diff --git a/chapter_4/exercises.md b/chapter_4/exercises.md index 2401ff3..1cb70a5 100644 --- a/chapter_4/exercises.md +++ b/chapter_4/exercises.md @@ -6493,7 +6493,7 @@ It then follows that $$ \lceil k^2 + k + \frac{1}{4} \rceil = k^2 + k + 1 $$ -Then by substution: +Then by substitution: $$ \frac{n^2 + 3}{4} = \frac{(2k + 1)^2 + 3}{4} $$ @@ -6613,23 +6613,164 @@ $1 < 2$ by \(c\) and dividing (d). Hence $\dfrac{x}{2}$ is a positive real number that is less than the least positive real number. This is a (e). _[Thus the supposition is false, and so there is no least positive real number.]_ +a. a contradiction + +b. a positive real number. + +c. $x$ + +d. both sides by $2$. + +e. contradiction + 2. Is $\dfrac{1}{0}$ an irrational number? Explain. +No. Since $\dfrac{1}{0}$ is undefined, it is not a number at all. + 3. Use proof by contradiction to show that for every integer $n$, $3n + 2$ is not divisible by $3$. +**Proof by contradiction:** + +Suppose not. That is, suppose that there is an integer $n$, such that $3n + 2$ +is divisible by 3. + +By definition of divisibility: + +$$ 3n + 2 = 3k $$ + +for some integer $k$. + +$$ 3n + 2 = 3k $$ + +$$ 2 = 3k - 3n $$ + +$$ 3k - 3n = 2 $$ + +$$ 3(k - n) = 2 $$ + +$$ k - n = \frac{2}{3} $$ + +Now, $k - n$ is an integer by the difference of integers, but $\dfrac{2}{3}$ is +not an integer. So $k - n$ is an integer and is not an integer. + +This is a contradiction. + +Q.E.D. + 4. Use proof by contradiction to show that for every integer $m$, $7m + 4$ is not divisible by $7$. +**Proof by contradiction:** + +Suppose not. That is, suppose that there is an integer $m$ such that $7m + 4$ is +divisible by $7$. + +By the definition of divisibility: + +$$ 7m + 4 = 7k $$ + +for some integer $k$. + +By the laws of algebra, this can be rewritten as: + +$$ 7m + 4 = 7k $$ + +$$ 7m - 7k = -4 $$ + +$$ 7k - 7m = 4 $$ + +$$ 7(k - m) = 4 $$ + +$$ k - m = \frac{4}{7} $$ + +Now, $k - m$ is an integer by the difference of integers, but $\dfrac{4}{7}$ is +not an integer. Therefore $k - m$ is an integer and not an integer. + +This is a contradiction. + +Q.E.D. + Carefully formulate the negations of each of the statements in 5-7. Then prove each statement by contradiction. 5. There is no greatest even integer. +Negation: There is some greatest even integer. + +**Proof by contradiction:** + +Suppose not. That is, suppose there is some greatest even integer $x$. + +Since $x$ is an even integer, $x = 2k$ for some integer $k$. Then suppose there +is number $y$, such that $y = x + 2$. + +By substitution: + +$$ y = 2k + 2 $$ + +$$ y = 2(k + 1) $$ + +Now, $k + 1$ is an integer by the sum of integers. Since $y$ is expressed in the +form of $2 \cdot (\text{an integer})$, $y$ is an integer by the product of +integers and an even integer by the definition of even. Since $y = x + 2$, $x$ +is not the greatest even integer, since $y > x$ and $y$ is even. So, $x$ is not +the greatest even integer and $x$ is the greatest even integer. + +This is a contradiction. + +Q.E.D. + 6. There is no greatest negative real number. +Negation: There is some greatest negative real number. + +**Proof by contradiction:** + +Suppose not. That is, suppose that there is some greatest negative real number +$y$. In other words, there exists some negative real number $y$ such that for +all negative real numbers $x$, $y \geq x$. + +Let $Y = \dfrac{y}{2}$. + +$\dfrac{y}{2} > y$ since $y$ is negative, $Y$ is a negative number that is +greater than $y$. So, $y$ is not the greatest negative real number and $y$ is +the greatest negative real number. + +This is a contradiction. + +Q.E.D. + 7. There is no least positive rational number. +Negation: There is some least positive rational number. + +**Proof by contradiction:** + +Suppose not. That 