Question

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

Answer 1

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

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

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