Question

Prove that ($4-5\sqrt{2}$) is an irrational number.

Answer 1

​Let x = $4-5\sqrt{2}$ be a rational number.

$$\begin{gathered} x=4-5\sqrt{2} \\ \Rightarrow x^2={\left(4-5\sqrt{2}\right)}^2 \\ \Rightarrow x^2={\left(4\right)}^2+{\left(5\sqrt{2}\right)}^2-2\left(4\right)\left(5\sqrt{2}\right) \\ \Rightarrow x^2=16+50-40\sqrt{2} \\ \Rightarrow x^2-66=-40\sqrt{2} \\ \Rightarrow \frac{66-x^2}{40}=\sqrt{2} \end{gathered}$$
Since x is a rational number, x² is also a rational number.
⇒ 66 − x² is a rational number
⇒ $\frac{66-x^2}{40}$ is a rational number
⇒ $\sqrt{2}$ is a rational number
But $\sqrt{2}$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.

Thus, ($4-5\sqrt{2}$) is an irrational number.