1. If x and y are arbitrary real numbers with x < y, prove that there is at least one real z satisfying x<z<y. 2. I'll be using this theorem: T 1.32 Let h be a given positive number and let S be a set of real numbers. (a) If S has a supremum, then for some x in S we have x > sup S - h. 3. The attempt at a solution 1. Let S be the set of all real x satisfying x < y for a given real y. 2. y = sup S. 3. z > y - h where z is in S. ( because of T 1.32 ) 4. Let h = y - x then z > x. 5. Because z is a member of S, then z must also be less than y. Therefore I have shown there exists a real number z, such that x < z < y. Is my logic sound? I am a little thrown off by the phrase, "If x and y are arbitrary reals.." I am not sure how to stick that construction into a set S, hence the wordiness of step 1. If so, how can I better formalize the proof, I would prefer some type of set/formal logic notation, at least for step 1. Thank you for your help.