Prove by contradiction that 2^(1/3) is an irrational number

Assume 2^(1/3) is rational, so can be written as p/q where p and q are integers with no common factors. p/q = 2^(1/3) (p^3)/(q^3) = 2 p^3 = 2q^3 Hence, p is even. Thus, p can be written as 2r, where r is an integer. p^3 = (2r)^3 = 2q^3 8r^3 = 2q^3 4r^3 = q^3 Hence, q is even. Therefore, p and q have common factor 2, which is a contradiction.