8s, suppose that there is some least positive rational number, +$y$. + +Since $y$ is a rational number, $y = \dfrac{a}{b}$ where both $a$ and $b$ are +integers and $b \neq 0$. + +Let $Y = \dfrac{y}{2}$. + +By substitution: + +$$ Y = \frac{\dfrac{a}{b}}{2} $$ + +$$ Y = \frac{a}{2b} $$ + +Now, $2b$ is an integer by the product of integers and $2b \neq 0$ by the zero +product property. Thus, $Y$ is a rational number since both $a$ and $2b$ are +integers and $Y$ has a nonzero denominator. $Y < y$ as +$\dfrac{a}{2b} < \dfrac{a}{b}$. Therefore $y$ is not the least positive rational +number and $y$ is the least positive rational number. + +This is a contradiction. + +Q.E.D. + 8. Fill in the blanks for the following proof that the difference of any rational number and any irrational number is irrational. @@ -6650,6 +6791,24 @@ $$ = \frac{ad}{bd} - \frac{bc}{bd} $$ $$ = \frac{ad - bc}{bd} \quad \text{ by algebra} $$ +a. some rational number + +b. some irrational number + +c. $\dfrac{a}{b}$ + +d. $\dfrac{c}{d}$ + +e. $\dfrac{a}{b} - \dfrac{c}{d}$ + +f. integers + +g. integers + +h. zero product property + +i. rational + Now both $ad - bc$ and $bd$ are integers and products and differences of (f) are (g). And $bd \neq 0$ by the (h). Hence $y$ is a ratio of integers with a nonzero denominator, and thus $y$ is (i) by definition of rational. We therefore have @@ -6664,9 +6823,43 @@ the difference of any irrational number and any rational number is rational." What is wrong with beginning the proof in this way? (_Hint:_ If needed, review the answer to exercise 11 in Section 3.2.) +The problem is that the negation of a universal is an existential, and the +student did not apply that. Instead the statement should be: + +"Suppose not. That is, suppose the difference of _some_ irrational number and +_some_ rational number is rational." + b. Prove that the difference of any irrational number and any rational number is irrational. +**Proof by contradiction:** + +Suppose not. That is, suppose there is some irrational number $x$ and some +rational number $y$ such that $x - y$ is rational. + +Since $y$ is rational and $x - y$ is rational, $y = \dfrac{a}{b}$ and +$x - y = \dfrac{c}{d}$ where $a$ $b$, $c$, and $d$ are integers and $b \neq 0$ +and $d \neq 0$. + +Then by substitution: + +$$ x - \dfrac{a}{b} = \dfrac{c}{d} $$ + +$$ x = \dfrac{c}{d} - \dfrac{a}{b} $$ + +$$ x = \dfrac{cb}{bd} - \dfrac{ad}{bd} $$ + +$$ x = \dfrac{cb - ad}{bd} $$ + +Now, $cb - ad$ is an integer by the difference and product of integers. $bd$ is +an integer by the product of integers and $bd \neq 0$ by the zero product +property. By the definition of rational numbers then, $x$ is a rational number. +Therefore, $x$ is a rational number and an irrational number. + +This is a contradiction. + +Q.E.D. + 10. Let $S$ be the statement: For all positive real numbers $r$ and $s$, $\sqrt{r + s} \neq \sqrt{r} + \sqrt{s}$. Statement $S$ is true, but the following "proof" is incorrect. Find the mistake. @@ -6688,6 +6881,21 @@ contradicts the supposition that $\sqrt{r + s} = \sqrt{r} + \sqrt{s}$. This contradiction shows that the supposition is false, and hence statement $S$ is true." +The mistake the student makes is in that they do not take the proper negation of +the universal statement. Instead of supposing the negation of the statement for +all $r$ and $s$, the student should have started with an existential quantifier +of the negating statement. It should have started with: + +"Suppose not. That is, suppose that there exists some positive real numbers $r$ +and $s$ such that $\sqrt{r + s} = \sqrt{r} + \sqrt{s}$." + +The student also then goes to provide specific number examples for $r$ and $s$, +which is a technique generally reserved for proof by counterexample, not +contradiction. By