diff --git a/chapter_4/exercises.md b/chapter_4/exercises.md index 9c96d84..07d3b9c 100644 --- a/chapter_4/exercises.md +++ b/chapter_4/exercises.md @@ -9,96 +9,380 @@ composite numbers. a. Is $-17$ an odd integer? +$$ -17 = 2(-9) + 1 $$ + +Let $k = -9$, so our expression becomes by substitution: + +$$ -17 = 2k + 1 $$ + +Since $-17$ can be represented by the form $2k + 1$ where $k = -9$ and $k$ is an +integer, by the definition of an odd number, $-17$ is an odd integer. + b. Is $0$ neither even nor odd? +No, $0$ can be represented as $0 = 2(0)$, and let $k = 0$, so $0 = 2k$, where +$k$ is an integer, and by definition of an even number, $0$ is even. + c. Is $2k - 1$ odd? +Yes, $2k - 1 = 2(k - 1) + 1$ where $k - 1$ is an integer by the difference of +integers. Let $m = k - 1$, so our expression becomes $2k - 1 = 2m + 1$, and +since $m$ is an integer, we can conclude that $2k - 1$ is an odd integer by +definition of odd integers. + 2. Assume that $c$ is a particular integer. a. Is $-6c$ an even integer? +Yes $-6c = 2(-3c)$, where $-3c$ is an integer by the product of integers, and +since $-6c$ can be expressed as $2 \cdot \text{ some integer}$, it is even by +the definition of even integers. + b. Is $8c + 5$ an odd integer? -c. Is $(c^1 + 1) - (c^2 - 1) - 2$ an even integer? +Yes, $8c + 5 = 8c + 4 + 1 = 2(4c + 2) + 1$ where $4c + 2$ is an integer by the +sum of products of integers. Since $8c + 5$ can be expressed as +$2(\text{some integer}) + 1$, we can conclude that $8c + 5$ is an odd integer by +the definition of odd integers. + +c. Is $(c^2 + 1) - (c^2 - 1) - 2$ an even integer? + +Yes, if evaluate the statement by laws of algebra, we get: + +$$ (c^2 + 1) - (c^2 - 1) - 2 = c^2 + 1 - c^2 + 1 - 2 = 1 + 1 - 2 = 0 $$ + +And as established in 1b, $0$ is an even integer, so $(c^2 + 1) - (c^2 - 1) - 2$ +can be expressed in the form of $2 \cdot \text{ some integer}$, so by the +definition of integers, $(c^2 + 1) - (c^2 - 1) - 2$ is an even integer. 3. Assume that $m$ and $n$ are particular integers? a. Is $6m + 8n$ even? +Yes, $6m + 8n = 2(3m + 4n)$. $3m + 4n$ is an integer by the sum of products of +integers. Since $6m + 8n$ can be expressed as $2 \cdot \text{ some integer}$, by +the definition of even integers, $6m + 8n$ is even. + b. Is $10mn + 7$ odd? +$10mn + 7 = 10mn + 6 + 1 = 2(5mn + 3) + 1$. $5mn + 3$ is an integer by the +product and sum of integers. Since $10mn + 7$ can be expressed as +$2(\text{some integer}) + 1$, $10mn + 7$ is an odd integer. + c. If $m > n > 0$, is $m^2 - n^2$ composite? +Not necessarily. Consider $m = 3$ and $n = 2$, then $m^2 - n^2 = 9 - 4 = 5$, +which is a prime number. + 4. Assume that $r$ and $s$ are particular integers. a. Is $4rs$ even? +Yes, $4rs = 2(2rs)$, where $2rs$ is an integer by the product of integers. Since +$4rs = 2(\text{ some integer})$, by the definition of an even integer, $4rs$ is +an even integer. + b. Is $6r + 4s^2 + 3$ odd? +$6r + 4s^2 + 3 = 6r + 4s^2 + 2 + 1 = 2(3r + 2s^2 + 1) + 1$. $3r + 2s^2 + 1$ is +an integer by product and sum of integers. Let $k = 3r + 2s^2 + 1$, and so +$6r + 4s^2 + 3 = 2k + 1$. By definition of an odd integer, $6r + 4s^2 + 3$ is an +odd integer. + c. If $r$ and $s$ are both positive, is $r^2 + 2rs + s^2$ composite? +Since $r^2 + 2rs + s^2 = (r + s)^2 = (r + s)(r + s)$ and $r + s \geq 2$. And +since $r + s > 1$, the product of $(r + s)(r + s)$ is composite. + Prove the statements in 5-11. 