Q: Prove or disprove statement (Set theory) How would I prove or disprove the following statement?
$ \forall a \in \mathbb{Z} \forall b \in \mathbb{N}$ , if $a < b$ then $a^2 < b^2$

A: Not true. Take $a=-2$ and $b=1$. Then $a<b$ but $a^2=4> 1=b^2$.
Generally, it is good practice to start with easy examples to get an idea of why something might or not be true. 

A: take $a=-1$ and $b=1$, then $a^2=1 \nless1=b^2$

A: If you choose $a$ as negative number and $b$ as a positive number with module less then $a$ you find a counter example.
In fact your statement is false $\forall a\in\mathbb{Z},a<0\forall b\in\mathbb{N}:|a|^2\geq b^2$
As @kevin said if you take $a=-5$ and $b=1$ the statement is false.

A: To disprove it, one simply provides a counterexample, as the other answers have addressed.
If I were to attempt to prove it, here's how I might begin--and how one might discover that it is false, and find a key to generating counterexamples.

Since $x<y$ is equivalent to $y-x$ being positive, then we want to show that $b^2-a^2$ is positive whenever $a\in\Bbb Z$ and $b\in\Bbb N$ are such that $b-a$ is positive.
Now, $b^2-a^2=(b-a)(b+a),$ and so since $b-a$ is positive, then for $b^2-a^2$ to be positive, we must show that $b+a$ is positive.

Here, though, we'd be stymied. We'd be okay so long as $|a|<|b|,$ but if $|a|\ge|b|,$ then we'd have to have $a\le -b,$ so that $b+a\le b+-b=0.$
Now, if we'd instead had $a\in\Bbb N$ and $b\in\Bbb Z,$ then we'd be just fine. In that case, we'd simply note that $-a\le a,$ so that $0<b-a\le b+a,$ as desired.