providing a specific example for their (incorrect) universal +statement, they are not even proving for all, nor are they proving there exists +_some_. They are just proving that the negation is a contradiction for a +singular example. + 11. Let $T$ be the statement: The sum of any two rational numbers is rational. Then @@ -6704,43 +6912,471 @@ Hence $4$ is both rational and not rational, which is a contradiction. This contradiction shows that the supposition is false, and hence statement $T$ is true. +The mistake the student makes is in that they do not take the proper negation of +the universal statement. Instead of supposing that the sum of _any_ two rational +numbers is not rational, the student should have started with an existential +quantifier of the negating statement. It should have started with: + +"Suppose not. That is, suppose that there are some rational numbers $x$ and $y$ +such that $x + y$ is not rational." + 12. Let $R$ be the statement: The square root of any irrational number is irrational. a. Write the negation for $R$. +Negation: The square root of some irrational number is rational. + b. Prove $R$ by contradiction. +**Proof by contradiction:** + +Suppose not. That is, suppose there is some irrational number $x$ such that +$\sqrt{x}$ is rational. + +Since $\sqrt{x}$ is rational, $\sqrt{x} = \dfrac{a}{b}$ where $a$ and $b$ are +integers and $b \neq 0$. + +Then by substitution: + +$$ \sqrt{x} = \frac{a}{b} $$ + +$$ (\sqrt{x})^2 = \left(\frac{a}{b}\right)^2 $$ + +$$ x = \frac{a^2}{b^2} $$ + +Now, $a^2$ is an integer by the product of integers. Additionally, $b^2$ is an +integer by the product of integers and $b^2 \neq 0$ by the zero product +property. Thus $x$ is a rational number. Therefore $x$ is a rational number and +$x$ is an irrational number. + +This is a contradiction. + +Q.E.D. + 13. Let $S$ be the statement: The product of any irrational number and any nonzero rational number is irrational. a. Write the negation for $S$. +The product of some irrational number and some nonzero rational number is +rational. + b. Prove $S$ by contradiction. +**Proof by contradiction:** + +Suppose not. That is, suppose that there exists some irrational number $x$ and +some nonzero rational number $y$ such that $xy$ is rational. + +Since $y$ is a nonzero rational number, $y = \dfrac{a}{b}$ where $a$ and $b$ are +integers and $a \neq 0$ and $b \neq 0$. Since $xy$ is rational, +$xy = \dfrac{c}{d}$ where $c$ and $d$ are integers and $d \neq 0$. + +Then, by susbstitution: + +$$ xy = x\left(\frac{a}{b}\right) = \frac{c}{d} $$ + +$$ x\left(\frac{a}{b}\right) = \frac{c}{d} $$ + +$$ x = \frac{c}{d}\left(\frac{b}{a}\right) $$ + +$$ x = \frac{cb}{ad} $$ + +Now, $cd$ is an integer by the product of integers. Additionally, $ad$ is an +integer by the product of integers and $ad \neq 0$ by the zero product property. +Thus, $x$ is a rational number. Therefore $x$ is a rational number and $x$ is an +irrational number. + +This is a contradiction. + +Q.E.D. + 14. Let $T$ be the statement: For every integer $a$, if $a \mod 6 = 3$ , then $a \mod 3 \neq 2$. a. Write a negation for $T$. +Negation: For some integer $a$, $a \mod 6 = 3$ and $a \mod 3 = 2$. + b. Prove $T$ by contradiction. +**Proof by contradiction:** + +Suppose not. That is, suppose for some integer $a$, $a \mod 6 = 3$ and +$a \mod 3 = 2$. + +Since $a \mod 6 = 3$, then by the quotient remainder theorem: + +$$ a = 6q + 3 $$ + +for some integer $q$. + +$$ a = 6q + 3 $$ + +Then, since $a \mod 3 = 2$, then by the quotient remainder theorem: + +$$ a = 3s + 2 $$ + +for some integer $s$. + +$$ 6q + 3 = 3s + 2 $$ + +$$ 6q - 3s = 2 - 3 $$ + +$$ 3(2q - s) = -1 $$ + +$$ (2q - s) = -\frac{1}{3} $$ + +Now, $2q - s$ is an integer by the product and difference of integers, but +$-\dfrac{1}{3}$ is not an integer. So $2q - s$ is an integer and $2q - s$ is not +an integer. + +This is a contradiction. + +Q.E.D. + 15. Do there exists integers $a$, $b$, and $c$ such that $a$, $b$, and $c$ are all odd and $a^2 + b^2 = c^2$? Prove your answer. +**Proof by contradiction:** + +Suppose not. That is, suppose that there exist odd integers $a$, $b$, and $c$ +such that $a^2 + b^2 = c^2$. + +Since $a$ and $b$ are odd, $a = 2m + 1$, $b = 2n + 1$ for some integers $m$ and +$n$. + +By substitution: + +$$ a^2 = (2m + 1)^2 = 4m^2 + 4m + 1 = 4m(m + 1) + 1 $$ + +$$ b^2 = (2n + 1)^2 = 4n^2 + 4n + 1 = 4n(n + 1) + 1 $$ + +Therefore: + +$$ a^2 \equiv 1 (\mod 4), b^2 \equiv 1 (\mod 4) $$ + +Adding these congruences leaves: + +$$ a^2 + b^2 \equiv 1 + 1 \equiv 2 (\mod 4) $$ + +So the left hand satisfies: + +$$ a^2 + b^2 \equiv 2 (\mod 4) $$ + +Now, consider $c$. Since $c$ is odd, $c = 2k + 1$ for some integer $k$. + +Then by substitution: + +$$ c^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1 $$ + +So: + +$$ c^2 \equiv 1 (\mod 4) $$ + +But we were given $a^2 + b^2 = c^2$, so this implies: + +$$ a^2 + b^2 = c^2 (\mod 4) $$ + +This is, + +$$ 2 \equiv 1 (\mod 4) $$ + +This is impossible. + +This is a contradiction. + +Q.E.D. + Prove each statement in 16-19 by contradiction. 16. For all odd integers $a$ and $b$, $b^2 - a^2 \neq 4$. (_Hint:_ $b^2 - a^2 = (b + a)(b - a)$ and the only way to factor $4$ is either $4 = 2 \cdot 2$ or $4 = 4 \cdot 1$.) +**Proof by contradiction:** + +Suppose not. That is, suppose that there exists some odd integers $a$ and $b$ +such that $b^2 - a^2 = 4$. + +Consider: + +$$ b^2 - a^2 = (b + a)(b - a) = 4 $$ + +By the sum and difference of integers, both $b + a$ and $b - a$ are integers. +The only integers where $4$ is factored by are $2 \cdot 2$ and $1 \cdot 4$. + +Thus there are only some cases to consider: + +_Case $(b + a) = 2$ and $(b - a) = 2$:_ + +$$ b + a = b - a $$ + +$$ b = b - 2a $$ + +$$ b - b = -2a $$ + +$$ 0 = -2a $$ + +$$ a = 0 $$ + +Since $a = 0$, $a$ is an even integer. Therefore $a$ is an even integer and $a$ +is an odd integer. This is a contradiction. + +_Case $(b + a) = 4$ and $(b - a) = 1$:_ + +$$ b + a = 4 $$ + +$$ b = 4 - a $$ + +$$ b - a = 1 $$ + +By substitution: + +$$ b - (4 - a) = 1 $$ + +$$ b - 4 + a = 1 $$ + +$$ b + a = 5 $$ + +So, $b + a = 5$ and $b + a = 4$. This is a contradiction. + +_Case $(b + a) = 1$ and $(b - a) = 4$:_ + +$$ b + a = 1 $$ + +$$ b = 1 - a $$ + +By substitution: + +$$ b - a = 4 $$ + +$$ b - (1 - a) = 4 $$ + +$$ b - 1 + a = 4 $$ + +$$ b + a = 5 $$ + +So $b + a = 1$ and $b + a = 5$. This is a contradiction. + +In all cases, this is a contradiction. + +Q.E.D. + 17. For all prime numbers $a$, $b$, and $c$, $a^2 + b^2 \neq c^2$. +**Proof by contradiction:** + +Suppose not. That is, suppose that for some prime numbers $a$, $b$, and $c$, +$a^2 + b^2 = c^2$. + +Consider that: + +$$ a^2 + b^2 = c^2 $$ + +$$ a^2 = c^2 - b^2 $$ + +$$ a^2 = (c - b)(c + b) $$ + +Since $a$, $b$, and $c$, are prime. By the unique prime factorization of +integers theorem, the only possible values for $c - b$ and $c + b$ are: + +$$ a^2 = (a^2)(1) = (c - b)(c + b) $$ + +$$ a^2 = (1)(a^2) = (c - b)(c + b) $$ + +$$ a^2 = (a)(a) = (c - b)(c + b) $$ + +Let's check these cases: + +__Case $c - b = a^2$ and $c + b = 1$:_ + +Add: + +$$ (c - b) + (c + b) = a^2 + 1 $$ + +$$ 2c = a^2 + 1 $$ + +$$ c = \frac{a^2 + 1}2 $$ + +Subtract: + +$$ (c - b) - (c + b) = a^2 - 1 $$ + +$$ -2b = a^2 - 1 $$ + +$$ b = -\frac{a^2 - 1}{2} $$ + +$$ b = \frac{1 - a^2}{2} $$ + +Now, $a$, $b$, and $c$ are prime number, but checking for small prime(s) shows: + +$$ a = 2 $$ + +$$ c = \frac{(2)^2 + 1}2 $$ + +$$ c = \frac{5}2 $$ + +$$ b = \frac{1 - a^2}{2} $$ + +$$ b = \frac{1 - (2)^2}{2} $$ + +$$ b = -\frac{3}{2} $$ + +So $b$ and $c$ are not prime numbers and $b$ and $c$ are not prime numbers. This +is a contradiction. + +__Case $c - b = a$ and $c + b = a$:_ + +Add: + +$$ (c - b) + (c + b) = a + a $$ + +$$ 2c = 2a $$ + +$$ c = a $$ + +Subtract: + +$$ (c - b) - (c - b) = a - a $$ + +$$ -2b = 0 $$ + +$$ b = 0 $$ + +So, $b$ is a prime number and $b$ is not a prime number. This is a +contradiction. + +__Case $c - b = 1$ and $c + b = a^2$:_ + +Add: + +$$ (c - b) + (c + b) = 1 + a^2 $$ + +$$ 2c = 1 + a^2 $$ + +$$ c = \frac{1 + a^2}{2} $$ + +Subtract: + +$$ (c + b) - (c - b) = 1 - a^2 $$ + +$$ c - b - c - b = 1 - a^2 $$ + +$$ -2b = 1 + a^2 $$ + +$$ b = -\frac{1 - a^2}{2} $$ + +$$ b = \frac{a^2 + 1}{2} $$ + +Now, $a$, $b$, and $c$ are prime number, but checking for small prime(s) shows: + +$$ a = 2 $$ + +$$ b = \frac{(2)^2 + 1}{2} $$ + +$$ b = \frac{5}{2} $$ + +$$ c = \frac{1 + (2)^2}{2} $$ + +$$ c = \frac{5}{2} $$ + +So $b$ and $c$ are not prime numbers and $b$ and $c$ are prime numbers. This is +a contradiction. + +In all cases, this is a contradiction. + +Q.E.D. + 18. If $a$ and $b$ are rational numbers, $b \neq 0$, and $r$ is an irrational number, then $a + br$ is irrational. +**Proof by contradiction:** + +Suppose not. That is, suppose that $a$ and $b$ are rational numbers, $b \neq 0$, +and $r$ is an irrational number and $a + br$ is rational. + +Since $a$, $b$ and $a + br$ are rational numbers, $a = \dfrac{u}{t}$, +$b = \dfrac{v}{w}$, $a + br = \dfrac{x}{y}$ where $u$, $t$, $v$, $w$, $x$ and +$y$ are integers and $t \neq 0$, $v \neq 0$, $w \neq 0$, and $y \neq 0$. + +Then by substitution: + +$$ a + br = \frac{x}{y} = \left(\frac{u}{t}\right) + \left(\frac{v}{w}\right)r $$ + +$$ \frac{x}{y} = \left(\frac{u}{t}\right) + \left(\frac{v}{w}\right)r $$ + +$$ \frac{x}{y} - \left(\frac{u}{t}\right) = \left(\frac{v}{w}\right)r $$ + +$$ \frac{xt}{ty} - \left(\frac{uy}{ty}\right) = \left(\frac{v}{w}\right)r $$ + +$$ \frac{xt - uv}{ty} = \left(\frac{v}{w}\right)r $$ + +$$ \left(\frac{xt - uv}{ty}\right)\left(\frac{w}{v}\right) = r $$ + +$$ \frac{w(xt - uv)}{tyv} = r $$ + +Now $w(xt - uv)$ is an integer by the product and difference of integers. +Additionally, $tyv$ is an integer by the product of integers and $tyv \neq 0$ by +the zero product property. Thus, by the definition of rational numbers, $r$ is +rational. Therefore $r$ is rational and $r$ is irrational. + +This is a contradiction. + +Q.E.D. + 19. For any integer $n$, $n^2 - 2$ is not divisible by $4$. +**Proof by contradiction:** + +Suppose not. That is, suppose there exists an integer $n$, such that $n^2 - 2$ +is divisible by $4$. + +Since $n^2 - 2$ is divisible by $4$, then $(n^2 - 2) \equiv 0 (\mod 4)$. + +_Case $n$ is odd:_ + +Since $n$ is odd, $n = 2k + 1$ for some integer $k$. + +Then: + +$$ n^2 - 2 = (2k + 1)^2 - 2 $$ + +$$ = 4k^2 + 4k + 1 - 2 $$ + +$$ = 4k^2 + 4k - 1 $$ + +$$ = 4(k^2 + k) - 1 $$ + +$$ 4(k^2 + k) - 1 \equiv -1 (\mod 4) $$ + +So $(n^2 - 2) \equiv -1 (\mod 4)$ and $(n^2 - 2) \equiv 0 (\mod 4)$. This is a +contradiction. + +_Case $n$ is even:_ + +Since $n$ is even, $n = 2k$ for some integer $k$. + +Then: + +$$ n^2 - 2 = (2k)^2 - 2 $$ + +$$ = 4k^2 - 2 $$ + +$$ = 4(k^2) - 2 $$ + +$$ 4(k^2) - 2 \equiv -2 (\mod 4) $$ + +So $(n^2 - 2) \equiv -2 (\mod 4)$ and $(n^2 - 2) \equiv 0 (\mod 4)$. This is a +contradiction. + +In all cases, this is a contradiction. + +Q.E.D. + +By definition of divisibility, there exists an integer $k$ so that $b = ak$. + 20. Fill in the blanks in the following proof by contraposition that for every integer $n$, if $5 \cancel{\mid} n^2$ then $5 \cancel{\mid} n$. @@ -6751,48 +7387,342 @@ integer $k$. By substitution, $n^2 = $ (d) $= 5(5k^2)$. But $5k^2$ is an integer because it is a product of integers. Hence $n^2 = 5 \cdot (\text{an