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