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I was doing a problem where I had to prove that $\sqrt5$ was irrational, and I went through with the usual method. Then, I was curious what would've happened if I tried to do this with a value whose square root was rational, such as $\sqrt4$.

Let's assume that we do not know that $\sqrt4$ is rational or not.

Let's assume $\sqrt4$ is rational.

$\sqrt4 = p/q$
$4 = p^2/q^2$
$4q^2 = p^2$

This means that 4 is a factor of p.

Let's assume p = 4x

This means that:

$4q^2 = 16x^2$
$q^2 = 4x^2$

This means q is also a factor of 4. But, here we are making the assumption that p and q are co-prime, so this contradiction occurs when we assume that $\sqrt4$ is rational, and hence $\sqrt4$ is irrational.

The contradiction method seems to suggest that $\sqrt4$ is irrational, even though we know that it equals 2 which is rational. Did I make a mistake somewhere? Or is the perhaps a problem in the contradiction method?

Tsar Asterov XVII
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    To find your mistake, since you know $\sqrt{4}=2$, you could go through each step with $p=2$ and $q=1$ to find the first claim which is not true. (The first answer has already pointed out the same spot.) – aschepler Dec 02 '22 at 17:15
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    For the future: I assume you know that $2$ is a (rational) square root of $4$. So why did you not take that solution (say $p=2, q=1$), and go through your "proof" line by line to see where your proof is off? Then you would have seen the first three lines are fine for you $p,q$, but the line "This means $4$ is a factor of $p$" would be the first wrong one for your $p=2$. And indeed that is the wrong line in the proof, as the answer shows. – Torsten Schoeneberg Dec 02 '22 at 17:17
  • @TorstenSchoeneberg I understand what you mean. Thanks a lot for clearing this misconception for me. – Tsar Asterov XVII Dec 02 '22 at 17:18
  • Ha, was typing why @aschepler was expressing the same idea. – Torsten Schoeneberg Dec 02 '22 at 17:18
  • @aschepler Thanks a lot to you too sir/ma'am! – Tsar Asterov XVII Dec 02 '22 at 17:19

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$4q^2=p^2$ means $4$ is a factor of $p^2$. That does not mean $4$ is a factor of $p$, since $4$ is not prime (it might be, it might not be). Assuming $p,q$ are coprime, you can only guarantee $2$ is a factor of $p$. If $p=2x$, then $4q^2=4x^2\implies x=q$ ($p,q$ can obviously be taken positive) so that $p/q=2$.

You conclude that, if $p/q$ is a coprime rational representation of $\sqrt{4}$, then $p/q=2$. That's ok, because $\sqrt{4}$ is indeed equal to $2$.

FShrike
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