Question

Prove that √3 is an irrational number

Answer 1

Let 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.

Answer 2

suppose 3–√3 is rational, then 3–√=ab3=ab for some (a,b)(a,b) suppose we have a/ba/b in simplest form.

3–√a2=ab=3b23=aba2=3b2

if b is even, then a is also even in which case a/b is not in simplest form.
if b is odd then a is also odd. Therefore:

ab(2n+1)24n2+4n+12n2+2n2(n2+n)=2n+1=2m+1=3(2m+1)2=12m2+12m+3=6m2+6m+1=2(3m2+3m)+1a=2n+1b=2m+1(2n+1)2=3(2m+1)24n2+4n+1=12m2+12m+32n2+2n=6m2+6m+12(n2+n)=2(3m2+3m)+1

Since (n^2+n) is an integer, the left hand side is even. Since (3m^2+3m) is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.