# The square root of 2 is irrational

One proof by reductio ad absurdum consists of first assuming that the square root of 2 can be written as a rational number. Thus, there is a pair of coprime integers $\displaystyle{ p }$ and $\displaystyle{ q }$ such that their ratio $\displaystyle{ \frac{p}q = \sqrt 2 }$ (Every rational number has an irreducible representation). And so, $\displaystyle{ \frac{p^2}{q^2} = 2 }$, and $\displaystyle{ p^2=2q^2 }$. As $\displaystyle{ p^2 }$ has a factor of $\displaystyle{ 2 }$, it is even, and thus $\displaystyle{ p }$ is also even. Thus $\displaystyle{ p }$ can be written as $\displaystyle{ 2k }$, where $\displaystyle{ k }$ is an integer. It then follows that $\displaystyle{ (2k)^2=2q^2 }$, $\displaystyle{ 4k^2=2q^2 }$, $\displaystyle{ 2k^2=q^2 }$, and thus $\displaystyle{ q^2 }$ also has a factor of $\displaystyle{ 2 }$, and thus $\displaystyle{ q }$ is also even. If $\displaystyle{ p }$ and $\displaystyle{ q }$ are both even, then they share a common factor of $\displaystyle{ 2 }$, and thus are not coprime, leading to a contradiction as we have already established that $\displaystyle{ p }$ and $\displaystyle{ q }$ are coprime. And so, no such $\displaystyle{ p,q }$ can exist, and the square root of 2 is irrational.