Prove root 3 is irrational numberLet us assume to the contrary that √3 is a rational number.It can be expressed in the form of p/qwhere p and q are co-primes and q≠ 0.⇒ √3 = p/q⇒ 3 = p2/q2 (Squaring on both the sides)⇒ 3q2 = p2………………………………..(1)It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.So we have p = 3rwhere r is some integer.⇒ p2 = 9r2………………………………..(2)from equation (1) and (2)⇒ 3q2 = 9r2⇒ q2 = 3r2We have two cases to consider now.Case ISuppose that r is even. Then r2 is even, and 3r2 is even which implies that q2 is even and so q is even, but this cannot happen. If both q and r are even then gcd(q,r)≥2 which is a contradiction.Case IINow suppose that r is odd. Then r2 is odd and 3r2 is odd which implies that q2 is odd and so q is odd. Since both q and r are odd, we can write q=2m−1 and r=2n−1 for some m,n∈N.Thereforeq2=3r2(2m−1)2=3(2n−1)24m2−4m+1=3(4n2−4n+1)4m2−4m+1=12n2−12n+34m2−4m=12n2−12n+22m2−2m=6n2−6n+12(m2−m)=2(3n2−3n)+1We note that the lefthand side of this equation is even, while the righthand side of this equation is odd, which is a contradiction. Therefore there exists no rational number r such that r2=3.Hence the root of 3 is an irrational number.Hence Proved​