[latexpage]
$\sqrt{2}$ is irrational
If $\sqrt{2}$ were rational, then there are two integers $a$ and $b$ where
\begin{displaymath}
\sqrt{2} = \frac{a}{b}
\end{displaymath}
Squaring both sides and splitting the variables:
\begin{align}
2 &= \frac{a^2}{b^2}\notag\
2b^2 &= a^2
\end{align}
Like all integers, $a$ and $b$ can be written as a product of primes:
\begin{displaymath}
\sqrt{2} = \frac{a}{b} = \frac{2^x3^y5^z\dots}{2^p3^q5^r\dots}
\end{displaymath}
And when you square that, you double the exponents of the primes:
\begin{displaymath}
2 = \frac{a^2}{b^2} = \frac{2^{2x}3^{2y}5^{2z}\dots}{2^{2p}3^{2q}5^{2r}\dots}
\end{displaymath}
From that equation, it’s clear that all the factors of $a$ cancel with all the factors of $b$ except for one factor of $2$. Which means that $2x-2p=1$. But that can never be — the difference of two even numbers cannot be just $1$.
Therefore, there is no prime factorization that will work for $a$ and $b$, so there is no pair which makes this fraction work.