Is Z complete...I think so....i'm slow

No.12053061 ViewReplyOriginalReport
Let S be a subset of Z. By the axiom of completeness there exists an x in R s.t. x is an upper bound of S.

By the Archimedean principle there exists an n >x. Then n is an upper bound of S.

Do I need to show that S has a Sup in Z to show it is complete? Confused.