Question

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

Answer 1

To prove : $\sqrt{3}+\sqrt{5}$ is irrational.

Let us assume it to be a rational number.

Rational numbers are the ones that can be expressed in $\frac{p}{q}$ form where $p,q$ are integers and $q$ isn't equal to zero.

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

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

squaring on both sides,

$3=\frac{p^2}{q^2}-2.\sqrt{5}\left(\frac{p}{q}\right)+5$

$\Rightarrow \frac{\left(2\sqrt{5}p\right)}{q}=5-3+\left(\frac{p^2}{q^2}\right)$

$\Rightarrow \frac{\left(2\sqrt{5}p\right)}{q}=\frac{2q^2-p^2}{q^2}$

$\Rightarrow \sqrt{5}=\frac{2q^2-p^2}{q^2}.\frac{q}{2p}$

$\Rightarrow \sqrt{5}=\frac{(2q^2-p^2)}{2pq}$

As $p$ and $q$ are integers RHS is also rational.

As RHS is rational LHS is also rational i.e $\sqrt{5}$ is rational.

But this contradicts the fact that$\sqrt{5}$ is irrational.

This contradiction arose because of our false assumption.

so,$\sqrt{3}+\sqrt{5}$ irrational.