---The statement: Let $m$ and $n$ be two integers. Prove that $mn+m$ is odd if and only if $m$ is odd and $n$ is even.
---Solution: First, we show that if $m$ is odd and $n$ is even, then $mn+m$ is odd. Assume that $m$ is odd and $n$ is even. Then $m = 2a+1$ and $n = 2b$ for integers $a$ and $b$. So $mn+m = (2a+1)(2b)+(2a+1)= 4ab +2b+2a+1=2(2ab+b+a)+1$. Since $2ab+b+a$ is an integer, $mn+m$ is odd. Next, we verify the converse, that is, if $mn + m$ is odd, then $m$ is odd and $n$ is even. Assume that it is not the case that $m$ is odd and $n$ is even. Then either $m$ is even or $n$ is odd. Therefore, either $m$ is even or $m$ and $n$ are both odd. We consider these two cases. Case 1. $m$ is even. Case 2. $m$ and $n$ are both odd.
---My question: Why didn't we consider a third case where $m$ is even and $n$ is odd
$\endgroup$ 43 Answers
$\begingroup$$mn+m=m(n+1)$ is odd iff both factors $m$ and $n+1$ are odd iff $m$ is odd and $n$ is even.
$\endgroup$ $\begingroup$It's mainly just an aesthetic choice how you split this into cases. As you suggested handling the cases
- m even and n odd
- m even and n even
- m odd and n odd
Is perfectly acceptable. You're basically just enumerating the possible parity of each pair of m and n. If you do this you will probably realize that you don't even need to consider the value of n for the 1st two cases. So instead it's a good idea to consider only the state of m
- m is even
- m is odd (this forces n to also be odd)
$mn+m$ is even $\iff$ $m$ is even $\lor$ $n$ is odd.
Therefore if and only if $m$ is odd $\land$ $n$ is even, is $mn+m$ odd.
$\endgroup$
