How to prove by contrapositive that for all integers $m$ and $n$, if $m - n$ is odd, then $m$ is odd or $n$ is odd?

Question

So far this is what I've come up with

$p:= m-n$ is odd $q:= m$ is odd or $n$ is odd

since the contrapositive is defined as $\neg q \implies \neg p$, I have assumed that $m$ is even and $n$ is even (after applying De Morgan's law) and I am now proving that $m-n$ is even. However, this is where I am stuck. Since any even number can be represented as $2$ multiplied by an integer, I have $m = 2a$ and $n = 2b$ and plugging that in gives me $2a - 2b$ which, in the case of $a = b$ equals zero. Is there a way to rewrite this statement so that I can say it holds true in the case that $a$ does not equal $b$?

Answer 1

$2a-2b=2(a-b)$ is an even number since $a-b$ is an integer.

Note that $0$ is an even number as $0=2(0)$.

Answer 2

Let $\mathbb{O}$ and $\mathbb{E}$ be the sets of odd and even integers, respectively.

You want to show that

$$m-n \in \mathbb{O} \implies (m \in \mathbb{O}) \lor (n \in \mathbb{O}) .$$

By De Morgan's laws, the negation of the right side is (where single quote means negation)

$((m \in \mathbb{O}) \lor (n \in \mathbb{O}))'\\ =(m \in \mathbb{O})' \land (n \in \mathbb{O})'\\ =(m \not\in \mathbb{O}) \land (n \not\in \mathbb{O})\\ =(m \in \mathbb{E}) \land (n \in \mathbb{E})\\ $

and the negation of the left side is

$(m-n \in \mathbb{O})'\\ =(m-n \not\in \mathbb{O})\\ =(m-n \in \mathbb{E}). $

The contrapositive is therefore

$((m \in \mathbb{E}) \land (n \in \mathbb{E})) \implies (m-n \in \mathbb{E}) $.

In words: if $m$ and $n$ are even then $m-n$ is even.

This question of mine might be relevant: Prove that no positive integer is both even and odd, and that all positive integers are either even or odd

Answer 3

Well the contrapositive would be that if $m$ and $n$ are even, then $m-n$ is even. Make $m=2a$ and $n=2b$ $$M-n= 2a-2b =2(a-b) =2k$$

make $k=a-b$. $2k$ is positive making the contrapositive true therefore making the original statement true.