Question

Prove that $3+2\sqrt{5}$ is irrational. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
procedure : 2 Marks

Let us assume, to the contrary, that $3+2\sqrt{5}$ is rational.

That is, we can find co-prime integers a and b$(b\ne 0)$ such that $3+2\sqrt{5}=\frac{a}{b}$

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

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

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

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

Since a and b are integers, we get $\frac{a}{2b}-\frac{3}{2}$ is rational, and so $\frac{a-3b}{2b}=\sqrt{5}$ is rational.

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

$\Rightarrow L.H.S\ne R.H.S$

This contradiction has arisen because of our incorrect assumption that $3+2\sqrt{5}$ is rational.

So, we conclude that $3+2\sqrt{5}$ is irrational.