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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?

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    $\begingroup$ Do you have some reason to believe that it is true? $\endgroup$ Commented May 22, 2015 at 9:57
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    $\begingroup$ 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
    Commented May 22, 2015 at 10:48
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    $\begingroup$ I would guess that the proof of Mills' constant might be useful here (assuming you can find it, and understand it). $\endgroup$ Commented May 22, 2015 at 11:05
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    $\begingroup$ 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 $\endgroup$ Commented May 22, 2015 at 11:23
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    $\begingroup$ 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. $\endgroup$ Commented May 22, 2015 at 17:10

2 Answers 2

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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 $$[\alpha_{n},\beta_{n}]\supset[\alpha_{n+1},\beta_{n+1}].$$

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  • $\begingroup$ In your statement $$[m_{n+1}+\dfrac{1}{3},m_{n+1}+\dfrac{2}{3}]\supset[a,b],$$ Should the $\supset$ sign be reversed? $\endgroup$ Commented May 22, 2015 at 12:01
  • $\begingroup$ ok,I have edit it. $\endgroup$
    – math110
    Commented May 22, 2015 at 12:07
  • $\begingroup$ A very intuitive solution. Thanks. $\endgroup$
    – Idonknow
    Commented May 23, 2015 at 7:07
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    $\begingroup$ Implementing the algorithm on a computer leads to the result $\alpha \approx 5.62144865272$. $\endgroup$
    – Fabian
    Commented May 28, 2015 at 14:14
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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

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    $\begingroup$ 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
    Commented Jun 5, 2015 at 23:54

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