Prove That At Least 1 Integer Divides Another w/ Discrete Math

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Homework Help Overview

The problem involves using mathematical induction to demonstrate that in any set of n+1 positive integers, none exceeding 2n, at least one integer divides another. The context is within discrete mathematics, focusing on properties of integers and divisibility.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to establish a base case for the induction and formulates an inductive hypothesis. Some participants discuss the implications of the largest integer in the set and consider different cases based on its value. Others raise questions about the distinctness of integers in the set and the nature of the induction method to be used.

Discussion Status

The discussion is ongoing, with participants exploring various aspects of the problem. Some guidance has been offered regarding the structure of the proof and the nature of the integers involved, but there is no explicit consensus on the approach or resolution of the problem.

Contextual Notes

Participants note that there is no requirement for the integers to be distinct, although some argue that the definition of a set implies uniqueness. The discussion also touches on the potential use of weak versus strong induction, with hints about how to relate larger sets back to smaller subsets.

VinnyCee
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Homework Statement



Use mathematical induction to show that given a set of n\,+\,1 positive integers, none exceeding 2\,n, there is at least one integer in this set that divides another integer in the set.



Homework Equations



Mathematical induction, others, I am not sure



The Attempt at a Solution



Let P(n) be the claim that there will be two integers (x,\,y) such that y divides x.

\left{1,\,2,\,\dots,\,2\,n\right} <----- Choose only n\,+\,1 of these.

Basis: Does P(1) hold?

\left{1,\,\dots,\,2\,n\right}\,=\,\left{1,\,2\right} <----- for n\,=\,1

2 is divisible by 1, so x = 2 and y = 1. P(1) is true.


Inductive hypothesis:

\left{1,\,2,\,\dots,\,2\,n,\,2\,n\,+\,1,\,2\,n\,+\,2\right} <----- Choose only n\,+\,2 of these

How do I prove that though?
 
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Here's the start of a proof...

Let m be the largest integer in the set of n+1 integers.

If m <= 2(n-1), there are n integers in the set less than 2(n-1) , so by induction P(n) is true.

Now think about the cases where m = 2n-1 or m = 2n.
 
By the way- there is no requirement that the n integers be distinct. In the n= 1 case, the two integers may be {1, 2}, {1, 1} or {2,2}. Of course, if they are NOT distinct, then obviously one of the duplicates divides the other so you really only need to consider the case that they are distinct- but you should say that.
 


Resurrecting old thread, I also really need help with this problem. I'm completely lost, I couldn't even come with what the OP got on my own.

Thanks anyway
 


By the way- there is no requirement that the n integers be distinct.

I believe there is that requirement, as the claim states that it is a set of positive integers. Sets contain unique elements.

Since Induction in general was part of your question, I would recommend looking at inductiveproofs.com -- they have a nice template for writing inductive proofs that you might find helpful.

Does the question hint at using weak or strong induction? (if you haven't heard of strong induction yet, that is helpful :) since we know it should use weak then!) The thing to think about is how can we get back to the smaller subset? AKA In the inductive step, we have a set of n+2 elements, all no higher than 2n+2, how do we talk about a subset of n+1 elements all no higher than 2n? What do we know about the other sets that don't contain that subset?
 

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