diff --git a/chapter_4/exercises.md b/chapter_4/exercises.md index 1ec5f67..2401ff3 100644 --- a/chapter_4/exercises.md +++ b/chapter_4/exercises.md @@ -6649,3 +6649,240 @@ $$ y = \text{(e)} \quad \text{ by substitution} $$ $$ = \frac{ad}{bd} - \frac{bc}{bd} $$ $$ = \frac{ad - bc}{bd} \quad \text{ by algebra} $$ + +Now both $ad - bc$ and $bd$ are integers and products and differences of (f) are +(g). And $bd \neq 0$ by the (h). Hence $y$ is a ratio of integers with a nonzero +denominator, and thus $y$ is (i) by definition of rational. We therefore have +both that $y$ is irrational and that $y$ is rational, which is a contradiction. +_[Thus the supposition is false and the statement to be proved is true.]_ + +9. + +a. When asked to prove that the difference of any irrational number and any +rational number is irrational, a student began, "Suppose not. That is, suppose +the difference of any irrational number and any rational number is rational." +What is wrong with beginning the proof in this way? (_Hint:_ If needed, review +the answer to exercise 11 in Section 3.2.) + +b. Prove that the difference of any irrational number and any rational number is +irrational. + +10. Let $S$ be the statement: For all positive real numbers $r$ and $s$, + $\sqrt{r + s} \neq \sqrt{r} + \sqrt{s}$. Statement $S$ is true, but the + following "proof" is incorrect. Find the mistake. + +**"Proof by contradiction:** Suppose not, that is, suppose that for all positive +real numbers $r$ and $s$, $\sqrt{r + s} = \sqrt{r} + \sqrt{s}$. This means that +the equation will be true no matter what positive real numbers are substituted +for $r$ and $s$. So let $r = 9$ and $s = 16$. Then $r$ and $s$ are positive real +numbers and + +$$ \sqrt{r + s} = \sqrt{9 + 16} = \sqrt{25} = 5 $$ + +whereas + +$$ \sqrt{r} + \sqrt{s} = \sqrt{9} + \sqrt{16} = 3 + 4 k 7 $$ + +Since $5 \neq 7$, we have that $\sqrt{r + s} \neq \sqrt{r} + \sqrt{s}$, which +contradicts the supposition that $\sqrt{r + s} = \sqrt{r} + \sqrt{s}$. This +contradiction shows that the supposition is false, and hence statement $S$ is +true." + +11. + +Let $T$ be the statement: The sum of any two rational numbers is rational. Then +$T$ is true, but the following "proof" is incorrect. Find the mistake. + +**"Proof by contradiction:** Suppose not. That is, suppose that the sum of any +two rational numbers is not rational. This means that no matter what two +rational numbers are chosen their sum is not rational. Now both $1$ and $3$ are +rational because $1 = \dfrac{1}{1}$ and $3 = \dfrac{3}{1}$, and so both are +ratios of integers with a nonzero denominator. Hence, by supposition, the sum of +$1$ and $3$, which is $4$, is not rational. But $4$ is rational because +$4 = \dfrac{4}{1}$, which is a ratio of integers with a nonzero denominator. +Hence $4$ is both rational and not rational, which is a contradiction. This +contradiction shows that the supposition is false, and hence statement $T$ is +true. + +12. Let $R$ be the statement: The square root of any irrational number is + irrational. + +a. Write the negation for $R$. + +b. Prove $R$ by contradiction. + +13. Let $S$ be the statement: The product of any irrational number and any + nonzero rational number is irrational. + +a. Write the negation for $S$. + +b. Prove $S$ by contradiction. + +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 $T$. + +b. Prove $T$ by contradiction. + +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. + +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 $4 = 4 \cdot 1$.) + +17. For all prime numbers $a$, $b$, and $c$, $a^2 + b^2 \neq c^2$. + +18. If $a$ and $b$ are rational numbers, $b \neq 0$, and $r$ is an irrational + number, then $a + br$ is irrational. + +19. For any integer $n$, $n^2 - 2$ is not divisible by $4$. + +20. Fill in the blanks in the following proof by contraposition that for every + integer $n$, if $5 \cancel{\mid} n^2$ then $5 \cancel{\mid} n$. + +**Proof (by contraposition):** _[The contrapositive is: For every integer $n$, +if $5 \mid n$ then $5 \mid n^2$.]_ Suppose $n$ is any integer such that (a). +_[We must show that (b).]