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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.

Martin Sleziak
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CrispyBacon
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  • What is your question? – Wojowu Dec 23 '15 at 22:01
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    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. – Mathmo123 Dec 23 '15 at 22:02
  • Proof verification – CrispyBacon Dec 23 '15 at 22:02
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    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 – Mathmo123 Dec 23 '15 at 22:05
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    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. – Noah Olander Dec 23 '15 at 22:05
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    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. – Thomas Andrews Dec 23 '15 at 22:07
  • see here http://math.stackexchange.com/questions/919681/square-root-of-2-irrational-alternative-proof – Dr. Sonnhard Graubner Dec 23 '15 at 22:11
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    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. – Dave L. Renfro Dec 23 '15 at 22:23
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    As said in other comments, I too suggest you read a proof on the rational root theorem. – YoTengoUnLCD Dec 23 '15 at 22:26
  • @DaveL.Renfro $\sqrt{2}$ $ \neq \pm1 $ or $\sqrt{2} \neq \pm 2$ – CrispyBacon Dec 23 '15 at 22:33
  • thanks for the feedback guys – CrispyBacon Dec 23 '15 at 22:42
  • Actually Wikipedia´s proof (“the square root of 2”) seems overly complicated. Instead assume $sqrt(2)=p/q$. Then $2q^2=p^2$ so that $p^2$ must contain the factor 2 and therefore the factor 4 being a square. $q^2$ must therefore contain a factor of 2 and therefore a factor 4 which leads to a number of equations $2q_n^2=p_n^2$ for any number n in infinite descent. – Mikael Jensen Dec 23 '15 at 22:58
  • Closely related: Is this proof that √2 is irrational correct? – MJD Dec 23 '15 at 23:28

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