Question

Show that 6 + √2 is an irrational number.

Answer 1

Let’s assume on the contrary that 6+√2 is a rational number. 

Then, there exist co prime positive integers a and b such that 

6 + √2 = \(\frac{a}{b}\) 

⇒ √2 = \(\frac{a}{b – 6}\) 

⇒ √2 = \(\frac{(a-6b)}{b }\)

⇒ √2 is rational [∵ a and b are integers ∴ \(\frac{(a-6b)}{b }\)is a rational number] 

This contradicts the fact that √2 is irrational. So, our assumption is incorrect. 

Hence, 6 + √2 is an irrational number.