Let A be a nonempty bounded above subset of Z. Prove that A has a largest element.
To prove A has a largest element does it mean to prove A is the least upper bound of Z, if so how do I do this?
Again let $\displaystyle \alpha=\sup(A)$.
If $\displaystyle \alpha\in\mathbb{Z}$ we are done.
If $\displaystyle \alpha\notin\mathbb{Z}$ then $\displaystyle \left\lfloor \alpha \right\rfloor < \alpha $.
That means that $\displaystyle \left( {\exists n \in A} \right)\left[ {\left\lfloor \alpha \right\rfloor < n < \alpha } \right]$.
That is a contradiction. WHY?
it's a contradction because it's bounded above, i.e there is an upper bound for A, A will converge to some number eventually, however, we don't need to show A has a least upper bound here, we just need to show that there is an element $\displaystyle S_0$. in A such that $\displaystyle S<=S_0$. for all S in A.
but why is this has to be in Z though?
That you understand is definitely arguable (and may indeed be argued by those already in the discussion).
Now, it appears that you are introducing a new question.
Consider the intervals [0, 1) and [0, 1] as subsets of ℝ
Both are bounded above. Both have the same least upper-bound (supremum).
One of these intervals, however, does not have a largest element. Indeed - to prove this, you could take and arbitary element and find one that is larger, and still in the interval.