Show that $\sqrt{2}$ is irrational using integer root theorem.

Let $P(x)=x^2-2$. Since $\sqrt{2}$ is a root of this polynomial, had it been a rational (suppose $\sqrt{2}=\frac{p}{q}$) no, by integer root theorem $q|1$. Hence $q=\pm1$. Moreover $p|-2$. So $p=\pm1$ or $p=\pm2$. Either way $\sqrt{2}=\pm1$ or $\pm2$ which is absurd.

Is it alright?

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    $\begingroup$ Alright, but slightly overdone I guess. The integer root theorem already says that the rational roots of $x^2-2$ are integers and divisors of $(-)2$. $\endgroup$ – Hagen von Eitzen Aug 6 '14 at 8:03
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    $\begingroup$ Please avoid creating overly specific tags like "rational root theorem". Nobody is going to subscribe to that, or use it for filtering. I replaced it with the tags (polynomials) and (roots). $\endgroup$ – user147263 Aug 6 '14 at 18:12
  • $\begingroup$ Here's the fifth post (fourth question) on this website: math.stackexchange.com/questions/5. One of its answers solves this problem: math.stackexchange.com/a/16596 $\endgroup$ – Jonas Meyer Aug 6 '14 at 18:20
  • $\begingroup$ math.stackexchange.com/questions/917983/… $\endgroup$ – user117644 Nov 20 '15 at 0:33

"Absurd" is not quite enough.

You should say that if $q=\pm1$ and $p=\pm1$ or $p=\pm2$ then $\left(\frac{p}{q}\right)^2=1$ or $4$. But $\sqrt{2} ^2=2$, so $\sqrt{2}$ cannot be written as $\frac{p}{q}$.

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