This seems that the set $$\left\{\frac{3^m}{\alpha^n}:\;m,n\in\mathbb Z\right\}$$ is dense in $\mathbb R_+$ (the set of positive real numbers), but I can not find the proof. How to prove this?
Edit # 2: Here $\alpha$ is any trancendental number.
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Consider $\alpha=\sqrt{3}$, is it dense? – Sungjin Kim Jun 20 '13 at 03:21
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okey, I assume addition that $\alpha\neq 3^r$ for any $r\in\mathbb Q$. – Hai Minh Jun 20 '13 at 03:26
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Then, density of the set in positive real numbers follows since $\log_3 \alpha$ is irrational. (Consider taking base 3 logarithm) – Sungjin Kim Jun 20 '13 at 03:29
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Related: http://math.stackexchange.com/questions/136665/how-to-show-this-is-a-dense-set – Sungjin Kim Jun 20 '13 at 03:33
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For $\alpha > 0$ I guess i707107's idea does the trick, because the irrationality of $\log_3(\alpha)$ will make the integer-linear combinations of $1$ and $\log_3(\alpha)$ generate a dense subset of $\mathbb R$, and then we can use continuity to get back density in $\mathbb R$ of the original form.
For $\alpha < 0$, we cannot take the log, but note that $\alpha^2$ satisfies the same constraints as $\alpha$, so we can take the subset $\{3^n/\alpha^{2m} \, | \, m,n \in \mathbb Z \}$ instead and it is still dense.
Hope that helps,
Patrick Da Silva
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$\alpha^2$ may be rational, but I could do the rest. Thanks both of you. – Hai Minh Jun 20 '13 at 03:44
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@user : I think I had in mind transcendental instead of irrational when I wrote this... :S did you deal with the case $\sqrt{p/q}$ separately? – Patrick Da Silva Jun 20 '13 at 04:18
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You are right, I really mean transcendental, and $\alpha^2$ never be rational. Sorry – Hai Minh Jun 20 '13 at 07:58
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@user80121 : as long as $\log_3(\alpha)$ is irrational in the positive case it should work, I don't think you need transcendental to be honest. I think you should work out the details (Which I didn't entirely) and see what happens. – Patrick Da Silva Jun 20 '13 at 08:47