Show that \(\frac{2}{\sqrt7}\) is irrational.

1.	Show that \(\frac{2}{\sqrt7}\) is irrational.
Answer» \(\frac{2}{\sqrt7}\) = \(\frac{2}{\sqrt7}\) x \(\frac{\sqrt7}{\sqrt7}\) = \(\frac{2}{7}\)\(\sqrt7\) Let \(\frac{2}{7}\)\(\sqrt7\) is a rational number. ∴ \(\frac{2}{7}\)\(\sqrt7\) = \(\frac{p}{q}\), where p and q are some integers and HCF(p,q) = 1 ….(1) ⇒ 2√7q = 7p ⇒(2√7q)² = (7p)² ⇒7(4q² ) = 49p² ⇒4q² = 7p² ⇒ q² is divisible by 7 ⇒ q is divisible by 7 …..(2) Let q = 7m, where m is some integer. ∴2√7q = 7p ⇒ [2√7 (7m)]² = (7p)² ⇒343(4m² ) = 49p² ⇒ 7(4m² ) = p² ⇒ p² is divisible by 7 ⇒ p is divisible by 7 ….(3) From (2) and (3), 7 is a common factor of both p and q, which contradicts (1). Hence, our assumption is wrong. Thus, \(\frac{2}{\sqrt7}\) is irrational.

Answer»

\(\frac{2}{\sqrt7}\) = \(\frac{2}{\sqrt7}\) x \(\frac{\sqrt7}{\sqrt7}\) = \(\frac{2}{7}\)\(\sqrt7\)

Let \(\frac{2}{7}\)\(\sqrt7\) is a rational number.

∴ \(\frac{2}{7}\)\(\sqrt7\) = \(\frac{p}{q}\), where p and q are some integers and HCF(p,q) = 1 ….(1)

⇒ 2√7q = 7p

⇒(2√7q)² = (7p)²

⇒7(4q² ) = 49p²

⇒4q² = 7p²

⇒ q² is divisible by 7

⇒ q is divisible by 7 …..(2)

Let q = 7m, where m is some integer.

∴2√7q = 7p

⇒ [2√7 (7m)]² = (7p)²

⇒343(4m² ) = 49p²

⇒ 7(4m² ) = p²

⇒ p² is divisible by 7

⇒ p is divisible by 7 ….(3)

From (2) and (3), 7 is a common factor of both p and q, which contradicts (1).

Hence, our assumption is wrong.

Thus, \(\frac{2}{\sqrt7}\) is irrational.

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