$
|
|
|
|
$$ \lfloor x - y \rfloor \neq \lfloor x \rfloor - \lfloor y \rfloor$ $.
|
|
|
|
Therefore for the given $x$ and $y$, this statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
Some of the statements in 15-22 are true and some are false. Prove each true
|
|
statement and find a counterexample for each false statement, but do not use
|
|
Theorem 4.6.1 in your proofs.
|
|
|
|
15. For every real number $x$, $\lfloor x - 1 \rfloor = \lfloor x \rfloor - 1$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $x$ is any real number.
|
|
|
|
By the definition of floor, $x$ can then be expressed as:
|
|
|
|
$$ \lfloor x \rfloor \Leftrightarrow n \leq x < n + 1 $$
|
|
|
|
for some integer $n$.
|
|
|
|
Subtracting $1$ from all sides of the inequality is:
|
|
|
|
$$ n - 1 \leq x - 1 < n $$
|
|
|
|
Since $n - 1$ is an integer by the difference of integers, by the definition of
|
|
floor:
|
|
|
|
$$ \lfloor x - 1 \rfloor = n - 1 $$
|
|
|
|
It follows by substitution then that:
|
|
|
|
$$ \lfloor x - 1 \rfloor = \lfloor x \rfloor - 1 $$
|
|
|
|
Q.E.D.
|
|
|
|
16. For every real number $x$,
|
|
$\left\lfloor x^2 \right\rfloor = \lfloor x \rfloor^2$
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $x = -\dfrac{1}{2}$
|
|
|
|
$$ x^2 = \frac{1}{4} $$
|
|
|
|
$$ \lfloor x^2 \rfloor = \lfloor \frac{1}{4} \rfloor = 0 $$
|
|
|
|
Now consider:
|
|
|
|
$$ \lfloor x \rfloor^2 = \lfloor -\frac{1}{2} \rfloor^2 = (-1)^2 = 1 $$
|
|
|
|
And then:
|
|
|
|
$$ 0 \neq 1 $$
|
|
|
|
$$ \lfloor x^2 \rfloor \neq \lfloor x \rfloor^2 $$
|
|
|
|
Then for the given $x$, $\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2$.
|
|
Therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
17. For every integer $n$,
|
|
|
|
$$
|
|
\left\lfloor \dfrac{n}{3} \right\rfloor =
|
|
\begin{cases}
|
|
\dfrac{n}{3} & \text{if } n \mod 3 = 0 \\
|
|
\dfrac{(n - 1)}{3} & \text{if } n \mod 3 = 1 \\
|
|
\dfrac{(n - 2)}{3} & \text{if } n \mod 3 = 2 \\
|
|
\end{cases}
|
|
$$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
_Case where $n \mod 3 = 0$:_
|
|
|
|
Since $n \mod 3 = 0$, then by the definition of mod:
|
|
|
|
Since $n \mod 3 = 0$, $n$ can be written as:
|
|
|
|
$$ n = 3q + 0 $$
|
|
|
|
for some integer $q$ by the definition of mod.
|
|
|
|
By substitution:
|
|
|
|
$$ \left\lfloor \frac{n}{3} \right\rfloor = \left\lfloor \frac{3q}{3} \right\rfloor $$
|
|
|
|
$$ = \lfloor q \rfloor $$
|
|
|
|
$$ = q \quad \text{ by the definition of floor} $$
|
|
|
|
Since $q = \dfrac{n}{3}$:
|
|
|
|
$$ \left\lfloor \frac{n}{3} \right\rfloor = \frac{n}{3} $$
|
|
|
|
_Case where $n \mod 3 = 1$:_
|
|
|
|
Since $n \mod 3 = 1$, then by the definition of mod:
|
|
|
|
Since $n \mod 3 = 1$, $n$ can be written as:
|
|
|
|
$$ n = 3q + 1 $$
|
|
|
|
for some integer $q$ by the definition of mod.
|
|
|
|
By substitution:
|
|
|
|
$$ \left\lfloor \frac{n}{3} \right\rfloor = \left\lfloor \frac{3q + 1}{3} \right\rfloor $$
|
|
|
|
$$ = \left\lfloor \frac{3q}{3} + \frac{1}{3} \right\rfloor $$
|
|
|
|
$$ = \left\lfloor q + \frac{1}{3} \right\rfloor $$
|
|
|
|
$$ = q \quad \text{ by the definition of floor} $$
|
|
|
|
Since $q = \dfrac{n - 1}{3}$:
|
|
|
|
$$ \left\lfloor \frac{n}{3} \right\rfloor = \frac{n - 1}{3} $$
|
|
|
|
_Case where $n \mod 3 = 2$:_
|
|
|
|
Since $n \mod 3 = 2$, $n$ can be written as:
|
|
|
|
$$ n = 3q + 2 $$
|
|
|
|
for some integer $q$ by the definition of mod.
|
|
|
|
By substitution:
|
|
|
|
$$ \left\lfloor \frac{n}{3} \right\rfloor = \left\lfloor \frac{3q + 2}{3} \right\rfloor $$
|
|
|
|
$$ = \left\lfloor \frac{3q}{3} + \frac{2}{3} \right\rfloor $$
|
|
|
|
$$ = \left\lfloor q + \frac{2}{3} \right\rfloor $$
|
|
|
|
$$ = q \quad \text{ by the definition of floor} $$
|
|
|
|
Since $q = \dfrac{n - 2}{3}$:
|
|
|
|
$$ \left\lfloor \frac{n}{3} \right\rfloor = \frac{n - 2}{3} $$
|
|
|
|
Q.E.D.
|
|
|
|
18. For all real numbers $x$ and $y$,
|
|
$\lceil x + y \rceil = \lceil x \rceil + \lceil y \rceil$.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $x = -\dfrac{1}{2}$ and $y = -\dfrac{3}{4}$.
|
|
|
|
Consider:
|
|
|
|
$$ \lceil x + y \rceil = \left\lceil -\frac{1}{2} + \left(-\frac{3}{4}\right) \right\rceil $$
|
|
|
|
$$ = -1 $$
|
|
|
|
Then:
|
|
|
|
$$ \lceil x \rceil + \lceil y \rceil = \left\lceil -\frac{1}{2} \right\rceil + \left\lceil -\frac{3}{4} \right\rceil $$
|
|
|
|
$$ = 0 + 0 $$
|
|
|
|
$$ = 0 $$
|
|
|
|
Then:
|
|
|
|
$$ -1 \neq 0 $$
|
|
|
|
$$ \lceil x + y \rceil \neq \lceil x \rceil + \lceil y \rceil $$
|
|
|
|
Therefore, for the given $x$ and $y$, the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
19. For every real number $x$, $\lceil x - 1 \rceil = \lceil x \rceil - 1$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $x$ is any real number.
|
|
|
|
By the definition of ceiling, $x$ can be expressed as:
|
|
|
|
$$ \lceil x \rceil = n \Leftrightarrow n - 1 < x \leq n $$
|
|
|
|
for some integer $n$.
|
|
|
|
If we then subtract $1$ from the inequality, we get:
|
|
|
|
$$ n - 2 < x - 1 \leq n - 1 $$
|
|
|
|
Since $n - 1$ is an integer by the difference of integers, by the definition of
|
|
ceiling:
|
|
|
|
$$ \lceil x - 1 \rceil = n - 1 $$
|
|
|
|
It follows by substitution then that:
|
|
|
|
$$ \lceil x - 1 \rceil = \lceil x \rceil - 1 $$
|
|
|
|
Q.E.D.
|
|
|
|
20. For all real numbers $x$ and $y$,
|
|
$\lceil xy \rceil = \lceil x \rceil \cdot \lceil y \rceil$.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $x = 2$ and $y = \dfrac{1}{2}$.
|
|
|
|
Then:
|
|
|
|
$$ \lceil xy \rceil = \left\lceil 2\left(\frac{1}{2}\right) \right\rceil = \lceil 1 \rceil = 1 $$
|
|
|
|
Then consider:
|
|
|
|
$$ \lceil x \rceil \cdot \lceil y \rceil = \lceil 2 \rceil \cdot \left\lceil \frac{1}{2} \right\rceil $$
|
|
|
|
$$ = 2 \cdot 1 $$
|
|
|
|
$$ = 2 $$
|
|
|
|
Since:
|
|
|
|
$$ 1 \neq 2 $$
|
|
|
|
Then:
|
|
|
|
$$ \lceil xy \rceil \neq \lceil x \rceil \cdot \lceil y \rceil $$
|
|
|
|
Thus it has been shown that for at least one given $x$ and one given $y$,
|
|
|
|
$$ \lceil xy \rceil \neq \lceil x \rceil \cdot \lceil y \rceil $$
|
|
|
|
Therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
21. For every odd integer $n$,
|
|
$\lceil \dfrac{n}{2} \rceil = \dfrac{(n - 1)}{2}$.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $n = 1$.
|
|
|
|
Consider:
|
|
|
|
$$ \left\lceil \dfrac{n}{2} \right\rceil = \left\lceil \dfrac{1}{2} \right\rceil $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Then:
|
|
|
|
$$ \frac{n - 1}{2} = \frac{1 - 1}{2} = \frac{0}{2} = 0 $$
|
|
|
|
Since $1 \neq 0$:
|
|
|
|
$$ \left\lceil \dfrac{n}{2} \right\rceil \neq \frac{n - 1}{2} $$
|
|
|
|
Thus it has been shown that there exists some value for $n$ such that:
|
|
|
|
$$ \left\lceil \dfrac{n}{2} \right\rceil \neq \frac{n - 1}{2} $$
|
|
|
|
Therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
22. For all real numbers $x$ and $y$,
|
|
$\lceil xy \rceil = \lceil x \rceil \cdot \lfloor y \rfloor$.
|
|
|
|
**Proof by Counterexample:**
|
|
|
|
Let $x = \dfrac{5}{4}$ and $y = \dfrac{1}{2}$.
|
|
|
|
Then:
|
|
|
|
$$ \lceil xy \rceil = \left\lceil \left(\frac{5}{4}\right)\left(\frac{1}{2}\right) \right\rceil $$
|
|
|
|
$$ = 1 $$
|
|
|
|
Then:
|
|
|
|
$$ \lceil x \rceil \cdot \lfloor y \rfloor = \left\lceil \frac{5}{4} \right\rceil \cdot \left\lfloor \frac{1}{2} \right\rfloor $$
|
|
|
|
$$ = 2 \cdot 0 = 0 $$
|
|
|
|
So:
|
|
|
|
$$ 1 \neq 0 $$
|
|
|
|
$$ \lceil xy \rceil \neq \lceil x \rceil \cdot \lfloor y \rfloor $$
|
|
|
|
So, it has been shown that there exists a value for $x$ and a value for $y$ such
|
|
that:
|
|
|
|
$$ \lceil xy \rceil \neq \lceil x \rceil \cdot \lfloor y \rfloor $$
|
|
|
|
Therefore the statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
Prove each of the following statements in 23-33.
|
|
|
|
23. For any real number $x$, if $x$ is not an integer, then
|
|
$\lfloor x \rfloor + \lfloor -x \rfloor = -1$.
|
|
|
|
Suppose $x$ is any real number where $x$ is not an integer.
|
|
|
|
By the definition of floor, since $x$ is not an integer, $x$ can be expressed
|
|
as:
|
|
|
|
$$ \lfloor x \rfloor = n \Leftrightarrow n < x < n + 1 $$
|
|
|
|
for $n$ is some integer.
|
|
|
|
Note that since $x$ is not an integer there is no "or equal to" here.
|
|
|
|
We can then multiply the inequality by $-1$:
|
|
|
|
$$ -n > -x > -n - 1 $$
|
|
|
|
Where $-n - 1$ is an integer by the product and difference of integers. Since
|
|
$-n - 1$ is an integer, it then follows:
|
|
|
|
$$ \lfloor -x \rfloor = -n - 1 $$
|
|
|
|
$$ \lfloor x \rfloor + \lfloor -x \rfloor = n + (-n - 1) = -1 $$
|
|
|
|
Therefore:
|
|
|
|
$$ \lfloor x \rfloor + \lfloor -x \rfloor = -1 $$
|
|
|
|
Q.E.D.
|
|
|
|
24. For any integer $m$ and any real number $x$, if $x$ is not an integer, then
|
|
$\lfloor x \rfloor + \lfloor m - x \rfloor = m - 1$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $x$ is any real number where $x$ is not an integer, and suppose $m$ is
|
|
any an integer.
|
|
|
|
Since $m$ is an integer and $x$ is not an integer, then $m - x$ is not an
|
|
integer by the difference of integers.
|
|
|
|
It follows then that:
|
|
|
|
$$ \lfloor m - x \rfloor = n \Leftrightarrow n < m - x < n + 1 $$
|
|
|
|
for some integer $n$.
|
|
|
|
Let's then subtract $m$ from the inequality:
|
|
|
|
$$ n - m < -x < n + 1 - m $$
|
|
|
|
And multiply the inequality by $-1$:
|
|
|
|
$$ m - n > x > m - n - 1 $$
|
|
|
|
Rewritten:
|
|
|
|
$$ m - n - 1 < x < m - n $$
|
|
|
|
Since $m - n - 1$ is an integer, this means that:
|
|
|
|
$$ \lfloor x \rfloor = m - n - 1 $$
|
|
|
|
By substitution then:
|
|
|
|
$$ \lfloor x \rfloor + \lfloor m - x \rfloor = (m - n - 1) + (n) $$
|
|
|
|
$$ \lfloor x \rfloor + \lfloor m - x \rfloor = m - 1 $$
|
|
|
|
Q.E.D.
|
|
|
|
25. For every real number $x$,
|
|
$\left\lfloor \dfrac{\left\lfloor \dfrac{x}{2}\right\rfloor}{2} \right\rfloor = \left\lfloor \dfrac{x}{4} \right\rfloor$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $x$ is any real number.
|
|
|
|
Let $n = \left\lfloor \dfrac{x}{2} \right\rfloor$.
|
|
|
|
Note that $n$ is automatically an integer due to floor always outputting an
|
|
integer.
|
|
|
|
By the definition of floor, since $n$ is an integer:
|
|
|
|
$$ \left\lfloor \frac{x}{2} \right\rfloor = n \Leftrightarrow n \leq \frac{x}{2} < n + 1 $$
|
|
|
|
__Case where $n$ is even:_
|
|
|
|
Since $n$ is even, $n = 2k$ for some integer $k$.
|
|
|
|
By substitution:
|
|
|
|
$$ \left\lfloor \frac{x}{2} \right\rfloor = 2k \Leftrightarrow 2k \leq \frac{x}{2} < 2k + 1 $$
|
|
|
|
We can then multiply out our inequality:
|
|
|
|
$$ 2(2k) \leq x < 2(2k) + 2 $$
|
|
|
|
$$ 4k \leq x < 4k + 2 $$
|
|
|
|
We then divide by $4$:
|
|
|
|
$$ k \leq \frac{x}{4} < k + \frac{1}{2} $$
|
|
|
|
By substitution then:
|
|
|
|
$$ \left\lfloor \frac{x}{4} \right\rfloor = k $$
|
|
|
|
__Case where $n$ is odd:_
|
|
|
|
Since $n$ is odd, $n = 2k + 1$ for some integer $k$.
|
|
|
|
By substitution:
|
|
|
|
$$ \left\lfloor \frac{x}{2} \right\rfloor = 2k + 1 \Leftrightarrow 2k + 1 \leq \frac{x}{2} < (2k + 1) + 1 $$
|
|
|
|
We can then multiply out our inequality:
|
|
|
|
$$ 2(2k + 1) \leq x < 2(2k + 1) + 2 $$
|
|
|
|
$$ 4k + 2 \leq x < 4k + 4 $$
|
|
|
|
We then divide by $4$:
|
|
|
|
$$ k + \frac{1}{2} \leq \frac{x}{4} < k + 1 $$
|
|
|
|
By substitution then:
|
|
|
|
$$ \left\lfloor \frac{x}{4} \right\rfloor = k $$
|
|
|
|
Thus in both cases we have shown the given statement to be true for all real
|
|
numbers $x$.
|
|
|
|
Q.E.D.
|
|
|
|
26. For every real number $x$, if $x - \lfloor x \rfloor < \dfrac{1}{2}$ then
|
|
$\lfloor 2x \rfloor = 2\lfloor x \rfloor$.
|
|
|
|
Suppose $x$ is any real number where $x - \lfloor x \rfloor < \dfrac{1}{2}$.
|
|
|
|
$$ x - \lfloor x \rfloor < \frac{1}{2} $$
|
|
|
|
Multiplying by $2$ gets us:
|
|
|
|
$$ 2x - 2\lfloor x \rfloor < 1 $$
|
|
|
|
Now add $2\lfoor x \rfloor$ to both sides:
|
|
|
|
$$ 2x < 2\lfloor x \rfloor + 1 $$
|
|
|
|
By definition of floor $\lfoor x \rfloor \leq x$. Thus:
|
|
|
|
$$ 2\lfloor x \rfloor \leq 2x $$
|
|
|
|
Putting these two inequalities together shows:
|
|
|
|
$$ 2\lfloor x \rfloor \leq 2x < 2\lfloor x \rfloor + 1 $$
|
|
|
|
By definition of floor, this means then that:
|
|
|
|
$$ \lfloor 2x \rfloor = 2\lfloor x \rfloor $$
|
|
|
|
Which is what was to be shown.
|
|
|
|
Q.E.D.
|
|
|
|
27. For every real number $x$, if $x - \lfloor x \rfloor \geq \dfrac{1}{2}$ then
|
|
$\lfloor 2x \rfloor = 2\lfloor x \rfloor + 1$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $x$ is any real number where $x - \lfloor x \rfloor \geq \dfrac{1}{2}$.
|
|
|
|
By the definition of floor, one can express some integer $n$ such that:
|
|
|
|
$$ \lfloor x \rfloor = n \Leftrightarrow n \leq x < n + 1 $$
|
|
|
|
Then subtract n from the inequality:
|
|
|
|
$$ 0 \leq x - n < 1 $$
|
|
|
|
We know that $x - \lfloor x \rfloor \geq \dfrac{1}{2}$.
|
|
|
|
$$ \frac{1}{2} \leq x - n < 1 $$
|
|
|
|
Multiply all sides by $2$:
|
|
|
|
$$ 1 \leq 2(x - n) < 2 $$
|
|
|
|
By substitution:
|
|
|
|
$$ \lfloor 2(x - n) \rfloor = 1 $$
|
|
|
|
_[To get to the form we want, we have to express $x$ as a further expression of
|
|
$n$]_
|
|
|
|
We can rewrite $x$ as $x = x + n - n$:
|
|
|
|
Since $n = \lfloor x \rfloor$:
|
|
|
|
$$ 2x = 2(n + (x - n)) $$
|
|
|
|
$$ 2x = 2n + 2(x - n) $$
|
|
|
|
Then take the floor:
|
|
|
|
$$ \lfloor 2x \rfloor = \lfloor 2n + 2(x - n) \rfloor $$
|
|
|
|
Since $2n$ is an integer:
|
|
|
|
$$ = 2n + \lfloor 2(x - n) \rfloor $$
|
|
|
|
And we know that $\lfloor 2(x - n) \rfloor = 1$, so substitute:
|
|
|
|
$$ = 2n + 1 $$
|
|
|
|
And we know $n = \lfoor x \rfloor$, so substitute again:
|
|
|
|
$$ = 2\lfoor x \rfloor + 1 $$
|
|
|
|
Therefore $\lfloor 2x \rfloor = 2\lfloor x \rfloor + 1$.
|
|
|
|
Q.E.D.
|
|
|
|
28. For any odd integer $n$,
|
|
|
|
$$ \left\lfloor \frac{n^2}{4} \right\rfloor = \left(\frac{n - 1}{2}\right)\left(\frac{n + 1}{2}\right) $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any odd integer.
|
|
|
|
Since $n$ is odd, $n = 2k + 1$ for some integer $k$.
