3013 lines
75 KiB
Markdown
3013 lines
75 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.
|