Question

prove that root 3 is a irrational no.

Answer 1

Say 3–√3 is rational. Then 3–√3 can be represented as abab, where a and b have no common factors.

So 3=a2b23=a2b2 and 3b2=a23b2=a2. Now a2a2 must be divisible by 33, but then so must aa (fundamental theorem of arithmetic). So we have 3b2=(3k)23b2=(3k)2 and 3b2=9k23b2=9k2 or even b2=3k2b2=3k2 and now we have a contradiction.

Answer 2

 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.
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