|
|
|
|
Then, by substitution:
|
|
|
|
$$ \left\lfloor \frac{n^2}{4} \right\rfloor = \left\lfloor \frac{(2k + 1)^2}{4} \right\rfloor $$
|
|
|
|
$$ = \left\lfloor \frac{(2k + 1)(2k + 1)}{4} \right\rfloor $$
|
|
|
|
$$ = \left\lfloor \frac{4k^2 + 4k + 1}{4} \right\rfloor $$
|
|
|
|
$$ = \left\lfloor \frac{4k^2 + 4k}{4} + \frac{1}{4} \right\rfloor $$
|
|
|
|
$$ = \left\lfloor k^2 + k + \frac{1}{4} \right\rfloor $$
|
|
|
|
By the definition of floor, since $k^2 + k$ is an integer:
|
|
|
|
$$ k^2 + k \leq k^2 + k + \frac{1}{4} < k^2 + k + 1 $$
|
|
|
|
It then follows that:
|
|
|
|
$$ = \left\lfloor k^2 + k + \frac{1}{4} \right\rfloor = k^2 + k $$
|
|
|
|
Now, also by substitution:
|
|
|
|
$$ \left(\frac{n - 1}{2}\right)\left(\frac{n + 1}{2}\right) = \left(\frac{(2k + 1) - 1}{2}\right)\left(\frac{(2k + 1) + 1}{2}\right) $$
|
|
|
|
$$ = \left(\frac{2k}{2}\right)\left(\frac{2k + 2}{2}\right) $$
|
|
|
|
$$ = k(k + 1) $$
|
|
|
|
$$ k^2 + k $$
|
|
|
|
We have thus shown that the two expressions are equal:
|
|
|
|
$$ \left\lfloor \frac{n^2}{4} \right\rfloor = \left(\frac{n - 1}{2}\right)\left(\frac{n + 1}{2}\right) $$
|
|
|
|
Q.E.D.
|
|
|
|
29. For any odd integer $n$,
|
|
|
|
$$ \left\lceil \frac{n^2}{4} \right\rceil = \frac{n^2 + 3}{4} $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any odd integer.
|
|
|
|
Since $n$ is an odd integer, $n = 2k + 1$ for some integer $k$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ \left\lceil \frac{n^2}{4} \right\rceil = \left\lceil \frac{(2k + 1)^2}{4} \right\rceil $$
|
|
|
|
$$ = \left\lceil \frac{(2k + 1)(2k + 1)}{4} \right\rceil $$
|
|
|
|
$$ = \left\lceil \frac{4k^2 + 4k + 1}{4} \right\rceil $$
|
|
|
|
$$ = \left\lceil k^2 + k + \frac{1}{4} \right\rceil $$
|
|
|
|
By the definition of ceiling:
|
|
|
|
$$ k^2 + k < k^2 + k + \frac{1}{4} \leq k^2 + k + 1 $$
|
|
|
|
It then follows that
|
|
|
|
$$ \lceil k^2 + k + \frac{1}{4} \rceil = k^2 + k + 1 $$
|
|
|
|
Then by substitution:
|
|
|
|
$$ \frac{n^2 + 3}{4} = \frac{(2k + 1)^2 + 3}{4} $$
|
|
|
|
$$ = \frac{(2k + 1)(2k + 1) + 3}{4} $$
|
|
|
|
$$ = \frac{4k^2 + 4k + 1 + 3}{4} $$
|
|
|
|
$$ = \frac{4k^2 + 4k + 4}{4} $$
|
|
|
|
$$ = k^2 + k + 1 $$
|
|
|
|
Thus we have shown that these two expressions are equal.
|
|
|
|
$$ \left\lceil \frac{n^2}{4} \right\rceil = \frac{n^2 + 3}{4} $$
|
|
|
|
Q.E.D.
|
|
|
|
30. For every integer $n$,
|
|
$\left\lfloor \dfrac{n}{2} \right\rfloor + \left\lceil \frac{n}{2} \right\rceil = n$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ is any integer.
|
|
|
|
_Case where $n$ is even:_
|
|
|
|
Since $n$ is even, $n = 2k$ for some integer $k$.
|
|
|
|
By substitution:
|
|
|
|
$$ \left\lfloor \frac{n}{2} \right\rfloor + \left\lceil \frac{n}{2} \right\rceil = \left\lfloor \frac{2k}{2} \right\rfloor + \left\lceil \frac{2k}{2} \right\rceil $$
|
|
|
|
$$ = \lfloor k \rfloor + \lceil k \rceil $$
|
|
|
|
Since $k$ is an integer, then we know that:
|
|
|
|
$$ \lfloor k \rfloor = k \quad \text{ and } \lceil k \rceil = k $$
|
|
|
|
So:
|
|
|
|
$$ = \lfloor k \rfloor + \lceil k \rceil = k + k = 2k = n $$
|
|
|
|
_Case where $n$ is odd:_
|
|
|
|
Since $n$ is odd, $n = 2k + 1$ for some integer $k$.
|
|
|
|
By substitution:
|
|
|
|
$$ \left\lfloor \frac{n}{2} \right\rfloor + \left\lceil \frac{n}{2} \right\rceil = \left\lfloor \frac{2k + 1}{2} \right\rfloor + \left\lceil \frac{2k + 1}{2} \right\rceil $$
|
|
|
|
$$ = \left\lfloor \frac{2k}{2} + \frac{1}{2} \right\rfloor + \left\lceil \frac{2k}{2} + \frac{1}{2} \right\rceil $$
|
|
|
|
$$ = \left\lfloor k + \frac{1}{2} \right\rfloor + \left\lceil k + \frac{1}{2} \right\rceil $$
|
|
|
|
By the definition of floor:
|
|
|
|
$$ k \leq k + \frac{1}{2} < k + 1 $$
|
|
|
|
So:
|
|
|
|
$$ \left\lfloor k + \frac{1}{2} \right\rfloor = k $$
|
|
|
|
By the definition of ceiling:
|
|
|
|
$$ k < k + \frac{1}{2} \leq k + 1 $$
|
|
|
|
So:
|
|
|
|
$$ \left\lceil k + \frac{1}{2} \right\rceil = k + 1 $$
|
|
|
|
Then:
|
|
|
|
$$ = \left\lfloor k + \frac{1}{2} \right\rfloor + \left\lceil k + \frac{1}{2} \right\rceil = k + k + 1 $$
|
|
|
|
$$ = 2k + 1 = n $$
|
|
|
|
In both cases we have shown that the given equation is true. Therefore we have
|
|
shown that the given equation is true for every integer $n$.
|
|
|
|
Q.E.D.
|
|
|
|
31. For every integer $n$,
|
|
$\left\lfloor \dfrac{\left\lceil \dfrac{n}{2} \right\rceil}{3} \right\rfloor = \left\lfloor \dfrac{n}{6} \right\rfloor$.
|
|
|
|
Omitted.
|
|
|
|
32. For every integer $n$,
|
|
$\left\lceil \dfrac{\left\lceil \frac{n}{2} \right\rceil}{3} \right\rceil = \left\lceil \dfrac{n}{6} \right\rceil$
|
|
|
|
Omitted.
|
|
|
|
33. A necessary and sufficient condition for an integer $n$ to be divisible by a
|
|
nonzero integer $d$ is that
|
|
$n = \left\lfloor \dfrac{n}{d} \right\rfloor \cdot d$. In other words, for
|
|
every integer $n$ and nonzero integer $d$,
|
|
|
|
a. if $d \mid n$, then $n = \left\lfloor \dfrac{n}{d} \right\rfloor \cdot d$.
|
|
|
|
b. if $n = \left\lfloor \dfrac{n}{d} \right\rfloor \cdot d$ then $d \mid n$.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
**Exercise Set 4.7**
|
|
|
|
Page 248
|
|
|
|
1. Fill in the blanks in the following proof by contradiction that there is no
|
|
least positive real number.
|
|
|
|
**Proof:** Suppose not. That is, suppose that there is a least positive real
|
|
number $x$. _[We must deduce (a)]._ Consider the number $\dfrac{x}{2}$. Since
|
|
$x$ is a positive real number, $\dfrac{x}{2}$ is also (b). In addition, we can
|
|
deduce that $\dfrac{x}{2} < x$ by multiplying both sides of the inequality
|
|
$1 < 2$ by \(c\) and dividing (d). Hence $\dfrac{x}{2}$ is a positive real
|
|
number that is less than the least positive real number. This is a (e). _[Thus
|
|
the supposition is false, and so there is no least positive real number.]_
|
|
|
|
a. a contradiction
|
|
|
|
b. a positive real number.
|
|
|
|
c. $x$
|
|
|
|
d. both sides by $2$.
|
|
|
|
e. contradiction
|
|
|
|
2. Is $\dfrac{1}{0}$ an irrational number? Explain.
|
|
|
|
No. Since $\dfrac{1}{0}$ is undefined, it is not a number at all.
|
|
|
|
3. Use proof by contradiction to show that for every integer $n$, $3n + 2$ is
|
|
not divisible by $3$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that there is an integer $n$, such that $3n + 2$
|
|
is divisible by 3.
|
|
|
|
By definition of divisibility:
|
|
|
|
$$ 3n + 2 = 3k $$
|
|
|
|
for some integer $k$.
|
|
|
|
$$ 3n + 2 = 3k $$
|
|
|
|
$$ 2 = 3k - 3n $$
|
|
|
|
$$ 3k - 3n = 2 $$
|
|
|
|
$$ 3(k - n) = 2 $$
|
|
|
|
$$ k - n = \frac{2}{3} $$
|
|
|
|
Now, $k - n$ is an integer by the difference of integers, but $\dfrac{2}{3}$ is
|
|
not an integer. So $k - n$ is an integer and is not an integer.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
4. Use proof by contradiction to show that for every integer $m$, $7m + 4$ is
|
|
not divisible by $7$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that there is an integer $m$ such that $7m + 4$ is
|
|
divisible by $7$.
|
|
|
|
By the definition of divisibility:
|
|
|
|
$$ 7m + 4 = 7k $$
|
|
|
|
for some integer $k$.
|
|
|
|
By the laws of algebra, this can be rewritten as:
|
|
|
|
$$ 7m + 4 = 7k $$
|
|
|
|
$$ 7m - 7k = -4 $$
|
|
|
|
$$ 7k - 7m = 4 $$
|
|
|
|
$$ 7(k - m) = 4 $$
|
|
|
|
$$ k - m = \frac{4}{7} $$
|
|
|
|
Now, $k - m$ is an integer by the difference of integers, but $\dfrac{4}{7}$ is
|
|
not an integer. Therefore $k - m$ is an integer and not an integer.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
Carefully formulate the negations of each of the statements in 5-7. Then prove
|
|
each statement by contradiction.
|
|
|
|
5. There is no greatest even integer.
|
|
|
|
Negation: There is some greatest even integer.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose there is some greatest even integer $x$.
|
|
|
|
Since $x$ is an even integer, $x = 2k$ for some integer $k$. Then suppose there
|
|
is number $y$, such that $y = x + 2$.
|
|
|
|
By substitution:
|
|
|
|
$$ y = 2k + 2 $$
|
|
|
|
$$ y = 2(k + 1) $$
|
|
|
|
Now, $k + 1$ is an integer by the sum of integers. Since $y$ is expressed in the
|
|
form of $2 \cdot (\text{an integer})$, $y$ is an integer by the product of
|
|
integers and an even integer by the definition of even. Since $y = x + 2$, $x$
|
|
is not the greatest even integer, since $y > x$ and $y$ is even. So, $x$ is not
|
|
the greatest even integer and $x$ is the greatest even integer.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
6. There is no greatest negative real number.
|
|
|
|
Negation: There is some greatest negative real number.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that there is some greatest negative real number
|
|
$y$. In other words, there exists some negative real number $y$ such that for
|
|
all negative real numbers $x$, $y \geq x$.
|
|
|
|
Let $Y = \dfrac{y}{2}$.
|
|
|
|
$\dfrac{y}{2} > y$ since $y$ is negative, $Y$ is a negative number that is
|
|
greater than $y$. So, $y$ is not the greatest negative real number and $y$ is
|
|
the greatest negative real number.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
7. There is no least positive rational number.
|
|
|
|
Negation: There is some least positive rational number.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That 8s, suppose that there is some least positive rational number,
|
|
$y$.
|
|
|
|
Since $y$ is a rational number, $y = \dfrac{a}{b}$ where both $a$ and $b$ are
|
|
integers and $b \neq 0$.
|
|
|
|
Let $Y = \dfrac{y}{2}$.
|
|
|
|
By substitution:
|
|
|
|
$$ Y = \frac{\dfrac{a}{b}}{2} $$
|
|
|
|
$$ Y = \frac{a}{2b} $$
|
|
|
|
Now, $2b$ is an integer by the product of integers and $2b \neq 0$ by the zero
|
|
product property. Thus, $Y$ is a rational number since both $a$ and $2b$ are
|
|
integers and $Y$ has a nonzero denominator. $Y < y$ as
|
|
$\dfrac{a}{2b} < \dfrac{a}{b}$. Therefore $y$ is not the least positive rational
|
|
number and $y$ is the least positive rational number.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
8. Fill in the blanks for the following proof that the difference of any
|
|
rational number and any irrational number is irrational.
|
|
|
|
**Proof (by contradiction):**
|
|
|
|
Suppose not. That is, suppose that there exist (a) $x$ and (b) $y$ such that
|
|
$x - y$ is rational. By definition of rational, there exist integers $a$, $b$,
|
|
$c$, and $d$ with $b \neq 0$ and $d \neq 0$ so that $x = $ \(c\) and $x - y =$
|
|
(d). By substitution,
|
|
|
|
$$ \frac{a}{b} - y = \frac{c}{d} $$
|
|
|
|
Adding $y$ and subtracting $\dfrac{c}{d}$ on both sides gives
|
|
|
|
$$ y = \text{(e)} \quad \text{ by substitution} $$
|
|
|
|
$$ = \frac{ad}{bd} - \frac{bc}{bd} $$
|
|
|
|
$$ = \frac{ad - bc}{bd} \quad \text{ by algebra} $$
|
|
|
|
a. some rational number
|
|
|
|
b. some irrational number
|
|
|
|
c. $\dfrac{a}{b}$
|
|
|
|
d. $\dfrac{c}{d}$
|
|
|
|
e. $\dfrac{a}{b} - \dfrac{c}{d}$
|
|
|
|
f. integers
|
|
|
|
g. integers
|
|
|
|
h. zero product property
|
|
|
|
i. rational
|
|
|
|
Now both $ad - bc$ and $bd$ are integers and products and differences of (f) are
|
|
(g). And $bd \neq 0$ by the (h). Hence $y$ is a ratio of integers with a nonzero
|
|
denominator, and thus $y$ is (i) by definition of rational. We therefore have
|
|
both that $y$ is irrational and that $y$ is rational, which is a contradiction.
|
|
_[Thus the supposition is false and the statement to be proved is true.]_
|
|
|
|
9.
|
|
|
|
a. When asked to prove that the difference of any irrational number and any
|
|
rational number is irrational, a student began, "Suppose not. That is, suppose
|
|
the difference of any irrational number and any rational number is rational."
|
|
What is wrong with beginning the proof in this way? (_Hint:_ If needed, review
|
|
the answer to exercise 11 in Section 3.2.)
|
|
|
|
The problem is that the negation of a universal is an existential, and the
|
|
student did not apply that. Instead the statement should be:
|
|
|
|
"Suppose not. That is, suppose the difference of _some_ irrational number and
|
|
_some_ rational number is rational."
|
|
|
|
b. Prove that the difference of any irrational number and any rational number is
|
|
irrational.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose there is some irrational number $x$ and some
|
|
rational number $y$ such that $x - y$ is rational.
|
|
|
|
Since $y$ is rational and $x - y$ is rational, $y = \dfrac{a}{b}$ and
|
|
$x - y = \dfrac{c}{d}$ where $a$ $b$, $c$, and $d$ are integers and $b \neq 0$
|
|
and $d \neq 0$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ x - \dfrac{a}{b} = \dfrac{c}{d} $$
|
|
|
|
$$ x = \dfrac{c}{d} - \dfrac{a}{b} $$
|
|
|
|
$$ x = \dfrac{cb}{bd} - \dfrac{ad}{bd} $$
|
|
|
|
$$ x = \dfrac{cb - ad}{bd} $$
|
|
|
|
Now, $cb - ad$ is an integer by the difference and product of integers. $bd$ is
|
|
an integer by the product of integers and $bd \neq 0$ by the zero product
|
|
property. By the definition of rational numbers then, $x$ is a rational number.
|
|
Therefore, $x$ is a rational number and an irrational number.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
10. Let $S$ be the statement: For all positive real numbers $r$ and $s$,
|
|
$\sqrt{r + s} \neq \sqrt{r} + \sqrt{s}$. Statement $S$ is true, but the
|
|
following "proof" is incorrect. Find the mistake.
|
|
|
|
**"Proof by contradiction:** Suppose not, that is, suppose that for all positive
|
|
real numbers $r$ and $s$, $\sqrt{r + s} = \sqrt{r} + \sqrt{s}$. This means that
|
|
the equation will be true no matter what positive real numbers are substituted
|
|
for $r$ and $s$. So let $r = 9$ and $s = 16$. Then $r$ and $s$ are positive real
|
|
numbers and
|
|
|
|
$$ \sqrt{r + s} = \sqrt{9 + 16} = \sqrt{25} = 5 $$
|
|
|
|
whereas
|
|
|
|
$$ \sqrt{r} + \sqrt{s} = \sqrt{9} + \sqrt{16} = 3 + 4 k 7 $$
|
|
|
|
Since $5 \neq 7$, we have that $\sqrt{r + s} \neq \sqrt{r} + \sqrt{s}$, which
|
|
contradicts the supposition that $\sqrt{r + s} = \sqrt{r} + \sqrt{s}$. This
|
|
contradiction shows that the supposition is false, and hence statement $S$ is
|
|
true."
|
|
|
|
The mistake the student makes is in that they do not take the proper negation of
|
|
the universal statement. Instead of supposing the negation of the statement for
|
|
all $r$ and $s$, the student should have started with an existential quantifier
|
|
of the negating statement. It should have started with:
|
|
|
|
"Suppose not. That is, suppose that there exists some positive real numbers $r$
|
|
and $s$ such that $\sqrt{r + s} = \sqrt{r} + \sqrt{s}$."
|
|
|
|
The student also then goes to provide specific number examples for $r$ and $s$,
|
|
which is a technique generally reserved for proof by counterexample, not
|
|
contradiction. By providing a specific example for their (incorrect) universal
|
|
statement, they are not even proving for all, nor are they proving there exists
|
|
_some_. They are just proving that the negation is a contradiction for a
|
|
singular example.
|
|
|
|
11.
|
|
|
|
Let $T$ be the statement: The sum of any two rational numbers is rational. Then
|
|
$T$ is true, but the following "proof" is incorrect. Find the mistake.
|
|
|
|
**"Proof by contradiction:** Suppose not. That is, suppose that the sum of any
|
|
two rational numbers is not rational. This means that no matter what two
|
|
rational numbers are chosen their sum is not rational. Now both $1$ and $3$ are
|
|
rational because $1 = \dfrac{1}{1}$ and $3 = \dfrac{3}{1}$, and so both are
|
|
ratios of integers with a nonzero denominator. Hence, by supposition, the sum of
|
|
$1$ and $3$, which is $4$, is not rational. But $4$ is rational because
|
|
$4 = \dfrac{4}{1}$, which is a ratio of integers with a nonzero denominator.
|
|
Hence $4$ is both rational and not rational, which is a contradiction. This
|
|
contradiction shows that the supposition is false, and hence statement $T$ is
|
|
true.
|
|
|
|
The mistake the student makes is in that they do not take the proper negation of
|
|
the universal statement. Instead of supposing that the sum of _any_ two rational
|
|
numbers is not rational, the student should have started with an existential
|
|
quantifier of the negating statement. It should have started with:
|
|
|
|
"Suppose not. That is, suppose that there are some rational numbers $x$ and $y$
|
|
such that $x + y$ is not rational."
|
|
|
|
12. Let $R$ be the statement: The square root of any irrational number is
|
|
irrational.
|
|
|
|
a. Write the negation for $R$.
|
|
|
|
Negation: The square root of some irrational number is rational.
|
|
|
|
b. Prove $R$ by contradiction.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose there is some irrational number $x$ such that
|
|
$\sqrt{x}$ is rational.
|
|
|
|
Since $\sqrt{x}$ is rational, $\sqrt{x} = \dfrac{a}{b}$ where $a$ and $b$ are
|
|
integers and $b \neq 0$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ \sqrt{x} = \frac{a}{b} $$
|
|
|
|
$$ (\sqrt{x})^2 = \left(\frac{a}{b}\right)^2 $$
|
|
|
|
$$ x = \frac{a^2}{b^2} $$
|
|
|
|
Now, $a^2$ is an integer by the product of integers. Additionally, $b^2$ is an
|
|
integer by the product of integers and $b^2 \neq 0$ by the zero product
|
|
property. Thus $x$ is a rational number. Therefore $x$ is a rational number and
|
|
$x$ is an irrational number.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
13. Let $S$ be the statement: The product of any irrational number and any
|
|
nonzero rational number is irrational.
