prove that root 3 is irrational number
Proof: Let us assume the contrary that root 3 is rational. Then √3 = p/q, where p, q are the integers i.e., p, q ∈ Z and co-primes, i.e., GCD (p,q) = 1. √3 = p/q ⇒ p = √3 q By squaring both sides, we get, p2 = 3q2 p2 / 3 = q2 ------- (1) (1) shows that 3 is a factor of p. (Since we know that by theorem, if a is a prime number and if a divides p2, then a divides p, where a is a positive integer) Here 3 is the prime number that divides p2, then 3 divides p and thus 3 is a factor of p. Since 3 is a factor of p, we can write p = 3c (where c is a constant). Substituting p = 3c in (1), we get, (3c)2 / 3 = q2 9c2/3 = q2 3c2 = q2 c2 = q2 /3 ------- (2) Hence 3 is a factor of q (from 2) Equation 1 shows 3 is a factor of p and Equation 2 shows that 3 is a factor of q. This is the contradiction to our assumption that p and q are co-primes. So, √3 is not a rational number. Therefore, the root of 3 is irrational.
