Proving by contrapositive: x and y are integers, and xy is even, then x is even or y is even I need to prove the following by contrapositive:
"$x$ and $y$ are integers, and $xy$ is even, then $x$ is even or $y$ is even"
I'm sure this question isn't very hard to solve, however my understanding of contraposition is very weak. I have only learned it recently and I do not feel like I am totally grasping the concept.
From my understanding I am almost trying to prove this by saying the opposite statement. But I feel like this is oversimplifying it. 
I know that based on previous proofs an even number is in a form like $z = 2a$ an odd is the same as the even except plus 1: $x = 2a + 1$. Based on knowing this I assume I am able to use contraposition. But where do I start to use this?
 A: What your are starting with is 
Let $x, y$ be integers.
($xy$ is even) $\implies (x$ is even or y is even)
The contrapositive is then $\lnot(x$ is even or $y$ is even)$\implies \lnot(xy$ is even).
This means we want to prove that if $x$ is odd AND $y$ is odd, then $xy$ is odd.
Start in the standard way: Let $x = 2a +1$ and let $y = 2b +1$ where $a, b \in \mathbb Z$.
Then calculate $xy$, and represent the product as an odd integer.
A: To prove this, you must rewrite the statement as:
If $x$ is odd AND $y$ is odd, then xy is odd. 
You can do this by letting $x = 2a + 1$ and $y = 2b + 1$, for all integers, $a,b$.
Then, try multiplying them and see what you can say about $xy$
A: There is a very important different between the converse of a statement and the contrapositive of a statement. If you have a statement $S$ that says "if $A$, then $B$" the converse/opposite of $S$ would be "if not $A$, then not $B$". As a general rule, $S$ and the converse of $S$ are not logically equivalent. The exception is that $A$ and $B$ are logically equivalent. You'll encounter this periodically in math and proofs of this nature are commonly called "if and only if" proofs. Anyway, what is equivalent to $S$ is the contrapositive of $S$. The contrapositive of $S$ is "if not $B$, then not $A$". Using logic notation, the following two statements are identical.

$$(A \implies B) \equiv (\neg B \implies \neg A)$$

In the context of what you need to prove, it might help to rephrase the sentence slightly. We can say "Let $x,y$ be integers. If $xy$ is even, then $x$ is even or $y$ is even". Pattern matching that to what is above, then $A \equiv $ "$xy$ is even" and $B \equiv $ " $x$ is even or $y$ is even." Hence the contrapositive would be:

If $\neg(x$ is even or $y$ is even$)$ then $\neg (xy$ is even$)$
$\equiv$ If $x$ is not even and $y$ is not even, then $xy$ is not even.
$\equiv$ If $x$ is odd and $y$ is odd then $xy$ is odd.

At this point you can proceed as you correctly guessed, letting $x = 2a+1$ and $y = 2b+1$ for some $a,b \in \Bbb{Z}$. You'll find that $xy = 2c+1$ for some $c \in \Bbb{Z}$.
