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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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    Bingo. Ex$a$ctly what we needed.2012-10-07

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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 $p$.2012-10-06
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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.