Let us assume to the contrary, that √5 is rational.

That is we can find integers a and b(≠0) such that √5=(a/b)

Suppose a and b have a common factor other than 1, then we can divide by the common factor and assume that a and b are co-prime.

Therefore b√5 = a

Sqaring on both sides 5b² = a².......(1)

The above implies that a² is divisible by 5 and also a is divisible by 5.

Therefore we can write that a =5c for some interger c.
Substituting in (1)

5b² = (5f)²

5b² = 25f² 

b² = 5f²

b² is divisible by 5 which means b is also divisible by 5.

Therefore a and b have 5 as a common factor.

This contradicts the fact that a and b are co prime.

We arrivied at the contradictory statement as above since our √5 is rational assumption 

is not correct. Hence we can conclude that √5 is irrational