What is wrong with the following proof that $\{ (\frac{3}{2})^n\mod 1: n\in\mathbb{N} \} $ is dense in $\ [0,1]\ $?

Solution 1:

As pointed out in the comments by Zerox, $\ (1),\ $

There exist $\ n_1,\ n_2,\ m,\ N,\ $ such that $ N +\gamma -\delta < {a_{n_1}}^m < N + \gamma < {b_{n_2}}^m < N + \gamma + \delta.\quad (1)$

has not been justified.

My idea when writing the proof attempt was that for any $\ \delta>0,\ $ no matter how small, $\ \exists\ \varepsilon>0, m,N\in\mathbb{N}\ $ such that $\ y\in \left(\frac{3}{2}-\varepsilon, \frac{3}{2}-\varepsilon\right) \implies N +\gamma -\delta < \left(\frac{3}{2}-\varepsilon\right)^m\ < N+\gamma < \left(\frac{3}{2} +\varepsilon\right)^m\ < N +\gamma +\delta. $

However, this only makes sense if we assume that $\ \lbrace{ \left(\frac{3}{2}\right)^n \mod 1: n\in\mathbb{N} \rbrace}\ $ is dense in $\ [0,1],\ $ which is the result we are trying to prove. [I think I did this without realising it was what I was doing].

I have tried to salvage $(1)$ but I don't see a way to do this.