- #1

mr_persistance

- 7

- 2

**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.

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.

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.

## 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.