# How prove or disprove $\lfloor a^m\rfloor \neq \lfloor b^n\rfloor$ for all $m,n\in\mathbb{N}$

Question:

prove or disprove: there exsit irrational $a>1,b>1$ such that for all positive integers $m,n$,

$$\lfloor a^m\rfloor \neq \lfloor b^n\rfloor$$

Now I can't prove this problem.

I know this following: $$a^m-\{a^m\}\neq b^n-\{b^n\}$$ since $$a^m-b^n\neq \{a^m\}-\{b^n\}\in (-1,2)$$

I only have this idea. if one can take example, Thank you

• I cannot completely follow: You want to find irrational $a,b$ so that $\lfloor a^m \rfloor \neq \lfloor b^n \rfloor$ for all $m,n \in \mathbb N$? What does your $\{\}$ notation mean? – flawr Sep 5 '14 at 9:37
• “for any” is slightly murky language, I think. It could be read as “for some” or “for all” (or “for every”). I assume you mean the latter. – Harald Hanche-Olsen Sep 5 '14 at 9:37
• I have to post a mea culpa here. I misread the problem, thinking that it asked for multiples of the irrational numbers rather than powers. Thanks to Bill Dubuque for pointing this out to me. I am going to withdraw my answer. Apologies for my mistake. – paw88789 Sep 5 '14 at 19:41
• Aargh, I can't delete an accepted answer. If OP wants to unaccept it, than I will delete it. – paw88789 Sep 5 '14 at 19:42