Prove that \(\frac{2}{\sqrt7}\) is an irrational number.

1.	Prove that \(\frac{2}{\sqrt7}\) is an irrational number.
Answer» Let’s assume on the contrary that \(\frac{2}{\sqrt7}\) is a rational number. Then, there exist co-prime positive integers a and b such that \(\frac{2}{\sqrt7}\) = \(\frac{a}{b}\) ⇒ √7 =\(\frac{2b}{a}\) ⇒ √7 is rational [∵ 2, a and b are integers ∴ \(\frac{2b}{a}\) is a rational number] This contradicts the fact that √7 is irrational. So, our assumption is incorrect. Hence, \(\frac{2}{\sqrt7}\) is an irrational number.

Answer»

Let’s assume on the contrary that \(\frac{2}{\sqrt7}\) is a rational number. Then, there exist co-prime positive integers a and b such that

\(\frac{2}{\sqrt7}\) = \(\frac{a}{b}\)

⇒ √7 =\(\frac{2b}{a}\)

⇒ √7 is rational [∵ 2, a and b are integers ∴ \(\frac{2b}{a}\) is a rational number]

This contradicts the fact that √7 is irrational. So, our assumption is incorrect.

Hence, \(\frac{2}{\sqrt7}\) is an irrational number.

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