Question

prove that 2-3√5 is an irrational number​

Answer 1

To prove: $2 - 3\sqrt{5}$ is an irrational number.

Proof: Let, to the contrary, $2 - 3\sqrt{5}$ be a rational number.

Then, $2 - 3\sqrt{5} = \frac{a}{b}$ where a and b are co - prime integers (HCF = 1).

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

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

$- 3\sqrt{5} = \frac{a - 2b}{b}$ (cross - Multiplying).

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

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

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

$\sqrt{5} = \frac{\cancel{-}(-a + 2b)}{\cancel{-}(3b)}$

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

Here, a and b are integers as assumed above. So, $\frac{-a + 2b}{3b}$ is a rational number. So that, $\sqrt{5}$ is also a rational number.

But $\sqrt{5}$ is an irrational number. So, our assumption is wrong.

This contradiction arises due to wrong assumption.

Therefore, the above said statement is true i.e., $2 - 3\sqrt{5}$ is an irrational number.

Hence, the proof.