|
|
|
|
a. Write the negation for $S$.
|
|
|
|
The product of some irrational number and some nonzero rational number is
|
|
rational.
|
|
|
|
b. Prove $S$ by contradiction.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that there exists some irrational number $x$ and
|
|
some nonzero rational number $y$ such that $xy$ is rational.
|
|
|
|
Since $y$ is a nonzero rational number, $y = \dfrac{a}{b}$ where $a$ and $b$ are
|
|
integers and $a \neq 0$ and $b \neq 0$. Since $xy$ is rational,
|
|
$xy = \dfrac{c}{d}$ where $c$ and $d$ are integers and $d \neq 0$.
|
|
|
|
Then, by susbstitution:
|
|
|
|
$$ xy = x\left(\frac{a}{b}\right) = \frac{c}{d} $$
|
|
|
|
$$ x\left(\frac{a}{b}\right) = \frac{c}{d} $$
|
|
|
|
$$ x = \frac{c}{d}\left(\frac{b}{a}\right) $$
|
|
|
|
$$ x = \frac{cb}{ad} $$
|
|
|
|
Now, $cd$ is an integer by the product of integers. Additionally, $ad$ is an
|
|
integer by the product of integers and $ad \neq 0$ by the zero product property.
|
|
Thus, $x$ is a rational number. Therefore $x$ is a rational number and $x$ is an
|
|
irrational number.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
14. Let $T$ be the statement: For every integer $a$, if $a \mod 6 = 3$ , then
|
|
$a \mod 3 \neq 2$.
|
|
|
|
a. Write a negation for $T$.
|
|
|
|
Negation: For some integer $a$, $a \mod 6 = 3$ and $a \mod 3 = 2$.
|
|
|
|
b. Prove $T$ by contradiction.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose for some integer $a$, $a \mod 6 = 3$ and
|
|
$a \mod 3 = 2$.
|
|
|
|
Since $a \mod 6 = 3$, then by the quotient remainder theorem:
|
|
|
|
$$ a = 6q + 3 $$
|
|
|
|
for some integer $q$.
|
|
|
|
$$ a = 6q + 3 $$
|
|
|
|
Then, since $a \mod 3 = 2$, then by the quotient remainder theorem:
|
|
|
|
$$ a = 3s + 2 $$
|
|
|
|
for some integer $s$.
|
|
|
|
$$ 6q + 3 = 3s + 2 $$
|
|
|
|
$$ 6q - 3s = 2 - 3 $$
|
|
|
|
$$ 3(2q - s) = -1 $$
|
|
|
|
$$ (2q - s) = -\frac{1}{3} $$
|
|
|
|
Now, $2q - s$ is an integer by the product and difference of integers, but
|
|
$-\dfrac{1}{3}$ is not an integer. So $2q - s$ is an integer and $2q - s$ is not
|
|
an integer.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
15. Do there exists integers $a$, $b$, and $c$ such that $a$, $b$, and $c$ are
|
|
all odd and $a^2 + b^2 = c^2$? Prove your answer.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that there exist odd integers $a$, $b$, and $c$
|
|
such that $a^2 + b^2 = c^2$.
|
|
|
|
Since $a$ and $b$ are odd, $a = 2m + 1$, $b = 2n + 1$ for some integers $m$ and
|
|
$n$.
|
|
|
|
By substitution:
|
|
|
|
$$ a^2 = (2m + 1)^2 = 4m^2 + 4m + 1 = 4m(m + 1) + 1 $$
|
|
|
|
$$ b^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 4n(n + 1) + 1 $$
|
|
|
|
Therefore:
|
|
|
|
$$ a^2 \equiv 1 (\mod 4), b^2 \equiv 1 (\mod 4) $$
|
|
|
|
Adding these congruences leaves:
|
|
|
|
$$ a^2 + b^2 \equiv 1 + 1 \equiv 2 (\mod 4) $$
|
|
|
|
So the left hand satisfies:
|
|
|
|
$$ a^2 + b^2 \equiv 2 (\mod 4) $$
|
|
|
|
Now, consider $c$. Since $c$ is odd, $c = 2k + 1$ for some integer $k$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ c^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1 $$
|
|
|
|
So:
|
|
|
|
$$ c^2 \equiv 1 (\mod 4) $$
|
|
|
|
But we were given $a^2 + b^2 = c^2$, so this implies:
|
|
|
|
$$ a^2 + b^2 = c^2 (\mod 4) $$
|
|
|
|
This is,
|
|
|
|
$$ 2 \equiv 1 (\mod 4) $$
|
|
|
|
This is impossible.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
Prove each statement in 16-19 by contradiction.
|
|
|
|
16. For all odd integers $a$ and $b$, $b^2 - a^2 \neq 4$. (_Hint:_
|
|
$b^2 - a^2 = (b + a)(b - a)$ and the only way to factor $4$ is either
|
|
$4 = 2 \cdot 2$ or $4 = 4 \cdot 1$.)
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that there exists some odd integers $a$ and $b$
|
|
such that $b^2 - a^2 = 4$.
|
|
|
|
Consider:
|
|
|
|
$$ b^2 - a^2 = (b + a)(b - a) = 4 $$
|
|
|
|
By the sum and difference of integers, both $b + a$ and $b - a$ are integers.
|
|
The only integers where $4$ is factored by are $2 \cdot 2$ and $1 \cdot 4$.
|
|
|
|
Thus there are only some cases to consider:
|
|
|
|
_Case $(b + a) = 2$ and $(b - a) = 2$:_
|
|
|
|
$$ b + a = b - a $$
|
|
|
|
$$ b = b - 2a $$
|
|
|
|
$$ b - b = -2a $$
|
|
|
|
$$ 0 = -2a $$
|
|
|
|
$$ a = 0 $$
|
|
|
|
Since $a = 0$, $a$ is an even integer. Therefore $a$ is an even integer and $a$
|
|
is an odd integer. This is a contradiction.
|
|
|
|
_Case $(b + a) = 4$ and $(b - a) = 1$:_
|
|
|
|
$$ b + a = 4 $$
|
|
|
|
$$ b = 4 - a $$
|
|
|
|
$$ b - a = 1 $$
|
|
|
|
By substitution:
|
|
|
|
$$ b - (4 - a) = 1 $$
|
|
|
|
$$ b - 4 + a = 1 $$
|
|
|
|
$$ b + a = 5 $$
|
|
|
|
So, $b + a = 5$ and $b + a = 4$. This is a contradiction.
|
|
|
|
_Case $(b + a) = 1$ and $(b - a) = 4$:_
|
|
|
|
$$ b + a = 1 $$
|
|
|
|
$$ b = 1 - a $$
|
|
|
|
By substitution:
|
|
|
|
$$ b - a = 4 $$
|
|
|
|
$$ b - (1 - a) = 4 $$
|
|
|
|
$$ b - 1 + a = 4 $$
|
|
|
|
$$ b + a = 5 $$
|
|
|
|
So $b + a = 1$ and $b + a = 5$. This is a contradiction.
|
|
|
|
In all cases, this is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
17. For all prime numbers $a$, $b$, and $c$, $a^2 + b^2 \neq c^2$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that for some prime numbers $a$, $b$, and $c$,
|
|
$a^2 + b^2 = c^2$.
|
|
|
|
Consider that:
|
|
|
|
$$ a^2 + b^2 = c^2 $$
|
|
|
|
$$ a^2 = c^2 - b^2 $$
|
|
|
|
$$ a^2 = (c - b)(c + b) $$
|
|
|
|
Since $a$, $b$, and $c$, are prime. By the unique prime factorization of
|
|
integers theorem, the only possible values for $c - b$ and $c + b$ are:
|
|
|
|
$$ a^2 = (a^2)(1) = (c - b)(c + b) $$
|
|
|
|
$$ a^2 = (1)(a^2) = (c - b)(c + b) $$
|
|
|
|
$$ a^2 = (a)(a) = (c - b)(c + b) $$
|
|
|
|
Let's check these cases:
|
|
|
|
__Case $c - b = a^2$ and $c + b = 1$:_
|
|
|
|
Add:
|
|
|
|
$$ (c - b) + (c + b) = a^2 + 1 $$
|
|
|
|
$$ 2c = a^2 + 1 $$
|
|
|
|
$$ c = \frac{a^2 + 1}2 $$
|
|
|
|
Subtract:
|
|
|
|
$$ (c - b) - (c + b) = a^2 - 1 $$
|
|
|
|
$$ -2b = a^2 - 1 $$
|
|
|
|
$$ b = -\frac{a^2 - 1}{2} $$
|
|
|
|
$$ b = \frac{1 - a^2}{2} $$
|
|
|
|
Now, $a$, $b$, and $c$ are prime number, but checking for small prime(s) shows:
|
|
|
|
$$ a = 2 $$
|
|
|
|
$$ c = \frac{(2)^2 + 1}2 $$
|
|
|
|
$$ c = \frac{5}2 $$
|
|
|
|
$$ b = \frac{1 - a^2}{2} $$
|
|
|
|
$$ b = \frac{1 - (2)^2}{2} $$
|
|
|
|
$$ b = -\frac{3}{2} $$
|
|
|
|
So $b$ and $c$ are not prime numbers and $b$ and $c$ are not prime numbers. This
|
|
is a contradiction.
|
|
|
|
__Case $c - b = a$ and $c + b = a$:_
|
|
|
|
Add:
|
|
|
|
$$ (c - b) + (c + b) = a + a $$
|
|
|
|
$$ 2c = 2a $$
|
|
|
|
$$ c = a $$
|
|
|
|
Subtract:
|
|
|
|
$$ (c - b) - (c - b) = a - a $$
|
|
|
|
$$ -2b = 0 $$
|
|
|
|
$$ b = 0 $$
|
|
|
|
So, $b$ is a prime number and $b$ is not a prime number. This is a
|
|
contradiction.
|
|
|
|
__Case $c - b = 1$ and $c + b = a^2$:_
|
|
|
|
Add:
|
|
|
|
$$ (c - b) + (c + b) = 1 + a^2 $$
|
|
|
|
$$ 2c = 1 + a^2 $$
|
|
|
|
$$ c = \frac{1 + a^2}{2} $$
|
|
|
|
Subtract:
|
|
|
|
$$ (c + b) - (c - b) = 1 - a^2 $$
|
|
|
|
$$ c - b - c - b = 1 - a^2 $$
|
|
|
|
$$ -2b = 1 + a^2 $$
|
|
|
|
$$ b = -\frac{1 - a^2}{2} $$
|
|
|
|
$$ b = \frac{a^2 + 1}{2} $$
|
|
|
|
Now, $a$, $b$, and $c$ are prime number, but checking for small prime(s) shows:
|
|
|
|
$$ a = 2 $$
|
|
|
|
$$ b = \frac{(2)^2 + 1}{2} $$
|
|
|
|
$$ b = \frac{5}{2} $$
|
|
|
|
$$ c = \frac{1 + (2)^2}{2} $$
|
|
|
|
$$ c = \frac{5}{2} $$
|
|
|
|
So $b$ and $c$ are not prime numbers and $b$ and $c$ are prime numbers. This is
|
|
a contradiction.
|
|
|
|
In all cases, this is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
18. If $a$ and $b$ are rational numbers, $b \neq 0$, and $r$ is an irrational
|
|
number, then $a + br$ is irrational.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that $a$ and $b$ are rational numbers, $b \neq 0$,
|
|
and $r$ is an irrational number and $a + br$ is rational.
|
|
|
|
Since $a$, $b$ and $a + br$ are rational numbers, $a = \dfrac{u}{t}$,
|
|
$b = \dfrac{v}{w}$, $a + br = \dfrac{x}{y}$ where $u$, $t$, $v$, $w$, $x$ and
|
|
$y$ are integers and $t \neq 0$, $v \neq 0$, $w \neq 0$, and $y \neq 0$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ a + br = \frac{x}{y} = \left(\frac{u}{t}\right) + \left(\frac{v}{w}\right)r $$
|
|
|
|
$$ \frac{x}{y} = \left(\frac{u}{t}\right) + \left(\frac{v}{w}\right)r $$
|
|
|
|
$$ \frac{x}{y} - \left(\frac{u}{t}\right) = \left(\frac{v}{w}\right)r $$
|
|
|
|
$$ \frac{xt}{ty} - \left(\frac{uy}{ty}\right) = \left(\frac{v}{w}\right)r $$
|
|
|
|
$$ \frac{xt - uv}{ty} = \left(\frac{v}{w}\right)r $$
|
|
|
|
$$ \left(\frac{xt - uv}{ty}\right)\left(\frac{w}{v}\right) = r $$
|
|
|
|
$$ \frac{w(xt - uv)}{tyv} = r $$
|
|
|
|
Now $w(xt - uv)$ is an integer by the product and difference of integers.
|
|
Additionally, $tyv$ is an integer by the product of integers and $tyv \neq 0$ by
|
|
the zero product property. Thus, by the definition of rational numbers, $r$ is
|
|
rational. Therefore $r$ is rational and $r$ is irrational.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
19. For any integer $n$, $n^2 - 2$ is not divisible by $4$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose there exists an integer $n$, such that $n^2 - 2$
|
|
is divisible by $4$.
|
|
|
|
Since $n^2 - 2$ is divisible by $4$, then $(n^2 - 2) \equiv 0 (\mod 4)$.
|
|
|
|
_Case $n$ is odd:_
|
|
|
|
Since $n$ is odd, $n = 2k + 1$ for some integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ n^2 - 2 = (2k + 1)^2 - 2 $$
|
|
|
|
$$ = 4k^2 + 4k + 1 - 2 $$
|
|
|
|
$$ = 4k^2 + 4k - 1 $$
|
|
|
|
$$ = 4(k^2 + k) - 1 $$
|
|
|
|
$$ 4(k^2 + k) - 1 \equiv -1 (\mod 4) $$
|
|
|
|
So $(n^2 - 2) \equiv -1 (\mod 4)$ and $(n^2 - 2) \equiv 0 (\mod 4)$. This is a
|
|
contradiction.
|
|
|
|
_Case $n$ is even:_
|
|
|
|
Since $n$ is even, $n = 2k$ for some integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ n^2 - 2 = (2k)^2 - 2 $$
|
|
|
|
$$ = 4k^2 - 2 $$
|
|
|
|
$$ = 4(k^2) - 2 $$
|
|
|
|
$$ 4(k^2) - 2 \equiv -2 (\mod 4) $$
|
|
|
|
So $(n^2 - 2) \equiv -2 (\mod 4)$ and $(n^2 - 2) \equiv 0 (\mod 4)$. This is a
|
|
contradiction.
|
|
|
|
In all cases, this is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
By definition of divisibility, there exists an integer $k$ so that $b = ak$.
|
|
|
|
20. Fill in the blanks in the following proof by contraposition that for every
|
|
integer $n$, if $5 \cancel{\mid} n^2$ then $5 \cancel{\mid} n$.
|
|
|
|
**Proof (by contraposition):** _[The contrapositive is: For every integer $n$,
|
|
if $5 \mid n$ then $5 \mid n^2$.]_ Suppose $n$ is any integer such that (a).
|
|
_[We must show that (b).]_ By definition of divisibility, $n =$ \(c\) for some
|
|
integer $k$. By substitution, $n^2 = $ (d) $= 5(5k^2)$. But $5k^2$ is an integer
|
|
because it is a product of integers. Hence $n^2 = 5 \cdot (\text{an integer})$,
|
|
and so (e) _[as was to be shown]._
|
|
|
|
a. $5 \mid n$
|
|
|
|
b. $5 \mid n^2$
|
|
|
|
c. $5k$
|
|
|
|
d. $(5k)^2$
|
|
|
|
e. $5 \mid n^2$
|
|
|
|
21. Consider the statement "For every integer $n$, if $n^2$ is odd then $n$ is
|
|
odd."
|
|
|
|
a. Write what you would suppose and what you would need to show to prove this
|
|
statement by contradiction.
|
|
|
|
**Supposition:** Suppose that for some integer $n$, $n^2$ is odd and $n$ is
|
|
even.
|
|
|
|
**Consequent:** Show that $n^2$ is even, a contradiction.
|
|
|
|
b. Write what you would suppose and what you would need to show to prove this
|
|
statement by contraposition.
|
|
|
|
**Supposition:** Suppose that $n$ is any even integer.
|
|
|
|
**Consequent:** Show that $n^2$ is even.
|
|
|
|
22. Consider the statement "For every real number $r$, if $r^2$ is irrational
|
|
then $r$ is irrational."
|
|
|
|
a. Write what you would suppose and what you would need to show to prove this
|
|
statement by contradiction.
|
|
|
|
**Supposition:** Suppose there is some real number $r$ such that $r^2$ is
|
|
irrational and $r$ is rational.
|
|
|
|
**Consequent:** Show that $r^2$ is rational, a contradiction.
|
|
|
|
b. Write what you would suppose and what you would need to show to prove this
|
|
statement by contraposition.
|
|
|
|
**Supposition:** Suppose $r$ is any real number such that $r$ is rational.
|
|
|
|
**Consequent:** Show that $r^2$ is rational.
|
|
|
|
Prove each of the statements in 23-25 in two ways: (a) by contraposition and (b)
|
|
by contradiction.
|
|
|
|
23. The negative of any irrational number is irrational.
|
|
|
|
a.
|
|
|
|
**Proof by contraposition:**
|
|
|
|
Suppose $x$ is any number such that $-x$ is rational.
|
|
|
|
Since $-x$ is rational, $-x = \dfrac{a}{b}$ where $a$ and $b$ are some integers
|
|
and $b \neq 0$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ -x = \dfrac{a}{b} $$
|
|
|
|
$$ x = -\dfrac{a}{b} $$
|
|
|
|
$$ x = \dfrac{-a}{b} $$
|
|
|
|
Now, $-a$ is an integer by the product of integers.
|
|
|
|
Therefore $x$ is a rational number.
|
|
|
|
Q.E.D.
|
|
|
|
b.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not, That is, suppose $x$ is some irrational number and $-x$ is
|
|
rational.
|
|
|
|
Since $-x$ is rational, $-x = \dfrac{a}{b}$ where $a$ and $b$ are some integers
|
|
and $b \neq 0$.
|
|
|
|
Then, by algebra:
|
|
|
|
$$ -x = \dfrac{a}{b} $$
|
|
|
|
$$ x = -\dfrac{a}{b} $$
|
|
|
|
$$ x = \dfrac{-a}{b} $$
|
|
|
|
Now, $-a$ is an integer by the product of integers. Therefore $x$ is a rational
|
|
number and $x$ is an irrational number.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
24. The reciprocal of any irrational number is irrational. (The **reciprocal**
|
|
of a nonzero real number $x$ is $\dfrac{1}{x}$.)
|
|
|
|
a.
|
|
|
|
**Proof by contraposition:**
|
|
|
|
Suppose $x$ is any number such that $\dfrac{1}{x}$ is rational.
|
|
|
|
Since $\dfrac{1}{x}$ is a rational number, $\dfrac{1}{x} = \dfrac{a}{b}$ where
|
|
$a$ and $b$ are some integers and $b \neq 0$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ \frac{1}{x} = \frac{a}{b} $$
|
|
|
|
$$ x = \frac{b}{a} $$
|
|
|
|
It has been established that $b$ is an integer. It has also been established
|
|
that $a$ is an integer, and since $\dfrac{1}{x}$ is rational, it follows that
|
|
$a \neq 0$.
|
|
|
|
Therefore $x$ is a rational number.
|
|
|
|
Q.E.D.
|
|
|
|
b.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is suppose there is an irrational number $x$ such that
|
|
$\dfrac{1}{x}$ is rational.
|
|
|
|
Since $\dfrac{1}{x}$ is a rational number, $\dfrac{1}{x} = \dfrac{a}{b}$ where
|
|
$a$ and $b$ are some integers and $b \neq 0$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ \frac{1}{x} = \frac{a}{b} $$
|
|
|
|
$$ x = \frac{b}{a} $$
|
|
|
|
It has been established that $b$ is an integer. It has also been established
|
|
that $a$ is an integer, and since $\dfrac{1}{x}$ is rational, it follows that
|
|
$a \neq 0$.
|
|
|
|
Therefore $x$ is a rational number and $x$ is an irrational number.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
25. For every integer $n$, if $n^2$ is odd then $n$ is odd.
|
|
|
|
a.
