Question

Prove that $\frac{3}{\sqrt{5}}$ is irrational, given that $\sqrt{5}$ is irrational.

Answer 1

Let us assume that $\frac{3}{\sqrt{5}}$ is a rational number.

Thus, $\frac{3}{\sqrt{5}}$ can be represented in the form of $\frac{p}{q},$ where p and q are integers, q ≠ 0, p and q are co-prime numbers.

$$\begin{gathered} \frac{3}{\sqrt{5}}=\frac{p}{q} \\ \Rightarrow p\sqrt{5}=3q \\ \Rightarrow \sqrt{5}=\frac{3q}{p} \\ \mathrm{Since},\frac{3q}{p}\mathrm{is}\mathrm{rational}\Rightarrow \sqrt{5}\mathrm{is}\mathrm{rational} \\ \mathrm{But},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\sqrt{5}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \\ \mathrm{Therefore},\mathrm{our}\mathrm{assumption}\mathrm{is}\mathrm{wrong}. \\ \mathrm{Hence},\frac{3}{\sqrt{5}}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \end{gathered}$$