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This is an exercise from Carothers Real Analysis: Let $p \geq 2$ an integer. "Count" the real numbers in $(0,1)$ that have an eventually repeating base $p$ decimal expansion.

For example, let $p=9$.

This is what I have tried:

First, we count real numbers eventually ending with repeating 8's. This is easy because these numbers are just numbers with finite decimal expansion. Such as $0.49999999999... = 0.5$. So these numbers are just rational numbers, they are countable.

But how to deal with numbers ending with other digits? Like $0.4777777777777$? How to count them?

Based on the comments, here is a useful link

Hamilton
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    A number which is rational in one base is rational in all bases, and the same argument that works for base $10$ shows that rational is equivalent to being eventually periodic in base $b$. – lulu Nov 09 '21 at 14:45
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    But it is easier to do the exercise as stated than to use @lulu 's assertion about rationals. A number with eventually repeating base $p$ expansion is: finitely many digits left of the ".", finitely may digits to the right of the ".", then a finite block of digits repeated indefinitely. So count how many of these there are. – GEdgar Nov 09 '21 at 14:50
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    But there is value in recognizing that all such "decimals" are rational numbers. For instance, if $N=.1237979797979\cdots $ in base $p$, we could observe that $p^3N=123.\overline {79}$ and $p^5N=12379.\overline {79}$ so $(p^5-p^3)N=12379-123$ which implies that $N$ is rational. – lulu Nov 09 '21 at 14:57
  • @lulu You are making the natural assumption (for example) that a base $(5)$ decimal number represented by $0.1\overline{0}$ is to be included in the OP's intent behind the phrase eventual repeating decimal expansion. This is a reasonable interpretation that might not be what the OP intends. In fact, if the OP does not intend this, then you have two situations to consider: [1] The OP is concerned about only one specific fixed prime $p$, in which case you have a problem. ...see next comment – user2661923 Nov 09 '21 at 15:34
  • [2] The OP is instead interested in any numbers that might be expressible without trailing zeroes, for any prime $p$. Under such an interpretation, it would then be sufficient to show (for example) that for any base $(5)$ decimal of form $0.1\overline{0}$, there exists some other prime $p$ such that it's decimal expression for the corresponding number has an eventual repeating expression other than $00000\cdots$. – user2661923 Nov 09 '21 at 15:37

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