Question

Prove that $\surd 3+\surd 5$ is irrational.

Answer 1

$\surd 3+\surd 5$ is an irrational number.

Let us assume that $\surd 3+\surd 5$ is a rational number.

So it can be written in the form $\frac{a}{b}$

$\surd 3+\surd 5=\frac{a}{b}$

Here $a$ and $b$ are coprime numbers and $b\ne 0$

$\surd 3+\surd 5=\frac{a}{b}$

On squaring both sides we get,

${\left(\surd 3+\surd 5\right)}^2={\left(\frac{a}{b}\right)}^2$

${\left(\surd 3\right)}^{²}+{\left(\surd 5\right)}^{²}+2(\surd 5)(\surd 3)=\frac{a^2}{b^2}$

$3+5+2\surd 15=\frac{a^2}{b^2}$

$8+2\surd 15=\frac{a^2}{b^2}$

$2\surd 15=\frac{a^2}{b^2}-8$

$\surd 15=\frac{a^2-8b^2}{2b^2}$

$a,b$ are integers then $\frac{a^2-8b^2}{2b^2}$ is a rational number.

Then $\surd 15$ also a rational number.

But this contradicts the fact that $\surd 15$ is an irrational number.

Our assumption is incorrect

$\surd 3+\surd 5$ is an irrational number.

Hence, it is proved that $\surd 3+\surd 5$ is irrational.