Question: Prove by showing where is the cut in with .
Let be Dedekind cut with
Note (this can be proved separately for cut ). Since , , . We show .
Case . Let . Since , by Archimedes principle, exists s.t. . Then, By density of , exists s.t. . So and thus . Since and , contradiction that is upper bound of .
Case . Let . Since , for some . By density of , exists s.t. so and hence for all and is thus upper bound smaller than supposed lowest upper bound , contradiction.
Hence so .