5. There are integers $m$ and $n$ such that $m > 1$ and $n > 1$ and $\dfrac{1}{m} + \dfrac{1}{n}$ is an integer. +For example, let $m = 2$ and $n = 2$, then $\dfrac{1}{2} + \dfrac{1}{2} = 1$, +and $1$ is an integer. + 6. There are distinct integers $m$ and $n$ such that $\dfrac{1}{m} + \dfrac{1}{n}$ is an integer. +For example, let $m = -2$, and $n = 2$, then $\dfrac{1}{-2} + \dfrac{1}{2} = 0$, +and $0$ is an integer. + 7. There are real numbers $a$ and $b$ such that $$ \sqrt{a + b} = \sqrt{a} + \sqrt{b} $$ +For example, let $a = 0$ and $b = 9$, then +$\sqrt{0 + 9} = \sqrt{9} = 3 = 0 + 3 = \sqrt{0} + \sqrt{9}$. + 8. There is an integer $n > 5$ such that $2^n - 1$ is prime. +For example, let $n = 7$, then $2^7 - 1 = 127$, and $127$ is prime. + 9. There is a real number $x$ such that $x > 1$ and $2^x > x^{10}$. +For example, let $x = \dfrac{1}{2}$, then $2^{\frac{1}{2}} \approx 1.414213562 > +0.0009765625 = \left(\frac{1}{2}\right)^{10}$ + **Definition:** An integer $n$ is called a **perfect square** if, and only if, $n = k^2$ for some integer $k$. 10. There is a perfect square that can be written as a sum of two other perfect squares. +Let $n = 4$ and $m = 3$, and let $l = k^2$ be the sum of their squares: + +$$ l = k^2 = n^2 + m^2 = (4)^2 + (3)^2 = 16 + 9 = 25 $$ + +$$ l = k^2 = 25 $$ + +So $l = 25$ can be written as $n^2 + m^2$ where $n = 4$ and $m = 3$, and since +both $n$ and $m$ are integers, we can say that $l$ is a perfect square by +definition of a perfect square and $l$ can be written as the sum of two other +perfect squares. + 11. There is an integer $n$ such that $2n^2 - 5n + 2$ is prime. +For example let $n = 3$, then +$2n^2 - 5n + 2 = 2(3)^2 - 5(3) + 2 = 2(9) - 15 + 2 = 18 - 15 + 2 = 3 + 2 = 5$, +and $5$ is prime. + In 12-13, (a) write a negation for the given statement, and (b) use a counterexample to disprove the given statement. Explain how the counterexample actually shows that the given statement is false. 12. For all real numbers $a$ and $b$, if $a < b$ the $a^2 < b^2$. +(a) + +Original: + +$$ \forall a \in \mathbb{R} \forall b \in \mathbb{R}((a < b) \to (a^2 < b^2)) $$ + +Negation: + +$$ \exists a \in \mathbb{R} \exists b \in \mathbb{R}((a < b) \wedge (a^2 \geq b^2)) $$ + +There exist real numbers $a$ and $b$ such that $a < b$ and $a^2 \geq b^2$. + +(b) + +Counterexample: + +Let $a = -2$ and let $b = -1$. The hypothesis $a < b$ holds as $-2 < -1$ is +true, but the conclusion of the original statement $a^2 < b^2$ is false as +$(-2)^2 = 4 \cancel{<} 1 = (-1)^2$. + +Since the original statement claims that the implication holds true for all real +numbers $a$ and $b$, a single counterexample is sufficient to show that the +statement is false. + 13. For every integer $n$, if $n$ is odd then $\dfrac{n - 1}{2}$ is