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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 $\sqrt{2}$?

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student010101
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  • 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
  • https://math.stackexchange.com/questions/5/how-can-you-prove-that-the-square-root-of-two-is-irrational – 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? – Student Oct 20 '20 at 14:31
  • @nls No, I found it on https://www.math.utah.edu/~pa/math/q1.html – student010101 Oct 20 '20 at 14:33
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    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

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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.

HyperPro
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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.

Paul Frost
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if p and q have common factors divide by their hcf and you have a new fraction with p and q as coprime. hence your assumption was wrong that p and q are coprime and √2 is irrational.

harsha vardhan
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