Let m N, and define m : Z Z/mZ by m(a) = a mZ (so m sends an integer to its congruence class modulo m). (a) Why is m surjective? (Hint: it is easy to overthink this, but there is nothing fancy involved.) (b) Does there exist an integer m N for which m is injective? Justify your answer. (c) Let R be a subset of Z having m elements such that, if r1, r2 R are distinct, then r1 r2 (mod m) (we say that the elements of R are mutually incongruent modulo m). The restriction of m to R, denoted m|R, is the function from R to Z/mZ given by the same formula as m, i.e., m|R(r) = r mZ for each r R. Prove that m|R is bijective, and deduce that R is a complete set of representatives for Z/mZ. (Hint: You may use the fact that, if S and T are finite sets having the same number of elements, then an injective function f : S T is necessarily surjective. How does bijectivity of m|R imply that R is a complete set of representatives?)