Prove root 3 is an irrational

let root 3 is rational number. root 3=p/q [where,p and q are integers,q is not equal to 0 and p and q are co-primes]. root 3q=p squaring both sides, 3q2 =p2...............[1] 3 divides p2 3 divides p.............[2] p2=3q2 p2=3m Put the value of p2 in eqn [1]. p2=3q2 [3m]2=3q2 9m2=3q2 3m2=q2 q2=3m2 3 divides q2 3 divides q [3] By eqn [2] and [3].... 3 is the common factor of both p and q . This is contradiction that and q are co-primes. Our assumption is wrong. Hence , root 3 is irrational number.

 let us assume that {tex}\sqrt 3{/tex} be a rational number.

{tex}\sqrt { 3 } = \frac { a } { b }{/tex}, where a and b are integers and co-primes and b{tex} \neq{/tex}0
Squaring both sides, we have
{tex}\frac { a ^ { 2 } } { b ^ { 2 } } = 3{/tex}
or, {tex}a ^ { 2 } = 3 b ^ { 2 }{/tex}--------(i)
a² is divisible by 3.
Hence a is divisible by 3..........(ii)
Let a = 3c ( where c is any integer)

squaring on both sides we get
(3c)² = 3b²
9c² = 3b²
b² = 3c²
so b² is divisible by 3
hence, b is divisible by 3..........(iii)
From equation(ii) and (iii), we have
3 is a factor of a and b which is contradicting the fact that a and b are co-primes.
Thus, our assumption that {tex}\sqrt 3{/tex} is rational number is wrong.
Hence, {tex}\sqrt 3{/tex} is an irrational number.