Prove that 3/(2√5) is an irrational number.

Question

Prove that \(\frac{3}{(2√5)}\) is an irrational number.

Answer 1

Let’s assume on the contrary that \(\frac{3}{(2√5)}\) is a rational number. Then, there exist co – prime positive integers a and b such that 

\(\frac{3}{(2√5)}\)= \(\frac{a}{b}\)

⇒ √5 = \(\frac{3b}{2a}\)

⇒ √5 is rational [∵ 3, 2, a and b are integers ∴ \(\frac{3b}{2a}\) is a rational number] 

This contradicts the fact that √5 is irrational. So, our assumption is incorrect.

Hence, \(\frac{3}{(2√5)}\) is an irrational number.