Question

Please help with parts a-d if possible. I am very confused.

14. Suppose that there exists integers a and b with no common divisors such that a2b2 (a) Explain why at most one of a and b is even. (b) Explain why a must be even. (c) Explain why part (b) implies that b must be even as well. (d) Explain the contradiction between parts (a), (b), and (c)

Answer 1

(a) Suppose a & b both are even then a, & b has a common divisor 2 but we are assuming that a & b has no common divisor . (b) a^2/b^2 = 2 = > a^2 = 2b^2 then a^2 has a factor 2 but if a is odd say a = 2k + 1 then a^2 = (2k 1)^2 = 4k^2 + 1 + 4k = 2(2k^2 + 2k) + 1 which is again odd thus a is even. (c) suppose a = 2k the a^2 = 4k^2 a^2 = 2b^2 = > 4k^2 = 2b^2 = > b^2 = 2k^2 thus by same logic as in (b) b is even as well (d) In part (a) we concluded that at most one of a & b could be even but in (b) & (c) we proved that a & b are even which are contradictions.