Question

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

Answer 1

Let's assume 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 integers and b is not equal to 0.
Subtract 3 from both sides, we get,
$2\sqrt{5}=\frac{a}{b}-3$
$2\sqrt{5}=\frac{(a-3b)}{b}$
Now divide by 2; we get,
$\sqrt{5}=\frac{(a-3b)}{b}$
Since a and b are integers, $\frac{(a-3b)}{2b}$ is a rational number, then $\sqrt{5}$ is also a rational number as $\frac{(a-3b)}{2b}$ is equal to $\sqrt{5}$ .
But, we know that $\sqrt{5}$ is an irrational number, so it contradicts our assumption.
Hence, $3+2\sqrt{5}$ is an irrational number.