Let’s assume that √5 is a rational number.

Hence, √5 can be written in the form a/b [where a and b (b ≠ 0) are co-prime (i.e. no common factor other than 1)]

⸫ √5 = a/b

⇒ √5 b = a

Squaring both sides,

⇒ (√5 b)² = a²

⇒ 5b² = a²

⇒ a²/5 = b²

Hence, 5 divides a²

By theorem, if p is a prime number and p divides a², then p divides a, where a is a positive number

So, 5 divides a too

Hence, we can say a/5 = c where, c is some integer

So, a = 5c

Now we know that,

5b² = a²

Putting a = 5c,

⇒ 5b² = (5c)²

⇒ 5b² = 25c²

⇒ b² = 5c²

⸫ b²/5 = c²

Hence, 5 divides b²

By theorem, if p is a prime number and p divides a², then p divides a, where a is a positive number

So, 5 divides b too

By earlier deductions, 5 divides both a and b

Hence, 5 is a factor of a and b

⸫ a and b are not co-prime.

Hence, the assumption is wrong.

⸫ By contradiction,

⸫ √5 is irrational