Prove that there is a real number $a$ such that $\frac{1}{3} \leq \{ a^n \} \leq \frac{2}{3}$ for all $n=1,2,3,...$ Here, $\{ x \}$ denotes the fractional part of $x$.

My attempt: Clearly $a$ cannot be an integer because $\{ a^n \}=0$ for all $n \in \mathbb{N}$. Also $a$ cannot be a rational number because $\{ x \}$ is the same as $a$ modulo $1$. So the fractional part of $a$, at some point, will be less than $\frac{1}{3}$.

I try to find irrational $a$ which satisfies the inequality but I got no luck.

Can anyone give some hint to this question?

  • 3
    $\begingroup$ Do you have some reason to believe that it is true? $\endgroup$ May 22 '15 at 9:57
  • 7
    $\begingroup$ You use the tag (contest-math) and since people have some doubts about the fact you're trying to prove : please give us a reference of this problem. Where did you find it? What contest? Which year? $\endgroup$
    – user37238
    May 22 '15 at 10:48
  • 1
    $\begingroup$ I would guess that the proof of Mills' constant might be useful here (assuming you can find it, and understand it). $\endgroup$ May 22 '15 at 11:05
  • 5
    $\begingroup$ This is chapter 1, problem 3.8 of "Selected Problems in Real Analysis". The solution is given on pages 153-154: books.google.com/books?id=WMs97jhcwS4C&pg=PA153 $\endgroup$ May 22 '15 at 11:23
  • 3
    $\begingroup$ I don't see how you proof with rationals works; $a$ mod $1$ and $\{a\}$ are the same regardless of whether $a$ is rational. It seems like you're suggesting that since $\{a\}^n$ is eventually less than $\frac{1}3$, then so is $\{a^n\}$, but this is clearly not the case in general. $\endgroup$ May 22 '15 at 17:10

There exists $\alpha \in\left[\dfrac{16}{3},\dfrac{17}{3}\right]$ with the required property. To see this, we will construct an interval sequence $$\left[\dfrac{16}{3},\dfrac{17}{3}\right]=[\alpha_{1},\beta_{1}]\supset [\alpha_{2},\beta_{2}]\supset\cdots\supset[\alpha_{n},\beta_{n}],$$ where $\alpha_{n}$ and $\beta_{n}$ are such that $$\alpha^n_{n}-\dfrac{1}{3}=\beta^n_{n}-\dfrac{2}{3}=m_{n}\in \Bbb N^{+},$$ so that, for any $x\in [\alpha_{n},\beta_{n}]$, we have $$\dfrac{1}{3}\le\{x^n\}\le\dfrac{2}{3}.$$

We construct the interval sequence by induction. Assume that we have $[\alpha_{n},\beta_{n}]$. Let $$a=\alpha^{n+1}_{n},\quad\quad b=\beta^{n+1}_{n}.$$It follows that $$ b-a=(m_{n}+\dfrac{2}{3})\beta_{n}-(m_{n}+\dfrac{1}{3})\alpha_{n}>\dfrac{\alpha_{n}}{3}>\dfrac{5}{3}.$$ Then there exists $m_{n+1}\in \Bbb N^{+}$ such that $$\left[m_{n+1}+\dfrac{1}{3},m_{n+1}+\dfrac{2}{3}\right]\subset[a,b].$$ We take $$\alpha_{n+1}=\sqrt[n+1]{m_{n+1}+\dfrac{1}{3}},\qquad\beta_{n+1}=\sqrt[n+1]{m_{n+1}+\frac{2}{3}}.$$ Now $$\alpha^{n+1}_{n}=a<\alpha^{n+1}_{n+1}=m_{n+1}+\dfrac{1}{3}<\beta^{n+1}_{n+1}=m_{n+1}+\dfrac{2}{3}<b=\beta^{n+1}_{n},$$and hence $\alpha_{n}\le\alpha_{n+1}<\beta_{n+1}<\beta_{n},$ or $$[\alpha_{n},\beta_{n}]\supset[\alpha_{n+1},\beta_{n+1}].$$

  • $\begingroup$ In your statement $$[m_{n+1}+\dfrac{1}{3},m_{n+1}+\dfrac{2}{3}]\supset[a,b],$$ Should the $\supset$ sign be reversed? $\endgroup$ May 22 '15 at 12:01
  • $\begingroup$ ok,I have edit it. $\endgroup$
    – math110
    May 22 '15 at 12:07
  • $\begingroup$ A very intuitive solution. Thanks. $\endgroup$
    – Idonknow
    May 23 '15 at 7:07
  • 2
    $\begingroup$ Implementing the algorithm on a computer leads to the result $\alpha \approx 5.62144865272$. $\endgroup$
    – Fabian
    May 28 '15 at 14:14

Obviously it depends how approximately a≈5.62144865272, but a quick check with windows calculator shows that a^14 ≈ 31468458796.31579.. or a^15 ≈ 176898325303.72424... for instance

  • 1
    $\begingroup$ Yeah, but just adjusting the final $2$ to a $3$ due to rounding, by $a^{14}$ that corresponds to a difference of $0.78370$, and for $a^{15}$ it's $4.720$, so it really does require more decimal places by then. $\endgroup$
    – Mark Hurd
    Jun 5 '15 at 23:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.