Question

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

Answer 1

To prove ${\displaystyle \frac{2+\sqrt{3}}{5}}$ is irrational, let us assume that ${\displaystyle \frac{2+\sqrt{3}}{5}}$ is rational.

${\displaystyle \frac{2+\sqrt{3}}{5}=\frac{\text{a}}{\text{b}};\text{b}\ne 0}$ and a and b are integers.

${\displaystyle \Rightarrow 2\text{b}+\sqrt{3}\text{b}=5\text{a}}$

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

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

Since a and b are integers so, 5a - 2b will also be an integer.

So, ${\displaystyle \frac{5\text{a}-2\text{b}}{\text{b}}}$ will be rational which means ${\displaystyle \sqrt{3}}$ is also rational.

But we know ${\displaystyle \sqrt{3}}$ is irrational (given).

Thus, a contradiction has risen because of incorrect assumptions. 

Thus, ${\displaystyle \frac{2+\sqrt{3}}{5}}$ is irrational