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I have self-taught myself undergraduate Real Analysis using S. Abbott's Understanding Analysis. I am now moving on to learning "graduate-level" Real Analysis using Stein & Shakarchi.

One thing, that I have seen come up a couple times now - not just in Stein & Shakarchi, but in some papers as well, is to create "base-n expansions" of numbers less than 1. Here is an example of doing this in base-3 (ternary) in an excercise.

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And a similar "binary" expansion for numbers less than 1 in a paper I was reading:

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I think I am just a bit confused on why these expansions are defined the way they are, and I'm even more confused by the fact that there are "ambiguities" - i.e ways to represent the same number with different expansions? I understand, mostly (although am not that comfortable with, honestly) creating base-n representations for natural numbers, but I am confused here. Can anyone explain intuitively why these numbers are defined the way they are?

Thanks!

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    $\begingroup$ the ambiguity is the same as 1 = 0.999... and for base 3, 1 = 0.222... It helps to think about there is no reason we use base 10, and this base is defined in precisely the same way. $\endgroup$
    – Nikolaj Pedersen
    Dec 20, 2021 at 13:15

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As @Nikolaj said the ambiquity is the same as in base $10$.

About why $x = \sum_{k - 1}^{\infty} \ a_k \ 3^{-k} \ \mid a_k \in \{0, 1, 2\}$:
How we represent a decimal number in base $10$ as a sum? Let $x = 0.15$, we can say:
$x = 0.1 + 0.05 = 1 \cdot 10^{-1} + 5 \cdot 10^{-2} + 0 \cdot 10^{-3} + \cdots \\ = \sum_{k - 1}^{\infty} \ a_k \ 10^{-k} \ \mid a = \{1, 5, 0, 0, \cdots\}$

The same happens in base $3$.

A example of the ambiguity in the sum:
$\sum_{k}^{\infty} \ a_k \ 10^{-k} \ \mid a = \{1, 0, 0, \cdots\} = \sum_{k - 1}^{\infty} \ a_k \ 10^{-k} \ \mid a = \{2, 2, 2, \cdots\} \\ \Rightarrow 1 = 0.\bar{2}$
This is true for any base $n$: $x = (x - 1).\overline{(n - 1)}$

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  • $\begingroup$ Thank you - that makes a lot of sense to me. I know there are many methods to quickly convert base 10 whole numbers to base n numbers, but is there an "algorithm" or process to do this for numbers less than one? $\endgroup$
    – DownstairsPanda
    Dec 20, 2021 at 14:41
  • $\begingroup$ @DownstairsPanda see this answer $\endgroup$
    – Tiago Cavalcante Trindade
    Dec 20, 2021 at 17:22

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