# Simple analysis proofs

1. Dec 21, 2007

### ice109

given x<y and x,y,z are elements of R prove there exists at least one z such that x<z<y.

proof:

x<z<y -> z>x and y>z

by the fact that the reals are unbounded there is definitely at least one z such that z>x

now either z>y,z<y, or z=y by the order axioms.

so... do i just let z<y and then call it a day?

2. Dec 21, 2007

### gimmickposter

No, because you are asserting that such a z < y exists without any proof (i.e. that the set of reals greater than x and the set of reals less than y share some element). You need to establish this fact first.

Hint: it's easier to construct a number between the two and show that it's a real than it is to look at the set of all reals and show that there's a number between the two.

Big hint: What happens if you average x and y? Can you show this number is between the two? And a real?

3. Dec 21, 2007

### ice109

let's say i don't want to take your hint and i want to work with the entire set of reals. the only way the set of reals greater than x and less than y don't share an element is if x>y which is contradictory to give x<y. is that right?

edit

how do i "establish" the fact that set <y and >x share elements?

Last edited: Dec 21, 2007
4. Dec 22, 2007

### HallsofIvy

What do you mean by "share an element"? Are you assuming the Dedekind Cut definition of the reals? If you are then what definition of "x< y" are you using?

5. Dec 22, 2007

### ice109

you're killing me halls i was just parroting gimmickposter. basically how do i prove that there are infinite z in between x and y cause i know there are.

6. Dec 22, 2007

### LukeD

Ice109: What you are saying about the sets of numbers greater than x and less than y not sharing a number being true only if x<y is true, but not obvious from the axioms of the real numbers.

What HallsofIvy meant is that there are certain ways of constructing the real numbers from other sets. In the Dedekind Cuts construction, what you have said is in fact obvious by the way the real numbers are defined. However, the Dedekind Cuts construction (and the obviousness of your statement) depend on the property being true for the rational numbers.

Really, it's a lot easier to prove if you just assume that x<y and combine them in some way to construct a number that is between the two.
Hint: Let's say that x=1 and y=2. How would you find a number between them? Now just adopt this method to any x and y

7. Dec 22, 2007

### neutrino

There's a thread here, related to this very problem from Apostol. I'll post the link if you desperately need it. (But the answer has already been provided, more or less.)

8. Dec 22, 2007

the problem is definitely that you are thinking too hard about the problem. You need to write down a candidate for a number that should be between x and y, and then prove it really is.

Hint: Start with the case x=0. Then try shifting this example over by x.

9. Dec 22, 2007

### HallsofIvy

Why did you say you didn't want to take gimmickposter's hint? It is certainly easy to show that if x and y are any two real numbers, then (x+y)/2 is a real number between them.

10. Dec 24, 2007

### ice109

i didn't want to take gimmickposter's hint because the point of doing the problem for me is not getting the right answer but learning how to prove. since he showed me how to prove that there exists at least one i would like to solve the harder problem that there exists infinite.

11. Dec 24, 2007

### LukeD

Well, you already know that there exists at least one between each pair of real numbers. From this it is fairly easy to show that there is at least a countably infinite amount between any two real numbers. (Think about what would happen if there were not)

If you want to show that there is an uncountably infinite amount between any 2 real numbers, you can either prove that there are 2 different rational numbers between any 2 real numbers and then try to repeat Cantor's diagonal argument with your real numbers bounded between those 2 rational numbers or you can make an injection from the real numbers to (a,b). Of course, the second idea doesn't follow straight from the axioms.