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**ScottO** If x and y are arbitrary real numbers with x < y, prove that there is at least one real z satisfying x < z < y.

At my disposal, the least-upper-bound axiom - every nonempty set S of reals bounded above has a supremum. And a handful of theorems including x > sup S - h where h > 0 and set S has a supremum.

Is my proof OK?

Let the set S be bounded above by y, so y = sup S. Let x $\displaystyle \in$ S and x + h = y = sup S. Then by the above theorem, there exists a z > sup S - h, therefore sup S = y > z > sup S - h = x. Q.E.D.

Thanks,

Scott