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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.

patha
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    The question isn't disproving $\sqrt 4$ is irrational. That's trivial! $\sqrt 4 = 2\in \mathbb Z\subset \mathbb Q$. The question is why does this proof for proving $\sqrt 4$ is irrational when on the surface it appears to be exactly like the proof that $\sqrt 2$ is irrational. The question is a matter of testing our understanding of the original proof so we can explain why these two situations are different and one of them is just wrong. No, we were never agnostic to what $\sqrt 4$ was, and we were never agnostic that $2$ is prime. The issue is why it is wrong. not that it is. – fleablood Aug 02 '22 at 04:54
  • @fleablood Thanks for the helpful comment! – patha Aug 02 '22 at 04:57
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    "Aren't we agnostic of what 4–√ is at the beginning of this disproof?" No. "seems like implying that we already knew that 2∗2=4" We did. "If so, why do we even need to disprove the proposition" We aren't trying to disprove the proposition. We are trying to find a specific error in the proof. It's not enough to say "we know it is wrong" (which is obvious). It's a matter of saying "your reasoning is incorrect here because you made this mistake", not "well you must have made a mistake somewhere because we know it is false". – fleablood Aug 02 '22 at 04:58

2 Answers2

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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

$$2>0\text{ and }2\times 2=4\Rightarrow \sqrt{4}=2\in\mathbb{Z}$$

QC_QAOA
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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!

Gary
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