|
|
|
|
**Proof by contraposition:**
|
|
|
|
Suppose $n$ is any integer such that $n$ is even.
|
|
|
|
Since $n$ is even, $n = 2k$ for some integer $k$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ n^2 = (2k)^2 $$
|
|
|
|
$$ = 4k^2 $$
|
|
|
|
$$ = 2(2k^2) $$
|
|
|
|
Now, $2k^2$ is an integer by the product of integers.
|
|
|
|
Therefore $n^2$ is even by the definition of even.
|
|
|
|
b.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that $n$ is some integer such that $n^2$ is odd
|
|
and $n$ is even.
|
|
|
|
Since $n$ is even, $n = 2k$ for some integer $k$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ n^2 = (2k)^2 $$
|
|
|
|
$$ = 4k^2 $$
|
|
|
|
$$ = 2(2k^2) $$
|
|
|
|
Now, $2k^2$ is an integer by the product of integers. Therefore $n^2$ is even
|
|
and $n^2$ is odd.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
Use any method to prove the statements in 26-29.
|
|
|
|
26. For all integers $a$, $b$, and $c$, if $a \cancel{\mid} bc$ then
|
|
$a \cancel{\mid} b$.
|
|
|
|
**Proof by contraposition:**
|
|
|
|
Suppose $a$, $b$, and $c$ are any integers such that $a \mid b$.
|
|
|
|
Since $a \mid b$, by the definition of divisiblity, $b = ak$ for some integer
|
|
$k$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ bc = (ak)c $$
|
|
|
|
$$ = a(kc) $$
|
|
|
|
Now, $kc$ is an integer by the product of integers. Since $bc = a(kc)$, it
|
|
follows that $a \mid a(kc)$.
|
|
|
|
Therefore $a \mid bc$ by definition of divisibility.
|
|
|
|
27. For all positive real numbers $r$ and $s$,
|
|
$\sqrt{r + s} \neq \sqrt{r} + \sqrt{s}$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that for some positive real numbers $r$ and $s$,
|
|
$\sqrt{r + s} = \sqrt{r} + \sqrt{s}$.
|
|
|
|
$$ \sqrt{r + s} = \sqrt{r} + \sqrt{s} $$
|
|
|
|
$$ (\sqrt{r + s})^2 = (\sqrt{r} + \sqrt{s})^2 $$
|
|
|
|
$$ r + s = (\sqrt{r} + \sqrt{s})(\sqrt{r} + \sqrt{s}) $$
|
|
|
|
$$ r + s = (\sqrt{r})^2 + 2(\sqrt{r})(\sqrt{s}) + (\sqrt{s})^2 $$
|
|
|
|
$$ r + s = r + 2\sqrt{r}\sqrt{s} + s $$
|
|
|
|
$$ 0 = 2\sqrt{r}\sqrt{s} $$
|
|
|
|
$$ 0 = \sqrt{r}\sqrt{s} $$
|
|
|
|
By the zero product property, either $\sqrt{r}$ or $\sqrt{s}$ must be $0$, which
|
|
means that either $r = 0$ or $s = 0$.
|
|
|
|
_Case where $r = 0$:_
|
|
|
|
If $r = 0$, then $r$ is both $0$ and a positive real number. This is a
|
|
contradiction.
|
|
|
|
_Case where $s = 0$:_
|
|
|
|
If $s = 0$, then $s$ is both $0$ and a positive real number. This is a
|
|
contradiction.
|
|
|
|
In both cases, this is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
28. For all integers $a$, $b$, and $c$, if $a \mid b$ and $a \cancel{\mid} c$,
|
|
then $a \cancel{\mid} (b + c)$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that $a$, $b$, and $c$ are some integers such that
|
|
$a \mid b$, $a \cancel{\mid} c$ and $a \mid (b + c)$.
|
|
|
|
Since $a \mid (b + c)$ and $a \mid b$, then $b + c = ak$ and $b = am$ for some
|
|
integers $k$ and $m$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ (am) + c = ak $$
|
|
|
|
$$ am + c = ak $$
|
|
|
|
$$ c = ak - am $$
|
|
|
|
$$ c = a(k - m) $$
|
|
|
|
Now, $k - m$ is an integer by the difference of integers. Thus by the definition
|
|
of divisibility, $a \mid c$. Therefore $a \mid c$ and $a \cancel{\mid} c$.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
29. For all integers $m$ and $n$, if $m + n$ is even then $m$ and $n$ are both
|
|
even or $m$ and $n$ are both odd.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that $m$ and $n$ are some integers such that
|
|
$m + n$ is even and either $m$ is even and $n$ is odd or $m$ is odd and $n$ is
|
|
even.
|
|
|
|
_Case where $m$ is even and $n$ is odd:_
|
|
|
|
Since $m$ is even and $n$ is odd, then $m = 2k$ and $n = 2p + 1$ for some
|
|
integers $k$ and $p$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ m + n = (2k) + (2p + 1) $$
|
|
|
|
$$ = 2k + 2p + 1 $$
|
|
|
|
$$ = 2(k + p) + 1 $$
|
|
|
|
Now, $k + p$ is an integer by the sum of integers. Thus $m + n$ is odd by the
|
|
definition of odd integers. Therefore $m + n$ is odd and $m + n$ is even.
|
|
|
|
This is a contradiction.
|
|
|
|
_Case where $m$ is odd and $n$ is even:_
|
|
|
|
Since $m$ is odd and $n$ is even, then $m = 2k + 1$ and $n = 2p$ for some
|
|
integers $k$ and $p$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ m + n = (2k + 1) + (2p) $$
|
|
|
|
$$ = 2k + 2p + 1 $$
|
|
|
|
$$ = 2(k + p) + 1 $$
|
|
|
|
Now, $k + p$ is an integer by the sum of integers. Thus $m + n$ is odd by the
|
|
definition of odd integers. Therefore $m + n$ is odd and $m + n$ is even.
|
|
|
|
This is a contradiction.
|
|
|
|
In both cases, this is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
30.
|
|
|
|
a. Let $n = 53$. Find an approximate value for $\sqrt{n}$ and write a list of
|
|
all the prime numbers less than or equal to $\sqrt{n}$. Is the following
|
|
statement true or false? When $n = 53$, $n$ is not divisible by any prime number
|
|
less than or equal to $\sqrt{n}$.
|
|
|
|
$$ \sqrt{53} \approx 7.280109889 $$
|
|
|
|
All prime numbers $\leq \sqrt{n}$:
|
|
|
|
$$ \{2, 3, 5, 7\} $$
|
|
|
|
Yes this is true, as when 53 is divided by any of these prime numbers in this
|
|
set, the result does not equal an integer.
|
|
|
|
b. Suppose $n$ is a fixed integer. Let $S$ be the statement, "$n$ is not
|
|
divisible by any prime number less than or equal to $\sqrt{n}$." The following
|
|
statement is equivalent to $S$:
|
|
|
|
$\forall$ prime number $p$, if $p$ is less than or equal to $\sqrt{n}$ then $n$
|
|
is not divisible by $p$.
|
|
|
|
Which of the following are negations for $S$?
|
|
|
|
(i) $\exists$ a prime number $p$ such that $p \leq \sqrt{n}$ and $n$ is
|
|
divisible by $p$.
|
|
|
|
Yes, this is a negation for $S$.
|
|
|
|
(ii) $n$ is divisible by every prime number less than or equal to $\sqrt{n}$.
|
|
|
|
No, this is not a negation for $S$.
|
|
|
|
(iii) $\exists$ a prime number $p$ such that $p$ is a multiple of $n$ and $p$ is
|
|
less than or equal to $\sqrt{n}$.
|
|
|
|
No, this is not a negation for $S$.
|
|
|
|
(iv) $n$ is divisible by some prime number that is less than or equal to
|
|
$\sqrt{n}$.
|
|
|
|
Yes, this is a negation for $S$.
|
|
|
|
(v) $\forall$ prime number $p$, if $p$ is less than or equal to $\sqrt{n}$, then
|
|
$n$ is divisible by $p$.
|
|
|
|
No, this is not a negation for $S$.
|
|
|
|
31.
|
|
|
|
a. Prove by contraposition: For all positive integers $n$, $r$, and $s$, if
|
|
$rs \leq n$, then $r \leq \sqrt{n}$ or $s \leq \sqrt{n}$. (_Hint:_ Use Theorem
|
|
T27 in Appendix A.)
|
|
|
|
**Proof by contraposition:**
|
|
|
|
Suppose $n$, $r$, and $s$ are some positive integers such that $r > \sqrt{n}$
|
|
and $s > \sqrt{n}$.
|
|
|
|
Theorem T27 States:
|
|
|
|
If $0 < a < c$ and $0 < b < d$, then $0 < ab < cd$.
|
|
|
|
With $a = \sqrt{n}$, $b = \sqrt{n}$, $c = r$, and $d = s$, it follows that:
|
|
|
|
If $0 < \sqrt{n} < r$ and $0 < \sqrt{n} < s$, then $0 < \sqrt{n}\sqrt{n} < rs$.
|
|
|
|
Therefore $rs > n$.
|
|
|
|
Q.E.D.
|
|
|
|
b. Prove: For each integer $n > 1$, if $n$ is not prime then there exists a
|
|
prime number $p$ such that $p \leq \sqrt{n}$ and $n$ is divisible by $p$.
|
|
(_Hint:_ Use the results of part (a), Theorems 4.4.1, 4.4.3, and 4.4.4, and the
|
|
transitive property of order.)
|
|
|
|
Omitted.
|
|
|
|
c. State the contrapositive of the result of part (b). The results of exercise
|
|
31 provide a way to test whether an integer is prime.
|
|
|
|
Omitted.
|
|
|
|
**Test for Primality**
|
|
|
|
Given an integer $n > 1$, to test whether $n$ is prime check to see if it is
|
|
divisible by a prime number less than or equal to its square root. If it is not
|
|
divisible by any of these numbers, then it is prime.
|
|
|
|
32. Use the test for primality to determine whether the following numbers are
|
|
prime or not.
|
|
|
|
a. 667
|
|
|
|
b. 557
|
|
|
|
c. 527
|
|
|
|
d. 613
|
|
|
|
33. The sieve of Eratosthenes, named after its inventor, the Greek scholar
|
|
Eratosthenes (276-194 B.C.E.), provides a way to find all prime numbers less
|
|
than or equal to some fixed number $n$. To construct it, write out all the
|
|
integers from $2$ to $n$. Cross out all multiples of $2$ except $2$ itself,
|
|
then all multiples of $3$ except $3$ itself, then all multiples of $5$
|
|
except $5$ itself, and so forth. Continue crossing out the multiples of each
|
|
successive prime number up to $\sqrt{n}$. The numbers that are not crossed
|
|
out are all the prime numbers from $2$ to $n$. Here is a sieve of
|
|
Eratosthenes that includes the numbers from $2$ to $27$. The multiples of
|
|
$2$ are crossed out with a /, the multiples of 3 with a \, and the multiples
|
|
of 5 with a -.
|
|
|
|
See image on Page 250.
|
|
|
|
Use the sieve of Eratosthenes to find all prime numbers less than $100$.
|
|
|
|
34. Use the test for primality and the result of exercise 33 to determine
|
|
whether the following numbers are prime.
|
|
|
|
a. 9,269
|
|
|
|
b. 9,103
|
|
|
|
c. 8,623
|
|
|
|
d. 7,917
|
|
|
|
35. Use proof by contradiction to show that every integer greater than 11 is a
|
|
sum of two composite numbers.
|
|
|
|
36. For all odd integers $a$, $b$, and $c$, if $z$ is a solution of
|
|
$ax^2 + bx + c = 0$ then $z$ is irrational. (In the proof, use the
|
|
properties of even and odd integers that are listed in Example 4.3.3.)
|
|
|
|
---
|
|
|
|
Page 256
|
|
|
|
**Exercise Set 4.8**
|
|
|
|
1. A calculator display shows that $\sqrt{2} = 1.414213562$. Because
|
|
$1.414213562 = \dfrac{1414213562}{1000000000}$, this suggests that $\sqrt{2}$
|
|
is a rational number, which contradicts Theorem 4.8.1. Explain the
|
|
discrepancy.
|
|
|
|
The reason for this discrepancy is due to mistaking
|
|
$\sqrt{2} \approx 1.41.4213562$ for $\sqrt{2} = 1.414213562$. The calculator
|
|
cannot display $\sqrt{2}$ finitely as it is an irrational number and its
|
|
non-repeating decimal goes on forever. Therefore, you cannot find equivalencies
|
|
for 1.414213562 and express $\sqrt{2}$ as a rational number as it is
|
|
"demonstrated" here.
|
|
|
|
2. Example 4.3.1(h) illustrates a technique for showing that any repeating
|
|
decimal number is rational. A calculator display shows the result of a
|
|
certain calculation as $40.72727272727$. Can you be sure that the result of
|
|
the calculation is a rational number? Explain.
|
|
|
|
Yes. The reason you can be sure that the result of the calculation is a rational
|
|
number is because repeating decimal places can always be expressed as a rational
|
|
number usually by subtracting the repeating decimal places from a larger number
|
|
with the same repeating decimal places. For the given example:
|
|
|
|
Let $x = 40.72727272727 \dots$, so $100x = 4072.727272727 \dots$. Then:
|
|
|
|
$$ 100x - x = 99x = 4072.727272727\dots - 40.72727272727\dots = 4032 $$
|
|
|
|
$$ 99x = 4032 $$
|
|
|
|
$$ x = \frac{4032}{99} $$
|
|
|
|
Which is an expression for 40.72727272727 in rational form.
|
|
|
|
3. Could there be a rational number whose first trillion digits are the same as
|
|
the first trillion digits of $\sqrt{2}$? Explain.
|
|
|
|
Yes, because the first trillion digits of $\sqrt{2}$ is potentially finite. In
|
|
that case it is rational. In another case where the first trillion digits are
|
|
then repeated, then we know by 4.3.1(h) that this form of a decimal is also
|
|
rational. Similarly, if even smaller portions of those first trillion digits are
|
|
then repeated, this same principle applies.
|
|
|
|
4. A calculator display shows that the result of a certain calculation is $0.2$.
|
|
Can you be sure that the result of the calculation is a rational number?
|
|
|
|
Yes. Since the decimal $0.2$ has a finite amount of decimal places, we can
|
|
simply express it as a fraction:
|
|
|
|
Let $x = 0.2$ and $10x = 2$.
|
|
|
|
Then:
|
|
|
|
$$ 10x = 2 $$
|
|
|
|
$$ x = \frac{2}{10} $$
|
|
|
|
$$ x = \frac{1}{5} $$
|
|
|
|
Where $1$ and $5$ are integers and $5 \neq 0$. This is a rational number.
|
|
|
|
5. Let $s$ be the statement: The cube root of every irrational number is
|
|
irrational. This statement is true, but the following "proof" is incorrect.
|
|
Explain the mistake.
|
|
|
|
**"Proof (by contradiction):**
|
|
|
|
Suppose not. Suppose the cube root of every irrational number is rational. But
|
|
$2\sqrt{2}$ is irrational because it is a product of a rational and an
|
|
irrational number, and the cube root of $2\sqrt{2}$ is $\sqrt{2}$, which is
|
|
irrational. This is a contradiction, and hence it is not true that the cube root
|
|
of every irrational number is rational. Thus the statement to be proved is
|
|
true."
|
|
|
|
This incorrect proof has two problems. One is that in its supposition, the
|
|
wording suggests that the cube root of _every_ irrational number is rational,
|
|
when the negation of the given statement would be that "Suppose the cube root of
|
|
_some_ irrational number is rational."
|
|
|
|
The author of this incorrect proof then goes onto use a specific example of
|
|
$2\sqrt{2}$ for their proof. While this is fine for a disproof by
|
|
counterexample, a proof by contradiction should be more general. It should
|
|
instead read as:
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. Suppose the cube root of some irrational number $x$, is rational.
|
|
|
|
Since the cube root of $x$ is rational, this means that
|
|
$\sqrt[3]{x} = \dfrac{a}{b}$ for some integers $a$ and $b$ where $b \neq 0$.
|
|
|
|
Then, by laws of algebra:
|
|
|
|
$$ \sqrt[3]{x} = \frac{a}{b} $$
|
|
|
|
$$ x = \left(\frac{a}{b}\right)^3 $$
|
|
|
|
$$ x = \frac{a^3}{b^3} $$
|
|
|
|
Now, $a^3$ and $b^3$ are integers by the product of integers, where $b^3 \neq 0$
|
|
by the zero product property. Thus $x$ is a rational number and an irrational
|
|
number.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
Determine which statements in 6-16 are true and which are false. Prove those
|
|
that are true and disprove those that are false.
|
|
|
|
6. $6 - 7\sqrt{2}$ is irrational.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. Suppose $6 - 7\sqrt{2}$ is rational. Then by definition of
|
|
rational,
|
|
|
|
$$ 6 - 7\sqrt{2} = \frac{a}{b} $$
|
|
|
|
For some integers $a$ and $b$ where $b \neq 0$.
|
|
|
|
It follows that:
|
|
|
|
$$ 6 - 7\sqrt{2} = \frac{a}{b} $$
|
|
|
|
$$ -7\sqrt{2} = \frac{a}{b} - 6 $$
|
|
|
|
$$ \sqrt{2} = \frac{6 - \dfrac{a}{b}}{7} $$
|
|
|
|
$$ \sqrt{2} = \frac{6b - a}{7b} $$
|
|
|
|
Now, since $6b - a$ and $7b$ are integers by the product and difference of
|
|
integers and $7b \neq 0$ by the zero product property. This means that
|
|
$\sqrt{2}$ is rational. The $\sqrt{2}$, however, is known to not be rational by
|
|
Theorem 4.8.1.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
7. $3\sqrt{2} - 7$ is irrational.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. Suppose $\sqrt{2} - 7$ is rational.
|
|
|
|
Since $\sqrt{2} - 7$ is rational, $\sqrt{2} - 7 = \dfrac{a}{b}$ for some
|
|
integers $a$ and $b$ where $b \neq 0$.
|
|
|
|
Then, by laws of algebra:
|
|
|
|
$$ \sqrt{2} - 7 = \frac{a}{b} $$
|
|
|
|
$$ \sqrt{2} = \frac{a}{b} + 7 $$
|
|
|
|
$$ \sqrt{2} = \frac{a + 7b}{b} $$
|
|
|
|
Now, $a + 7b$ is an integer by the product and sum of integers. Thus $\sqrt{2}$
|
|
is rational. We know, however, by Theorem 4.8.1 that $\sqrt{2}$ is irrational.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
8. $\sqrt{4}$ is irrational.
|
|
|
|
This is false. $\sqrt{4} = 2 = \dfrac{2}{1}$, which is rational.
|
|
|
|
9. $\dfrac{\sqrt{2}}{6}$ is irrational.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. Suppose $\dfrac{\sqrt{2}}{6}$ is rational. Then by the definition
|
|
of rational:
|
|
|
|
$$ \frac{\sqrt{2}}{6} = \frac{a}{b} $$
|
|
|
|
for some integers $a$ and $b$ where $b \neq 0$.
|
|
|
|
Then, by laws of algebra:
|
|
|
|
$$ \sqrt{2} = \frac{6a}{b} $$
|
|
|
|
Now, $6a$ is an integer by the product of integers. Thus $\sqrt{2}$ is a
|
|
rational number. We know by Theorem 4.8.1 that $\sqrt{2}$ is irrational.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
10. The sum of any two irrational numbers is irrational.
|
|
|
|
**Proof by counterexample:**
|
|
|
|
Let $a = \sqrt{2}$, and let $b = -\sqrt{2}$. Then, their sum is:
|
|
|
|
$$ a + b = \sqrt{2} + (-\sqrt{2}) = 0 = \frac{0}{1} $$
|
|
|
|
Which is rational. This statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
11. The difference of any two irrational numbers is irrational.
|
|
|
|
**Proof by counterexample:**
|
|
|
|
Let $a = \sqrt{2}$, and let $b = \sqrt{2}$. Then, their sum is:
|
|
|
|
$$ a + b = \sqrt{2} - \sqrt{2}) = 0 = \frac{0}{1} $$
|
|
|
|
Which is rational. This statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
12. The positive square root of a positive irrational number is irrational.
|
|
|
|
$$ \forall x \in \mathbb{R} (I(x) \to I(\sqrt{x})) $$
|
|
|
|
Contrapositive:
|
|
|
|
$$ \forall x \in \mathbb{R} (\neg I(\sqrt{x}) \to \neg I(x)) $$
|
|
|
|
$$ \forall x \in \mathbb{R} (R(\sqrt{x}) \to R(x)) $$
|
|
|
|
**Proof by contraposition:**
|
|
|
|
Suppose $r$ is any positive real number such that $\sqrt{r}$ is rational.
