Discrete Mathematics Proof If m is an even integer and n is an odd integer, then m+n is odd.
Proof:
Suppose m is an even integer and n is an odd integer.
  Then $m=2k$ and $n=2j+1$ for some integers k and j.
  So $$m+n = 2k + 2j + 1\\
         = (2k+2j) + 1\\
         = 2(k+j) + 1 $$
Hence, $m+n$ is odd. 
Converse:
Suppose $m+n$ is odd. 
  Let $m+n = 2(x+y) + 1$
I am not sure if how I am starting the converse proof is correct, and if it is correct, I am a little stuck on where to go from here.
 A: I would do the converse by contradiction if I were you. Let 
$m+n = 2k + 1$
and suppose both of $m,n$ are even. You immediately get a contradiction by applying the definition of evenness to $m$ and $n$. Likewise, if you suppose they are both odd, you immediately get a contradiction by applying the definition of oddness to both $m$ and $n$. Those two cases cover everything, so you'll be done after demonstrating the contradictions. Your forward direction is correct.
Actually, a slightly simpler way to do it would be to suppose first that $m$ is even, and prove then that $n$ is odd. Next, suppose that $m$ is odd, and prove then that $n$ is even. That covers everything, and proves the result (you're renaming the variables but it doesn't incur any loss of generality).
A: Do it by cases:
If $m+n$ is odd what can $m$ and $n$ be?
Case 1:  $m$ and $n$ are even.... no, because then $m+n$ is even.
Case 2: $m$ and $n$ are odd... no, because then $m+n$ is even.
Case 3: One is odd the other is even, and this works!  Because there are no other possibilities we conclude this is what $must$ be true.
