Question

Prove that for any prime positive integer p, √p is an irrational number.

Answer 1

Let’s assume on the contrary that √p is a rational number. Then, there exist positive integers a and b such that 

√p = \(\frac{a}{b}\) where, a and b, are co-primes 

⇒ (√p)² = (\(\frac{a}{b}\))² 

⇒ p = \(\frac{a^2}{b^2}\)

⇒ pb² = a² 

⇒ p | a² [∵ p|pb² and pb² = a²] 

⇒ p | a ………… (ii) 

⇒ a = pc for some integer c. 

Now, b²p = a² 

⇒ b²p = p²c² 

⇒ b² = pc² 

⇒ p | b² [∵ p|pc² and pc² = b²] 

⇒ p | b ……… (i) 

From (i) and (ii), we can infer that p is a common factor of a and b. But, this contradicts the fact that a and b are co-primes. So, our assumption is incorrect. 

Hence, √p is an irrational number.