Question

Prove that √5 is an irrational number.

Answer 1

Let us assume to the contrary, that √5 is rational. Then we can find a and b ( ≠ 0) such that √5 = a/b (assuming that a and b are co-primes). 

So, a = √5 b ⇒ a² = 5b² 

Here 5 is a prime number that divides a² then 5 divides a also 

(Using the theorem, if a is a prime number and if a divides p² , then a divides p, where a is a positive integer) 

Thus 5 is a factor of a 

Since 5 is a factor of a, we can write a = 5c (where c is a constant). Substituting a = 5c 

We get (5c)² = 5b² ⇒ 5c² = b² 

This means 5 divides b² so 5 divides b also (Using the theorem, 

if a is a prime number and if a divides p² , then a divides p, where a is a positive integer). 

Hence a and b have at least 5 as a common factor. 

But this contradicts the fact that a and b are coprime. This is the contradiction to our assumption that p and q are co-primes. 

So, √5 is not a rational number. Therefore, the √5 is irrational.