Question

Prove that $(2\sqrt{3}+\sqrt{5})$ is an irrational number. Also check whether $(2\sqrt{3}+\sqrt{5})(2\sqrt{3}-\sqrt{5})$ is rational or irrational

Answer 1

a) Let us assume that $2\sqrt{3}+\sqrt{5}$ is rational number.

Let $P=2\sqrt{3}+\sqrt{5}$ is rational

on squaring both sides we get

$P^2=(2\sqrt{3}+\sqrt{5}{)}^2=(2\sqrt{3}{)}^2+(\sqrt{5}{)}^2+2\times 2\sqrt{3}\times \sqrt{5}$

$P^2=12+5+4\sqrt{15}$

$P^2=17+4\sqrt{15}$

$\frac{P^2-17}{4}=\sqrt{15}$ ………..$\left(1\right)$

Since $P$ is rational no. therefore $P^2$ is also rational & $\frac{P^2-17}{4}$ is also rational.

But $\sqrt{15}$ is irrational & in equation$\left(1\right)$

$\frac{P^2-17}{4}=\sqrt{15}$

Rational $\ne$ irrational

Hence our assumption is incorrect & $2\sqrt{3}+\sqrt{5}$ is irrational number.

b) $P=(2\sqrt{3}+\sqrt{5})(2\sqrt{3}-\sqrt{5})$

$P=12-5=7$

Hence P is rational as $\frac{p}{q}=\frac{7}{1}$ & both $p$ & $q$ are coprime numbers.