Proof that a quantity is greater than 1/2

  • Context: Undergrad 
  • Thread starter Thread starter Mr Davis 97
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary

Discussion Overview

The discussion revolves around proving that the quantity ##\displaystyle 1 - \frac{N}{n}## is greater than ##1/2## under the condition that ##n > N##. The scope includes mathematical reasoning and exploration of convergence concepts.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant proposes using the definition of convergence to show that ##1 - \frac{N}{n} > 1/2## as ##n## approaches infinity.
  • Another participant challenges this by providing a specific example with ##N = 7## and ##n = 8##, demonstrating that in this case, the quantity evaluates to ##1/8##, which is less than ##1/2##.
  • A later reply questions the relevance of this discussion to another thread involving a different mathematical sequence.
  • One participant clarifies that the current question is not related to the other thread and indicates they have figured out what they want to show.

Areas of Agreement / Disagreement

Participants do not appear to agree on the validity of the initial claim, as one participant provides a counterexample that contradicts the assertion. The discussion remains unresolved regarding the proof of the inequality.

Contextual Notes

The discussion highlights the dependence on the values of ##N## and ##n##, as well as the conditions under which the inequality might hold. There is also an indication of missing assumptions regarding the behavior of the quantity as ##n## varies.

Mr Davis 97
Messages
1,461
Reaction score
44
I'm looking at the quantity ##\displaystyle 1 - \frac{N}{n}##, and trying to prove that it is greater than ##1/2##, given that ##n> N##. I thought that since ##\lim_{n \to \infty} 1 - \frac{N}{n} = 1##, we could use the definition of convergence to get this inequality, for suitable ##\epsilon##. For example, from the definition of convergence we can see that ##|(1 - N/n) - 1| < \epsilon## which implies ##N/n < \epsilon##, and so ##1-N/n > 1- \epsilon##. So if we let ##\epsilon = 1/2##, we get our result. Is that how I would do it?
 
Physics news on Phys.org
Suppose N = 7 and n = 8. Then n > N. Then 1 - N/n = 1 - 7/8 = 1/8 < 1/2. So I don't think you will have any luck proving the statement as you've stated it.
 
  • Like
Likes   Reactions: member 587159
Is this related to your other thread with ##a_n = 2^{(-1)^n}## and still open? What do you want to show?
 
fresh_42 said:
Is this related to your other thread with ##a_n = 2^{(-1)^n}## and still open? What do you want to show?
This is not related, but I actually figured out what I wanted to show for this question.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 48 ·
2
Replies
48
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K