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

In summary, the given proposition states that for any real numbers a and b such that a < b, there exists a sequence of real numbers x1, x2, x3, ..., xn that satisfies the conditions a < x1 < x2 < x3 < ... < xn < b. The hint provided suggests recursively defining the sequence xi as x1 = (a+b)/2 and x(i+1) = (xi +b)/2. This creates a sequence that starts at the midpoint of a and b and converges towards b. Using induction, it can be shown that this sequence satisfies the conditions a < x1 < x2 < x3 < ... < xn < b for any natural number n.
  • #1
snes_nerd
13
0
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 doesn't really help me at all but I think I have an idea of what the question is asking. Let's 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.
 
Physics news on Phys.org
  • #2


Okay so the hint doesn't really help me at all

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


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 don't 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 that's greater then x1. Cant really connect this to the proof
 
  • #4


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


snes_nerd said:
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 doesn't really help me at all but I think I have an idea of what the question is asking. Let's 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 there.
...
For your example in which a=0 and b= 10: What are the values you get for x1, x2, x3, x4, x5, etc. ?
 
  • #6


So x3=(x2+b)/2 implies that a < x1 < x2 < x3 < b?.
 
  • #7


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


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


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


snes_nerd said:
...

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.
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.
 

1. How do you prove that a

To prove that a

2. What does it mean for a number to be "natural"?

In mathematics, a natural number is a positive integer (excluding 0) that is used for counting and ordering objects. It is denoted by the symbol "n" and can be written as n = 1, 2, 3, 4, ...

3. Can you prove this statement for all possible values of n?

Yes, we can prove this statement for all possible values of n by using mathematical induction. This method allows us to prove that a statement is true for all natural numbers by showing that it is true for the first value of n (usually n = 1), and then showing that if it is true for any arbitrary value of n, it must also be true for the next value of n.

4. Is this statement true for all real numbers a and b?

No, this statement is only true for real numbers a and b such that a

5. Can you provide an example to illustrate this statement?

Sure, for example, if a = 2 and b = 5, then a

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
907
  • Programming and Computer Science
Replies
14
Views
496
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
455
  • Calculus and Beyond Homework Help
Replies
3
Views
761
  • Calculus and Beyond Homework Help
Replies
8
Views
6K

Back
Top