Solution:
Let us assume, to the contrary, that √5 is rational.
So, we can find coprime integers a and b(b ≠ 0)
such that √5 = a/b
=> √5b = a
=> 5b2 = a2 ….(i) [Squaring both the sides]
=> a2 is divisible by 5
=> a is divisible by 5
So, we can write a = 5c for some integer c.
=> a2 = 25c2 ….[Squaring both the sides]
=> 5b2 = 25c2 ….[From (i)]
=> b2 = 5c2
=> b2 is divisible by 5
=> b is divisible by 5
=> 5 divides both a and b.
=> a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction arises because we have assumed that √5 is rational.
Hence, √5 is irrational.
