Question

Prove that $\left(2+3\sqrt{5}\right)$ is an irrational number, given that $\sqrt{5}$ is an irrational number.

Answer 1

Let us assume that $\left(2+3\sqrt{5}\right)$ is a rational number.

Thus, $\left(2+3\sqrt{5}\right)$ 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.

$$\begin{gathered} 2+3\sqrt{5}=\frac{p}{q} \\ \Rightarrow 3\sqrt{5}=\frac{p}{q}-2 \\ \Rightarrow 3\sqrt{5}=\frac{p-2q}{q} \\ \Rightarrow \sqrt{5}=\frac{p-2q}{3q} \\ \mathrm{Since},\frac{p-2q}{3q}\mathrm{is}\mathrm{rational}\Rightarrow \sqrt{5}\mathrm{is}\mathrm{rational} \\ \mathrm{But},\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\sqrt{5}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \\ \mathrm{Therefore},\mathrm{our}\mathrm{assumption}\mathrm{is}\mathrm{wrong}. \\ \mathrm{Hence},2+3\sqrt{5}\mathrm{is}\mathrm{an}\mathrm{irrational}\mathrm{number}. \end{gathered}$$