Question

Prove that is `sqrt3` irrational number.

Answer 1

Let us assume, to contrary, that  is rational. That is, we can find integers a and b (≠0) such that `sqrt2=a/b`

Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b`sqrt3`= a

Squaring on both sides, and rearranging, we get 3b² = a².

Therefore, a² is divisible by 3, and by Theorem, it follows that a is also divisible by 3.

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

Substituting for a, we get 3b² = 9c², that is, b² = 3c².

This means that b² is divisible by 3, and so b is also divisible by 3 (using Theorem with p = 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 contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that `sqrt3` is rational.

So, we conclude that `sqrt3` is irrational.

Answer 2

Let `sqrt3` is rational

∴`sqrt3 = "p"/"q"` where p and q are co-prime integers and q ≠ 0.

⇒ `sqrt3"q" = "p"`

⇒ 3q² = p²..........(1)

⇒ 3 divides p² 

⇒ 3 divides p   .....(A)

Let p = 3c where c is an integer

⇒  p²  = 9c²
⇒ 3q² = 9c²   [from (1)]
⇒ q² = 3c²
⇒ 3 divides q² 
⇒ 3 divides q   .....(B)

From statements (A) and (B), 3 divides p and q both that means p and q are not co-prime which contradicts our assumption.
So, our assumption is wrong.
Hence `sqrt3` is irrational.