So I understand how to prove that $\sqrt 3$ is irrational. However, I think I'm missing something, conceptually.
So we assume that $\sqrt 3$ is rational, and thus can be expressed as $\frac{p}q$, $p \in \mathbb Z, q \in \mathbb Z^*$.
Thus, $(\frac{p}q)^2$ = 3, which means that $p^2 = 3q^2$, which means that $q^2$ is a multiple of $3$, which means that $q$ is a multiple of 3. How do we know this? I've read that it comes from the Fundamental Theorem of Arithmetic, but I just don't see how it follows?
We then go on and express $q$ as $3r, r \in \mathbb Z$, thus $3r = q^2$. Therefore, $p$ and $q$ are multiples of each other, and we have a contradiction.
My question is, why coudln't we use the exact same logic for proving that $\sqrt 4$ is irrational (which of course it isn't)? Does the Fundamental Theorem of Arthmetic imply something for our $\sqrt 3$ that it doesn't for the $\sqrt 4$ proof? This feels like a stupid questions seeing as $\sqrt 4$ is obviously not irrational. I just don't understand how the logic doesn't tranpose.
Any help is appreciated, thanks.
First - I think you meant to write $$ \left( \frac{p}{q} \right)^2 = 3 $$ - squared, not cubed.
The argument then continues by claiming that if $p^2 = 3q^2$ then $3$ must divide $p$. That follows from the fact that if a prime (in this case $3$) divides a product it must divide one of the factors (here they are $p$ and $p$).
You can't assert that if $p^2 = 4q^2$ then $4$ must divide $p$. If $p$ were just singly even then $4$ would still divide $p^2$.