The set {${m^m}/{n^n}: m,n\in \mathbb{N}$} density in ${\mathbb{Q}_+}$

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.

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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?

Henry
November 08, 2018 22:41 PM

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 $$z$$.

Mike
November 08, 2018 22:50 PM