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Please help me to find (if they exist): the least upper bound, greatest lower bound, the smallest and the largest elements of these sets. $$\left\{ \frac{p-q}{p+q},\space p \text{ and } q\in \mathbb{N},\space p>q\right\}$$ and $$\{x\in \mathbb{Q},\space x^2\leq3 \}$$

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It's considered rude on this site to post in the imperative, rather than asking a question. Also, what have you tried? – Brad Oct 6 '12 at 22:19
For the first problem, we can find all sorts of sequences going to different things. For example, take $p=n+1, q=n$. Then $p+q = 2n+1, p-q =1$. Then $\frac{p-q}{p+q} = \frac{1}{2n+1}\rightarrow0$. Similarly, if $p = n+1, q= -n$, we get $p+q = 1, p-q = 2n+1$, so that $\frac{p-q}{p+q} =\frac{2n+1}{1}\rightarrow \infty$. You can figure out your question from here. – Euler....IS_ALIVE Oct 6 '12 at 22:28
Do you mean the least upper bound and greatest lower bound? – Christopher A. Wong Oct 6 '12 at 22:29
I figure out that $1\geq\frac{p-q}{p+q}>0$, so the problem is how can I be sure that $0$ is the smalest one? – Mohamez Oct 6 '12 at 22:36
It isn't the smallest one, because it isn't in the set. You can show that it's the greatest lower bound, though. – Cameron Buie Oct 6 '12 at 22:39
up vote 2 down vote accepted

A way to approach the first problem is in two cases, $q=0$ and $q\ne0$. If $q=0$ then $\frac{p-q}{p+q} = 1$. If $q\ne0$ then $0 \lt \frac{p-q}{p+q} \lt 1$.

A way to be sure that $\frac{p-q}{p+q} \gt 0$ is to consider when $p$ and $q$ are the closest they can be, $1$ apart. i.e. $p = q +1$. Then $\frac{p-q}{p+q} = \frac{1}{2q+1} $ and $\lim_{q \to \infty} \frac{1}{2q+1} = 0 $

So an upper bound would be $1$ and a lower bound $0$. The set contains $1$, $1$ is also the upper bound, so $1$ is the largest element of the set. The lower bound is $0$, but $\frac{p-q}{p+q}$ can never be $0$, so the set does not have a smallest element.

note: If you don't take $0 \in \mathbb{N}$ then there is no largest element, because $\frac{p-q}{p+q}\ne 1$.

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I edited that bit for clarity, hope it helps! But when you say "0 is the smallest one" it's important to realize that it is not the smallest element in the set, just a lower bound for the set. Specifically in your statement with $\alpha$ assume that $p$ and $q$ are such that $\frac{p-q}{p+q} = \alpha$ Now add $1$ to $p$. – seth Oct 6 '12 at 22:54

For the first one, let $r = p/q$. The fraction is, dividing numerator and denominator by $q$ and calling it $f$, $f = (r-1)/(r+1) = 1 - 2/(r+1)$.

Since $p > q \ge 1$, $r > 1$ and $r$ can be arbitrarily large, so $f > 0$ (for $r$ close to $1$), and $f < 1$ (for $r$ large).

There is no largest or smallest element, since the bounds ($0$ and $1$) are never reached.

For the second, since $\sqrt{3}$ is irrational, the upper bound is not in the set.

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