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I get that this has to be solved as a by conditional. So first I would just solve directly: $m$ is even and $n$ is odd, so $m = 2k$ and $n = 2j+1$. Therefore $mn+n = 2(2kj+k+j)+1$, which is odd.

This is where I'm getting confused. Next when you solve the the converse, by way of contrapositive you would try to prove that $mn+m$ is even if $m$ is even OR $n$ is odd.

The book I'm using says that there are two cases to consider:

case one: $m$ is even
case two: $m$ and $n$ are both odd

I don't understand why you need to do multiple cases to begin with since you are dealing with an OR state. One case being true should suffice, right?

Also in case one, if $m$ is even, then what is $n$ supposed to be? Does the parity of $n$ even matter?

Théophile
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user3915992
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    When is $mn+m$ odd? If $m$ is even, then obviously $mn$ and $m$ are even. Thus $m$ must be odd. Then if $n$ is odd, both $m$ and $mn$ are odd and the sum is even. Thus $n$ has to be even: then $mn$ is even and $m$ is odd, and the sum is odd, bingo. – Jean-Claude Arbaut Oct 08 '15 at 18:27
  • I tried to correct the title, but would you please check to see if I did it correctly. – user84413 Oct 08 '15 at 18:28
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    Other way to see this: $mn+m=m(n+1)$ is odd iff both factors are odd, that is $m$ is odd and $n$ is even. – Jean-Claude Arbaut Oct 08 '15 at 18:29
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    About "one case being true should suffice right?": this is important, because for an iff (if and only if) proof, you must be sure that the LHS and the RHS are equivalent in all circumstances. It's not just "if $m$ is odd and $n$ is even, then it's ok", because you have to check what happens if they are not. – Jean-Claude Arbaut Oct 08 '15 at 18:31

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$mn+n=n(m+1)$. Take mod 2 for both sides.

$ mn+n$ is odd iff $n(m+1)=1$ iff $n=1, m=0 $ mod $2 $ iff $n$ is odd and $m$ is even.

Nameless
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Adelafif
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