Question

Prove that $\sqrt{3}$ is irrational.

Answer 1

Let us assume, to the contrary, that $\sqrt{3}$ is rational.

That is, we can find integers a and b $(\ne 0)$ such that

$\sqrt{3}=\frac{a}{b}$.

Suppose a and b have a common factor other than $1$, then we can divide by the common factor, and assume that a and b are coprime.

So, $b\sqrt{3}=a$.

Squaring on both sides, and rearranging, we get $3b^2=a^2$.

Therefore, $a^2$ is divisible by $3$, and by Theorem $1.3$, it follows

that a is also divisible by $3$.

So, we can write a $=3c$ for some integer c.

Substituting for a, we get $3b^2=9c^2$, that is, $b^2=3c^2$.

This means that $b^2$ is divisible by $3$, and so b is also divisible by $3$(using Theorem $1.3$ with $p=3$).

Therefore, a and b have at least $3$ as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational. So, we conclude that $\sqrt{3}$ is irrational.