Prove that is `sqrt3` irrational number.

#### Solution 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 3b^{2} = a^{2}.

Therefore, a^{2} 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 3b^{2} = 9c^{2}, that is, b^{2} = 3c^{2}.

This means that b^{2} 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.

#### Solution 2

Let `sqrt3` is rational

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

⇒ `sqrt3"q" = "p"`

⇒ 3q^{2} = p^{2}..........(1)

⇒ 3 divides p^{2 }

⇒ 3 divides p .....(A)

Let p = 3c where c is an integer

⇒ p^{2 } = 9c^{2}

⇒ 3q^{2} = 9c^{2} [from (1)]

⇒ q^{2} = 3c^{2}

⇒ 3 divides q^{2}

⇒ 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.