How do I prove the following statement? How do I prove the following statement?

If $x^2$ is irrational, then $x$ is irrational. The number $y = π^2$ is irrational. Therefore, the number $x = π$ is irrational

 A: The statement 
"$x^2$ irrational $\implies$ $x$ irrational" 
is logically equivalent to 
"$x$ not irrational $\implies x^2$ not irrational".
A: Hint
$$(A\implies B)\iff (\neg B\implies \neg A)$$
A: To prove the first assertion, we can use a proof by contradiction.
Suppose to the contrary that $x$ is rational:
$$\exists (a, b) : b\ne 0\land x=\frac{a}{b}$$
$$\implies x^2=\frac{a^2}{b^2}$$
$$\therefore x^2 \text{ is rational. }$$
This contradicts the given fact that $x^2$ is irrational.
Added: If you are only asking for how one goes from the first (universal) assertion to the particular case about $\pi$, it is by rule often called Universal Instantiation.
A: As written, it doesn't appear that you have a statement that needs proved, but an argument (consisting of multiple statements) that needs validated.
It appears that you have a modus ponens argument. All such arguments are valid, so you're done, as far as I can tell.
In order to prove that this particular argument is sound, you would need to prove that both the premises are true. The answers above suggest how you may show that the first premise is true. Showing that $\pi^2$ is irrational, though, is likely not something that you're being asked to do. Hence, I strongly suspect that a proof isn't what's required, here.
A: $$\;(A\implies B)\equiv(\neg B\implies\neg A)\;$$
and this means that
($\;x^2\;$ is irrational $\;\implies\;x\;$ is irrational)$\;\equiv (x\;$ is rational $\;\implies\;x^2\;$ is rational) , which seems way simpler to prove.
