# Given any real numbers a and b such that a<b, prove that for any natural number n

1. Dec 3, 2011

### snes_nerd

Given any real numbers a and b such that a < b, prove that for any natural number n, there are real numbers x1, x2, x3, ... , xn such that a < x1 < x2 < x3 < ... < xn < b.

The hint I was given says : Define xi recursively by x1 = (a+b)/2 and x(i+1) = (xi +b)/2. Prove that xi < xi + 1 < b, and use this result to prove by induction that a < x1 < x2 < x3 < ... < xn < b for any n in the natural numbers.

Okay so the hint doesnt really help me at all but I think I have an idea of what the question is asking. Lets suppose a = 1 and b= 10. Then there are natural numbers between these such that 1 < .... < 10. So going back to the original proposition I could say that there are real numbers x1, x2, x3, ..., xn such that a < x1 < x2 < x3 < .. < xn < x(n+1) < b. Not really sure where to go from their.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 3, 2011

### Office_Shredder

Staff Emeritus
Re: Given any real numbers a and b such that a<b, prove that for any natural number n

The hint is a fairly explicit statement of how to solve the problem to be honest. It's worth going over it and identifying what is confusing you about it

3. Dec 3, 2011

### snes_nerd

Re: Given any real numbers a and b such that a<b, prove that for any natural number n

I am sure it is very straightforward once I actually know what it means. In fact, its probably really easy. I feel kind of dumb not seeing it, and I am sure once I see it, I will feel even more dumb.

Maybe its because I dont understand why x1 = (a+b)/2 and (xi+b)/2. (a+b)/2 tells me a number between a and b, and (xi+b)/2 tells me another number between a and b thats greater then x1. Cant really connect this to the proof

4. Dec 3, 2011

### Office_Shredder

Staff Emeritus
Re: Given any real numbers a and b such that a<b, prove that for any natural number n

So if x1=(a+b)/2 and x2=(x1_b)/2 we have:

a<x1<x2<b

and if x3=(x2+b)/2 what do we have?

5. Dec 3, 2011

### SammyS

Staff Emeritus
Re: Given any real numbers a and b such that a<b, prove that for any natural number n

For your example in which a=0 and b= 10: What are the values you get for x1, x2, x3, x4, x5, etc. ?

6. Dec 3, 2011

### snes_nerd

Re: Given any real numbers a and b such that a<b, prove that for any natural number n

So x3=(x2+b)/2 implies that a < x1 < x2 < x3 < b?.

7. Dec 4, 2011

### SammyS

Staff Emeritus
Re: Given any real numbers a and b such that a<b, prove that for any natural number n

Be more specific.

Using the hint, and a=0, b=10:
x1 = (a+b)/2 = 5

x2 = (x1+b)/2 = 7.5

x3 = (x2+b)/2 = 8.75

...​

8. Dec 4, 2011

### HallsofIvy

Re: Given any real numbers a and b such that a<b, prove that for any natural number n

If a< b then a+ b< 2b so (a+ b)/2< b.

If a< b then 2a< a+ b so a< (a+ b)/2.

a< (a+ b)/2< b.

Yes, that's the whole point of an "average"- it lies between the numbers.

9. Dec 4, 2011

### snes_nerd

Re: Given any real numbers a and b such that a<b, prove that for any natural number n

Proof

If a < b, then a + b < 2b (axiom or reasoning I can assume this?)

Then with simple algebra, (a+b)/2 < b (definition of multiplicative inverse)

If a < b, then 2a < a + b (axiom or reasoning I can assume this?)

Then a < (a+b)/2 ( definition of multiplicative inverse

Thus, a < (a+b)/2 < b

Okay that makes sense but this problem has to be solved with induction. And Sammy 0 and 10 is not in the actual problem. I was just throwing an example out. I still have to prove with induction that xi < xi+1 < b. It does make sense this way though although I HAVE to show it with induction for any n in the natural numbers.

10. Dec 4, 2011

### SammyS

Staff Emeritus
Re: Given any real numbers a and b such that a<b, prove that for any natural number n

And I was trying to point out how to use your example to help you understand how the hint that was given could be useful.