Question

prove that $3+\sqrt{5}$ is an irrational number.

Answer 1

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

Now,

$3+\sqrt{5}=\frac{a}{b}$ [Here a and b are co-prime numbers]

$\sqrt{5}=[\left(\frac{a}{b}\right)-3]$

$\sqrt{5}=\left[\left(\frac{a-3b}{b}\right)\right]$

Here, $\left[\left(\frac{a-3b}{b}\right)\right]$ is a rational number.

But we know that $\sqrt{5}$ is an irrational number.

So, $\left[\left(\frac{a-3b}{b}\right)\right]$ is also a irrational number.

So, our assumption is wrong.

$3+\sqrt{5}$ is an irrational number.

Hence, proved.