# Proving a property of real numbers

1. Jul 3, 2007

### neutrino

1. The problem statement, all variables and given/known data

Given x<y for some real numbers x and y. Prove that there is at least
one real z satisfying x<z<y

2. Relevant equations

This is an exercise from Apostol's Calculus Vol. 1. The usual laws of
algebra, inequalities, a brief discussion on supremum, infimum and
the Archimedean property preceeded this exercise.

3. The attempt at a solution

I'm trying to solve as many problems from the first chapter, which I had
just glossed over during a first read. Proving theorems in a rigorous
fashion is not exactly my forte. It becomes all the more difficult when
I'm asked to prove something "intuitively obvious."

I think I have the solution in fragments.

Let S be the set of all numbers greater than x. x is a lower bound for
this set, and therefore S has a greatest lower bound - inf S.

$$inf S \geq x$$

[The following argument assumes that x = inf S]

Consider some $$y \in S$$. I can always find at least one $$z \in S$$ such that z < y, because if there are no z's satisfying that inequality then y would be inf S.

The only problem here seems to be to prove that inf S = x. The first choice is a proof by contradiction.

Assume inf S > x...:uhh:

Am I on the right track?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 3, 2007

3. Jul 3, 2007

### neutrino

Yes, thank you...I really don't know why I was thinking about lower bounds and stuff.

Was my earlier "proof" correct? It will be complete when I prove x = inf S, right?

4. Jul 3, 2007

### Dick

The thing is that I think you might find this property useful for proving x=inf S.

5. Jul 3, 2007

### neutrino

I thought you were suggesting this...

x<(x+y)/2
y>(x+y)/2

Therefore x<(x+y)/2<y - this being a much shorter solution than what I had written down earlier. I'm not sure how (x+y)/2 relates to inf S.

6. Jul 3, 2007

### Dick

Well, x is a lower bound of S. Suppose there is another lower bound y and y>x. Knowing there is a z such that x<z<y gives you a problem since then z is in S. Trying to prove inf S=x leads you to prove things that are an awful lot like what you are trying to prove to begin with.

7. Jul 3, 2007

### neutrino

Ah...
Assume inf S > x. This implies that inf S belongs to S.

But then x<(inf S + x)/2 < inf S and it also belongs to S, which would imply inf S is not the infimum. A contradiction!

But trying to prove that x = inf S this way seems redundant since it relies on the fact that for all real x and y, x<(x+y)/2<y, which was what I set to prove in the first place.

And that's what I think you tried to convey in your last post.

8. Jul 3, 2007

### Dick

Yes. If there were a gap in the real numbers between x and y, then inf S would be y. So you want to show there are no gaps before you prove inf S=x.

9. Jul 3, 2007

### neutrino

Got it now. Thank you.

10. Jul 4, 2007

### HallsofIvy

Staff Emeritus
?? No, it isn't. If y is a lower bounds on S, x< y, then x is NOT in s and neither is the z above.

In fact, "infimum" cannot have anything to do with this property: If x and y are rational numbers then there exist a rational number between them (again (x+y)/2 is a rational number) but bounded sets of rational numbers do not necessarily have rational infimum.

Last edited: Jul 4, 2007
11. Jul 4, 2007

### Dick

I didn't say x in is S. S is defined to be the set of all y such that y>x. I find it difficult to accept that z is not in S if x<z<y.