odd. +(a) + +Original: + +Let $P(n) = n \text{ is odd}$ + +Let $Q(m) = m \text{ is odd}$ + +$$ \forall n \in \mathbb{Z}(P(n) \to Q\left(\frac{n - 1}{2}\right)) $$ + +Negation: + +$$ \exists n \in \mathbb{Z}(P(n) \wedge \neg Q\left(\frac{n - 1}{2}\right)) $$ + +There exists some integer $n$ such that $n$ is odd and $\dfrac{n - 1}{2}$ is not +odd. + +(b) + +Counterexample: + +Let $n = 1$. $n$ is odd as $1$ can be expressed as $n = 1 = 2(k) + 1$, where +$k = 0$. This means that $1$ is odd by the definition of an odd integer, and the +hypothesis of the original statement is true. The conclusion of the original +statement, however, is false, as +$\dfrac{n - 1}{2} = \dfrac{1 - 1}{2} = \dfrac{0}{2} = 0$, and $0$ is not odd. + +Since the original statement claims that the implication holds true for all +integers $n$, a single counterexample is sufficient to show that the statement +is false. + Disprove each of the statements in 14-16 by giving a counterexample. In each case explain how the counterexample actually disproves the statement. 14. For all integers $m$ and $n$, if $2m + n$ is odd then $m$ and $n$ are both odd. +Let $m = 2$ and let $n = 1$, the hypothesis $2m + n$ is odd is true as +$2(2) + 1 = 5$, and $5$ is odd, but the conclusion that both $m$ and $n$ are odd +is false, as $m$ is even. + 15. For every integer $p$, if $p$ is prime then $p^2 - 1$ is even. +Let $p = 2$. The hypothesis holds true as $2$ is prime, but the conclusion +"$p^2 - 1$ is even" is false for this $p$ as $(2)^2 - 1 = 4 - 1 = 3$, and $3$ is +not even. + 16. For every integer $n$, if $n$ is even then $n^2 + 1$ is prime. +Let $n = 0$, the hypothesis "$n$ is even" holds true for this $n$ as $0$ is +even. The conclusion "$n^2 + 1$ is prime" fails for this $n$ as +$0^2 + 1 = 0 + 1 = 1$, and $1$ is not prime. + In 17-20, determine whether the property is true for all integers, true for no integers, or true for some integers and false for other integers. Justify your answers. 17. $(a + b)^2 = a^2 + b^2$ +This property is true for some integers and not others. + +For example where it is true, consider $a = 0$ and $b = 1$, then +$(0 + 1)^2 = 1^2 = 1 = 0 + 1 = (0)^2 + (1)^2$ holds true for at least two +integers. + +For example where it is false, consider $a = 1$ and $b = 1$, then +$(1 + 1)^2 = 2^2 = 4 \neq 2 = 1 + 2 = (1)^2 + (1)^2$. Since this provides a +counterexample, this property cannot hold true for all integers. + +Therefore, this property holds true for some integers and not others. + 18. $\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{a + c}{b + d}$ +This is true for $a = c = 0$ and $b = d = 1$as: + +$$ \frac{0}{1} + \frac{0}{1} = 0 + 0 = 0 = \frac{0}{2} = \frac{0 + 0}{1 + 1} $$ + +This is false for $a = b = c = d = 1$, as: + +$$ \frac{1}{1} + \frac{1}{1} = 1 + 1 = 2 \neq 1 = \frac{2}{2} = \frac{1 + 1}{1 + 1} $$ + +Therefore, this property holds true for some integers and not others. + 19. $-a^n = (-a)^n$ +This is true for $a = -1$ and $n = 1$. + +$$ -(-1)^1 = (-(-1))^1 $$ + +$$ -(-1) = (-(-1)) $$ + +$$ 1 = 1 $$ + +This is false for $a = -1$ and $n = 2$. + +$$ -(-1)^2 = (-(-1))^2 $$ + +$$ -(1) = (1)^2 $$ + +$$ -1 \neq 1 $$ + +Therefore, this property holds true for some integers and not others. + 20. The average of any two odd integers is odd. +Let $m$ and $n$ be odd integers. Let $m = 2k + 1$ and $n = 2p + 1$ where $k$ and +$p$ are any integers. + +We are asserting that $\dfrac{m + n}{2}$ is odd. By substitution, we can express +this as: + +$$ \frac{m + n}{2} = \frac{(2k + 1) + (2p + 1)}{2} = \frac{2k + 2p + 2}{2} = \frac{2(k + p + 1)}{2} = k + p + 1 $$ + +In order to prove that $k + p + 1$ is odd, we need to be able to express it in +the form of $2(\text{some integer}) + 1$ by the definition of an odd integer. + +An example where this is true is if $k = 2$ and $p = 4$, then +$k + p + 1 = 2 + 4 + 1 = 7$, and $7$ is an odd integer. + +A counterexample where this is false is if $k = 3$ and $p = 4$, then +$k + p + 1 = 3 + 4 + 1 = 8$, and $8$ is not an odd integer. + +Therefore, this property holds true for some integers and not others. + Prove the statement in 21 and 22 by the method of exhaustion. 21. Every positive even integer less than 26 can be expressed as a sum of three - of fewer perfect squares. (For instance, $10 = 1^2 + 3^2$ and $16 = 4^2$.) + or fewer perfect squares. (For instance, $10 = 1^2 + 3^2$ and $16 = 4^2$.) -22. For each integer $n$ with $1 \leq n \leq 10$, $n^2 -n + 11$ is a prime +Let's first establish all positive even integers less than 26: + +$$ \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} $$ + +$$ 2 = 1^2 + 1^2 $$ + +$$ 4 = 2^2 $$ + +$$ 6 = 2^2 + 1^2 + 1^2 $$ + +$$ 8 = 2^2 + 2^2 $$ + +$$ 10 = 3^2 + 1^2 $$ + +$$ 12 = 2^2 + 2^2 + 2^2 $$ + +$$ 14 = 3^2 + 2^2 + 1^2 $$ + +$$ 16 = 4^2 $$ + +$$ 18 = 4^2 + 1^2 + 1^2 $$ + +$$ 20 = 4^2 + 2^2 $$ + +$$ 22 = 3^2 + 3^2 + 2^2 $$ + +$$ 24 = 4^2 + 2^2 + 2^2 $$ + +22. For each integer $n$ with $1 \leq n \leq 10$, $n^2 - n + 11$ is a prime number. +Let's establish all possible values for $n$: + +$$ \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} $$ + +$n = 1$: + +$$ (1)^2 - (1) + 11 = 1 - 1 + 11 = 11 \text{ is prime} $$ + +$n = 2$: + +$$ (2)^2 - (2) + 11 = 4 - 2 + 11 = 13 \text{ is prime} $$ + +$n = 3$: + +$$ (3)^2 - (3) + 11 = 9 - 3 + 11 = 17 \text{ is prime} $$ + +$n = 4$: + +$$ (4)^2 - (4) + 11 = 16 - 4 + 11 = 23 \text{ is prime} $$ + +$n = 5$: + +$$ (5)^2 - (5) + 11 = 25 - 5 + 11 = 31 \text{ is prime} $$ + +$n = 6$: + +$$ (6)^2 - (6) + 11 = 36 - 6 + 11 = 41 \text{ is prime} $$ + +$n = 7$: + +$$ (7)^2 - (7) + 11 = 49 - 7 + 11 = 53 \text{ is prime} $$ + +$n = 8$: + +$$ (8)^2 - (8) + 11 = 64 - 8 + 11 = 67 \text{ is prime} $$ + +$n = 9$: + +$$ (9)^2 - (9) + 11 = 81 - 9 + 11 = 83 \text{ is prime} $$ + +$n = 10$: + +$$ (10)^2 - (10) + 11 = 100 - 10 + 11 = 101 \text{ is prime} $$ + Each of the statements in 23-26 is true. For each, (a) rewrite the statement with the quantification implicit as If _____, then _____, and (b) write the first sentence of a proof (the "starting point") and the last sentence of a @@ -107,12 +391,45 @@ the statements in order to be able to do these exercises.) 