Is this proof that $\sqrt 2$ is irrational correct? Suppose $\sqrt 2$ were rational.  Then we would have integers $a$ and $b$ with $\sqrt 2 = \frac ab$ and $a$ and $b$ relatively prime.
Since $\gcd(a,b)=1$, we have $\gcd(a^2, b^2)=1$, and the fraction $\frac{a^2}{b^2}$ is also in lowest terms.
Squaring both sides,
$2 = \frac 21 = \frac{a^2}{b^2}$.
Lowest terms representations of rational numbers are unique, so we have $a^2 = 2$ and $b^2=1$.
But there is no such integer $a$, and therefore we have a contradiction and $\sqrt 2$ is irrational.
I am not interested in the pedagogical value of this purported proof; I am only interested in whether the logic is sound.

Hhmmm ... Here is a short one.
We all know that $\sqrt{2}$ is a root of $x^2-2 = 0$
Apply Rational Root Theorem and you're done
It goes something like this : if $\sqrt{2} = \dfrac{p}{q}$, then $q=1$ and $p$ is a factor of $2$. Hence the only possible rational root of $x^2-2=0$ would be $-2,-1,1,-2$.
A quick check shows that this is not the case. Therefore $\sqrt{2}$ is irrational.
A bit further now : using the same Theorem, shiw that $\sqrt{n^2+1}$ is irrational.