|
|
|
|
Since $\sqrt{r}$ is rational, $\sqrt{r} = \dfrac{a}{b}$ where $a$ and $b$ are
|
|
integers and $b \neq 0$.
|
|
|
|
Then:
|
|
|
|
$$ \sqrt{r} = \frac{a}{b} $$
|
|
|
|
$$ r = \left(\frac{a}{b}\right)^2 $$
|
|
|
|
$$ r = \frac{a^2}{b^2} $$
|
|
|
|
Now, $a^2$ and $b^2$ are both integers by the product of integers and
|
|
$b^2 \neq 0$ by the zero product property.
|
|
|
|
Therefore $r$ is a rational number.
|
|
|
|
Q.E.D.
|
|
|
|
13. If $r$ is any rational number and $s$ is any irrational number, then
|
|
$\dfrac{r}{s}$ is irrational.
|
|
|
|
**Proof by counterexample:**
|
|
|
|
Let $r = 0$ and $s = \sqrt{2}$, then
|
|
$\dfrac{r}{s} = \dfrac{0}{\sqrt{2}} = 0 = \dfrac{0}{1}$ which is rational.
|
|
|
|
Therefore, this statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
14. The sum of any two positive irrational numbers is irrational.
|
|
|
|
**Proof by counterexample:**
|
|
|
|
Let $x = \sqrt{2}$ and $y = 2 - \sqrt{2}$, then:
|
|
|
|
$$ x + y = \sqrt{2} + (2 - \sqrt{2}) = 2 = \dfrac{2}{1} $$
|
|
|
|
Thus, $x + y$ is a rational number.
|
|
|
|
Therefore, this statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
15. The product of two irrational numbers is irrational.
|
|
|
|
**Proof by counterexample:**
|
|
|
|
$$ \sqrt{2} \cdot \sqrt{2} = (\sqrt{2})^2 = 2 = \frac{2}{1} $$
|
|
|
|
Which is a rational number.
|
|
|
|
This statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
16. If an integer greater than $1$ is a perfect square, then its cube root is
|
|
irrational.
|
|
|
|
**Proof by counterexample:**
|
|
|
|
Consider $64 = 8^2$, then $64 > 1$ and $64$ is a perfect square. Then consider
|
|
$\sqrt[3]{64} = 4 = \dfrac{4}{1}$. Thus $\sqrt[3]{64}$ is rational.
|
|
|
|
Thus there is at least one integer greater than $1$ that is a perfect square and
|
|
its cube root is rational.
|
|
|
|
This statement is false.
|
|
|
|
Q.E.D.
|
|
|
|
17. Consider the following sentence: If $x$ is rational then $\sqrt{x}$ is
|
|
irrational. Is this sentence always true, sometimes true and sometimes
|
|
false, or always false? Justify your answer.
|
|
|
|
This statement is sometimes true and sometimes false. Consider when $x = 2$,
|
|
then $\sqrt{2}$ is irrational. This is the case when the statement is true. Then
|
|
consider when $x = 9$ then $\sqrt{9} = 3 = \dfrac{3}{1}$ is rational.. This is a
|
|
case when the statement is false. Therefore this statement is sometimes true and
|
|
sometimes false.
|
|
|
|
18.
|
|
|
|
a. Prove that for every integer $a$, if $a^3$ is even then $a$ is even.
|
|
|
|
**Proof by contrapositive:**
|
|
|
|
Suppose $a$ is any integer such that $a$ is odd.
|
|
|
|
Since $a$ is odd, $a = 2k + 1$ for some integer $k$.
|
|
|
|
Then:
|
|
|
|
$$ a^3 = (2k + 1)^3 $$
|
|
|
|
$$ a^3 = 8k^3 + 12k^2 + 6k + 1 $$
|
|
|
|
$$ a^3 = 2(4k^3 + 6k^2 + 3k) + 1 $$
|
|
|
|
Now, $4k^3 + 6k^2 + 3k$ is an integer by the product and sum of integers.
|
|
Therefore $a^3$ is odd by definition of odd integers.
|
|
|
|
Q.E.D.
|
|
|
|
b. Prove that $\sqrt[3]{2}$ is irrational.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose $\sqrt[3]{2}$ is rational. Then, by definition of
|
|
rational:
|
|
|
|
$$ \sqrt[3]{2} = \frac{a}{b} $$
|
|
|
|
for some integers $a$ and $b$ where $b \neq 0$ and $\dfrac{a}{b}$ is written in
|
|
lowest terms.
|
|
|
|
Then:
|
|
|
|
$$ \sqrt[3]{2} = \frac{a}{b} $$
|
|
|
|
$$ 2 = \left(\frac{a}{b}\right)^3 $$
|
|
|
|
$$ 2 = \frac{a^3}{b^3} $$
|
|
|
|
$$ 2b^3 = a^3 $$
|
|
|
|
Now, by the definition of even integers, we know that $a^3$ is even. We also
|
|
know by part (a), that if $a^3$ is even, then $a$ is even.
|
|
|
|
Since $a$ is even, $a = 2k$ for some integer $k$. Then:
|
|
|
|
$$ 2b^3 = (2k)^3 $$
|
|
|
|
$$ 2b^3 = 8k^3 $$
|
|
|
|
$$ 2b^3 = 8k^3 $$
|
|
|
|
$$ b^3 = 4k^3 $$
|
|
|
|
$$ b^3 = 2(2k^3) $$
|
|
|
|
Now, $2k^3$ is an integer by the product of integers. Additionally, $b^3$ is
|
|
even by the definition of even integers. Additionally, by part (a) $b$ is even
|
|
since $b^3$ is even.
|
|
|
|
Since both $a$ and $b$ and even, $\sqrt[3]{2}$ is not in lowest terms, so
|
|
$\sqrt[3]{2}$ is not rational, which contradicts the supposition.
|
|
|
|
Q.E.D.
|
|
|
|
19.
|
|
|
|
a. Use proof by contradiction to show that for any integer $n$, it is impossible
|
|
for $n$ to equal both $3q_1 + r_1$ and $3q_2 + r_2$, where $q_1$, $q_2$, $r_1$,
|
|
and $r_2$ are integers, $0 \leq r_1 < 3$, $0 \leq r_2 < 3$, and $r_1 \neq r_2$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose $n$ is some integer such
|
|
$n = 3q_1 + r_1 = 3q_2 + r_2$, where $q_1$, $q_2$, $r_1$, and $r_2$ are some
|
|
integers such that $0 \leq r_1 < 3$ and $0 \leq r_2 < 3$ and $r_1 \neq r_2$.
|
|
|
|
Then:
|
|
|
|
$$ 3q_1 + r_1 = 3q_2 + r_2 $$
|
|
|
|
$$ 3q_1 - 3q_2 = r_2 - r_1 $$
|
|
|
|
$$ 3(q_1 - q_2) = r_2 - r_1 $$
|
|
|
|
This means that $3 \mid (r_2 - r_1)$.
|
|
|
|
Since $0 \leq r_1,r_2 < 3$, it follows that:
|
|
|
|
$$ -2 \leq r_2 - r_1 \leq 2 $$
|
|
|
|
This means that $3 \cancel{\mid} (r_2 - r_1)$, which contradicts the earlier
|
|
finding that $3 \mid (r_2 - r_1)$.
|
|
|
|
Q.E.D.
|
|
|
|
b. Use proof by contradiction, the quotient-remainder theorem, division into
|
|
cases, and the result of part (a) to prove that for every integer $n$, if $n^2$
|
|
is divisible by $3$ then $n$ is divisible by $3$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose $n$ is some integer such that $n^2$ is divisible
|
|
by $3$ and $n$ is not divisible by $3$.
|
|
|
|
Since $n$ is not divisible by $3$, then $n = 3k + 1$ or $n = 3k + 2$ for some
|
|
integer $k$.
|
|
|
|
_Case where $n = 3k + 1$:_
|
|
|
|
$$ n^2 = (3k + 1)^2 $$
|
|
|
|
$$ n^2 = 9k^2 + 6k + 1 $$
|
|
|
|
$$ n^2 = 3(3k^2 + 2k) + 1 $$
|
|
|
|
By the quotient-remainder theorem, this means that $3 \cancel{\mid} n^2$. This
|
|
contradicts the supposition.
|
|
|
|
_Case where $n = 3k + 2$:_
|
|
|
|
$$ n^2 = (3k + 2)^2 $$
|
|
|
|
$$ n^2 = 9k^2 + 12k + 4 $$
|
|
|
|
$$ n^2 = 9k^2 + 12k + 3 + 1 $$
|
|
|
|
$$ n^2 = 3(3k^2 + 4k + 1) + 1 $$
|
|
|
|
By the quotient-remainder theorem, this means that $3 \cancel{\mid} n^2$. This
|
|
contradicts the supposition.
|
|
|
|
In both cases, the supposition is contradicted.
|
|
|
|
Q.E.D.
|
|
|
|
c. Prove that $\sqrt{3}$ is irrational.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. Suppose that $\sqrt{3}$ is rational.
|
|
|
|
Since $\sqrt{3}$ is rational, $\sqrt{3} = \dfrac{a}{b}$ for some integers $a$
|
|
and $b$ where $b \neq 0$ and $\dfrac{a}{b}$ is in lowest terms.
|
|
|
|
Then:
|
|
|
|
$$ \sqrt{3} = \frac{a}{b} $$
|
|
|
|
$$ 3 = \left(\frac{a}{b}\right)^2 $$
|
|
|
|
$$ 3 = \frac{a^2}{b^2} $$
|
|
|
|
$$ 3b^2 = a^2 $$
|
|
|
|
This means that $3 \mid a^2$. By part (b), we then know that $3 \mid a$. This
|
|
means that $a = 3k$ for some integer $k$. Then:
|
|
|
|
$$ 3b^2 = (3k)^2 $$
|
|
|
|
$$ 3b^2 = 9k^2 $$
|
|
|
|
$$ b^2 = 3k^2 $$
|
|
|
|
This means that $3 \mid b^2$. Then, however, $\dfrac{a}{b}$ is not in lowest
|
|
terms since $3 \mid a$ and $3 \mid b$. This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
20. Give an example to show that if $d$ is not prime and $n^2$ is divisible by
|
|
$d$, then $n$ need not be divisible by $d$.
|
|
|
|
We need to find a non prime number $d$ and a real number $n$ such that:
|
|
|
|
$$ d \mid n^2 $$
|
|
|
|
$$ d \cancel{\mid} n $$
|
|
|
|
Let $d = 4$ and $n = 2$.
|
|
|
|
Then:
|
|
|
|
$$ 4 \mid 4 $$
|
|
|
|
$$ 4 \cancel{\mid} 2 $$
|
|
|
|
21. The quotient-remainder theorem says not only that there exist quotients and
|
|
remainders but also that the quotient and remainder of a division are
|
|
unique. Prove the uniqueness. That is, prove that if $a$ and $d$ are
|
|
integers with $d > 0$ and if $q_1$, $r_1$, $q_2$, and $r_2$ are integers
|
|
such that
|
|
|
|
$$ a = dq_1 + r_1 \quad \text{ where } 0 \leq r_1 < d $$
|
|
|
|
and
|
|
|
|
$$ a = dq_2 + r_2 \quad \text{ where } 0 \leq r_2 < d $$
|
|
|
|
then
|
|
|
|
$$ q_1 = q_2 \quad \text{ and } r_1 = r_2 $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$, $d$ are any integers and $q_1$, $q_2$, $r_1$, and $r_2$ are some
|
|
integers such that $a = dq_1 + r_1$ where $0 \leq r_1 < d$ and $a = dq_2 + r_2$
|
|
where $0 \leq r_2 < d$.
|
|
|
|
Then:
|
|
|
|
$$ dq_1 + r_1 = dq_2 + r_2 $$
|
|
|
|
$$ dq_1 - dq_2 = r_2 - r_1 $$
|
|
|
|
$$ d(q_1 - q_2) = r_2 - r_1 $$
|
|
|
|
Since $0 \leq r_1,r_2 < d$, this means that:
|
|
|
|
$$ -(d - 1) \leq r_2 - r_1 \leq d - 1 $$
|
|
|
|
And this means:
|
|
|
|
$$ |r_2 - r_1| < d $$
|
|
|
|
Since $d(q_1 - q_2) = r_2 - r_1$ means $d \mid (r_2 - r_1)$, but we also know
|
|
that $|r_2 - r_1| < d$, then it follows that $r_2 - r_1 = 0$. Thus, it then
|
|
follows that $r_1 = r_2$
|
|
|
|
Then by substitution:
|
|
|
|
$$ d(q_1 - q_2) = r_2 - r_1 $$
|
|
|
|
$$ q_1 - q_2 = 0 $$
|
|
|
|
Thus it follows that $q_1 = q_2$.
|
|
|
|
Therefore, it has been shown that $q_1 = q_2$ and $r_1 = r_2$.
|
|
|
|
Q.E.D.
|
|
|
|
22. Prove that $\sqrt{5}$ is irrational.
|
|
|
|
**Lemma 22:**
|
|
|
|
If $5$ divides $a^2$, then $5$ divides $a$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not, that is suppose $a$ is some integer such that $5 \mid a^2$ and
|
|
$5 \cancel{\mid} a$.
|
|
|
|
Since $5 \cancel{\mid} a$, this means that $a = 5q + r$ for some unique integers
|
|
$q$ and $r$, such that $1 \leq r < 5$.
|
|
|
|
Then, by substitution:
|
|
|
|
$$ a^2 = (5q + r)^2 $$
|
|
|
|
$$ a^2 = 25q^2 + 10qr + r^2 $$
|
|
|
|
$$ a^2 = 5(5q^2 + 2qr) + r^2 $$
|
|
|
|
So:
|
|
|
|
$$ a^2 \equiv r^2 (\mod 5) $$
|
|
|
|
Then by cases:
|
|
|
|
$$ r = 1 \to r^2 (\mod 5) = 1 $$
|
|
|
|
$$ r = 2 \to r^2 (\mod 5) = 4 $$
|
|
|
|
$$ r = 3 \to r^2 (\mod 5) = 4 $$
|
|
|
|
$$ r = 4 \to r^2 (\mod 5) = 1 $$
|
|
|
|
So in all cases $r^2 \cancel{\equiv} 0 (\mod 5)$
|
|
|
|
Therefore $5 \cancel{\mid} a^2$, which contradicts the supposition.$
|
|
|
|
Q.E.D.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. Suppose that $\sqrt{5}$ is rational.
|
|
|
|
Since $\sqrt{5}$ is rational, $\sqrt{5} = \dfrac{a}{b}$ for some integers $a$
|
|
and $b$ where $b \neq 0$ and $\dfrac{a}{b}$ is in lowest terms.
|
|
|
|
Then:
|
|
|
|
$$ \sqrt{5} = \frac{a}{b} $$
|
|
|
|
$$ 5 = \left(\frac{a}{b}\right)^2 $$
|
|
|
|
$$ 5 = \frac{a^2}{b^2} $$
|
|
|
|
$$ 5b^2 = a^2 $$
|
|
|
|
Thus we know that $5 \mid a^2$, and by Lemma 22, we then know that $5 \mid a$.
|
|
|
|
So, since $5 \mid a$, $a = 5k$ for some integer $k$.
|
|
|
|
Then by substitution:
|
|
|
|
$$ 5b^2 = (5k)^2 $$
|
|
|
|
$$ 5b^2 = 25k^2 $$
|
|
|
|
$$ b^2 = 5k^2 $$
|
|
|
|
So then we know that $5 \mid b^2$, and by Lemma 22, we then know that
|
|
$5 \mid b$. Then, however, $\dfrac{a}{b}$ is not in lowest terms. Therefore
|
|
$\sqrt{5}$ is irrational, which contradicts the supposition.
|
|
|
|
Q.E.D.
|
|
|
|
23. Prove that for any integer $a$, $9 \cancel{\mid} (a^2 - 3)$.
|
|
|
|
_Hint:_ This statement is true. If $a^2 - 3 = 9b$, then
|
|
$a^2 = 9b + 3 = 3(3b + 1)$, and so $a^2$ is divisible by $3$. Hence, by exercise
|
|
19(b), $a$ is divisible by $3$. Thus $a^2 =(3c)^2$ for some integer $c$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that for some integer $a$, $9 \mid (a^2 - 3)$.
|
|
|
|
Since $9 \mid (a^2 - 3)$, then $a^2 - 3 = 9b$ for some integer $b$. This then
|
|
becomes:
|
|
|
|
$$ a^2 - 3 = 9b $$
|
|
|
|
$$ a^2 = 9b + 3 $$
|
|
|
|
$$ a^2 = 3(3b + 1) $$
|
|
|
|
So $3 \mid a^2$, by 9(b), we then know that $3 \mid a$. Since $3 \mid a$,
|
|
$a = 3c$ for some integer $c$. By substitution:
|
|
|
|
$$ (3c)^2 = 3(3b + 1) $$
|
|
|
|
$$ 9c^2 = 3(3b + 1) $$
|
|
|
|
$$ 3c^2 = 3b + 1 $$
|
|
|
|
$$ 3c^2 - 1 = 3b $$
|
|
|
|
But since $ 3 \cancel{\mid} 3c^2 - 1$ and $3 \mid 3b$, this statement can never
|
|
hold for $b$ and $c$. This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
24. An alternative proof of the irrationality of $\sqrt{2}$ counts the number of
|
|
$2$'s on the two sides of the equation $2n^2 = m^2$ and uses the unique
|
|
factorization of integers theorem to deduce a contradiction. Write a proof
|
|
that uses this approach.
|
|
|
|
Statement: $\sqrt{2}$ is irrational.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose $\sqrt{2}$ is rational. Then there are some
|
|
integers $m$ and $n$ where $n \neq 0$ such that:
|
|
|
|
$$ \sqrt{2} = \frac{m}{n} $$
|
|
|
|
where $\dfrac{m}{n}$ have no common factors. Then:
|
|
|
|
$$ \sqrt{2} = \frac{m}{n} $$
|
|
|
|
$$ 2 = \frac{m^2}{n^2} $$
|
|
|
|
$$ 2n^2 = m^2 $$
|
|
|
|
By the unique factorization theorem, we can express $n$ as the product of prime
|
|
numbers as:
|
|
|
|
$$ n = 2^a \cdot p_1^{e_1} \dots \cdot p_k^{e_k} $$
|
|
|
|
Then, we can express $n^2$ as:
|
|
|
|
$$ n^2 = 2^{2a} $$
|
|
|
|
Since the exponent is even, we know the number of $2s$ in $n^2$ is even. It then
|
|
follows that $2n^2$ adds one more $2$:
|
|
|
|
$$ 2n^2 = 2^{2a + 1} $$
|
|
|
|
So $2n^2$ has an odd number of $2$s.
|
|
|
|
We can then apply the same logic to $m^2$.
|
|
|
|
$$ m = 2^b \cdot p_1^{e_1} \dots \cdot p_q^{e_q} $$
|
|
|
|
$$ m^2 = 2^{2b} $$
|
|
|
|
So $m^2$ has an even number of $2$s. Thus $2n^2$ has an odd number of $2$s and
|
|
$m^2$ has an even number of $2$s, which contradicts the statement $2n^2 = m^2$.
|
|
|
|
Q.E.D.
|
|
|
|
25. Use the proof technique illustrated in exercise 24 to prove that if $n$ is
|
|
any positive integer that is not a perfect square, then $\sqrt{n}$ is
|
|
irrational.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose $n$ is some positive integer such that $n$ is not
|
|
a perfect square and $\sqrt{n}$ is rational.
|
|
|
|
Since $\sqrt{n}$ is rational, $\sqrt{n} = \dfrac{a}{b}$ for some integers $a$
|
|
and $b$, where $b \neq 0$ and where $\dfrac{a}{b}$ is in lowest terms.
|
|
|
|
Then:
|
|
|
|
$$ \sqrt{n} = \frac{a}{b} $$
|
|
|
|
$$ n = \left(\frac{a}{b}\right)^2 $$
|
|
|
|
$$ n = \frac{a^2}{b^2} $$
|
|
|
|
$$ nb^2 = a^2 $$
|
|
|
|
By the unique factorization theorem, $a$ can be written as a product of prime
|
|
numbers:
|
|
|
|
$$ a = p_1^{e_1}p_2^{e_2} \cdot \dots \cdot p_k^{e_k} $$
|
|
|
|
And:
|
|
|
|
$$ a^2 = p_1^{2e_1}p_2^{2e_2} \cdot \dots \cdot p_k^{2e_k} $$
|
|
|
|
This means that all prime exponents in $a^2$ are even.
