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I want to prove that $M=\{2^a 3^b: a,b \in\mathbb Z\}$ is dense $[0,\infty)$

Therefore I want to show that $\overline{M}=[0,\infty)$

$\overline{M}\subseteq[0,\infty)$ because $M\subseteq[0,\infty)$

For the other direction I have an element $x\in[0,\infty)$, now I need to show that $\exists (x_n)\in\overline{M}:x_n\rightarrow x$, i.e to show $\exists (a_n)(b_n)\in\mathbb Z:x_n=2^{a_n}3^{b_n}\rightarrow x$.

Could you help me with that direction?

EDIT: As discussed in the comments, of course there are different ways proving this result, but I would like to be most interested in the continuation of my approach. If you know different solution approaches it is also interesting to know them and discuss them.

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I'm not sure but I think you only need to prove that your set is dense in $\Bbb Q_+$ since $\Bbb Q_+$ is dense in $\Bbb R_+$. – xavierm02 Mar 27 '13 at 11:37
up vote 5 down vote accepted

In general if $x,y\in\mathbb N, \ y>1$ and there is a prime $p$ such that $p\mid x$ and $p\nmid y$ then the set $M=\{x^a y^b: a,b \in\mathbb Z\}$ is dense in $[0,\infty)$.

Kronecker's theorem can be stated as:
If $\xi$ is irrational then the set $\{n\xi+m:n,m\in\mathbb Z\}$ is dense in $\mathbb R$.
Now use the fact that $\frac{\ln x}{\ln y}$ is irrational.

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I know what you mean, I can also consider $A=\{n\log2+m\log3: m,n\in\mathbb Z\}$ and this is by Kronecker dense in $\mathbb R$ and because of $e^A$ the set $M$ is dense in $[0,\infty)$ but I want to do it in a different way. – Alexander Mar 27 '13 at 11:41
Why do you want to do it in a different way? Is this a homework? – P.. Mar 27 '13 at 11:43
No it is not a homework, I want to try it for myseld in a different way because I think the sequence approach is more general than using Kronecker. – Alexander Mar 27 '13 at 11:44
It is inherently a Diophantine approximation problem. Doubt one can avoid the ideas. – André Nicolas Mar 27 '13 at 11:45
"The sequence approach is more general"? Note that $x_{m,n}:=2^m3^n$ over $\mathbb{Z}^2$ is the set of values of a sequence as $\mathbb{Z}^2\simeq \mathbb{N}$. – 1015 Mar 27 '13 at 12:39

Let $x>0$ and let $\varepsilon>0$. As in P.'s proof, which looks fine to me, we can choose $a,b\in\mathbb Z$ so that

$$x< 2^a 3^b <x+\varepsilon.$$

To see this, simply take the logarithm and divide through by $\log(3)$ to show that your inequality is equivalent to

$$\frac{\log(x)}{\log(3)} < a\frac{\log(2)}{\log(3)}+b < \frac{\log(x+\varepsilon)}{\log(3)}.$$

Again, as in P.'s answer, the existence of $a$ and $b$ follows from Kronecker's theorem.

Finally, if you really want to construct a sequence of such numbers that converges to $x$, simply choose $a_n$ and $b_n$ so that

$$x< 2^{a_n} 3^{b_n} <x+\frac{1}{n}.$$

Then the sequence defined by $x_n = 2^{a_n} 3^{b_n}$ satisfies your requirement.

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Thank you, how would you choose $a_n,b_n$ explicitly? Or is it enough to conlcude the proof using the log argument? – Alexander Mar 29 '13 at 13:58
@Alexander I think that P.'s proof (or the log argument, as you call it), is clearly sufficient. The point I'm trying to make is that the existence of your desired sequence follows easily from the inequality here. – Mark McClure Mar 29 '13 at 14:13

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