prove using contraposition that if a and b are integers:

$a^2-4b-2\neq0$

So the contrapositive:

$a^2-4b-2=0$ implies that at least one of either a or b are not integers

Working from the above, I got:

$a^2=4b+2$

$a^2=2(2b+1)$

If both a and b are integers, $2(2b+1)$ is a perfect square.

In other words:

$\sqrt{2(2b+1)}=k$

Where k is an integer

I also realised that 2b+1 is an odd number so:

$k=\sqrt{2}\sqrt{2b+1}$

I ended my proof by saying that since 2b+1 is odd, it does not have a multiple of 2 so k will always have multiple or $\sqrt{2}$ which means that it cannot be an integer.