Question

show that root 3 is irrational

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

let us assume that root 3 is rational
therefore it can be written in the fprm p/q where p and q are integers
so
root 3=p/q
we will square on both the sides and get
3=p^2/q^2
now by cross multiply 3q^2 will be equal to p^2
since q^2 is divided by 3 q is also divided
and p and q has a common factor 3 now but it opposes the fact that p and q are co primes

our assumption is wrong and root 3 is irrational