Least Upper Bounds/Supremum Proof

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Homework Help Overview

The discussion revolves around proving that the least upper bound (supremum) of the set {1/2, 2/3, 3/4, 4/5,..., n/(n+1),...} is 1, based on the formal definition of least upper bounds.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to assume that 1 is not the least upper bound and seeks to find a contradiction by identifying a number in the set that exceeds a chosen real number p, which is less than 1. Some participants suggest manipulating inequalities to find suitable n values that satisfy the conditions.

Discussion Status

Participants are exploring various mathematical manipulations and reasoning to address the original poster's approach. There is an ongoing exchange of ideas regarding how to find an n that meets the necessary conditions, but no consensus has been reached yet.

Contextual Notes

The discussion includes attempts to clarify the implications of inequalities and the conditions under which n must be chosen, with some participants questioning the effectiveness of the proposed methods.

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Homework Statement


Give a formal proof based on the definition of least upper bounds that the least upper bound of the set {1/2,2/3,3/4,4/5,...,n/(n+1),...} is 1.


Homework Equations


None.


The Attempt at a Solution


Basically, the best attempt at a solution that I have worked out is as follows:
assume 1 is not the least upper bound of the set. then, let p be a real number such that p<1 and p>k/(k+1) for some k. (i.e. let p be the "least upper bound"). now, i need to find a number that would somehow bring up a contradiction proving p>1 or that there is a number of the form n/(n+1)>p thus showing p cannot be the "least upper bound". this is where i get stuck. i can't seem to determine a number greater than p which is in the set. i have also tried to start with the trivial fact that n+1>n and try to make something appear, but i get nowhere.

if anyone could give some pointers as to how to continue along with the approach i have already used or lead me towards any approach, it would be greatly appreciated.

thanks.
 
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You want to find n such that
\frac{n}{n+1} &gt; p

Take the reciprocal of both sides. This is equivalent to finding n such that

\frac{n+1}{n}&lt; \frac{1}{p}

which is the same as finding n such that

1+\frac{1}{n} &lt; \frac{1}{p}

What can you do with that?
 
I don't really see how this how this could help the problem. Could you please elaborate?

Thanks a lot.
 
We can change this last line to be

\frac{1}{n} &lt; \frac{1}{p} - 1

But we know \frac{1}{p}-1&gt;0 so we can look at its reciprocal. So finding an n equivalent to what you want is finding an n such that (taking the reciprocals of the first inequality in this post

n &gt; \frac{1}{ \frac{1}{p}-1}

And we know the right hand side is a positive real number
 
so basically taking any n greater than 1/(1/p - 1) should do the trick right? from what I understand of your suggestion is that taking any n greater than 1/(1/p - 1) would make p>1. am i right?
 

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