|
|
|
|
Similarly:
|
|
|
|
$$ b = q_1^{f_1}q_2^{f_2} \cdot \dots \cdot q_m^{f_m} $$
|
|
|
|
And:
|
|
|
|
$$ b^2 = q_1^{2e_1}q_2^{2e_2} \cdot \dots \cdot q_k^{2e_m} $$
|
|
|
|
This means that all prime exponents in $b^2$ are even.
|
|
|
|
Since $nb^2 = a^2$, all prime exponents in $nb^2$ are even since all the prime
|
|
exponents in $a^2$ are even. Therefore $n$ is a perfect square
|
|
|
|
This contradicts the supposition that $n$ is not a perfect square.
|
|
|
|
Q.E.D.
|
|
|
|
26. Prove that $\sqrt{2} + \sqrt{3}$ is irrational.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is suppose that $\sqrt{2} + \sqrt{3}$ is rational.
|
|
|
|
Since $\sqrt{2} + \sqrt{3}$ is rational, then
|
|
$\sqrt{2} + \sqrt{3} = \dfrac{a}{b}$ where $a$ and $b$ are some integers and
|
|
$b \neq 0$ and $\dfrac{a}{b}$ has no common factors.
|
|
|
|
Then, by substitution:
|
|
|
|
$$ \sqrt{2} + \sqrt{3} = \frac{a}{b} $$
|
|
|
|
$$ \left(\sqrt{2} + \sqrt{3}\right)^2 = \left(\frac{a}{b}\right)^2 $$
|
|
|
|
$$ \left(\sqrt{2} + \sqrt{3}\right)\left(\sqrt{2} + \sqrt{3}\right) = \frac{a^2}{b^2} $$
|
|
|
|
$$ 2 + 2\sqrt{2}\sqrt{3} + 3 = \frac{a^2}{b^2} $$
|
|
|
|
$$ 2\sqrt{6} + 5 = \frac{a^2}{b^2} $$
|
|
|
|
$$ 2\sqrt{6} = \frac{a^2}{b^2} - 5 $$
|
|
|
|
$$ \sqrt{6} = \frac{\dfrac{a^2}{b^2} - 5}{2} $$
|
|
|
|
$$ \sqrt{6} = \frac{a^2 - 5b^2}{2b^2} $$
|
|
|
|
By the proof given in exercise 25, we know that $\sqrt{6}$ is irrational.
|
|
|
|
Now, $a^2 - 5b^2$ is an integer by the product and difference of integers. Also
|
|
$2b^2$ is an integer by the product of integers and $2b^2 \neq 0$ by the zero
|
|
product property.
|
|
|
|
Therefore $\sqrt{6}$ is a rational number which contradicts what we derived from
|
|
the proof given in exercise 25 that $\sqrt{6}$ is irrational.
|
|
|
|
Q.E.D.
|
|
|
|
27. Prove that $\log_5(2)$ is irrational. (_Hint:_ Use the unique factorization
|
|
of integers theorem.)
|
|
|
|
Omitted.
|
|
|
|
28. Let $N = 2 \cdot 3 \cdot 5 \cdot 7 + 1$. What remainder is obtained when $N$
|
|
is divided by $2$?$3$?$5$?$7$? Justify your answer.
|
|
|
|
$$ N = 2(3 \cdot 5 \cdot 7) + 1 $$
|
|
|
|
$$ N \mod 2 = 1 $$
|
|
|
|
$$ N = 3(2 \cdot 5 \cdot 7) + 1 $$
|
|
|
|
$$ N \mod 3 = 1 $$
|
|
|
|
$$ N = 5(2 \cdot 3 \cdot 7) + 1 $$
|
|
|
|
$$ N \mod 5 = 1 $$
|
|
|
|
$$ N = 7(2 \cdot 3 \cdot 5) + 1 $$
|
|
|
|
$$ N \mod 7 = 1 $$
|
|
|
|
29. Suppose $a$ is an integer and $p$ is a prime number such that $p \mid a$ and
|
|
$p \mid (a + 3)$. What can you deduce about $p$? Why?
|
|
|
|
Since $p \mid a$ and $p \mid (a + 3)$, then $p \mid ((a + 3) - a)$. It follows
|
|
then that $p \mid 3$, and the only prime number that divides $3$ is $3$ itself.
|
|
|
|
$$ p = 3 $$
|
|
|
|
30. Let $p_1, p_2, p_3, \dots$ be a list of all prime numbers in ascending
|
|
order. Here is a table of the first six:
|
|
|
|
| $p_1$ | $p_2$ | $p_3$ | $p_4$ | $p_5$ | $p_6$ |
|
|
| ----- | ----- | ----- | ----- | ----- | ----- |
|
|
| $2$ | $3$ | $5$ | $7$ | $11$ | $13$ |
|
|
|
|
a. Let
|
|
$N_1 = p_1, N_2 = p_1 \cdot p_2, N_3 = p_1 \cdot p_2 \cdot p_3, \dots N_6 = p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot p_5 \cdot p_6$.
|
|
Calculate $N_1$, $N_2$, $N_3$, $N_4$, $N_5$, and $N_6$.
|
|
|
|
| $N_1$ | $N_2$ | $N_3$ | $N_4$ | $N_5$ | $N_6$ |
|
|
| ----- | ----- | ----- | ----- | ------ | ------- |
|
|
| $2$ | $6$ | $30$ | $210$ | $2310$ | $30030$ |
|
|
|
|
b. For each $i = 1, 2, 3, 4, 5, 6$, find whether $N_i$ is itself prime or just
|
|
has a prime factor less than itself. (_Hint:_ Use the test for primality from
|
|
exercise 31 in Section 4.7 to determine your answers.)
|
|
|
|
**Test for Primality**
|
|
|
|
Given an integer $n > 1$, to test whether $n$ is prime check to see if it is
|
|
divisible by a prime number less than or equal to its square root. If it is not
|
|
divisible by any of these numbers, then it is prime.
|
|
|
|
$N_1$ is prime.
|
|
|
|
$N_2$ is not prime.
|
|
|
|
$N_3$ is not prime.
|
|
|
|
$N_4$ is not prime.
|
|
|
|
$N_5$ is not prime.
|
|
|
|
$N_6$ is not prime.
|
|
|
|
For exercises 31 and 32, use the fact that for ever integer $n$,
|
|
|
|
$$ n! = n(n - 1) \dots 3 \cdot 2 \cdot 1 $$
|
|
|
|
31. An alternative proof of the infinitude of the prime numbers begins as
|
|
follows:
|
|
|
|
**Proof:** Suppose there are only finitely many prime numbers. Then one is the
|
|
largest. Call it $p$. Let $M = p! + 1$. We will show that there is a prime
|
|
number $q$ such that $q > p$. Complete this proof.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is suppose there are only finitely many prime numbers. Then
|
|
one is the largest. Call it $p$. Let $M = p! + 1$.
|
|
|
|
Let $q$ be a prime number such that $q \leq p$.
|
|
|
|
Since $q \leq p$, $q \mid p!$. This means that:
|
|
|
|
$$ p! \equiv 0 (\mod q) $$
|
|
|
|
It then follows that:
|
|
|
|
$$ M = p! + 1 \equiv 1 (\mod q) $$
|
|
|
|
So, $q \cancel{\mid} M$.
|
|
|
|
Since $q$ is a prime number that cannot divide $M$, either $M$ is prime itself
|
|
or $M$ has a prime factor greater than $p$, which contradicts the supposition.
|
|
|
|
Q.E.D.
|
|
|
|
We will show that there is a prime number $q$ such that $q > p$.
|
|
|
|
32. Prove that for every integer $n$, if $n > 2$ then there is a prime number
|
|
$p$ such that $n < p < n!$.
|
|
|
|
Omitted.
|
|
|
|
33. Prove that if $p_1, p_2, \dots$, and $p_n$ are distinct prime numbers with
|
|
$p_1 = 2$ and $n > 1$, then $p_1, p_2, \dots, p_n + 1$ can be written in the
|
|
form $4k + 3$ for some integer $k$.
|
|
|
|
Omitted.
|
|
|
|
34.
|
|
|
|
a. Fermat's last theorem says that for every integer $n > 2$, the equation
|
|
$x^n + y^n = z^n$ has no positive integer solution (solution for which $x$, $y$,
|
|
and $z$ are positive integers). Prove the following: If for every prime number
|
|
$p > 2$, $x^p + y^p = z^p$ has no positive integer solution, then for any
|
|
integer $n > 2$ that is not a power of $2$, $x^n + y^n = z^n$ has no positive
|
|
integer solution.
|
|
|
|
Omitted.
|
|
|
|
b. Fermat proved that there are no integers $x$, $y$, and $z$ such that
|
|
$x^4 + y^4 = z^4$. Use this result to remove the restriction in part (a) that
|
|
$n$ not be a power of $2$. That is, prove that if $n$ is a power of $2$ and
|
|
$n > 4$, then $x^n + y^n = z^n$ has no positive integer solution.
|
|
|
|
Omitted.
|
|
|
|
For exercises 35-38 note that to show there is a unique object with a certain
|
|
property, show that (1) there is an object with the property and (2) if objects
|
|
$A$ and $B$ have the property, then $A = B$.
|
|
|
|
35. Prove that there exists a unique prime number of the form $n^2 - 1$, where
|
|
$n$ is an integer that is greater than or equal to $2$.
|
|
|
|
Omitted.
|
|
|
|
36. Prove that there exists a unique prime number of the form $n^2 + 2n - 3$,
|
|
where $n$ is a positive integer.
|
|
|
|
Omitted.
|
|
|
|
37. Prove that there is at most one real number $a$ with the property that
|
|
$a + r = r$ for every real number $r$. (Such a number is called an _additive
|
|
identity_.)
|
|
|
|
Omitted.
|
|
|
|
38. Prove that there is at most one real number $b$ with the property that
|
|
$br = r$ for every real number $r$. (Such a number is called a
|
|
_multiplicative identity_.)
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
**Exercise Set 4.9**
|
|
|
|
Page 265
|
|
|
|
In 1 and 2 find the degree of each vertex and the total degree of the graph.
|
|
Check that the number of edges equals one-half of the total degree.
|
|
|
|
1. See page 265.
|
|
|
|
$\text{deg(v_1)} = 3$
|
|
|
|
$\text{deg(v_2)} = 2$
|
|
|
|
$\text{deg(v_3)} = 4$
|
|
|
|
$\text{deg(v_4)} = 2$
|
|
|
|
$\text{deg(v_5)} = 1$
|
|
|
|
$\text{deg(v_6)} = 0$
|
|
|
|
$\text{deg}(\text{total}) = 12$
|
|
|
|
$\text{total edges} = 6$
|
|
|
|
2. See page 265.
|
|
|
|
$\text{deg(v_1)} = 1$
|
|
|
|
$\text{deg(v_2)} = 5$
|
|
|
|
$\text{deg(v_3)} = 4$
|
|
|
|
$\text{deg(v_4)} = 4$
|
|
|
|
$\text{deg(v_5)} = 1$
|
|
|
|
$\text{deg(v_6)} = 3$
|
|
|
|
$\text{deg}(\text{total}) = 18$
|
|
|
|
$\text{total edges} = 9$
|
|
|
|
3. A graph has vertices of degrees 0, 2, 2, 3, and 9. How many edges does the
|
|
graph have?
|
|
|
|
$$ \frac{1}{2}(0 + 2 + 2 + 3 + 9) = 8 \text{ edges} $$
|
|
|
|
4. A graph has vertices of degrees 1, 1, 4, 4, and 6. How many edges does the
|
|
graph have?
|
|
|
|
$$ \frac{1}{2}(1 + 1 + 4 + 4 + 6) = 8 \text{ edges} $$
|
|
|
|
In each of 5-13 either draw a graph with the specified properties or explain why
|
|
no such graph exists.
|
|
|
|
5. Graph with five vertices of degrees 1, 2, 3, 3, and 5.
|
|
|
|
6. Graph of four vertices of degrees 1, 2, 3, and 3.
|
|
|
|
Not possible. It has an odd number of total degrees, 9. By 4.9.2, the total
|
|
degree of a graph must be even.
|
|
|
|
7. Graph with four vertices of degrees 1, 1, 1, and 4.
|
|
|
|
Not possible. It has an odd number of total degrees, 7. By 4.9.2, the total
|
|
degree of a graph must be even.
|
|
|
|
8. Graph with four vertices of degrees 1, 2, 3, and 4.
|
|
|
|
9. Simple graph with four vertices of degrees 1, 2, 3, and 4.
|
|
|
|
No such graph. The vertex of degree 4 would have to be connected by edges to 4
|
|
distinct vertices other than itself. This is not possible in a simple graph
|
|
since it cannot loop back on itself.
|
|
|
|
10. Simple graph with five vertices of degrees 2, 3, 3, 3, and 5.
|
|
|
|
Not possible, as the vertex of degree 5 would have to loop back on itself in a
|
|
graph of 5 vertices, which contradicts the definition of a simple graph.
|
|
|
|
11. Simple graph with five vertices of degrees 1, 1, 1, 2, and 3.
|
|
|
|
12. Simple graph with six edges and all vertices of degree 3.
|
|
|
|
13. Simple graph with nine edges and all vertices of degree 3.
|
|
|
|
14. At a party attended by a group of people, two people knew one other person
|
|
before the party, and five people knew two other people before the party.
|
|
The rest of the people knew three other people before the party. A total of
|
|
15 pairs of people knew each other before the party.
|
|
|
|
a. How many people attending the party knew three other people before the party?
|
|
|
|
$$ (2 \cdot 1) + (5 \cdot 2) + 3x = 2 + 10 + 3x = 12 + 3x $$
|
|
|
|
$$ 12 + 3x = 2(15) $$
|
|
|
|
$$ 12 + 3x = 30 $$
|
|
|
|
$$ 3x = 18 $$
|
|
|
|
$$ \boxed{x = 6} $$
|
|
|
|
b. How many people attended the party?
|
|
|
|
$$ 2 + 5 + 6 = \boxed{13} $$
|
|
|
|
15. A small social network contains three people who are network friends with
|
|
six other people in the network, one person who is network friend with five
|
|
other people in the network, and five people who are network friends with
|
|
four other people in the network. The rest are network friends with three
|
|
other people in the network. The network contains 41 pairs of network
|
|
friends.
|
|
|
|
a. How many people are network friends with three other people in the network?
|
|
|
|
$$ (3 \cdot 6) + (1 \cdot 5) + (5 \cdot 4) + 3x = 41(2) $$
|
|
|
|
$$ 43 + 3x = 82 $$
|
|
|
|
$$ 3x = 39 $$
|
|
|
|
$$ \boxed{x = 13} $$
|
|
|
|
b. How many people are in the network?
|
|
|
|
$$ 3 + 1 + 5 + 13 = \boxed{22} $$
|
|
|
|
16.
|
|
|
|
a. In a group of 15 people, is it possible for each person to have exactly 3
|
|
friends? Justify your answer. (Assume that friendship is a symmetric
|
|
relationship: If $x$ is a friend of $y$, then $y$ is a friend of $x$.)
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose that, in a group of 15 people, each person had exactly three friends.
|
|
Then you could draw a graph representing each person by a vertex and connecting
|
|
two vertices by an edge if the corresponding people were friends. But such a
|
|
graph would have 15 vertices, each of degree 3, for a total of 45. This would
|
|
contradict the fact that the total degree of any graph is even. Hence the
|
|
supposition must be false, and in a group of 15 people it is not possible for
|
|
each to have exactly three friends.
|
|
|
|
b. In a group of 4 people, is it possible for each person to have exactly 3
|
|
friends? Justify your answer.
|
|
|
|
**Proof:**
|
|
|
|
Suppose that, in a group of 4 people, each person has exactly 3 friends. Then
|
|
you could draw a graph representing each person by a vertex and connecting two
|
|
vertices by an edge if the corresponding people were friends. Such a graph would
|
|
have 4 vertices, each of degree 3, for a total of 12. The total degree is even,
|
|
and therefore it is possible for each person to have exactly 3 friends.
|
|
|
|
17. In a group of 25 people, is it possible for each to shake hands with exactly
|
|
3 other people? Justify your answer.
|
|
|
|
No $25 \cdot 3 = 75$ is odd number of total degrees.
|
|
|
|
18. Is there a simple graph, each of whose vertices has even degree? Justify
|
|
your answer.
|
|
|
|
Yes, a minimum number of vertices would be 3. Each vertex would have a degree of
|
|
2 that would reach out to the other two vertices.
|
|
|
|
19. Suppose that $G$ is a graph with $v$ vertices and $e$ edges and that the
|
|
degree of each vertex is at least $d_{\text{min}}$ and at most
|
|
$d_{\text{max}}$. Show that
|
|
|
|
$$ \frac{1}{2}d_{\text{min}} \cdot v \leq e \leq \frac{1}{2}d_{\text{max}} \cdot v $$
|
|
|
|
**Proof:**
|
|
|
|
Let $e$ be the total number of edges, let $t$ be the total degree of the graph,
|
|
let $d_{\text{min}}$ be the minimum degree of any vertex in $G$, and let
|
|
$d_{\text{max}}$ be the maximum degree of any vertex in $G$.
|
|
|
|
The total degree of $G$ is greater than or equal to the minimum degree times the
|
|
total amount of vertices.
|
|
|
|
$$ d_{\text{min}} \cdot v \leq t $$
|
|
|
|
Also the total degree of $G$ is less than or equal to the maximum degree times
|
|
the total amount of vertices.
|
|
|
|
$$ d_{\text{min}} \cdot v \leq t \leq d_{\text{max}} \cdot v $$
|
|
|
|
The total degree of $G$ is $2e$.
|
|
|
|
$$ d_{\text{min}} \cdot v \leq 2e \leq d_{\text{max}} \cdot v $$
|
|
|
|
$$ \frac{1}{2}d_{\text{min}} \cdot v \leq e \leq \frac{1}{2}d_{\text{max}} \cdot v $$
|
|
|
|
20.
|
|
|
|
a. Draw $K_6$, a complete graph on six vertices.
|
|
|
|
b. Use the result of Example 4.9.9 to show that the number of edges of a simple
|
|
graph with $n$ vertices is less than or equal to $\dfrac{n(n - 1)}{2}$.
|
|
|
|
Prove that for any positive integer $n$, the number of edges of a simple graph
|
|
with $n$ vertices is less than or equal to $\dfrac{n(n - 1)}{2}$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $n$ and $e$ are any positive integers such that a simple graph $K_n$ has
|
|
$n$ vertices and $e$ edges.
|
|
|
|
By Example 4.9.9, we know the number of edges of a complete graph, $K_m$ is:
|
|
|
|
$$ \text{the number of edges of } K_m = \frac{m(m - 1)}{2} $$
|
|
|
|
A simple graph is a graph that does not have any loops or parallel edges, while
|
|
a complete graph is a simple graph with $m$ vertices and exactly one edge
|
|
connecting each pair of distinct vertices.
|
|
|
|
It follows then that the total number of edges for the simple graph $K_n$ could
|
|
only have at most the total number of edges for a complete graph of $n$
|
|
vertices.
|
|
|
|
Therefore we have proven that:
|
|
|
|
$$ e \text{ for } K_n \leq \frac{n(n - 1)}{2} $$
|
|
|
|
Q.E.D.
|
|
|
|
21.
|
|
|
|
a. In a simple graph, must every vertex have degree that is less than the number
|
|
of vertices in the graph? Why?
|
|
|
|
Yes. Let $G$ be a simple graph with $n$ vertices and let $v$ be a vertex of $G$.
|
|
Since $G$ has no parallel edges, $v$ can be joined by at most a single edge to
|
|
each of the $n - 1$ other vertices of $G$, and since $G$ has no loops, $v$
|
|
cannot be joined to itself. Therefore, the maximum degree of $v$ is $n - 1$.
|
|
|
|
b. Can there be a simple graph that has four vertices all of different degrees?
|
|
Why?
|
|
|
|
No. Suppose there is a simple graph with four vertices, all of which have
|
|
different degrees. By part (a), no vertex can have a degree greater than three,
|
|
and of course, no vertex can have a degree less than $0$. Therefore, the only
|
|
possible degrees of the vertices are 0, 1, 2, and 3. Since all four vertices
|
|
have different degrees, there is one vertex with each degree. But then the
|
|
vertex of degree 3 is connected to all other vertices, which contradicts the
|
|
fact that one of the vertices has degree 0. Hence the supposition is false, and
|
|
there is no simple graph with four vertices each of which has a different
|
|
degree.
