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@ -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.)