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I am wondering if $f: \mathbb{R}_+ \rightarrow (0,1)$ is an injection if $f$ just moves the decimal point to the left of each number an equal amount of times as how far the decimal point is from the right of the first digit of the real number which has the decimal point featured to the furthest point right from its first digit, adding a $0$ every time we move the decimal point left and there's nothing there: e.g. if we had $3.14, 314.1, 34567.2$ which $f$ was mapping from, $f$ would output $0.0000314, 0.00314,$ and $0.34567$ respectively.

I am unsure because I don't know whether there is a well defined number which constitutes the greatest number of spots to the right which the decimal point is featured in a real number. If this isn't an injection, is there a way to use a similar idea to define an injection from $\mathbb{R}_+$ to $(0,1)$?

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    $\begingroup$ What's $f({1\over 3})$? $\endgroup$ Commented Sep 24, 2023 at 0:38
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    $\begingroup$ I don't understand the definition of $f$. It seems to only be defined for terminating decimals. I have no guess how to define $f\left(\frac 13\right)$, for instance. If you think it's clear, what is the value? $\endgroup$
    – lulu
    Commented Sep 24, 2023 at 0:43
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    $\begingroup$ @Shmuel OK, so then what is $f({1\over 3})$? It's possible I'm misunderstanding your idea; I'll be honest, your description " moves the decimal point to the left of each number an equal amount of times as how far the decimal point is from the right of the first digit of the real number which has the decimal point featured to the furthest point right from its first digit" is not very clear to me. $\endgroup$ Commented Sep 24, 2023 at 0:57
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    $\begingroup$ The fact that you can't simply tell us what $f\left(\frac 13\right)$ is suggests that you don't understand your definition either. Frankly, your description is extremely hard to follow. $\endgroup$
    – lulu
    Commented Sep 24, 2023 at 1:01
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    $\begingroup$ Or, the answer is "no, this is not an injective function from $\mathbb{R}_+$ to $(0,1)$ because it is not a function with domain $\mathbb{R}_+$ at all". $\endgroup$
    – aschepler
    Commented Sep 24, 2023 at 1:02

1 Answer 1

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Your proposal is rather unclear, but per your comments it does not give a function with codomain $\mathbb{R}$ at all; this is because things like

[a sequence] of the form 0.0000000000…33333333… where [the] amount of zeroes preceding the first 3 is some infinite number

is not actually (a valid representation of) a real number in the first place.


However, you may be interested to know that something similar to your idea does actually work! Specifically, letting $d(\alpha)$ be the number of digits in the integer part of $\alpha$ (so e.g. $d(\pi)=1$ and $d(12.3)=2$), the function $$g:\alpha\mapsto \alpha\cdot 10^{-2d(\alpha)}$$ gives an injection from $\mathbb{R}_{>1}$ to $(0,1)$. Some example values include:

  • $g(31.234)=0.0031234$.

  • $g(3.3333...)=g({10\over 3})=0.0333333...={1\over 30}$.

  • $g(103)=0.000103$.

The injectivity of $g$ comes from the fact that we can "read off" of $g(\alpha)$ the size of the integer part of $\alpha$, by looking at the number of zeroes in the decimal expansion (a bit of care is needed here around numbers with two decimal expansions, so I'm leaving the detailed proof of injectivity as an exercise).

Note that there's a crucial asymmetry here: the "left-of-decimal-point part" of a real number is required to be finite. This is why the above actually results in something well-defined, while your proposal runs into trouble with things like $1\over 3$.

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