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

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

Answer»

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.

Discussion