|
|
|
|
c. For any integer $n \geq 5$, can there be a simple graph that has $n$ vertices
|
|
all of different degrees? Why?
|
|
|
|
No, let $n$ be an integer such that $n \geq 5$, and let $G$ be a simple graph
|
|
with $n$ vertices. By part (a) no vertex can have a degree greater than $n - 1$.
|
|
Since $n \geq 5$, then $n - 1 \geq 4$. Therefore the only possible degrees of
|
|
the vertices are $0, 1, \dots, n - 1$.
|
|
|
|
Since there are $n$ vertices and $n$ possible degree values, each degree must
|
|
occur exactly once. In particular there must be a vertex of degree $0$ and a
|
|
vertex of degree $n - 1$.
|
|
|
|
However, a vertex of degree $n - 1$ is adjacent to every other vertex in the
|
|
graph, including the vertex of degree $0$. This contradicts the fact that a
|
|
vertex of degree $0$ is adjacent to no vertices.
|
|
|
|
22. In a group of two or more people, must there always be at least two people
|
|
who are acquainted with the same number of people within the group? Why?
|
|
|
|
Yes.
|
|
|
|
Let the group contain $n \geq 2$ people. Model the situation with a simple graph
|
|
$G$ where each person is represented by a vertex. Two vertices are connected by
|
|
an edge if the corresponding people are acquainted.
|
|
|
|
Then the degree of a vertex is the number of people in the group that person is
|
|
acquainted with.
|
|
|
|
By Exercise 21, a simple graph with $n$ vertices cannot have all $n$ vertices of
|
|
different degrees. Equivalently, there must be at least two vertices with the
|
|
same degree.
|
|
|
|
Therefore, there must be at least two people who are acquainted with the same
|
|
number of people within the group.
|
|
|
|
23. Recall that $K_{m, n}$ denotes a complete bipartite graph on $(m, n)$
|
|
vertices.
|
|
|
|
a. Draw $K_{4, 2}$.
|
|
|
|
b. Draw $K_{1, 3}$.
|
|
|
|
c. Draw $K_{3, 4}$.
|
|
|
|
d. How many vertices of $K_{m, n}$ have degree $m$? degree $n$?
|
|
|
|
Vertices that have degree $m$ are all vertices in $\{n\}$.
|
|
|
|
Vertices that have degree $n$ are all vertices in $\{m\}$.
|
|
|
|
e. What is the total degree of $K_{m, n}$?
|
|
|
|
$$ (m \cdot n) + (n \cdot m) = 2mn $$
|
|
|
|
f. Find a formula in terms of $m$ and $n$ for the number of edges of $K_{m, n}$.
|
|
Justify your answer.
|
|
|
|
$$ e = mn $$
|
|
|
|
Justification omitted.
|
|
|
|
24. A (general) **bipartite graph** $G$ is a simple graph whose vertex set can
|
|
be partitioned into two disjoint nonempty subsets $V_1$ and $V_2$ such that
|
|
vertices in $V_1$ may be connected to vertices in $V_2$, but no vertices in
|
|
$V_1$ and no vertices in $V_2$ are connected to other vertices in $V_2$. For
|
|
example, the bipartite graph $G$ illustrated in (i) can be redrawn as shown
|
|
in (ii). From the drawing in (ii), you can see that $G$ is bipartite with
|
|
mutually disjoint vertex sets $V_w = \{v_1, v_3, v_5\}$ and
|
|
$V_2 = \{v_2, v_4, v_6\}$.
|
|
|
|
(i) See Page 266
|
|
|
|
(ii) See Page 266
|
|
|
|
Find which of the following graphs are bipartite. Redraw the bipartite graphs so
|
|
that their bipartite nature is evident.
|
|
|
|
See Page 266.
|
|
|
|
25. Suppose $r$ and $s$ are any positive integers. Does there exist a graph $G$
|
|
with the property that $G$ has vertices of degrees $r$ and $s$ and no other
|
|
degrees? Explain.
|
|
|
|
Omitted.
|
|
|
|
---
|
|
|
|
**Exercise Set 4.10**
|
|
|
|
Page 278
|
|
|
|
Find the value of $z$ when each of the algorithm segments in 1 and 2 is
|
|
executed.
|
|
|
|
1.
|
|
|
|
$i := 2\\ \text{\textbf{if }} (i > 3 \text{ or } i \leq 0)\\ \ \ \ \ \text{\textbf{then }} z := 1\\ \ \ \ \ \text{\textbf{else }} z := 0$
|
|
|
|
$z = 0$
|
|
|
|
2.
|
|
|
|
$i := 3\\ \text{\textbf{if }} (i \leq 3 \text{ or } i > 6)\\ \ \ \ \ \text{\textbf{then }} z := 2\\ \ \ \ \ \text{\textbf{else }} z := 0$
|
|
|
|
$z = 2$
|
|
|
|
3. Consider the following algorithm segment:
|
|
|
|
$\text{\textbf{if }} x \cdot y > 0 \text{\textbf{ then do }} y := 3 \cdot x\\ \ \ \ \ x := x + 1 \text{\textbf{end do}}\\ \ \ \ \ z := x \cdot y$
|
|
|
|
Find the value of $z$ if prior to execution $x$ and $y$ have the values given
|
|
below.
|
|
|
|
a. $x = 2, y = 3$
|
|
|
|
$y = 3 \cdot 3 = 9$, and $x = 2 + 1 = 3$, and $z = 9 \cdot 3 \cdot = 27$
|
|
|
|
b. $x = 1, y = 1$
|
|
|
|
$y = 3 \cdot 1 = 3$, and $x = 1 + 1 = 2$, and $z = 3 \cdot 2 = 6$
|
|
|
|
Find the values of $a$ and $e$ after execution of the loops in 4 and 5 by first
|
|
making trace tables for them.
|
|
|
|
4.
|
|
|
|
$a := 2\\ \text{\textbf{for }} i := 1 \text{\textbf{ to }} 3\\ \ \ \ \ a:= 3a + 1\\ \text{\textbf{next }} i$
|
|
|
|
| | 0 | 1 | 2 | 3 |
|
|
| --- | - | - | -- | -- |
|
|
| $a$ | 2 | 7 | 22 | 67 |
|
|
| $i$ | 1 | 2 | 3 | 4 |
|
|
|
|
After execution, $a = 67$.
|
|
|
|
5.
|
|
|
|
$e := 2, f := 0\\ \text{\textbf{for }} k := 1 \text{\textbf{ to }} 3\\ \ \ \ \ e := e \cdot k\\ \ \ \ \ f := e + f\\ \text{\textbf{next }} k$
|
|
|
|
| | 0 | 1 | 2 | 3 |
|
|
| --- | - | - | - | -- |
|
|
| $e$ | 2 | 2 | 4 | 12 |
|
|
| $f$ | 0 | 2 | 6 | 18 |
|
|
| $k$ | 1 | 2 | 3 | 4 |
|
|
|
|
After execution, $e = 12$, $f = 18$.
|
|
|
|
Make a trace table to trace the action of Algorithm 4.10.1 for the input
|
|
variables given in 6 and 7.
|
|
|
|
6. $a = 26, d = 7$
|
|
|
|
| | 0 | 1 | 2 | 3 |
|
|
| --- | -- | -- | -- | -- |
|
|
| $a$ | 26 | 26 | 26 | 26 |
|
|
| $d$ | 7 | 7 | 7 | 7 |
|
|
| $r$ | 26 | 19 | 12 | 5 |
|
|
| $q$ | 0 | 1 | 2 | 3 |
|
|
|
|
After execution, $q = 3$, and $r = 5$.
|
|
|
|
7. $a = 59, d = 13$
|
|
|
|
| | 0 | 1 | 2 | 3 | 4 |
|
|
| --- | -- | -- | -- | -- | -- |
|
|
| $a$ | 59 | 59 | 59 | 59 | 59 |
|
|
| $d$ | 13 | 13 | 13 | 13 | 13 |
|
|
| $r$ | 59 | 46 | 33 | 20 | 7 |
|
|
| $q$ | 0 | 1 | 2 | 3 | 4 |
|
|
|
|
After execution, $q = 4$, $r = 7$.
|
|
|
|
8. The following algorithm segment makes change; given an amount of money $A$
|
|
between 1¢ and 99¢, it determines a breakdown of $A$ into quarters $(q)$,
|
|
dimes $(d)$, nickels $(n)$, and pennies $(p)$.
|
|
|
|
$$
|
|
q := A \text{div } 25 \\
|
|
A := A \mod 25 \\
|
|
d := A \text{div } 10 \\
|
|
A := A \mod 10 \\
|
|
n := A \text{div } 5 \\
|
|
p := A \mod 5
|
|
$$
|
|
|
|
a. Trace this algorithm segment for $A = 69$.
|
|
|
|
| | | | | |
|
|
| --- | -- | -- | - | - |
|
|
| $A$ | 69 | 19 | 9 | |
|
|
| $q$ | 2 | | | |
|
|
| $d$ | | 1 | | |
|
|
| $n$ | | | 1 | |
|
|
| $p$ | | | | 4 |
|
|
|
|
b. Trace this algorithm segment for $A = 87$.
|
|
|
|
| | | | | |
|
|
| --- | -- | -- | - | - |
|
|
| $A$ | 87 | 12 | 2 | |
|
|
| $q$ | 3 | | | |
|
|
| $d$ | | 1 | | |
|
|
| $n$ | | | 0 | |
|
|
| $p$ | | | | 0 |
|
|
|
|
Find the greatest common divisor of each of the pairs of integers in 9-12. (Use
|
|
any method you wish.)
|
|
|
|
9. $27$ and $72$
|
|
|
|
$$ \text{gcd}(27, 72) = 9 $$
|
|
|
|
10. $5$ and $9$
|
|
|
|
$$ \text{gcd}(5, 9) = 1 $$
|
|
|
|
11. $7$ and $21$
|
|
|
|
$$ \text{gcd}(7, 21) = 7 $$
|
|
|
|
12. $48$ and $54$
|
|
|
|
$$ \text{gcd}(54, 48) = \text{gcd}(48, 6) = \text{gcd}(6, 0) = 6 $$
|
|
|
|
Use the Euclidean algorithm to hand-calculate the greatest common divisors of
|
|
each of the pairs of itnegers in 13-16.
|
|
|
|
13. $1,188$ and $385$
|
|
|
|
$$ \text{gcd}(1188, 385) = \text{gcd}(385, 33) = \text{gcd}(33, 22) = \text{gcd}(22, 11) = \text{gcd}(11, 0) = 11 $$
|
|
|
|
14. $509$ and $1,177$
|
|
|
|
$$ \text{gcd}(1177, 509) = \text{gcd}(509, 159) = \text{gcd}(159, 32) = \text{gcd}(32, 31) = \text{gcd}(31, 1) = \text{gcd}(1, 0) = 1 $$
|
|
|
|
15. $832$ and $10,933$
|
|
|
|
$$ \text{gcd}(10933, 832) = \text{gcd}(832, 117) = \text{gcd}(117, 13) = \text{gcd}(13, 0) = 13 $$
|
|
|
|
16. $4,131$ and $2,431$
|
|
|
|
$$ \text{gcd}(4131, 2431) = \text{gcd}(2431, 1700) = \text{gcd}(1700, 731) = \text{gcd}(731, 238) = \text{gcd}(238, 17) = \text{gcd}(17, 0) = 17 $$
|
|
|
|
Make a trace table to trace the action of Algorithm 4.10.2 for the input
|
|
variables given in 17-19.
|
|
|
|
17. $1,001$ and $871$
|
|
|
|
| | | | | | | | |
|
|
| ------------ | ---- | --- | --- | -- | -- | -- | -- |
|
|
| $A$ | 1001 | | | | | | |
|
|
| $B$ | 871 | | | | | | |
|
|
| $a$ | 1001 | 871 | 130 | 91 | 39 | 13 | |
|
|
| $b$ | 871 | 130 | 91 | 39 | 13 | 0 | |
|
|
| $r$ | 871 | 130 | 91 | 39 | 13 | 0 | |
|
|
| $\text{gcd}$ | | | | | | | 13 |
|
|
|
|
After execution, $\text{gcd} = 13$.
|
|
|
|
18. $5,859$ and $1,232$
|
|
|
|
| | | | | | | | | |
|
|
| ------------ | ---- | ---- | --- | --- | -- | -- | - | - |
|
|
| $A$ | 5859 | | | | | | | |
|
|
| $B$ | 1232 | | | | | | | |
|
|
| $a$ | 5859 | 1232 | 931 | 301 | 28 | 21 | 7 | |
|
|
| $b$ | 1232 | 931 | 301 | 28 | 21 | 7 | 0 | |
|
|
| $r$ | 1232 | 931 | 301 | 28 | 21 | 7 | 0 | |
|
|
| $\text{gcd}$ | | | | | | | | 7 |
|
|
|
|
After execution, $\text{gcd} = 7$.
|
|
|
|
19. $1,570$ and $488$
|
|
|
|
| | | | | | | | | | |
|
|
| ------------ | ---- | --- | --- | -- | -- | -- | -- | - | - |
|
|
| $A$ | 1570 | | | | | | | | |
|
|
| $B$ | 488 | | | | | | | | |
|
|
| $a$ | 1570 | 488 | 106 | 64 | 42 | 22 | 20 | 2 | |
|
|
| $b$ | 488 | 106 | 64 | 42 | 22 | 20 | 2 | 0 | |
|
|
| $r$ | 488 | 106 | 64 | 42 | 22 | 20 | 2 | 0 | |
|
|
| $\text{gcd}$ | | | | | | | | | 2 |
|
|
|
|
After execution, $\text{gcd} = 2$.
|
|
|
|
**Definition: Integers $a$ and $b$ are said to be **relatively prime** if, and
|
|
only if, their greatest common divisor is $1$.
|
|
|
|
In 20 and 21 trace the action of Algorithm 4.10.2 to determine whether the
|
|
integers are relatively prime.
|
|
|
|
20. $4,167$ and $2,563$
|
|
|
|
| | | | | | | | | | | |
|
|
| ------------ | ---- | ---- | ---- | --- | --- | --- | -- | - | - | - |
|
|
| $A$ | 4167 | | | | | | | | | |
|
|
| $B$ | 2563 | | | | | | | | | |
|
|
| $a$ | 4167 | 2563 | 1604 | 959 | 645 | 314 | 17 | 8 | 1 | |
|
|
| $b$ | 2563 | 1604 | 959 | 645 | 314 | 17 | 8 | 1 | 0 | |
|
|
| $r$ | 2563 | 1604 | 959 | 645 | 314 | 17 | 8 | 1 | 0 | |
|
|
| $\text{gcd}$ | | | | | | | | | | 1 |
|
|
|
|
After execution, $\text{gcd} = 1$, and because of this, $4,167$ and $2,563$ are
|
|
_relatively_ prime.
|
|
|
|
21. $34,391$ and $6,728$
|
|
|
|
| | | | | | | | | | |
|
|
| ------------ | ----- | ---- | --- | --- | -- | - | - | - | - |
|
|
| $A$ | 34391 | | | | | | | | |
|
|
| $B$ | 6728 | | | | | | | | |
|
|
| $a$ | 34391 | 6728 | 751 | 720 | 31 | 7 | 3 | 1 | |
|
|
| $b$ | 6728 | 751 | 720 | 31 | 7 | 3 | 1 | 0 | |
|
|
| $r$ | 6728 | 751 | 720 | 31 | 7 | 3 | 1 | 0 | |
|
|
| $\text{gcd}$ | | | | | | | | | 1 |
|
|
|
|
After execution, $\text{gcd} = 1$, and because of this, $34,391$ and $6,728$ are
|
|
_relatively_ prime.
|
|
|
|
22. Prove that for all positive integers $a$ and $b$, $a \mid b$ if, and only
|
|
if, $\text{gcd}(a, b) = a$. (Note that to prove "$A$ if, and only if, $B$,"
|
|
you need to prove "if $A$ then $B$" and "if $B$ then $A$.")
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$ and $b$ are any positive integers such that $a \mid b$.
|
|
|
|
By the definition of divisibility, any number is divisible by at least $1$ and
|
|
itself. It follows then that $a \mid a$. This makes $a$ a common divisor of $a$
|
|
and $b$. Thus, by definition of the common divisor, $a$ is less than or equal to
|
|
the greatest common divisor of $a$ and $b$, $a \leq \text{gcd}(a, b)$.
|
|
|
|
Similarly, since $\text{gcd}(a, b)$ is the greatest common divisor of $a$ and
|
|
$b$, it must also divide $a$, $\text{gcd}(a, b) \mid a$. Thus, by Theorem 4.4.1,
|
|
$\text{gcd}(a, b) \leq a$.
|
|
|
|
We know that when $a \leq \text{gcd}(a, b)$ and $\text{gcd}(a, b) \leq a$ are
|
|
both true, this means that $\text{gcd}(a, b) = a$.
|
|
|
|
Therefore $\text{gcd}(a, b) = a$. _[as was to be shown.]_
|
|
|
|
Then, suppose $a$ and $b$ are any positive integers such that their greatest
|
|
common divisor is $\text{gcd}(a, b) = a$.
|
|
|
|
By definition of a common divisor, $a \mid \text{gcd}(a, b)$ and
|
|
$\text{gcd}(a, b) \mid b$.
|
|
|
|
Therefore $a \mid b$. _[as was to be shown.]_
|
|
|
|
Q.E.D.
|
|
|
|
23.
|
|
|
|
a. Prove that if $a$ and $b$ are integers, not both zero, and
|
|
$d = \text{gcd}(a, b)$, then $\dfrac{a}{d}$ and $\dfrac{b}{d}$ are integers with
|
|
no common divisor that is greater than $1$.
|
|
|
|
**Proof by contradiction:**
|
|
|
|
Suppose not. That is, suppose that $a$ and $b$ are any nonzero integers and
|
|
there is some positive integer $d$, where $d = \text{gcd}(a, b)$ and
|
|
$\dfrac{a}{d}$ and $\dfrac{b}{d}$ are integers with a common divisor that is
|
|
greater than $1$.
|
|
|
|
Since $d = \text{gcd}(a, b)$. By the definition of a common divisor, $d \mid a$
|
|
and $d \mid b$.
|
|
|
|
Since $d \mid a$ and $d \mid b$, by the definition of divisibility, $a = dk$ and
|
|
$b = dm$ for some integers $k$ and $m$. By algebra:
|
|
|
|
$$ a = dk \to \frac{a}{d} = k $$
|
|
|
|
$$ b = dm \to \frac{b}{d} = m $$
|
|
|
|
Since $\dfrac{a}{d}$ and $\dfrac{b}{d}$ are integers with a common divisor that
|
|
is greater than $1$. This means that $k$ and $m$ have some common divisor, $c$
|
|
such that $c > 1$. Because $c > 1$, this means that $dc > d$. It follows that
|
|
$dc$ would then be a common divisor of $a$ and $b$ where $dc > d$, which means
|
|
that $d$ is not the greatest common divisor of $a$ and $b$,
|
|
$d \neq \text{gcd}(a, b)$.
|
|
|
|
This is a contradiction.
|
|
|
|
Q.E.D.
|
|
|
|
b. Write an algorithm that accepts the numerator and denominator of a fraction
|
|
as input and produces as output the numerator and denominator of that fraction
|
|
written in lowest terms. (The algorithm may call upon the Euclidean algorithm as
|
|
needed.)
