Question

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

Answer 1

Let us suppose that $\sqrt{3}$ is a rational number.

Then there are positive integers $a$ and $b$ such that $\sqrt{3}=\frac{a}{b}$, where $a$ and $b$ are co-prime, meaning their HCF is $1$.

$$\begin{gathered} \sqrt{3}=\frac{a}{b} \\ \Rightarrow 3=\frac{a^2}{b^2} \\ \Rightarrow 3b^2=a^2 \\ \Rightarrow 3divides{a}^2[\because 3divides3b^2] \\ \Rightarrow 3dividesa.....................\left(i\right) \\ \Rightarrow a=3cforsomeintegerc \\ \Rightarrow a^2=9c^2 \\ \Rightarrow 3b^2=9c^2[\because a^2=3b^2] \\ \Rightarrow b^2=3c^2 \\ \Rightarrow 3divides{b}^2[\because 3divides3c^2] \\ \Rightarrow 3dividesb..............................\left(ii\right) \end{gathered}$$

We can see that $a$ and $b$ share at least $3$ as a common factor from $\left(i\right)$ and $\left(ii\right)$.

Because of the fact that$a$ and $b$ are co-prime, however, contradicts this and indicates that our hypothesis is incorrect.

Hence, $\sqrt{3}$​ is an irrational number.