Prove that√3 is irrational

Let us assume that √3 is a rational number.

That is, we can find integers a and b (≠ 0) such that √3 = (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 coprime.

√3b = a

⇒ 3b²=a² (Squaring on both sides) → (1)

Therefore, a² is divisible by 3

Hence ‘a’ is also divisible by 3.

So, we can write a = 3c for some integer c.

Equation (1) becomes,

3b² =(3c)²

⇒ 3b² = 9c²

∴ b² = 3c²

This means that b² is divisible by 3, and so b is also divisible by 3.

Therefore, a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that √3 is rational.
So, we conclude that √3 is irrational.

Now
√2 = 1.4142...
√3 = 1.7321...

1.45, 1.5, 1.55, 1.6, 1.65, 1.7 lies between √2 and √3.

Hence the rational numbers between √2 and √3 are:

145/100, 15/10, 155/100, 16/10, 165/100 and 17/10

Let root3 is a rational no and let its simplest form a/b ,but b is not equal to 0 Root 3 =a/b => 3=a^2/b^2 (sq.both sides) => 3b=a^2 (3 divides 3 b^2) 3 divides a Let a=3c for c is some integer 3b^2 =9c^2 => b^2 = 3c^2 => 3 divides b^2 So,root 3 is an irrational. PROVED