# 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

You may be interested to look at information on Beatty sequences and Rayleigh's Theorem: http://en.wikipedia.org/wiki/Beatty_sequence#Rayleigh_theorem

• Is you answer meant to imply that you are certain that the result follows by the linked theorem, or, rather, that you guess that the theorem might be useful for this problem? – Bill Dubuque Sep 5 '14 at 16:56
• I would say that the link points you to information that solves the problem. – paw88789 Sep 5 '14 at 17:32
• If you know a way to uee the theorem to give a complete proof, then please provide further details. Otherwise the answer reads like "maybe this will help, maybe not...". – Bill Dubuque Sep 5 '14 at 17:58
• @BillD After studying your comment and taking another look, I see that I misread the problem. I thought it asked for multiples of the irrational numbers rather than powers. I plan to post a comment to the question and withdraw my answer. Thanks for helping me notice the mistake I made. – paw88789 Sep 5 '14 at 19:40