Question

Prove that is `sqrt2` irrational number.

Answer 1

Let us assume, to the contrary, that `sqrt2` is rational. So, we can find integers r and s (≠ 0) such that `sqrt2="r"/"s"`.

 Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get `sqrt2="a"/"b"` 

where a and b are coprime.

So,  b`sqrt2` = a.

Squaring on both sides and rearranging, we get 2b² = a².

Therefore, 2 divides a². Now, by Theorem it following that 2 divides a.

So, we can write a = 2c for some integer c.

Substituting for a, we get 2b² = 4c², that is, b² = 2c².

This means that 2 divides b², and so 2 divides b (again using Theorem with p = 2).

Therefore, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that `sqrt2` is rational.

So, we conclude that `sqrt2` is irrational.

Answer 2

Let `sqrt(2)` be rational.

∴ `sqrt(2)  = "p"/"q"` where p and q are co-prime integers and, q ≠ 0
Implies that `sqrt(2"q")  = "p"`

2q² - p²          ...(i) 

⇒ 2 divides p²

⇒ 2 divides p       ...(A)

Let p = 2C for some integer c

p² = 4c²

⇒ 2q² = 4C²

⇒ q² = 2c²

⇒ 2 divides q²

⇒ 2 divides q        ...(B)

From (A) and (B), we get 

∴ 2 is common factor of both p and q. But this contradicts the fact that p and q have no common factor other than 1.

∴  Our supposition is wrong Hence, `sqrt(2)` is an irrational number.