find $u < r$, $v < s$ such that $p< uv < rs$

Question

Let $p < rs$ where both $p, r, s$ are positive rational numbers.

I want to find rationals $u < r$, $v<s$ such than $p < uv < rs$ holds.

It seems obvious to me since for sufficient small $\epsilon$, $u = r - \epsilon$, $v = s - \epsilon$ is my desired.

But how can I construct such $\epsilon$ exactly?

Answer 1

Addendum added to correct an analytical error in my answer.

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Without loss of generality, $u,v$ are both positive. That is, since $0 < p < uv,$ then $u$ and $v$ must have the same sign. This means that $(|u|,|v|)$ is a satisfying ordered pair if and only if $(-|u|,-|v|)$ is also a satisfying ordered pair.

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Edit
Actually, the above paragraph contains an analytical error. See the Addendum.

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Further, with $u,v$ both assumed positive,
$u < r, v < s$ automatically implies $uv < rs.$

So, the question reduces to finding positive $u,v$ such that

• $p < uv.$
• $u < r, v < s.$

One way of doing this is to establish a range of values of $u$ and then, for each value of $u$, establishing a range of values for $v$.

The easiest way of doing this is to temporarily pretend that the constraints are:

• $p \leq uv.$
• $u \leq r, v \leq s.$

So, under the pretend constraints above, the largest value for $u$ is $r$. This implies that the smallest value for $v$ is given by $\dfrac{p}{r}.$

Similarly, the largest value for $v$ is $s$, and the smallest value for $u$ is given by $\dfrac{p}{s}.$

So, you have that $\dfrac{p}{s} \leq u \leq r.$

Then, you have that for any value of $u$, you must have that $\dfrac{p}{u} \leq v \leq s.$

Putting this all together, and reverting back to the actual constraints, the complete set of all (positive) $(u,v)$ is given by

$$\{ ~(u,v) ~: ~ \frac{p}{s} < u < r, ~\frac{p}{u} < v < s ~\}.$$

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Addendum

It is true that if $(|u|,|v|)$ is a satisfying ordered pair, then so is $(-|u|, -|v|).$

However, the converse is false.

For example, take $r,s$ = 2,3, respectively.

Take $p = 1.$

Then $\displaystyle ~(u,v) = \left( ~-10^6, ~-\frac{1}{10^6 - 1} ~\right)$ is a satisfying ordered pair.

So, when considering satisfying ordered pairs $(u,v)$, where $0 > u,v$, the only relevant constraints are that $p < uv < rs.$

So, $u$ can be any negative number.

Then, the set of all satisfying (negative) values of $v$ is given by

$$\frac{p}{|u|} < |v| < \frac{rs}{|u|} \implies \frac{p}{u} > v > \frac{rs}{u}.$$

Answer 2

• Find any $k_1$ satisfying $\quad 2^{k_1} > \frac{r+s}{rs-p}\qquad (1)$
• Find any $k_2$ satisfying $\quad 2^{k_2} > \frac{1}{r+s}\qquad (2)$
• Let $k = \max\{k_1,k_2\}.\ $ Then $\ k\ $ (in place of $k_1$ and $k_2$) satisfies both $(1)$ and $(2).$
• Then $u = r - 2^{-k},\ v = s - 2^{-k}\ $ will do, since:

$(i)\quad(1)\implies rs - (r+s)2^{-k} > p,\implies uv=rs - (r+s)2^{-k} + 2^{-2k} > p,\ $ and

$(ii)\quad (2) \implies r+s > 2^{-k}, \implies 2^{-k}(r+s) > 2^{-2k},\implies 2^{-k}(r+s) - 2^{-2k} > 0,\implies rs - (2^{-k}(r+s) - 2^{-2k}) < rs.$

Answer 3

The $\epsilon$ requested by the OP has to satisfy 2 conditions.

The first condition is:

$$rs-(r-\epsilon)(s-\epsilon) \gt 0$$

$$\epsilon(r+s-\epsilon) \gt 0$$

Clearly any $\epsilon$ satisfying

$$0 \lt \epsilon \leq \frac{r+s}{2} \tag{1}$$

meets the requirement.

The 2nd condition is

$$(r-\epsilon)(s-\epsilon)-p \gt 0$$

$$rs-p - \epsilon(r+s)+ \epsilon^2 \gt 0$$

This condition is satisfied when

$$0 \lt \epsilon \leq \frac{rs-p}{2(r+s)} \tag{2}$$

Combining (1), (2), we may let

$$\epsilon = \min \left\{ \frac{r+s}{2}, \frac{rs-p}{2(r+s)} \right\}$$

Answer 4

In this answer, we will construct a closed form example for the rational pair $(u,v)$.

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Your exact problem is:

I want to find rationals $u < r$, $v<s$ such than $p < uv < rs$ holds.

Let $u,v>0$ and $\alpha,\beta >1$, then our first move is to set up the following system of equations:

$$\begin{align}\begin{cases}u=\dfrac r\alpha \wedge v=\dfrac s\beta \\ uv=\dfrac {p+rs}{2}\end{cases}\end{align}$$

This implies that,

$$\begin{align}&\frac {rs}{\alpha\beta}=\frac {p+rs}{2}\\ \implies &\alpha\beta=\frac {2rs}{p+rs}\\ \implies &\alpha=\frac {2rs}{\beta({p+rs})}\end{align}$$

where $1<\beta<\frac {2rs}{p+rs}$, since $2rs>p+rs$ and $\alpha>1$.

Therefore, we can take:

$$\begin{align}\beta=\frac{1+\frac {2rs}{p+rs}}{2}=\frac{p+3rs}{2(p+rs)}\end{align}$$

This leads to:

$$\begin{align}\alpha=\frac {4rs}{p+3rs},~\beta=\frac{p+3rs}{2(p+rs)}\end{align}$$

Finally, the closed form example we want to construct is:

$$\begin{align}\bbox[5px,border:2px solid #C0A000]{u=\frac {p+3rs}{4s},~v=\frac {2s(p+rs)}{p+3rs}}\end{align}$$