Question

Prove that $\sqrt{5}$ is an irrational number.

Answer 1

We will show that $\sqrt{5}$ is irrational by contradiction.
Let us assume that $\sqrt{5}$ is rational.

Therefore we can represent it in the form of $\frac{a}{b}$.
$\sqrt{5}$ = $\frac{a}{b}$
Let us assume that $\frac{a}{b}$ is in its lowest form, i.e. a and b are co-prime.
$$\begin{gathered} \sqrt{5}=\frac{a}{b} \\ \Rightarrow a=\sqrt{5}b \\ \Rightarrow a^2=5b^2-----\left(1\right) \end{gathered}$$

Therefore, 5 is the factor of $a^2$ . If 5 is a factor of $a^2$, 5 will also be the factor of a.

So ,we can write a = 5c.

$$\begin{gathered} {\left(5c\right)}^2=5b^2 \\ \Rightarrow 25c^2=5b^2 \\ \Rightarrow b^2=5c^2 \end{gathered}$$

Therefore b² is divisible by 5.
Since, it is divisible by 5, b will also be divisible by 5.

Therefore, a and b have at least 5 as a common factor.

But this contradicts that a and b are co-prime.

This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.

Therefore, $\sqrt{5}$ is irrational.