_ By definition of divisibility, $n =$ \(c\) for some +integer $k$. By substitution, $n^2 = $ (d) $= 5(5k^2)$. But $5k^2$ is an integer +because it is a product of integers. Hence $n^2 = 5 \cdot (\text{an integer})$, +and so (e) _[as was to be shown]._ + +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. + +b. Write what you would suppose and what you would need to show to prove this +statement by contraposition. + +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. + +b. Write what you would suppose and what you would need to show to prove this +statement by contraposition. + +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. + +24. The reciprocal of any irrational number is irrational. (The **reciprocal** + of a nonzero real number $x$ is $\dfrac{1}{x}$.) + +25. For every integer $n$, if $n^2$ is odd then $n$ is odd. + +Use any method to prove the statements in 26-29. + +26. For all integers $a$, $b$, and $c$, if $a \cancel{\mid} bc$ then + $a \cancel{\mid} b$. + +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 $a \cancel{\mid} c$, + then $a \cancel{\mid} (b + c)$. + +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. + +30. + +a. Let $n = 53$. Find an approximate value for $\sqrt{n}$ and write a list of +all the prime numbers less than or equal to $\sqrt{n}$. Is the following +statement true or false? When $n = 53$, $n$ is not divisible by any prime number +less than or equal to $\sqrt{n}$. + +b. Suppose $n$ is a fixed integer. Let $S$ be the statement, "$n$ is not +divisible by any prime number less than or equal to $\sqrt{n}$." The following +statement is equivalent to $S$: + +$\forall$ prime number $p$, if $p$ is less than or equal to $\sqrt{n}$ then $n$ +is not divisible by $p$. + +Which of the following are negations for $S$? + +(i) $\exists$ a prime number $p$ such that $p \leq \sqrt{n}$ and $n$ is +divisible by $p$. + +(ii) $n$ is divisible by every prime number less than or equal to $\sqrt{n}$. + +(iii) $\exists$ a prime number $p$ such that $p$ is a multiple of $n$ and $p$ is +less than or equal to $\sqrt{n}$. + +(iv) $n$ is divisible by some prime number that is less than or equal to +$\sqrt{n}$. + +(v) $\forall$ prime number $p$, if $p$ is less than or equal to $\sqrt{n}$, then +$n$ is divisible by $p$. + +31. + +a. Prove by contraposition: For all positive integers $n$, $r$, and $s$, if +$rs \leq n$, then $r \leq \sqrt{n}$ or $s \leq \sqrt{n}$. + +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.) + +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. + +**Test for Primality** + +Given an integer $n > 1$, to test whether $n$ is prime check to see if it is +divisible by a prime number less than or equal to its square root. If it is not +divisible by any of these numbers, then it is prime. + +32. Use the test for primality to determine whether the following numbers are + prime or not. + +a. 667 + +b. 557 + +c. 527 + +d. 613 + +33. The sieve of Eratosthenes, named after its inventor, the Greek scholar + Eratosthenes (276-194 B.C.E.), provides a way to find all prime numbers less + than or equal to some fixed number $n$. To construct it, write out all the + integers from $2$ to $n$. Cross out all multiples of $2$ except $2$ itself, + then all multiples of $3$ except $3$ itself, then all multiples of $5$ + except $5$ itself, and so forth. Continue crossing out the multiples of each + successive prime number up to $\sqrt{n}$. The numbers that are not crossed + out are all the prime numbers from $2$ to $n$. Here is a sieve of + Eratosthenes that includes the numbers from $2$ to $27$. The multiples of + $2$ are crossed out with a /, the multiples of 3 with a \, and the multiples + of 5 with a -. + +See image on Page 250. + +Use the sieve of Eratosthenes to find all prime numbers less than $100$. + +34. Use the test for primality and the result of exercise 33 to determine + whether the following numbers are prime. + +a. 9,269 + +b. 9,103 + +c. 8,623 + +d. 7,917 + +35. Use proof by contradiction to show that every integer greater than 11 is a + sum of two composite numbers. + +36. For all odd integers $a$, $b$, and $c$, if $z$ is a solution of + $ax^2 + bx + c = 0$ then $z$ is irrational. (In the proof, use the + properties of even and odd integers that are listed in Example 4.3.3.)