Disproving that $\sqrt{4}$ is irrational with the same logic proving that $\sqrt{2}$ is irrational I have looked up another question regarding this disproof, but I got confused.
If I understood it correctly, the disproof flows like this:
Just like we have shown that $\sqrt{2}$ is irrational by contradiction,
'Assuming' that $\sqrt{4}$ is rational,
$\sqrt{4}$ = $\frac{p}{q}$ for $p,q \in \mathbb{Z}$
$4$ = $\frac{p^2}{q^2}$
$4 * q^2$ = $p^2$
Here, we claim that $p$ is not necessarily a multiple of 4 since the same condition can be sufficed when p is a multiple of 2.
This, to me, feels like a circular reasoning.
Aren't we agnostic of what $\sqrt{4}$ is at the beginning of this disproof?
Bringing up 2(which is an integer obviously) here seems like implying that we already knew that $2 * 2 = 4$
If so, why do we even need to disprove the proposition at the first place rather than stating that $\sqrt{4} = 2 \in \mathbb{Z}$?
I think I am misinterpreting something here, but I cannot see what that is.
Thanks.
 A: In the linked proof, they are not assuming $4=2\times 2$. Rather, they are assuming
$$4\text{ is even }\Rightarrow\ 2|4\ \Rightarrow\ 2|p^2\ \Rightarrow\ 2|p\Rightarrow p=2r$$
(since $2$ is prime). Of course, in the linked question they do not prove that the square root of $4$ is rational, only that the same method for proving $\sqrt{2}$ is irrational cannot work for proving that the $\sqrt{4}$ is irrational. This isn't really a big deal, for many $n$ we wouldn't use this reasoning to show $\sqrt{n}$ is irrational and it shouldn't worry us that it primarily works for $\sqrt{2}$. To show that $\sqrt{4}$ is rational, it suffices to say that
$$2>0\text{ and }2\times 2=4\Rightarrow \sqrt{4}=2\in\mathbb{Z}$$
A: The proof that $\sqrt 2$ is irrational works for $2$ because 2 is a prime number. If we know that a number $p$ is prime, then we can show that $\sqrt p$ is irrational.
The proof is given by contradiction as we assume $\sqrt p$ to be rational, that is, equal to $\frac ab$, where a and b are positive integers and co-prime. Then we square both sides to get $p=\frac {a^2}{b^2}$ to get $pb^2=a^2$. Now, we can say that "$p$ divides $a^2$, so $p$ also divides $a$" only when $p$ is prime, which we know it is, as we began with $p$ being a prime(I will not explain why it is so, as I'm assuming you already know the reason for that. Basically, you factorize a number into its prime factors and square that number, and notice that only those prime factors are the prime factors of its square as well). Ultimately we show that a and b have a common factor $p$ and so they are not co-prime. Otherwise, if you don't know whether a number is prime, you can't use the same line of reasoning as we did for $\sqrt 2$.  Besides, it is much easier to check rationality of the square root of a number than it is to prove it!
