Question

Can anyone prove (by contradiction) that z+ is not bounded above?

Answer 1

Assume, on the contrary, that
there exists an integer m that is the upper bound of Z+.
Is m+1 >m? yes.
Is m+1 in Z+? yes.
Contradiction since it was assumed that m was upper bound/max
therefore assumption that m was an upper bound fails
therefore there is no upper bound in Z+

Answer 2

Hey, to be fair, the upper bound is not necessarily an integer. Assume there is an upper bound m for Z+
Then we use the ceiling function:
Let
\[\huge u = \lceil m \rceil\]\[\huge u \ge m\]\[\huge u+1>m\]\[\huge u+1 \in \ Z^{+}\]
contradiction