Prove that root 3 is irrational …

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q 

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side 

we get,

(√3q)² = p²

3q² = p² ........ ( i )
So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m² 

putting the value of p² in equation ( i )

3q² = p² 

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q
Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

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.