Is the set
$$ \left\{ 2^a \cdot 3^b \; : \; a, b \in \mathbb{Z} \right\} $$
dense in $[0, \infty)$? Seems like the answer is yes, but what's a simple proof?
Asked
Active
Viewed 179 times
3
-
sorry, the previous answer was very,very,very wrong. i am sorry thrice, I'll try and write a better answer. – Sarvesh Ravichandran Iyer Mar 12 '16 at 06:42
-
2it would be equivalent to prove that ${a\ln 2 + b\ln 3:a,b\in\mathbb Z}$ is dense in $(-\infty,\infty)$. Similarly, that ${a + b\log_2 3:a,b\in\mathbb Z}$ is dense in $(-\infty,\infty)$. Isn't it true in general that ${a + b p:a,b\in\mathbb Z}$ is dense in $(-\infty,\infty)$ if $p$ is irrational? – Mirko Mar 12 '16 at 07:01
-
2Continuing Mirko's comment, see For every irrational $\alpha$, the set ${a+b\alpha: a,b\in \mathbb{Z}}$ is dense in $\mathbb R$ – Bart Michels Mar 12 '16 at 07:23
-
1Yes. Because $\Bbb{Z}$ is a UFD, $\log_23$ is irrational, and that older thread saves the day as per Mirko's argument. – Jyrki Lahtonen Mar 12 '16 at 07:33
1 Answers
2
Yes, but it's kinda tricky to prove. Here is my strategy: First prove that you can get a number of the form $2^a \cdot 3^b$ arbitrarily close to any number in [0,1]. Then observe that you can form another number of the same form that exceeds any given number $n$ (just make the exponent(s) large enough). Let $k$ denote a number of the form $2^a \cdot 3^b$ exceeding $n$. Then $n$ can be expressed as $r \cdot k$ where $r \in [0,1]$, thus by our first part of the proof, we have shown that there is a number of the form $2^a \cdot 3^b$ arbitrarily close to $r$, and call such a number $m$. The product $m \cdot k$ is a product of two numbers of the form $2^a \cdot 3^b$, and so by exponent properties and the closure of $\mathbb{Z}$ under addition, we have that $m \cdot k$ is of the form $2^a \cdot 3^b$.
Justin Benfield
- 3,992
-
1I was trying to figure out why I could simplify to $[0,1]$ but everything I was coming up with was too complicated. Your strategy is very nice :) – Eric Stucky Mar 12 '16 at 06:59
-
1At first I was actually skeptical the statement was true, then once I started thinking about it more carefully I realized that it very well could be. The key insight (which took me a few minutes to come up with), was that I could express $n$ as $r \cdot k$. Once I recognized that the rest was straightforward. – Justin Benfield Mar 12 '16 at 07:05