integer})$, and so (e) _[as was to be shown]._ +a. $5 \mid n$ + +b. $5 \mid n^2$ + +c. $5k$ + +d. $(5k)^2$ + +e. $5 \mid n^2$ + 21. Consider the statement "For every integer $n$, if $n^2$ is odd then $n$ is odd." a. Write what you would suppose and what you would need to show to prove this statement by contradiction. +**Supposition:** Suppose that for some integer $n$, $n^2$ is odd and $n$ is +even. + +**Consequent:** Show that $n^2$ is even, a contradiction. + b. Write what you would suppose and what you would need to show to prove this statement by contraposition. +**Supposition:** Suppose that $n$ is any even integer. + +**Consequent:** Show that $n^2$ is even. + 22. Consider the statement "For every real number $r$, if $r^2$ is irrational then $r$ is irrational." a. Write what you would suppose and what you would need to show to prove this statement by contradiction. +**Supposition:** Suppose there is some real number $r$ such that $r^2$ is +irrational and $r$ is rational. + +**Consequent:** Show that $r^2$ is rational, a contradiction. + b. Write what you would suppose and what you would need to show to prove this statement by contraposition. +**Supposition:** Suppose $r$ is any real number such that $r$ is rational. + +**Consequent:** Show that $r^2$ is rational. + Prove each of the statements in 23-25 in two ways: (a) by contraposition and (b) by contradiction. 23. The negative of any irrational number is irrational. +a. + +**Proof by contraposition:** + +Suppose $x$ is any number such that $-x$ is rational. + +Since $-x$ is rational, $-x = \dfrac{a}{b}$ where $a$ and $b$ are some integers +and $b \neq 0$. + +Then by substitution: + +$$ -x = \dfrac{a}{b} $$ + +$$ x = -\dfrac{a}{b} $$ + +$$ x = \dfrac{-a}{b} $$ + +Now, $-a$ is an integer by the product of integers. + +Therefore $x$ is a rational number. + +Q.E.D. + +b. + +**Proof by contradiction:** + +Suppose not, That is, suppose $x$ is some irrational number and $-x$ is +rational. + +Since $-x$ is rational, $-x = \dfrac{a}{b}$ where $a$ and $b$ are some integers +and $b \neq 0$. + +Then, by algebra: + +$$ -x = \dfrac{a}{b} $$ + +$$ x = -\dfrac{a}{b} $$ + +$$ x = \dfrac{-a}{b} $$ + +Now, $-a$ is an integer by the product of integers. Therefore $x$ is a rational +number and $x$ is an irrational number. + +This is a contradiction. + +Q.E.D. + 24. The reciprocal of any irrational number is irrational. (The **reciprocal** of a nonzero real number $x$ is $\dfrac{1}{x}$.) +a. + +**Proof by contraposition:** + +Suppose $x$ is any number such that $\dfrac{1}{x}$ is rational. + +Since $\dfrac{1}{x}$ is a rational number, $\dfrac{1}{x} = \dfrac{a}{b}$ where +$a$ and $b$ are some integers and $b \neq 0$. + +Then by substitution: + +$$ \frac{1}{x} = \frac{a}{b} $$ + +$$ x = \frac{b}{a} $$ + +It has been established that $b$ is an integer. It has also been established +that $a$ is an integer, and since $\dfrac{1}{x}$ is rational, it follows that +$a \neq 0$. + +Therefore $x$ is a rational number. + +Q.E.D. + +b. + +**Proof by contradiction:** + +Suppose not. That is suppose there is an irrational number $x$ such that +$\dfrac{1}{x}$ is rational. + +Since $\dfrac{1}{x}$ is a rational number, $\dfrac{1}{x} = \dfrac{a}{b}$ where +$a$ and $b$ are some integers and $b \neq 0$. + +Then by substitution: + +$$ \frac{1}{x} = \frac{a}{b} $$ + +$$ x = \frac{b}{a} $$ + +It has been established that $b$ is an integer. It has also been established +that $a$ is an integer, and since $\dfrac{1}{x}$ is rational, it follows that +$a \neq 0$. + +Therefore $x$ is a rational number and $x$ is an irrational number. + +This is a contradiction. + +Q.E.D. + 25. For every integer $n$, if $n^2$ is odd then $n$ is odd. +a. + +**Proof by contraposition:** + +Suppose $n$ is any integer such that $n$ is