# Proof the indices

1. Apr 2, 2010

### DorumonSg

Let {x1, x2, x3, x4, x5} be distinct real numbers. Prove there are indices a, b with 0< xa-xb<1+xaxb.

Seriously I have no idea how to even start...

I tried subbing random numbers in... but nope...

Can anyone give a hint?

Hey wait, the sets do not need to be ordered right? Can I do this?

Direct proof:

{x1, x2, x3, x4, x5} = {-1, -2, -3, -4, -5}

proven.

Last edited: Apr 2, 2010
2. Apr 3, 2010

### sutupidmath

Try to use the fact that we can put an order on the set, say x1<x2<..<x5.

Last edited: Apr 3, 2010
3. Apr 3, 2010

### DorumonSg

Then I dun see any combination that can prove it...

4. Apr 3, 2010

### sutupidmath

You might want to start with xi-xj>0, for some i,j. You can make such a choice. I would be inclined to consider three cases separately: when the set consists of all positive real nr., all negative and a combination of both.

5. Apr 3, 2010

### DorumonSg

Erm... how is tat possible...

Example for xi - xj, if they are an ordered set...

For a positive case... a smaller number - a bigger number I will get negative... which is smaller than 0

For a negative case... a smaller number - a bigger number I will get negative... which is smaller than 0

if I mix them up... which I can't...because if u say I should do the sets in order the set would look sumting like... {-2, -1, 0, 1, 2} <- juz an example, its not the right anwer. and since i and j have to be consective elements in the set... halfway thru' the set they would unmixed... and besides, I still get a negative number...

6. Apr 3, 2010

### sutupidmath

The question is asking only two show that such indices exist, not to show that such a thing is true for all of them. So, given any two elements from your set, say x_i, x_j, since they are distinct, then precisely one of the following should hold: xi<xj or xj<xi.

None of them holds a special status, so we can choose either one, say the secon holds, then certainly xi-xj>0.

7. Apr 3, 2010

### DorumonSg

Okay... but I still don't see how it can be done if the set is ordered... like I alreadi explained above.

8. Apr 3, 2010

### sutupidmath

I am working under the assumption that R has the simple order (the natural order).

9. Apr 3, 2010

### DorumonSg

Yes I understand ur saying if u mean natural order as ...-1,-2,0,1,2,3,4...

But I dun see how the logic can be proven true using this order?

Do u have a set of number in mind tat can work? Coz' I can't see ani.

10. Apr 3, 2010

### sutupidmath

{1,2,3,4,5}={x1,x2,x3,x4,x5}

0<x3-x1=2<1+3*1=1+x3*x1

However, the point here is to show that this works for any such set.

11. Apr 3, 2010

### DorumonSg

hold on a sec!!

when it says indices, a, b.

I am pretty sure it doesn't meant any 2 elements in the set.

But 2 consecutive elements in the set?

Fine, I mistyped abit, they actually stated i, j instead of a, b. if i, j actualli means anything...

12. Apr 3, 2010

### Staff: Mentor

No, they don't have to be consecutive. Here is a revision of the problem description you posted at the beginning of this thread.

I interpret this to mean that the inequality doesn't have to hold for all choices of i and j, but it has to hold for at least one choice of i and j. It doesn't say anything about i and j being consecutive values.