Question

prove that root 3 is an irrational number​

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.

Answer 2

Step-by-step explanation:

firstly we will assume that root 3 is a rational number and then prove that root 3 has common factor as 3 show so it can't be a rational number because rational number has only two factors one and itself that prove root 3 AS irrational number.

in the attachment I have taken 2 equations and from that two equations we got that P and Q have common factor as 3

now REFER TO ATTACHMENT