Show that if S is a subring with identity of the real numbers then S contains Z (integers).?

Answer 1

Since S is a subring with identity, 1 is in S. Since S is a subring, the additive identity is in S, so 0 is in S. Since S is closed under addition, so 1 + 1 = 2 is in S. Continue by induction, and prove S contains all the positive integers. Finally, since S is a subring, it contains the additive inverses of the positive integers, so all the negative integers are is S. Thus, S contains Z.