Question

Prove that √3 is irrational and hence solve that 2-√3 is irrational

Answer 1

Sol:
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.

2-√3=a/b
-√3=a/b-2
-√3=a-2b/b
√3=-(a-2b/b)
here -(a-2b/b) is rational where as √3 is irrational so 2-√3 is irrational
please mark it as a brainlist answer
please mark it as a brainlist answer
please mark it as a brainlist answer