Question

Prove that 3 + 2${\displaystyle \sqrt{5}}$ is irrational.

Answer 1

If possible let a = 3 + 2${\displaystyle \sqrt{5}}$ be a rational number.

We can find two co-prime integers a and b such that ${\displaystyle 3+2\sqrt{5}=\frac{a}{b},}$ where b ≠ 0

${\displaystyle \frac{a-3b}{b}}$

= ${\displaystyle 2\sqrt{5}}$

= ${\displaystyle \frac{a-3b}{2b}}$

= ${\displaystyle \sqrt{5}}$

∵ a and b are integers,

∴ ${\displaystyle \frac{a-3b}{2b}}$

= ${\displaystyle \frac{\text{Integer - 3 (interger)}}{\text{2 interger}}}$

= ${\displaystyle \frac{a-3b}{2b}}$ is rational.

= From (1), ${\displaystyle \sqrt{5}}$ is rational.

= But this contradicts the fact that ${\displaystyle \sqrt{5}}$ is rational.

∴ Our supposition is wrong.

Hence, ${\displaystyle 3+2\sqrt{5}}$ is irrational.