# Least Upper Bounds/Supremum Proof

1. Sep 21, 2009

### jmirenzi

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
None.

3. 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.

2. Sep 21, 2009

### Office_Shredder

Staff Emeritus
You want to find n such that
$$\frac{n}{n+1} > p$$

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

$$\frac{n+1}{n}< \frac{1}{p}$$

which is the same as finding n such that

$$1+\frac{1}{n} < \frac{1}{p}$$

What can you do with that?

3. Sep 21, 2009

### jmirenzi

I dont really see how this how this could help the problem. Could you please elaborate?

Thanks a lot.

4. Sep 21, 2009

### Office_Shredder

Staff Emeritus
We can change this last line to be

$$\frac{1}{n} < \frac{1}{p} - 1$$

But we know $\frac{1}{p}-1>0$ 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

$$n > \frac{1}{ \frac{1}{p}-1}$$

And we know the right hand side is a positive real number

5. Sep 21, 2009

### jmirenzi

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?