Question

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

Answer 1

Let , $2 \div \sqrt{3}$ +5 be a rational number

$\frac{2}{ \sqrt{3} } + 5 = \frac{p}{q} \\ \\ = > \frac{2}{ \sqrt{3} } = \frac{p - 5q}{q} \\ \\ = > \sqrt{3} = \frac{2q}{p - 5q}$

But

$\sqrt{3}$ is irrational and $\frac{2q}{p - 5q}$

therefore, $\sqrt{3}$ ≠ $\frac{2q}{p - 5q}$

therefore, $\frac{2}{ \sqrt{3} } + 5$ is an irrational number

Answer 2

hye!!!!gud afternoon

assume 2/root 3 +5 is rational

2/ roo3 +5 =p/q

root 3= 2q/p-5q

but not eqaul

hence, it is irrational