Show that is irrational.

Answer:

is a rational number.

p and q are some integers and HCF(p,q) = 1

2√7q = 7p

(2√7q)² = (7p) ²

7(4q²) = 49p²

4q² = 7p²

q² is divisible by 7

q is divisible by 7

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

The assumption is wrong.

Hence, 2/√7 is irrational.