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Let $x,y\in\ [0,1]$ a real numbers. Suppose that $$x=\sum_{n=1}^{+\infty}\frac{a_n}{3^n},\quad y=\sum_{n=1}^{+\infty} \frac{b_n}{3^n},\;a_n,b_n\in\{0,1,2\}$$ they are their ternary expansions.

Suppose we know only the ternary expressions and not actually the real number they represent.

Question Can I compare $x$ and $y$ working exclusively in base 3? Or rather, how can I determine if $x\le y$ or $y\le x$?

Thanks!

Jack J.
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1 Answers1

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Yes you can, except for one small problem : some numbers have TWO different ternary developments.

Simple rule : if $a_n=b_n$ for all $n<n_0$, and $a_{n_0}<b_{n_0}$, then $x\le y$.

They can still be the same if $a_{n_0}=b_{n_0}-1$, and $a_n=2$ and $b_n=0$ for all $n>n_0$.

Nicolas FRANCOIS
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  • Thanks for the answer. But if I already know that $x\le y$ can I say about $a_n$ and $b_n$? – Jack J. Jul 25 '19 at 08:54
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    Just what I said above : consider the development of $y-x$ : if it is greater than $3^{-n_0}$, then some $a_n$ is strictly smaller then $b_n$ for $n\le n_0$. – Nicolas FRANCOIS Jul 25 '19 at 08:56
  • For example $\frac{1}{3}=0,00\overline{2}_3$ and $1/3=0,0\overline{1}_3$, what can I say about the digits of the numbers that are less than $1/3$? – Jack J. Jul 25 '19 at 08:56
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    $\frac13=0,1_3$, or $0,0\overline1_3$. If $x<\frac13$, consider the development of $\frac13-x$ : it's a strictly positive number, so it has to have some strictly positive coefficients. Consider the one with the smaller index. – Nicolas FRANCOIS Jul 25 '19 at 08:59
  • Then if $x<\frac{1}{3}$, we can say with certainty that the first digit $a_1$ of $x$ is zero, if were $a_1=1$, then may be other digits other different from zero and then $x>\frac{1}{3}$ or $x=\frac{1}{3}$. Moreover if were $a_1=2$, then $x>\frac{1}{3}$. In any case we would get an absurdity. Quite right? – Jack J. Jul 25 '19 at 09:57