Does there exist real $a$ such that for every positive integer $c$ there exists integer $b$ such that $\lfloor{a^b}\rfloor = c$?
Asked
Active
Viewed 44 times
3
-
1You mean a single $a$ for all $b,c$? – gammatester Nov 27 '18 at 10:09
-
I've edited the question, sorry for my mistake. Yes, one $a$. – user4201961 Nov 27 '18 at 10:11
1 Answers
2
No.
The simplest way to see why the answer is no is the fact that, for $a>1$, you have
$$\lim_{b\to\infty} (a^b-a^{b-1}) = \infty$$
(which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)
The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $\lfloor a^{b-1}\rfloor$ and $\lfloor a^b\rfloor$ (that is, $\lfloor a^{b-1}\rfloor<n<\lfloor a^b\rfloor$. Couple that with the fact that $\lfloor a^b\rfloor$ is an increasing sequence, and you are done, because $$\forall b: \lfloor a^b\rfloor\neq n$$
5xum
- 123,496
- 6
- 128
- 204