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A friend asked me this problem but I couldn't solve. Not homework.

Given reals $x>1$, $\epsilon>0$, does there always exist $i,n \in \mathbb N$ such that $|x^i -n |<\epsilon$. If the answer is in the negative, is there a nontrivial subset $S \subset (1,\infty)$ such that all $x\in S$ satisfy this?

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1 Answer 1

up vote 4 down vote accepted

We are going to find a sequence of nested intervals $(a_i,b_i)$ such that if $x\in[a_i,b_i]$ then $x^i\in [n_i+\frac{1}{3},n_i+\frac 2 3]$ for some integer $i$.

Then you'd have a counter-example by taking the limit.

Let $a_1=9\frac 1 3$ and $b_1=9\frac 2 3$.

Now, assume you have $(a_i,b_i)$ and $n_i$ such that $a_i^i = n_i + \frac{1}{3}$ and $b_i^i= n_i +\frac 2 3$.

Then look at the interval $[a_i^{i+1},b_i^{i+1}]$. By choose $a_1>9$, we see that this interval is at least $9$ times the length of the $[a_i^i,b_i^i]$, which is length $\frac{1}{3}$, so, in particular, this interval must contain some full integer interval, $[n_{i+1},n_{i+1}+1]$. Now, choose $a_{i+1}=\sqrt[i+1]{n_{i+1}+\frac{1}3}$ and $b_{i+1}=\sqrt[i+1]{n_{i+1}+\frac 2 3}$. Then $[a_{i+1},b_{i+1}]\subset [a_i,b_i]$ and, for any value $x\in [a_{i+1},b_{i+1}]$, $x^i$ is at least $\frac 1 3$ away from the nearest integer.

Now, if $x\in \cap_i [a_i,b_i]$, then $|x^i-n|>\frac{1}{3}$ for all $i>0$ and all integers $n$.

Therefore the answer is in the negative.

I think it is fairly easy to use a variation of this argument to prove that the set of counter-examples are dense in the set $(1,\infty)$.

I think a similar argument can show that the set of reals satisfying your condition is also dense.

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What real number is in the sequence of nested open intervals $(0,1),(1/2,1),(2/3,1),(3/4,1),\dots$? –  Gerry Myerson Apr 25 '12 at 1:53
Ah, okay, you can take closed interval then. @GerryMyerson :) –  Thomas Andrews Apr 25 '12 at 3:09
@GerryMyerson Changed to use closed intervals, but not really necessary, because you can ensure that $a_{i+1}>a_i$ and $b_{i+1}<b_i$. –  Thomas Andrews Apr 25 '12 at 3:12

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