Question

Prove that : $4-5\sqrt{3}$ is irrational.

Answer 1

Let us assume that $4-5\sqrt{3}$ be rational

$⟹4-5\sqrt{3}=\frac{a}{b}$ where $a,bϵZ$

$⟹-5\sqrt{3}=\frac{a}{b}-4$

$⟹\sqrt{3}=\frac{4}{5}-\frac{a}{b}$

L.H.S = irrational number as $\sqrt{3}$ is a irrational number.

R.H.S = rational number

$asRHS\ne LHS$

Hence, our assumption is wrong and $4-5\sqrt{3}$ is a irrational number.