Question

Prove that ($5-2\sqrt{3}$) is an irrational number.

Answer 1

​​Let x = $5-2\sqrt{3}$ be a rational number.

$$\begin{gathered} x=5-2\sqrt{3} \\ \Rightarrow x^2={\left(5-2\sqrt{3}\right)}^2 \\ \Rightarrow x^2={\left(5\right)}^2+{\left(2\sqrt{3}\right)}^2-2\left(5\right)\left(2\sqrt{3}\right) \\ \Rightarrow x^2=25+12-20\sqrt{3} \\ \Rightarrow x^2-37=-20\sqrt{3} \\ \Rightarrow \frac{37-x^2}{20}=\sqrt{3} \end{gathered}$$
Since x is a rational number, x² is also a rational number.
⇒ 37 − x² is a rational number
⇒ $\frac{37-x^2}{20}$ is a rational number
⇒ $\sqrt{3}$ is a rational number
But $\sqrt{3}$ is an irrational number, which is a contradiction.
Hence, our assumption is wrong.

Thus, ($5-2\sqrt{3}$) is an irrational number.