Question

Show that³√6 is irrational number.

Answer 1

Assume cube root 6 is rational. Then let cube root 6 = a/b ( a & b are co-prime and b not = 0) Cubing both sides : 6=a^3/b^3 a^3 = 6b^3 a^3 = 2(3b^3) Therefore, 2 divides a^3 or a^2 * a . By Euclid's Lemma if a prime number divides the product of two integers then it must divide one of the two integers. Since all the terms here are the same we conclude that 2 divides a. Now there exists an integer k such that a=2k Substituting 2k in the above equation 8k^3 = 6b^3 b^3 = 2{(2k^3) / 3)} Therefore, 2 divides b^3. Using the same logic as above. 2 divides b. Hence 2 is common factor of both a & b. But this is a contradiction of the fact that a & b are co-prime. Therefore, the initial assumption is wrong. cube root 6 is irrational