Question

Prove that $3+2\sqrt{5}$ is an irrational number.

Answer 1

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

So we can write this number as $3+2\sqrt{5}$ = $\frac{a}{b}$

Here $a$ and $b$ are two co prime number and $b$ is not equal to $0$

Subtract $3$ both sides we get

$2\sqrt{5}$ = $\frac{a}{b}-3$

$2\sqrt{5}$ = $\frac{a-3b}{b}$

Now divide by $2$, we get

$\sqrt{5}$ = $\frac{a-3b}{2b}$

Here a and b are integer so $\frac{a-3b}{2b}$ is a rational number so

$\sqrt{5}$ should be a rational number but $\sqrt{5}$ irrational

number so it contradicts.

Hence, $3+2\sqrt{5}$ is a irrational number.