Chapter 1.5, Problem 12E

Assign a grade of A (correct), C (partially correct), or F (failure) to each. Justify assignments of grades other than A.
(a) Suppose m is an integer.
Claim. If m2 is odd, then m is odd.
“Proof.” Assume that $m^{\text{2}}$ is not odd. Then $m^{\text{2}}$ is even and $m^{\text{2}}=\text{2}k$ for some integer k. Thus 2k is a perfect square; that is, $\sqrt{\text{2}k}$ is an integer. If $\sqrt{\text{2}k}$ is odd, then $\sqrt{\text{2}k}=\text{2}n+\text{1}$ for some integer n, which means $m^{\text{2}}=\text{2}k={(\text{2}n+\text{1})}^{\text{2}}=\text{4}{n}^{\text{2}}+\text{4}n+\text{1}=\text{2}(\text{2}{n}^{\text{2}}+\text{2}n)+\text{1}$ . Thus $m^{\text{2}}$ is odd, contrary to our assumption. Therefore, $\sqrt{\text{2}k}=m$ must be even. Thus if $m^{\text{2}}$ is not odd, then m is not odd. Hence if $m^{\text{2}}$ is odd, then m
is odd.
(b) Suppose t is a real number.
Claim. If t is irrational, then 5t is irrational.
“Proof.” Suppose 5t is rational. Then $\text{5}t=p/q$ , where p and q are integers and $q\ne 0$ . Therefore, $t=p/(\text{5}q)$ , where p and 5q are integers and $\text{5}q\ne 0$ , so t is rational. Therefore, if t is irrational, then 5t is irrational.
(c) Suppose x and y are integers.
Claim. If x and y are even, then $x+y$ is even.
“Proof.” Suppose x and y are even but $x+y$ is odd. Then, for some integer k, $x+y=\text{2}k+\text{1}$ . Therefore, $x+y+(-\text{2})k=\text{1}$ . The left side of the equation is even because it is the sum of even numbers. However, the right side, 1, is odd. Because an even cannot equal an odd, we have a contradiction. Therefore, $x+y$ is even.
(d) Suppose a, b, and c are integers.
Claim. If a divides both b and c, then a divides $b+c$ .
“Proof.” Assume that a does not divide $b+c$ . Then there is no integer k such that $ak=b+c$ . However, a divides b, so $am=b$ for some integer m; and a divides c, so $an=c$ for some integer n. Thus $am+an=a(m+n)=b+c$ . Therefore, $k=m+n$ is an integer satisfying $ak=b+c$ . Thus the assumption that a does not divide $b+c$ is false, and a does divide $b+c$ .
(e) Suppose m and n are integers.
Claim. If $m^{\text{2}}+n^{\text{2}}$ is even, then m and n have the same parity.
“Proof.” Suppose $m^{\text{2}}+n^{\text{2}}$ is even, and m and n have opposite parity. We may assume that m is odd and n is even. Then for some integers j and k, $m=\text{2}j+\text{1}$ and $n=\text{2}k$ , so $m^{\text{2}}+n^{\text{2}}=\text{4}{j}^{\text{2}}+\text{4}j+\text{1}+\text{4}{k}^{\text{2}}=\text{2}(\text{2}{j}^{\text{2}}+\text{2}j+\text{2}{k}^{\text{2}})+\text{1}$ , which is odd. This is a contradiction.
(f) Suppose a and b are positive integers.
Claim. If $a+\text{1}$ divides b and b divides $b+\text{3}$ , then $a=\text{2}$ and $b=\text{3}$ .
“Proof.” Assume $a+\text{1}$ divides b and b divides $b=\text{3}$ . Then $b=(a+\text{1})k$ and $b+\text{3}=bj$, for some integers k and j. Choose $k=\text{1}$ and $j=\text{2}$ . Then $b+\text{3}=\text{2}b$ , so $\text{2}b-b=\text{3}$ . Thus $b=\text{3}$ . From $b=(a+\text{1})k$ , we have $\text{3}=(a+\text{1})\text{1}$ , so $a=\text{2}$ . Therefore, $a=\text{2}$ and $b=\text{3}$ .