Prove that √3 is irrational. (CBSE 2008, 2009)

Let us assume that √3 is rational.

Hence, √3 can be written in the form

where a and b (≠ 0) are co-prime (no common factor other than 1).

Hence, √3 =

So, b√3 = a.

Squaring on both sides, we get

3b² = a²

= b²

Hence, 3 divides a².

By theorem: Let p is a prime number and p divides a² , then p divides a, where a is a positive integer,

3 divides a also …(1)

Hence, we can say a = 3c for some integer c

Now, we know that 3b² = a²

Putting a = 3c

3b² = (3c)²

3b² = 9c²

b² = 3c²

Hence, 3 divides b²

By theorem: Let p is a prime number and p divides a², then p divides a, where a is a positive integer,

So, 3 divides b also …(2)

By (1) and (2)

3 divides both a and b

Hence, 3 is a factor of a and b

So, a and b have a factor 3

Therefore, a and b are not co-prime.

Hence, our assumption is wrong

Therefore, by contradiction √3 is irrational.