# Proof of why $\sqrt{2}$ is an irrational number [duplicate]

I am studying the proof by contradiction below. But I am confused on why the proof is valid. It first assumes that $$p, q$$ have no common factor, and then arrives at a conclusion where $$p, q$$ are both divisible by $$2$$, and hence they do have a common factor, contradicting the earlier assumption. But I am confused on how this leads to the conclusion that $$\sqrt{2}$$ is irrational.

To me, it seems all it says is that we can't assume that $$p, q$$ have no common factors. How does it prove the case where $$p, q$$ have a common factor, but doesn't result in $$\sqrt{2}$$?

• If $p,q$ have common factors, cancel their gcd. You get a new fraction, without common factors, which is equal to $\sqrt{2}$. – N. S. Oct 20 '20 at 14:26
• math.stackexchange.com/questions/5/… – Lion Heart Oct 20 '20 at 14:26
• @N.S. I understand that you can cancel their gcd, but I still don't understand why this proof by contradiction works. All it seems to do is tell me that their original assumption that $p, q$ have no common factors is wrong, but not that $\sqrt{2}$ is irrational. – student010101 Oct 20 '20 at 14:29
• @student010101 Is this from an IB book? – nls Oct 20 '20 at 14:31
• Well if $\sqrt{2}$ is rational, then it can be written as a reduced fraction, which you just showed it is wrong. If Something implies wrong, what does it mean? – N. S. Oct 20 '20 at 14:55

To me, it seems all it says is that we can't assume that p,q have no common factors.

That would mean that you could reduce the fraction. But you can not reduce a fraction infinitely many times. At some point, $$p$$ and $$q$$ have to coprime, what means that they have no common factor.
That means, you can assume that you can write $$\sqrt 2$$ as a fully reduced fraction, just because you can fully reduce any fraction. But given such a fraction that proof shows, that you can still reduce it, and that is the contradicting.
If you can write a number as a fraction, you can always write it as a fraction with coprime $$p,q$$. It follows that the assumption that there even exists such a fraction must be wrong.
You know that it is a proof by contradiction. The assumption is that $$\sqrt 2$$ is not irrational, i.e. $$\sqrt 2$$ is rational. This means $$\sqrt 2 = p/q$$ for some integers $$p,q$$. There are many such representations as a fraction, but we may cancel the gcd of $$p, q$$ and get a representation in which $$p, q$$ do not have a common factor. The proof shows then that both $$p, q$$ must have a factor $$2$$ which contradicts the fact that they do not have a common factor. Thus is assumption was wrong, and $$\sqrt 2$$ is irrational.