Representing any number in $R_+$ by means of two numbers

Question

I have the following question:

Let $R_+$ denote the nonnegative reals. Let $0<a<1$ and $b>0$, and set $p = a^n b^m$. Is it possible to find a $n,m\in N$ such that $p$ can approximate any number in $R_+$ arbitrarily close?

Answer 1

Yes as long as $\frac {\log a}{\log b}$ is irrational. Given a number $r$ to approximate, we have $\log r \approx n \log a + m \log b$ You can use the equidistribution theorem to show you can get as close as you want.

Answer 2

This cannot be true in general, because a simple counter-example:

Set $a=\frac{1}{2}$ and $b=2$. Now, the only number that can be arbitrarily approximate is $0$, as is easily seen.

But I believe that it could be true when $\frac{a}{b}$ is irrational.

Answer 3

I give here a proof that it's possible if either $\frac{\ln(a)}{\ln(b)}$ or $\frac{\ln(b)}{\ln(a)}$ is a normal number (and $b>1$) :

https://en.wikipedia.org/wiki/Normal_number

First notice that there is equivalence between "$p(n,m)$ can approximate any number in $R_+$ arbitrarily close", and "$\ln(p(n,m))$ can approximate any number in $R$" (this is a consequence of the continuity of $\ln$)

$$\ln(p(n,m)) = n\ln(a) + m\ln(b)$$

Now take an arbitrary $x \in R$. If $\gamma = \frac{\ln(a)}{\ln(b)} < 0$ is a normal number, then the k-th first decimal of $\frac{x}{\ln(b)}$ are the decimal of $\gamma$ between $n_k$ and $n_k+k$ for a certain $n_k$ that exist by normality of $\gamma$. You have then :

$$ -10^{-k} < 10^{n_k} \gamma - \lfloor 10^{n_k} \gamma \rfloor +\lfloor \frac{x}{\ln(b)} \rfloor - \frac{x}{\ln(b)} < 10^{-k}$$

Multiply this by $\ln(b)$, and you get

$$ -\ln(b) 10^{-k} < 10^{n_k} \ln(a) + \left( - \lfloor 10^{n_k} \gamma \rfloor + \lfloor \frac{x}{\ln(b)} \rfloor \right) \ln(b) - x < \ln(b) 10^{-k}$$

Now, notice that $m=- \lfloor 10^{n_k} \gamma \rfloor + \lfloor \frac{x}{\ln(b)} \rfloor $ is a positive integer if $n_k$ is taken big enough, and this is possible.

So you indeed have

$$\left| n \ln(a) + m\ln(b) - x \right| < \ln(b) 10^{-k}$$

And x can be approximed arbitrarily close.

Now, notice that I only proved that it's a sufficent condition, not a necessarily one.