**Question change √5 is irrational.**

Assume contrary that √5 is a rational number.

Then √5 = a/b, where a and b are co-prime and b ≠ 0.

∴ HCF (a, b) = 1

⇒ a = √5 b

**Now, a**^{2} = 5b^{2} (By squaring both sides)

⇒ 5 divides a^{2}

⇒ 5 divides a (∵ If any prime number divides a^{2} then that prime number must divide a)

⇒ a = 5m

⇒ a^{2} = 25 m^{2}

⇒ 25 m^{2} = 5b^{2} (∵ a^{2} = 5b^{2})

⇒ b^{2 }= 5 m^{2}

⇒ 5 divides b^{2}

⇒ 5 divides b

Since, 5 divides both a and b.

Therefore, HCF (a, b) ≠ 1, it is a multiple of 5.

Which is contradicts the fact that a and b are co-prime numbers.

Therefore, our assumption is wrong.

**i.e., √5 is an irrational number.**