If possible, find two positive integers $\displaystyle m,n$ such that: $\displaystyle 8\pi - 2 < m \pi - n < e^{\pi}.$

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- Dec 24th 2008, 03:50 PMNonCommAlgAn inequality from Alex Lupas
If possible, find two positive integers $\displaystyle m,n$ such that: $\displaystyle 8\pi - 2 < m \pi - n < e^{\pi}.$

- Dec 24th 2008, 11:33 PMbkarpuz
Do you exactly want the numbers or prove existence of them?

If you want the existence, for any $\displaystyle \lambda\in\mathbb{R}\backslash\mathbb{Q}$ the set $\displaystyle S:=\{m\lambda+n:m,n\in\mathbb{Z}\}$ is dense in $\displaystyle \mathbb{R}$, hence for any interval of type $\displaystyle (p,q)$ with $\displaystyle {\color{red}{p<q}}$, there exists at least one $\displaystyle r\in S$ such that $\displaystyle r\in(p,q)$. - Dec 25th 2008, 12:13 AMIsomorphism
Hey bkarpuz, I think he expects us to "find" m and n.

The inequality is pretty tight. I think m = 3391, n = 10630 works. But I did that the inelegant way (using MATLAB) (Crying)

I am wondering if a "Elementary and Middle School Math" technique exists for this (Thinking) - Dec 25th 2008, 12:15 AMNonCommAlg
- Dec 25th 2008, 05:59 AMbkarpuz
- Dec 26th 2008, 10:08 AMOpalg
According to my little pocket calculator (which may not be quite up to the job),

$\displaystyle \begin{aligned}8\pi-2&\approx23.132741,\\

16612\pi-52165&\approx23.137161,\\

e^\pi&\approx23.140693.\end{aligned}$

If $\displaystyle 8\pi-2<m\pi-n<e^\pi$ then $\displaystyle 0<(m-8)\pi-(n-2) < e^\pi-(8\pi-2)\approx23.140693 - 23.132741 = 7.952\times10^{-3}$. Therefore $\displaystyle 0<\pi - \frac{n-2}{m-8}< \frac{7.952\times10^{-3}}{m-8}$.

So we are looking for a rational approximation to π that is less than π, but close enough that the difference is of the order of one-thousandth of the reciprocal of its denominator. Among the known rational approximations to π, the first one that seems to fit these conditions is $\displaystyle \frac{52163}{16604}$. So take $\displaystyle n-2=52163,\;\;m-8=16604$. Unless my little calculator is misleading me, that appears to work.

Those are bigger numbers than**Isomorphism**gets, but I don't have MATLAB to numbercrunch for me. - Dec 26th 2008, 12:35 PMNonCommAlg
let $\displaystyle \{x \}=x-\lfloor x \rfloor.$ it's known that if $\displaystyle \theta \notin \mathbb{Q},$ then the sequence $\displaystyle \{\{k \theta \}: \ \ k \in \mathbb{N} \}$ is dense in the unit interval. so there exists $\displaystyle k \in \mathbb{N}$ such that: $\displaystyle \left \{\frac{k}{\pi} \right \} > 9 - \frac{e^{\pi} + 2}{\pi}. \ \ \ \ \ (*)$

now let $\displaystyle m=\left \lfloor \frac{k}{\pi} \right \rfloor + 9,$ and $\displaystyle n=k+2. $ then it's easy to see that $\displaystyle 8\pi - 2 < m \pi - n < e^{\pi}.$ this proves the existence of $\displaystyle m,n.$ to find $\displaystyle m,n,$ we need to find a solution for $\displaystyle (*).$ (Wondering)