Question

Prove that `sqrt3` is an irrational number.

Answer 1

Let us assume, to the contrary, that is rational. That is, we can find integers a and b (≠ 0) such that `sqrt3="a"/"b"`.

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

So `bsqrt3` = a

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

Therefore, a² is divisible by 3, and by the 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 the theorem with p = 3).

Therefore, a and b have at least 3 as a common factor.

However, this contradicts the principle that a and b are coprime.

This contradicts the concept of coprimity between a and b.

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), we can see that 3 divides both p and q. This implies that p and q are not coprime, which contradicts our assumption.

So, our assumption is wrong.

Hence `sqrt3` is irrational.