by EeMm
Last Updated November 08, 2018 23:20 PM

I missed this topic (density) , so need help.

Is the set {${m^m}/{n^n}: m,n\in \mathbb{N}$} everywhere dense in positive rational numbers set?

I think maybe first of all here we must try with, for example $m=1, n=2; m=2, n=3;\ldots$ But need help to end this idea and proof.

Hint:

try to find values between $\dfrac{1}{2}$ and $2$

$\dfrac{n^n}{n^n}=1$ is a possibility. Are there any others? If so, what values do they take? If not, why not?

No, this set is not dense.

If $m > n$ then

$\frac{m^m}{n^n}$ is at least $m$ which (as $m,n \in \mathbb{N}$) must be at least 2, and

$\frac{n^n}{m^m}$ is no larger than $\frac{1}{m}$ which must be no larger than $\frac{1}{2}$.

In fact, at any positive integer $z$ one can show that there are at most $z^2$ ordered pairs $(m,n)$ that satisfy $m^m/n^n \in [z,z+1]$. Indeed, $m$ must satisfy $m>n$ but $m$ cannot be larger than $z$.

- ServerfaultXchanger
- SuperuserXchanger
- UbuntuXchanger
- WebappsXchanger
- WebmastersXchanger
- ProgrammersXchanger
- DbaXchanger
- DrupalXchanger
- WordpressXchanger
- MagentoXchanger
- JoomlaXchanger
- AndroidXchanger
- AppleXchanger
- GameXchanger
- GamingXchanger
- BlenderXchanger
- UxXchanger
- CookingXchanger
- PhotoXchanger
- StatsXchanger
- MathXchanger
- DiyXchanger
- GisXchanger
- TexXchanger
- MetaXchanger
- ElectronicsXchanger
- StackoverflowXchanger
- BitcoinXchanger
- EthereumXcanger