23. For every integer $m$, if $m > 1$ then $0 < \dfrac{1}{m} < 1$. +(a) If an integer is greater than 1, then its reciprocal is between 0 and 1. + +(b) + +Starting Point: Suppose $m$ is any integer such that $m > 1$. + +To Show: $0 < \dfrac{1}{m} < 1$ + 24. For every real number $x$, if $x > 1$ then $x^2 > x$. +(a) If a real number is greater than 1, then it's square is greater than itself. + +(b) + +Starting Point: Suppose $x$ is any real number such that $x > 1$. + +To Show: $x^2 > x$. + 25. For all integers $m$ and $n$, if $mn = 1$ then $m = n = 1$ or $m = n = -1$. +(a) If the product of any two integers is equal to 1, then both integers either +equal 1 or -1. + +(b) + +Starting Point: Suppose $m$ and $n$ are any integers such that $mn = 1$. + +To Show: $m = n = 1$ or $m = n = -1$. + 26. For every real number $x$, if $0 < x < 1$ then $x^2 < x$. +(a) If a real number is between 0 and 1, then its square is less than itself. + +(b) + +Starting Point: Suppose $x$ is any real number such that $0 < x < 1$. + +To Show: $x^2 < x$. + 27. Fill in the blanks in the following proof. **Theorem:** For every odd integer $n$, $n^2$ is odd. @@ -134,6 +451,14 @@ Because we have not assumed anything about $n$ except that it is an odd integer, it follows from the principle of ___ (d) ___ that for _every_ odd integer $n$, $n^2$ is odd. +a. odd integer. + +b. $2k + 1$ + +c. $n^2$ + +d. universal generalization + In each of 28-31: a. Rewrite the theorem in three different ways: @@ -170,6 +495,26 @@ 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. @@ -193,6 +538,26 @@ Let $r = -k$. Then $r$ is an integer because $(-1)$ and $k$ are integers and __ 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. @@ -206,6 +571,27 @@ $$ m + n = \text{\_\_ (c) \_\_} = 2(r + s) + 1 $$ 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. @@ -231,3 +617,23 @@ 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 diff --git a/chapter_4/test_yourself.md b/chapter_4/test_yourself.md index 572dfd7..13be4a0 100644 --- a/chapter_4/test_yourself.md +++ b/chapter_4/test_yourself.md @@ -4,16 +4,31 @@ Page 194 1. An integer is even if, and only if, ______. +it equals twice some integer. + 2. An integer is odd if, and only if, ______. +it equals twice some integer plus 1. + 3. An integer $n$ is prime if, and only if, ______. +$n$ is greater than $1$ and if $n$ equals the product of any two positive +integers, then one of the integers equals $1$ and the other equals $n$. + 4. The most common way to disprove a universal statement is to find ______. +a counterexample. + 5. According to the method of generalizing from the generic particular, to show that every element of a set satisfies a certain property, suppose $x$ is a ______, and show that ______. +particular but arbitrarily chosen element of the set; $x$ satisfies the given +property. + 6. To use the method of direct proof to prove a statement of the form, "For every $x$ in a set $D$, if $P(x)$ then $Q(x)$," one supposes that ______ and one shows that ______. + +$x$ is a particular but arbitrarily chosen element of the set $D$ that makes the +hypothesis $P(x)$ true; $x$ makes the conclusion $Q(x)$ true.