# Homework Help: Infimim and supremum proof

1. Feb 2, 2010

### Hotsuma

The problem statement, all variables and given/known data

Find the infimum and supremum of each of the following sets; state whether the infimum and supremum belong to the set E.

$$\item 1. ~~~~E={p/q \in \mathbf{Q} | p^2 < 5q^2 \mbox{ and } p,q >0}. \mbox{ Prove your result. } \item 2. ~~~~E={2-(-1)^n/n^2|n \in \mathbf{N}. \mbox { Just list your answers}.$$

2. Relevant equations

$$sup E - \epsilon < a \leq sup E.$$

3. The attempt at a solution

$$\item 1. ~~~~ \mbox{Here I claim the supremum of E is }\frac{1}{\sqrt{5}}, \mbox{ in E, and the infimum of E is 0, which is not in Q}. \item \mbox{Proof: Assume p, q greater than 0. Then } p^2 < 5q^2 \Leftrightarrow p < \sqrt{5} q \Leftrightarrow s = sup E = p/q = \frac{1}{\sqrt{5}} \mbox{how do I prove the infimum part. Hmm...} \item 2. ~~~~ \mbox{inf E = 0, sup E = 2 (assume 0 in N). }$$

2. Feb 2, 2010

p/q < √5, not 1/√5.

3. Feb 2, 2010

### Hotsuma

Thanks, it is late. Woo silly mistakes. How do I prove this result, or is that sufficient? I mean, should I just plug in the typical supremum proof for this result?

Thanks by the way

4. Feb 2, 2010

Yes, you can prove this from the definition of the supremum and by using the fact that between every two distinct real numbers you can find a rational number.

5. Feb 2, 2010

### HallsofIvy

You mean $\sqrt{5}$ not, $1/\sqrt{5}$, but, more importantly, $\sqrt{5}$ is NOT in E since it is not a rational number. 0, on the other hand, is in E: take p= 0, q= 1. And it is the infimum because it is a lower bound (there are no negative numbers in E) while for any n, 1/n, is in E (take p= 1, q= n).

You had better not "assume 0 in N" because then you are dividing by 0 when n=0.

When n= 1, 2-(-1)1/12= 3 so 2 is not an upper bound on this set.

Last edited by a moderator: Feb 2, 2010