Proof - The Square-Root of 2 is Irrational

Relevant Courses: EECS 203, Math 297

Proof by Contradiction

First, recall that a Rational Number is a number that can be written as the quotient of an Integer and a non-zero Natural Number. Furthermore, every rational number has a “simplified” form, where the quotient and divisor have no common factors (and thus are Relatively Prime).

Assume for contradiction that $\sqrt{2}$ is rational. That means that there exists an integer $p\in \mathbb{Z}$ and nonzero $q\in {\mathbb{Z}}_{>0}$ such that $\sqrt{2}=\frac{p}{q}$. If we square both sides, we get that $2=\frac{p^2}{q^2}$, which implies that $2q^2=p^2$. Therefore, $p^2$ is even. This can only be true if $p$ is even. But this means that $p^2$ is actually divisible by 4, hence $q^2$ must also be even, and so therefore $q$ must be even. So that means both $p$ and $q$ are even, which is a contradiction to our assumption that they have no common factors.

Therefore, our initial assumption must be false, so $\sqrt{2}$ is not rational, which means it must be irrational.