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Can anyone check if this proof is correct. Thank you.

Proof that $\sqrt{2}$ is irrational.

Let $x = \sqrt{2}$

then $x^2=2$

and $x^2-2=0$

By the Rational Root Theorem, we have:

the number $1$ that is the coefficient of $x^2$

and the number $(-2)$ that is the constant of the polynomial.

Let assume that $\sqrt{2}$ is rational then $\sqrt{2}=\displaystyle\frac{p}{q}$.

By the rational root theorem, $\sqrt{2}$ is a root of the equation then $p|2$ and $q|1$.

  1. $p|(-2)$ with $p=\pm1$ or $p=\pm2$
  2. $q|1$ with $q=1$

Let list all $\displaystyle\frac{p}{q}$ : $\pm\displaystyle\frac{1}{1}$ or $\pm\displaystyle\frac{2}{1}$.

Since there is no such $p$ and $q$ that satisfy $\sqrt{2}=\displaystyle\frac{p}{q}$ we conclude that $\sqrt{2}$ is irrational.

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    $\begingroup$ This is correct. However, the standard proof of the rational root theorem uses an argument very similar to the standard argument that $\sqrt 2$ is irrational, so you're not gaining much with this proof. If anything, you may be hiding what is really going on. $\endgroup$
    – Mathmo123
    Commented Dec 23, 2015 at 22:02
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    $\begingroup$ The question was obvious given the tags, and the it has now been edited. Since proof verification is on topic, I don't think the downvotes are justified $\endgroup$
    – Mathmo123
    Commented Dec 23, 2015 at 22:05
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    $\begingroup$ Looks good! You should think a little about the proof of the rational root theorem because if you add that into your argument, you get back a pretty standard proof that $\sqrt{2}$ is rational. $\endgroup$
    – Noah Olander
    Commented Dec 23, 2015 at 22:05
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    $\begingroup$ The rational root theorem is basically much more general than the theorem that $\sqrt{2}$ is irrational, and proving the rational root theorem is actually a generalization of the standard proof that $\sqrt{2}$ is irrational, so you are correct, but essentially hiding the details of the same proof in the shout out to RRT. $\endgroup$
    – Thomas Andrews
    Commented Dec 23, 2015 at 22:07
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    $\begingroup$ I think "Since there is no such $p$ and $q$ that satisfy $\sqrt{2}=\displaystyle\frac{p}{q}$" requires more explanation. Specifically, the rational root theorem gives you specific rational number possibilities for $\sqrt{2}.$ What are these rational numbers? And how do you know that $\sqrt{2}$ is not equal to one of these rational numbers? Also, this handout of mine might be of help. $\endgroup$
    – Dave L. Renfro
    Commented Dec 23, 2015 at 22:23

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