|
|
|
|
**Algorithm**
|
|
|
|
_[Given a rational number $\dfrac{N}{D}$, output both $N$ and $D$ in lowest
|
|
terms.]_
|
|
|
|
_[Note that $EA()$ here represents calling the Euclidean Algorithm]_
|
|
|
|
**Input:** $\dfrac{N}{D}$ _[a rational number.]_
|
|
|
|
**Algorithm Body:**
|
|
|
|
$d := EA(N, D)\\ n := \dfrac{N}{d}\\ m := \dfrac{D}{d}$
|
|
|
|
**Output:** $n$ and $m$.
|
|
|
|
24. Complete the proof of Lemma 4.10.2 by proving the following: If $a$ and $b$
|
|
are any integers with $b \neq 0$ and $q$ and $r$ are any integers such that
|
|
|
|
$$ a = bq + r $$
|
|
|
|
then
|
|
|
|
$$ \text{gcd}(b, r) \leq \text{gcd}(a, b) $$
|
|
|
|
**Continuation of Lemma 4.10.2:**
|
|
|
|
2. $\text{\textbf{gcd}}(b, r) \leq \text{\textbf{gcd}}(a, b)$:
|
|
|
|
a. _[We will first show that any common divisor of $b$ and $r$ is also a common
|
|
divisor of $a$ and $b$.]_
|
|
|
|
Let $b$ and $r$ be integers where $b \neq 0$, and let $c$ be a common divisor of
|
|
$b$ and $r$. Then $c \mid b$ and $c \mid r$ and so, by definition of
|
|
divisibility, $b = nc$ and $r = mc$, for some integers $n$ and $m$. Substitute
|
|
into the equation
|
|
|
|
$$ a = bq + r $$
|
|
|
|
to obtain
|
|
|
|
$$ a = (nc)q + mc $$
|
|
|
|
$$ a = (nq + m)c $$
|
|
|
|
Now, $nq + m$ is an integer, and so, by definition of divisibility, $c \mid a$.
|
|
Because we already know that $c \mid b$, we can conclude that $c$ is a common
|
|
divisor of $a$ and $b$ _[as was to be shown]._
|
|
|
|
b. _[Next, we show that $\text{gcd}(b, r) \leq \text{gcd}(a, b)$.]_
|
|
|
|
Now the greatest common divisor of $b$ and $r$ is defined because $b$ and $r$
|
|
are not both zero. Also, by part (a), every common divisor of $b$ and $r$ is a
|
|
common divisor of $a$ and $b$, and so the greatest common divisor of $b$ and $r$
|
|
is a common divisor of $a$ and $b$. But then $\text{gcd}(b, r)$ (being one of
|
|
the common divisors of $a$ and $b$) is less than or equal to the greatest common
|
|
divisor of $a$ and $b$:
|
|
|
|
$$ \text{gcd}(b, r) \leq \text{gcd}(a, b) $$
|
|
|
|
25.
|
|
|
|
a. Prove: If $a$ and $d$ are positive integers and $q$ and $r$ are integers such
|
|
that $a = dq + r$ and $0 \leq r < d$, then
|
|
|
|
$$ -a = d(-(q + 1)) + (d - r) $$
|
|
|
|
and
|
|
|
|
$$ 0 < d - r \leq d $$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$ and $d$ are positive integers, and that $q$ and $r$ are integers
|
|
such that $a = dq + r$ and $0 \leq r < d$.
|
|
|
|
By algebra:
|
|
|
|
$$ a = dq + r $$
|
|
|
|
$$ -a = -(dq + r) $$
|
|
|
|
$$ -a = -(dq + d - d + r) $$
|
|
|
|
$$ -a = -(d(q + 1) - d + r) $$
|
|
|
|
$$ -a = -d(q + 1) + d - r $$
|
|
|
|
$$ -a = d(-(q + 1)) + (d - r) $$
|
|
|
|
Then, also by algebra:
|
|
|
|
$$ 0 \leq r < d $$
|
|
|
|
$$ 0 \geq -r > -d $$
|
|
|
|
Then add $d$ to all sides:
|
|
|
|
$$ 0 + d \geq -r + d > -d + d $$
|
|
|
|
$$ d \geq -r + d > 0 $$
|
|
|
|
Rearranged:
|
|
|
|
$$ 0 < d - r \leq d $$
|
|
|
|
Therefore $-a = d(-(q + 1)) + (d - r)$ and $0 < d - r \leq d$.
|
|
|
|
Q.E.D.
|
|
|
|
b. Indicate how to modify Algorithm 4.10.1 to allow for the input $a$ to be
|
|
negative.
|
|
|
|
**Algorithm 4.10.1 Division Algorithm**
|
|
|
|
_[Given an integer $a$ and a positive integer $d$, the aim of the algorithm is
|
|
to find integers $q$ and $r$ that satisfy the conditions $a = dq + r$ and
|
|
$0 \leq r < d$. This is done by subtracting $d$ repeatedly from $a$ until the
|
|
result is less than $d$ but is nonnegative._
|
|
|
|
$$ 0 \leq a - d - d - d - \dots - d = a - dq < d$$
|
|
|
|
_The total number of $d$'s that are subtracted is the quotient $q$. The quantity
|
|
$a - dq$ equals the remainder $r$.]_
|
|
|
|
**Input:** _$a$ [an integer], $d$ [a positive integer]_
|
|
|
|
**Algorithm Body:**
|
|
|
|
$\text{\textbf{if }}(a \geq 0) \text{\textbf{then}}\\ \ \ r := a, q := 0\\ \ \ \text{\textbf{while }}(r \geq d)\\ \ \ \ \ r := r - d\\ \ \ \ \ q := q + 1\\ \ \ \text{\textbf{end while}}\\ \text{\textbf{else}}\\ \ \ r := -a, q := 0\\ \ \ \text{\textbf{while }}(r \geq d)\\ \ \ \ \ r := r - d\\ \ \ \ \ q := q + 1\\ \ \ \text{\textbf{end while}}\\ \ \ \text{\textbf{if }}(r == 0) \text{\textbf{ then}}\\ \ \ \ \ q := -q\\ \ \ \text{\textbf{else}}\\ \ \ \ \ q := -(q + 1)\\ \ \ \ \ r := d - r\\ \ \ \text{\textbf{end if}}\\ \text{\textbf{end if}}$
|
|
|
|
**Output:** $q$, $r$
|
|
|
|
26.
|
|
|
|
a. Prove that if $a$, $d$, $q$, and $r$ are integers such that $a = dq + r$ and
|
|
$0 \leq r < d$, then
|
|
|
|
$$ q = \left\lfloor \frac{a}{d} \right\rfloor \quad \text{ and } r = a - \left\lfloor \frac{a}{d} \right\rfloor \cdot d$$
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$, $d$, $q$, and $r$ are any integers such that $a = dq + r$ and
|
|
$0 \leq r < d$.
|
|
|
|
By algebra:
|
|
|
|
$$ a = dq + r $$
|
|
|
|
$$ r = a - dq $$
|
|
|
|
Then by substitution:
|
|
|
|
$$ 0 \leq r < d $$
|
|
|
|
$$ 0 \leq a - dq < d $$
|
|
|
|
$$ dq \leq a < dq + d $$
|
|
|
|
$$ dq \leq a < d(q + 1) $$
|
|
|
|
$$ q \leq \frac{a}{d} < q + 1 $$
|
|
|
|
Thus, by the definition of floor, $q = \left\lfloor \dfrac{a}{d} \right\rfloor$.
|
|
|
|
Then, by substitution:
|
|
|
|
$$ a = dq + r $$
|
|
|
|
$$ a = d\left\lfloor \frac{a}{d} \right\rfloor + r $$
|
|
|
|
$$ r = a - d\left\lfloor \frac{a}{d} \right\rfloor $$
|
|
|
|
$$ r = a - \left\lfloor \frac{a}{d} \right\rfloor \cdot d $$
|
|
|
|
Therefore $q = \left\lfloor \dfrac{a}{d} \right\rfloor$ and
|
|
$r = a - \left\lfloor \dfrac{a}{d} \right\rfloor \cdot d$.
|
|
|
|
Q.E.D.
|
|
|
|
b. In a computer language with a built-in floor function, $\text{div}$ and
|
|
$\text{mod}$ can be calculated as follows:
|
|
|
|
$$ a \text{ div } d = \left\lfloor \frac{a}{d} \right\rfloor \quad \text{ and } \quad a \mod d = a - \left\lfloor \frac{a}{d} \right\rfloor \cdot d $$
|
|
|
|
Rewrite the steps of Algorithm 4.10.2 for a computer language with a built-in
|
|
floor function but without $\text{div}$ and $\text{mod}$.
|
|
|
|
**Algorithm Body:**
|
|
|
|
$a := A, b := B, r:= B$
|
|
|
|
_[If $b \neq 0$, compute $a \mod b$, the remainder of the integer division of
|
|
$a$ by $b$, and set $r$ equal to this value. Then repeat the process using $b$
|
|
in place of $a$ and $r$ in place of $b$.]_
|
|
|
|
$\text{\textbf{while }} (b \neq 0)\\ \ \ \ \ r := a - \left\lfloor \dfrac{a}{b} \right\rfloor \cdot b$
|
|
|
|
_[The value of $a \mod b$ can be obtained by calling the division algorithm.]_
|
|
|
|
$\ \ \ \ a := b\\ \ \ \ \ b := r\\ \text{\textbf{end while}}$
|
|
|
|
_[After execution of the $\text{\textbf{while}}$ loop, $\text{gcd}(A, B) = a$.]_
|
|
|
|
$\text{gcd} := a$
|
|
|
|
27. An alternative to the Euclidean algorithm uses subtraction rather than
|
|
division to compute greatest common divisors. (After all, division is
|
|
repeated subtraction.) It is based on the following lemma.
|
|
|
|
**Lemma 4.10.3**
|
|
|
|
**Algorithm 4.10.3 Computing gcd's by Subtraction**
|
|
|
|
_[Given two positive integers $A$ and $B$, variables $a$ and $b$ are set equal
|
|
to $A$ and $B$. Then a repetitive process begins. If $a \neq 0$, and $b \neq 0$,
|
|
then the larger of $a$ and $b$ is set equal to
|
|
$a - b (\text{if } a \geq b) \text{ or to } b - a(\text{if } a < b)$, and the
|
|
smaller of $a$ and $b$ is left unchanged. This process is repeated over and over
|
|
until eventually $a$ or $b$ becomes $0$. By Lemma 4.10.3, after each repetition
|
|
of the process,_
|
|
|
|
$$ \text{gcd}(A, B) = \text{gcd}(a, b) $$
|
|
|
|
_After the last repetition,_
|
|
|
|
$$ \text{gcd}(A, B) = \text{gcd}(a, 0) \quad \text{ or } \quad \text{gcd}(A, B) = \text{gcd}(0, b) $$
|
|
|
|
_depending on whether $a$ or $b$ is nonzero. But by Lemma 4.10.1,_
|
|
|
|
$$ \text{gcd}(a, 0) = a \quad \text{ and } \quad \text{gcd}(0, b) = b $$
|
|
|
|
_Hence, after the last repetition,_
|
|
|
|
$$ \text{gcd}(A, B) = a \text{ if } a \neq 0 \quad \text{ or } \quad \text{gcd}(A, B) = b \text{ if } b \neq 0 $$
|
|
|
|
**Input:** $A, B$ _[positive integers]_
|
|
|
|
**Algorithm Body:**
|
|
|
|
$a := A, b := B\\ \text{\textbf{while }} (a \neq 0 \text{ and } b \neq 0)\\ \ \ \ \ \text{\textbf{if }} a \geq b \text{\textbf{ then }} a := a - b\\ \ \ \ \ \ \ \ \ \text{\textbf{else }} b := b - a\\ \text{\textbf{end while}}\\ \ \ \ \ \text{\textbf{if }} a = 0 \text{\textbf{ then }} gcd := b\\ \ \ \ \ \text{\textbf{else }} gcd := a$
|
|
|
|
_[After execution of the **if-then-else** statement,
|
|
$\text{gcd} = \text{gcd}(A, B)$.]_
|
|
|
|
**Output:** $\text{gcd}$ _[a positive integer]_
|
|
|
|
a. Prove Lemma 4.10.3.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$ and $b$ are any integers such that $a \geq b > 0$.
|
|
|
|
Let $c$ be some integer such that $c$ is common divisor of both $a$ and $b$ such
|
|
that $c \mid a$ and $c \mid b$.
|
|
|
|
Since $c \mid a$ and $c \mid b$, $a = nc$ and $b = mc$ for some integers $n$ and
|
|
$m$.
|
|
|
|
By substitution:
|
|
|
|
$$ a - b = nc - mc $$
|
|
|
|
$$ a - b = c(n - m) $$
|
|
|
|
Now, $n - m$ is an integer by the difference of integers, and so by the
|
|
definition of divisibility, $c \mid (a - b)$.
|
|
|
|
Thus, it follows that $\text{gcd}(a, b) \leq \text{gcd}(b, a - b)$.
|
|
|
|
Let $k$ be some integer such that $k$ is common divisor of both $a - b$ and $b$
|
|
such that $k \mid (a - b)$ and $k \mid b$.
|
|
|
|
Since $k \mid (a - b)$ and $k \mid b$, $a - b = nk$ and $b = mk$ for some
|
|
integers $n$ and $m$.
|
|
|
|
By substitution:
|
|
|
|
$$ a - b = nk $$
|
|
|
|
$$ a = nk + b $$
|
|
|
|
$$ a = nk + mk $$
|
|
|
|
$$ a = k(n + m) $$
|
|
|
|
Since $m + n$ is an integer by the sum of integers, the definition of
|
|
divisibility tells us that $k \mid a$.
|
|
|
|
Since we already know that $k \mid b$, it follows that $k$ is a common divisor
|
|
of both $a$ and $b$.
|
|
|
|
Thus it follows that $\text{gcd}(a, b) \geq \text{gcd}(b, a - b)$.
|
|
|
|
It has now been established that $\text{gcd}(a, b) \leq \text{gcd}(b, a - b)$
|
|
and also that $\text{gcd}(a, b) \geq \text{gcd}(b, a- b)$.
|
|
|
|
Therefore $\text{gcd}(a, b) = \text{gcd}(b, a - b)$
|
|
|
|
Q.E.D.
|
|
|
|
b. Trace the execution of Algorithm 4.10.3 for $A = 630$ and $B = 336$.
|
|
|
|
| | | | | | | | | | | |
|
|
| ------------ | --- | --- | --- | --- | --- | --- | --- | -- | -- | -- |
|
|
| $A$ | 630 | | | | | | | | | |
|
|
| $B$ | 336 | | | | | | | | | |
|
|
| $a$ | 630 | 294 | 294 | 252 | 210 | 168 | 126 | 84 | 42 | 0 |
|
|
| $b$ | 336 | 336 | 42 | 42 | 42 | 42 | 42 | 42 | 42 | 42 |
|
|
| $\text{gcd}$ | | | | | | | | | | 42 |
|
|
|
|
c. Trace the execution of Algorithm 4.10.3 for $A = 768$ and $B = 348$.
|
|
|
|
| | | | | | | | | | | | | | |
|
|
| ------------ | --- | --- | --- | --- | --- | --- | -- | -- | -- | -- | -- | -- | -- |
|
|
| $A$ | 768 | | | | | | | | | | | | |
|
|
| $B$ | 348 | | | | | | | | | | | | |
|
|
| $a$ | 768 | 420 | 72 | 72 | 72 | 72 | 72 | 12 | 12 | 12 | 12 | 12 | 0 |
|
|
| $b$ | 348 | 348 | 348 | 276 | 204 | 132 | 60 | 60 | 48 | 36 | 24 | 12 | 12 |
|
|
| $\text{gcd}$ | | | | | | | | | | | | | 12 |
|
|
|
|
Exercises 28-32 refer to the following definition.
|
|
|
|
**Definition:** The **least common multiple** of two nonzero integers $a$ and
|
|
$b$, denoted $\text{\textbf{lcm}}(a, b)$, is the positive integer $c$ such that
|
|
|
|
a. $a \mid c$ and $b \mid c$
|
|
|
|
b. for all positive integers $m$, if $a \mid m$ and $b \mid m$, then $c \leq m$.
|
|
|
|
28. Find
|
|
|
|
a. $\text{lcm}(12, 18)$
|
|
|
|
$\text{lcm}(12, 18) = 36$
|
|
|
|
b. $\text{lcm}(2^2 \cdot 3 \cdot 5, 2^3 \cdot 3^2)$
|
|
|
|
$$ \text{lcm}(12, 18) = 2^3 \cdot 3^2 \cdot 5 = 360 $$
|
|
|
|
c. $\text{lcm}(2800, 6125)$
|
|
|
|
$$ 2800 = 100 \cdot 28 = 10^2 \cdot 7 \cdot 4 = 2^2 \cdot 5^2 \cdot 7 \cdot 2^2 = 2^4 \cdot 5^2 \cdot 7 $$
|
|
|
|
$$ 6125 = 25 \cdot 245 = 5^2 \cdot 5 \cdot 49 = 5^3 \cdot 7^2 $$
|
|
|
|
$$ 2800 = 2^4 \cdot 5^2 \cdot 7 $$
|
|
|
|
$$ 6125 = 5^3 \cdot 7^2 $$
|
|
|
|
$$ \text{lcm}(2800, 6125) = 2^4 \cdot 5^3 \cdot 7^2 = 98000 $$
|
|
|
|
29. Prove that for all positive integers $a$ and $b$,
|
|
$\text{gcd}(a, b) = \text{lcm}(a, b)$ if, and only if, $a = b$.
|
|
|
|
**Proof:**
|
|
|
|
Suppose $a$ and $b$ are any positive integers such that
|
|
$\text{gcd}(a, b) = \text{lcm}(a, b)$.
|
|
|
|
Let $k$ be some integer such that $k = \text{gcd}(a, b) = \text{lcm}(a, b)$.
|
|
|
|
Since $k = \text{gcd}(a, b)$, $k \mid \text{gcd}(a, b)$, and thus $k \leq a$ and
|
|
$k \leq b$.
|
|
|
|
And since $k = \text{lcm}(a, b)$, $\text{lcm}(a, b) \mid k$ and thus $a \leq k$
|
|
and $b \leq k$.
|
|
|
|
Since $k \leq a$ and $a \leq k$, this means that $k = a$.
|
|
|
|
And since $k \leq b$ and $b \leq k$, this means that $k = b$.
|
|
|
|
By the law of transitivity, $a = k = b$.
|
|
|
|
Thus $a = b$.
|
|
|
|
Now, suppose $a$ and $b$ are any positive integers such that $a = b$.
|
|
|
|
By the definition of greatest common divisor, $\text{gcd}(a, a) = a$.
|
|
|
|
By the definition of least common multiple, $\text{lcm}(a, a) = a$.
|
|
|
|
This means that $\text{gcd}(a, a) = \text{lcm}(a, a)$.
|
|
|
|
Since $a = b$, $b$ can be substituted for $a$:
|
|
|
|
$$ \text{gcd}(a, b) = \text{lcm}(a, b) $$
|
|
|
|
Thus $\text{gcd}(a, b) = \text{lcm}(a, b)$
|
|
|
|
Therefore it has been shown that if $\text{gcd}(a, b) = \text{lcm}(a, b)$, then
|
|
$a = b$, and it has also been shown that if $a = b$, then
|
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$\text{gcd}(a, b) = \text{lcm}(a, b)$.
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Q.E.D.
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30. Prove that for all positive integers $a$ and $b$, $a \mid b$ if, and only
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if, $\text{lcm}(a, b) = b$.
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**Proof:**
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Suppose $a$ and $b$ are any positive integers such that $a \mid b$.
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By the definition of a multiple, $b \mid b$ means that $b$ is a multiple of $b$.
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This means that since $a \mid b$ and $b \mid b$, that $b$ is a common multiple
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of both $a$ and $b$.
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Let $m$ be some positive integer such that $b \mid m$ and $a \mid m$. Since $b$
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and $m$ are positive integers, and since $b \mid m$, this means that $b \leq m$.
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Thus $\text{lcm}(a, b) = b$.
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Then suppose $a$ and $b$ are any positive integers such that
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$\text{lcm}(a, b) = b$.
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By the definition of a common multiple, we know that $a \mid \text{lcm}(a, b)$.
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Since we also know that $\text{lcm}(a, b) = b$, by substitution:
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$$ a \mid b $$
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Thus $a \mid b$.
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Therefore it has been shown that if $a \mid b$, then $\text{lcm}(a, b) = b$ and
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it has also been shown that if $\text{lcm}(a, b) = b$, then $a \mid b$.
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Q.E.D.
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31. Prove that for all integers $a$ and $b$,
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$\text{gcd}(a, b) \mid \text{lcm}(a, b)$.
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**Proof:**
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Suppose $a$ and $b$ are any integers.
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Since $a$ is an integer, this means that $\text{gcd}(a, b) \mid a$ and
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$a \mid \text{lcm}(a, b)$.
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By the transitive property of divisibility, this means that:
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$$ \text{gcd}(a, b) \mid \text{lcm}(a, b) $$
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Therefore $\text{gcd}(a, b) \mid \text{lcm}(a, b)$.
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Q.E.D.
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32. Prove that for all positive integers $a$ and $b$,
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$\text{gcd}(a, b) \cdot \text{lcm}(a, b) = ab$.
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Omitted.
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