This is inspired by a recent math.SE question.

Given that mathematicians like to come up with theoretical constructs which do not necessarily always have any practical purpose (but sometimes provide lots of fun anyway): to generalize the above question, let's define a "distance function" on reals greater than $1$ by $$d(x,y):=\inf_\limits{m,n\in\mathbb N}|x^m-y^n|.$$ As $0\not\in\mathbb N$, this function is supposedly rather pathological. I think that certain values of $d(x,y)>0$ can at least be found as minima, e.g. if $x=\frac{\sqrt{5}+1}2$, then $x^n$ is close to a Lucas number, so if $y$ then is an integer (or a $k$th root of an integer), the question of finding $d(x,y)$ essentially boils down to the question whether Lucas numbers can be infinitely often odd powers (which is very probably not the case).

Maybe there are any other fun facts that can be said about the function $d$, starting with the question:

Is it possible that $d$ is (against intuition) continuous in each variable?

Note that one could also ask:

Does it satisfy the triangle inequality?

Now for that, we would obviously need to have $d(x,y)=d(x^k,y)=d(x,y^k)$ for all $x,y>1$ and all $k\in\mathbb N$, but as soon as there is a $d(x,y)>0$ given by a minimum (as probably for the case above), this fails.

Any other ideas?


1 Answer 1


The answer to both of your questions is no.

Notice that $d(2, 3) = 1$. However, for any $\varepsilon > 0$, there exists a number $2 - \varepsilon < x < 2$ such that $d(x, 3) = 0$ and thus it is not a continuous function. To see this, take $n$ to be some large positive integer and $m$ the unique integer satisfying $$2^{m - 1} < 3^n < 2^m$$ or equivalently $$\left( 2^{1 - \frac{1}{m}} \right)^m < 3^n < 2^m$$ therefore there is a number $2^{1 - \frac{1}{m}} < x < 2^m$ such that $x^m = 3^n$ and therefore $d(x, 3) = 0$. Clearly, taking $n \to \infty$ we can let $x$ be arbitrarily close to $2$. In fact, this argument shows that for any $y > 1$, the set of $x > 1$ such that $d(x, y) = 0$ is dense in the set of numbers that are larger than $1$.

As for the triangle inequality, it is easy to see that $d(2, 3) = 1$ and $d(2, 16) = 0$. Furthermore, $d(3, 16) > 1$ because the only solutions to the equation $$2^m - 3^n = \pm 1$$ in positive integers $m, n$ are $(m, n) = (1, 1), (2, 1), (3, 2)$, which shows that there are no solutions to $$16^m - 3^n = \pm 1$$ and, thus, $d(3, 16) > d(3, 2) + d(2, 16)$.

  • $\begingroup$ By the way, a more direct way to prove that $16^m - 3^n = \pm 1$ has no solutions is that looking modulo 8 we see that $n$ must be even, but then our expression is a difference of two non-zero squares which is never equal to $\pm 1$. $\endgroup$
    – Random
    Aug 5, 2021 at 21:44

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