Let $x \in \mathbb{Z}$. Prove that $5x-11$ is even IFF $x$ is odd. Let $x \in \mathbb{Z}$. Prove that $5x-11$ is even IFF $x$ is odd.
I know that for an IFF proof we prove it directly, and then prove it again by the contrapositive.
I don't know how to prove this directly, so in a normal direct proof I would use the contrapositive since that is logically equivalent to the direct proof.
The contrapositive would be:
Assume $x$ is even. By definition $x=2k$ for $k \in\mathbb{Z}$. Therefore we can see $5x-11=5(2k)-11$. This becomes $5x-11= 2(5k-6)+1$. Since $5k-6 \in\mathbb{Z}$, then $5x-11$ is odd. Which is true for the contrapositive.
I am confused on how to prove this as an IFF proof.
 A: To prove an if and only if statement you would prove the statement and the converse. If the statement is true, the contraposition is always going to be true.
So here by proving the contrapositive, you proved the if direction. i.e. $ \forall x \in \mathbb{Z}~5x - 11\text{ even} \implies x \text{ odd}.$ You need to prove the other direction
$$ \forall x \in \mathbb{Z}~5x - 11 \text{ even}\impliedby x \text{ odd}.$$
And the other direction is easy. $x$ odd implies that $x=2k+1$ for some $k\in \mathbb{Z}$. So $5x - 11 = 10k - 6 = 2(5k -3)$, which is even.
A: You could prove $A\iff B$ by proving $A\implies B$ and the inverse $\lnot A \implies \lnot B$.  (Since the inverse is the contrapositive of the converse, the inverse $\lnot A \implies \lnot B$ and converse $B \implies A$ are logically equivalent to each other.)
So, to prove what you want, it suffices to show that, if $x$ is odd, then $5x-11$ is even, and, if $x$ is not odd (i.e., $x$ is even), then $5x-11$ is not even (i.e., $5x-1$ is odd).  Those proofs should be fairly straightforward.
