I'm just trying to make sure I have this right:
(b) Give a proof by contradiction of: “If n is an odd integer, then n
2
is odd.”
$n = 2k-1$
$n^2 = (2k-1)^2$
$a = n$ is odd
$b = n^2$ is odd
Since any integer $k$ multiplied by $2$ is even, and we subtract $1$, we get an odd result. That odd result, when squared, is always odd. Therefore, to assume that $n^2$ is even is actually a false statement, giving the following:
$a \rightarrow \neg b$
$TRUE \rightarrow FALSE$
$FALSE$
(c) Give a proof by contraposition of: “If n is an odd integer, then n + 2 is odd.”
$n = 2k-1$
$n+2 = 2k+1$
$a = n$ is odd
$b = n+2$ is odd
Any integer $k$ multiplied by 2 is even, and subtracting 1 is always an odd result. Any odd result + 2 is still odd, so $a$ and $b$ are true. Assuming n+2 is even and n as even would be false statements, so we get the following:
$\neg b \rightarrow \neg a$
$FALSE \rightarrow FALSE$
$TRUE$
Best Answer
Why are you using a contradiction to prove something? $$(2k+1)(2k+1)= \text{ even+even+even}+1!$$ that's the simple way.