Given a real number a, define S≡{x|x in Q, x < a}. Prove that a = sup S using dedekind cuts.

Here is a summary of the proof I used:

If ∃ B|B is an upper bound of S and B < a

then (B+a)/2 is not in S.

But (B+a)/2 < a, so we have proof by contradiction.

But I don't know how dedekind cuts would be used in this, since I don't really understand what they are and they aren't used in my text book.

If anyone could point me in the right direction, I would appreciate it.