even. + +Since $n$ is even, $n = 2k$ for some integer $k$. + +Then by substitution: + +$$ n^2 = (2k)^2 $$ + +$$ = 4k^2 $$ + +$$ = 2(2k^2) $$ + +Now, $2k^2$ is an integer by the product of integers. + +Therefore $n^2$ is even by the definition of even. + +b. + +**Proof by contradiction:** + +Suppose not. That is, suppose that $n$ is some integer such that $n^2$ is odd +and $n$ is even. + +Since $n$ is even, $n = 2k$ for some integer $k$. + +Then by substitution: + +$$ n^2 = (2k)^2 $$ + +$$ = 4k^2 $$ + +$$ = 2(2k^2) $$ + +Now, $2k^2$ is an integer by the product of integers. Therefore $n^2$ is even +and $n^2$ is odd. + +This is a contradiction. + +Q.E.D. + Use any method to prove the statements in 26-29. 26. For all integers $a$, $b$, and $c$, if $a \cancel{\mid} bc$ then $a \cancel{\mid} b$. +**Proof by contraposition:** + +Suppose $a$, $b$, and $c$ are any integers such that $a \mid b$. + +Since $a \mid b$, by the definition of divisiblity, $b = ak$ for some integer +$k$. + +Then by substitution: + +$$ bc = (ak)c $$ + +$$ = a(kc) $$ + +Now, $kc$ is an integer by the product of integers. Since $bc = a(kc)$, it +follows that $a \mid a(kc)$. + +Therefore $a \mid bc$ by definition of divisibility. + 27. For all positive real numbers $r$ and $s$, $\sqrt{r + s} \neq \sqrt{r} + \sqrt{s}$. -28. For all integers $a$, $c$, and $c$, if $a \mid b$ and $a \cancel{\mid} c$, +**Proof by contradiction:** + +Suppose not. That is, suppose that for some positive real numbers $r$ and $s$, +$\sqrt{r + s} = \sqrt{r} + \sqrt{s}$. + +$$ \sqrt{r + s} = \sqrt{r} + \sqrt{s} $$ + +$$ (\sqrt{r + s})^2 = (\sqrt{r} + \sqrt{s})^2 $$ + +$$ r + s = (\sqrt{r} + \sqrt{s})(\sqrt{r} + \sqrt{s}) $$ + +$$ r + s = (\sqrt{r})^2 + 2(\sqrt{r})(\sqrt{s}) + (\sqrt{s})^2 $$ + +$$ r + s = r + 2\sqrt{r}\sqrt{s} + s $$ + +$$ 0 = 2\sqrt{r}\sqrt{s} $$ + +$$ 0 = \sqrt{r}\sqrt{s} $$ + +By the zero product property, either $\sqrt{r}$ or $\sqrt{s}$ must be $0$, which +means that either $r = 0$ or $s = 0$. + +_Case where $r = 0$:_ + +If $r = 0$, then $r$ is both $0$ and a positive real number. This is a +contradiction. + +_Case where $s = 0$:_ + +If $s = 0$, then $s$ is both $0$ and a positive real number. This is a +contradiction. + +In both cases, this is a contradiction. + +Q.E.D. + +28. For all integers $a$, $b$, and $c$, if $a \mid b$ and $a \cancel{\mid} c$, then $a \cancel{\mid} (b + c)$. +**Proof by contradiction:** + +Suppose not. That is, suppose that $a$, $b$, and $c$ are some integers such that +$a \mid b$, $a \cancel{\mid} c$ and $a \mid (b + c)$. + +Since $a \mid (b + c)$ and $a \mid b$, then $b + c = ak$ and $b = am$ for some +integers $k$ and $m$. + +Then by substitution: + +$$ (am) + c = ak $$ + +$$ am + c = ak $$ + +$$ c = ak - am $$ + +$$ c = a(k - m) $$ + +Now, $k - m$ is an integer by the difference of integers. Thus by the definition +of divisibility, $a \mid c$. Therefore $a \mid c$ and $a \cancel{\mid} c$. + +This is a contradiction. + +Q.E.D. + 29. For all integers $m$ and $n$, if $m + n$ is even then $m$ and $n$ are both even or $m$ and $n$ are both odd. +**Proof by contradiction:** + +Suppose not. That is, suppose that $m$ and $n$ are some integers such that +$m + n$ is even and either $m$ is even and $n$ is odd or $m$ is odd and $n$ is +even. + +_Case where $m$ is even and $n$ is odd:_ + +Since $m$ is even and $n$ is odd, then $m = 2k$ and $n = 2p + 1$ for some +integers $k$ and $p$. + +Then by substitution: + +$$ m + n = (2k) + (2p + 1) $$ + +$$ = 2k + 2p + 1 $$ + +$$ = 2(k + p) + 1 $$ + +Now, $k + p$ is an integer by the sum of integers. Thus $m + n$ is odd by the +definition of odd integers. Therefore $m + n$ is odd and $m + n$ is even. + +This is a contradiction. + +_Case where $m$ is odd and $n$ is even:_ + +Since $m$ is odd and $n$ is even, then $m = 2k + 1$ and $n = 2p$ for some +integers $k$ and $p$. + +Then by substitution: + +$$ m + n = (2k + 1) + (2p) $$ + +$$ = 2k + 2p + 1 $$ + +$$ = 2(k + p) + 1 $$ + +Now, $k + p$ is an integer by the sum of integers. Thus $m + n$ is odd by the +definition of odd integers. Therefore $m + n$ is odd and $m + n$ is even. + +This is a contradiction. + +In both cases, this is a contradiction. + +Q.E.D. + 30. a. Let $n = 53$. Find an approximate value for $\sqrt{n}$ and write a list of @@ -6800,6 +7730,15 @@ all the prime numbers less than or equal to $\sqrt{n}$. Is the following statement true or false? When $n = 53$, $n$ is not divisible by any prime number less than or equal to $\sqrt{n}$. +$$ \sqrt{53} \approx 7.280109889 $$ + +All prime numbers $\leq \sqrt{n}$: + +$$ \{2, 3, 5, 7\} $$ + +Yes this is true, as when 53 is divided by any of these prime numbers in this +set, the result does not equal an integer. + b. Suppose $n$ is a fixed integer. Let $S$ be the statement, "$n$ is not divisible by any prime number less than or equal to $\sqrt{n}$." The following statement is equivalent to $S$: @@ -6812,30 +7751,64 @@ Which of the following are negations for $S$? (i) $\exists$ a prime number $p$ such that $p \leq \sqrt{n}$ and $n$ is divisible by $p$. +Yes, this is a negation for $S$. + (ii) $n$ is divisible by every prime number less than or equal to $\sqrt{n}$. +No, this is not a negation for $S$. + (iii) $\exists$ a prime number $p$ such that $p$ is a multiple of $n$ and $p$ is less than or equal to $\sqrt{n}$. +No, this is not a negation for $S$. + (iv) $n$ is divisible by some prime number that is less than or equal to $\sqrt{n}$. +Yes, this is a negation for $S$. + (v) $\forall$ prime number $p$, if $p$ is less than or equal to $\sqrt{n}$, then $n$ is divisible by $p$. +No, this is not a negation for $S$. + 31. a. Prove by contraposition: For all positive integers $n$, $r$, and $s$, if -$rs \leq n$, then $r \leq \sqrt{n}$ or $s \leq \sqrt{n}$. +$rs \leq n$, then $r \leq \sqrt{n}$ or $s \leq \sqrt{n}$. (_Hint:_ Use Theorem +T27 in Appendix A.) + +**Proof by contraposition:** + +Suppose $n$, $r$, and $s$ are some positive integers such that $r > \sqrt{n}$ +and $s > \sqrt{n}$. + +Theorem T27 States: + +If $0 < a < c$ and $0 < b < d$, then $0 < ab < cd$. + +With $a = \sqrt{n}$, $b = \sqrt{n}$, $c = r$, and $d = s$, it follows that: + +If $0 < \sqrt{n} < r$ and $0 < \sqrt{n} < s$, then $0 < \sqrt{n}\sqrt{n} < rs$. + +Therefore $rs > n$. + +Q.E.D. b. Prove: For each integer $n > 1$, if $n$ is not prime then there exists a prime number $p$ such that $p \leq \sqrt{n}$ and $n$ is divisible by $p$. (_Hint:_ Use the results of part (a), Theorems 4.4.1, 4.4.3, and 4.4.4, and the transitive property of order.) +Omitted. + c. State the contrapositive of the result of part (b). The results of exercise 31 provide a way to test whether an integer is prime. +Omitted. + +RESUME HERE. + **Test for Primality** Given an integer $n > 1$, to test whether $n$ is prime check to see if it is diff --git a/chapter_4/test_yourself.md b/chapter_4/test_yourself.md index 42d8d80..8bfd070 100644 --- a/chapter_4/test_yourself.md +++ b/chapter_4/test_yourself.md @@ -210,10 +210,16 @@ Page 248 1. To prove a statement by contradiction, you suppose that ______ and you show that ______. +the statement is false; this supposition leads to a contradiction + 2. A proof by contraposition of a statement of the form "$\forall x \in D, \text{ if } P(x) \text{ then } Q(x)$" is a direct proof of ______. +$\forall x \in D, \text{ if } \neg Q(x) \text{ then } \neg P(x)$ + 3. To prove a statement of the form "$\forall x \in D, \text{ if } P(x) \text{ then } Q(x)$" by contraposition, you suppose that ______ and you show that ______. + +$Q(x)$ is false; $P(x)$ is false.