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Consider a function f that takes all the supraunitary digits of the real number and assign them on the odd positions of the natural number and all the subunitary digits in reverse order and assign them on the even positions of the natural number. This mapping is unique.

How would Cantor's diagonal argument work on this mapping? Surely for any real number r, constructed using this argument, f would construct a unique natural number, which is part of the mapping, contradicting that r isn't mapped.

A more formal definition for the proposed function and its inverse:

Let $f\colon\mathbb{R}\to\mathbb{N}$ such as for any real number with the decimal representation $a_n ... a_2 a_1. b_1 b_2 ... b_m$ with $a_n \neq 0$ and $b_m \neq 0$: \begin{equation} f(a_n ... a_2 a_1. b_1 b_2 ... b_m)= \begin{cases} b_m 0 \dots b_2 a_2 b_1 a_1& \text{if}\ n<m\\ b_n a_n \dots b_2 a_2 b_1 a_1& \text{if}\ n=m\\ a_n 0 \dots b_2 a_2 b_1 a_1& \text{if}\ n>m\\ \end{cases}\label{fdef} \end{equation}

Let $f^{-1}\colon\mathbb{N}\to\mathbb{R}$ such as for any natural number with the decimal representation $\dots x_6 x_5 x_4 x_3 x_2 x_1$: \begin{equation} f^{-1}(\dots x_6 x_5 x_4 x_3 x_2 x_1)= \dots x_5 x_3 x_1. x_2 x_4 x_6 \dots\label{finvdef} \end{equation}

ctapus
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  • How would it work? You have the literal mapping. Show how it works. – Asaf Karagila Jan 05 '17 at 15:34
  • Also, last time I checked, the integral part of a real number had only finitely many decimals. I might be wrong about that, though. – Asaf Karagila Jan 05 '17 at 15:35
  • But every natural number has only finite number of digts. – Zoran Loncarevic Jan 05 '17 at 15:35
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    "Surely for any real number r, constructed using this argument, f would construct a unique natural number." Sentences like that are exactly where to look for the mistake - in particular, $f$ does not in general yield a natural number, as stated above and below. When you're trying to prove something, always focus special attention on the claims that are "too obvious" to warrant explicit proof (and this doesn't just apply to you; I've been nailed by this more times than I am comfortable admitting). – Noah Schweber Jan 05 '17 at 15:42
  • @AsafKaragila The mapping would work like this: 1 is mapped to 1; 0,1 is mapped to 10; 10 is mapped to 100; 123.456 is mapped to 415263; etc. – ctapus Jan 05 '17 at 22:43
  • @ZoranLoncarevic The set of natural numbers is infinite, so there are naturals number with an infinity of digits. – ctapus Jan 05 '17 at 22:47
  • @NoahSchweber By its definition f will yield a string of digits without any decimals, eg f(1.2)=21; f(12)=102; why are those not natural numbers? – ctapus Jan 05 '17 at 22:53
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    @ctapus "The set of natural numbers is infinite, so there are naturals number with an infinity of digits." That's nonsense. Every natural number is finite - so every natural number has finitely many digits. Just because the set of natural numbers is infinite, doesn't mean any individual natural number is infinite. (Incidentally, there are number systems where "infinite leftward expansions" make sense - look at the $p$-adics, for instance - but $\mathbb{N}$ isn't one of them.) – Noah Schweber Jan 05 '17 at 22:55
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    @ctapus Re: your comment to me, those are - but something like "$...3030303030$" isn't. (Incidentally, if you insist on such things being in your number system, how do you compare them? Which is larger: $...121212$ or $...212121$? Note that they "keep switching": $2>1$, $21>12$, $212>121$, .... So even if you insist on a number system with such weird objects (which, to reiterate, will not be $\mathbb{N}$), things are going to get ugly.) – Noah Schweber Jan 05 '17 at 23:02
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    I hope you agree that every non-empty set of natural numbers has a minimal element. What is the smallest natural number to have infinitely many digits? – Asaf Karagila Jan 06 '17 at 07:43
  • @NoahSchweber "Every natural number is finite - so every natural number has finitely many digits." Thanks for this insight, it would be great if you could provide me with a resource on this topic. I've seen this statement in few other topics on mathflow, but they don't go into much details. However, going back to Cantor's argument, he's also allowing numbers with infinitely many digits. Or am I reading this wrong? https://en.wikipedia.org/wiki/Cantor's_diagonal_argument – ctapus Jan 07 '17 at 12:44
  • @AsafKaragila This is interesting, but the fact that you can only think about something only in abstract terms doesn't mean it doesn't exist? I can ask for example what's the next number after 1/3 (or 0.33... to stay in decimal notation). You can't express it but it does exist. – ctapus Jan 07 '17 at 12:47
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    I have negative idea what you're talking about. Note that 1/3 is *not* a natural number, nor its integral part (which is 0, by the way) has infinitely many non-zero digits. What is the next number after 1/3, by the way? You got me curious. – Asaf Karagila Jan 07 '17 at 12:50
  • @AsafKaragila Like I said, it can't be expressed, but you also agree that it exists ( http://math.stackexchange.com/questions/696043/why-there-is-not-the-next-real-number ) I was just pointing to the fact that for real numbers we are happy to accept them having infinitely many digits, even if we are unable to use some of the number system's properties. This comment was made also in regards with the assertions that natural numbers have to have a finite number of digits, otherwise we are not able to make use of some of their properties (like ordering). – ctapus Jan 10 '17 at 13:43
  • I have no idea why you say that I agree that it exists. I literally explained that it does not exist. And my point is that the "number" that you map 1/3 will have infinitely many digits in its integral part. Which is therefore not a natural number. – Asaf Karagila Jan 10 '17 at 13:44
  • Probably a confirmation bias when I read this: <<So if we endow the set of real numbers with a well-ordering then there is a notion of "the next real number".>> And from wikipedia (https://en.wikipedia.org/wiki/Well-order#Reals) "[...] it is consistent with ZFC that a definable well ordering of the reals exists". – ctapus Jan 11 '17 at 14:08

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The problem is that almost all real numbers will have infinitely many nonzero digits $b_i$ following their decimal point. These numbers are not mapped to natural numbers by your scheme, since they have an infinite number of digits, while all natural numbers only have a finite number of digits.

For instance $1/3$ gets mapped to $\dots 30303030$ with an infinite number of copies of $30$, which is not a natural number.

Lothar Narins
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  • Why 30303030... is not a natural number? Every natural number has a successor, which means that there are infinitely big natural numbers, meaning there are natural numbers with an infinity of digits. And wouldn't Cantor's proof for countability of rational numbers would suffer from the same issue? Given the fact that there are infinitely many prime numbers, in his ordering schema you will get to a rational number that is the ratio of two infinite prime numbers. – ctapus Jan 05 '17 at 23:00
  • @ctapus "Every natural number has a successor, which means that there are infinitely big natural numbers, meaning there are natural numbers with an infinity of digits." That's incorrect. And incidentally, the "number" $f$ gives on input ${1\over 3}$ isn't $303030...$ but rather $...303030$: it has a last term, but no first term. – Noah Schweber Jan 05 '17 at 23:04
  • @NoahSchweber Thank you for the insight, Noah. If I may draw a similarity with the proof of countability of Q, why can't real numbers be ordered by their number of digits? So first you will have real numbers with 1 digit (0..9), followed by numbers with 2 digits(10..99, with 0.1 .. 0.9) followed by real numbers with 3 digits, etc? Is it be because of numbers like 0.(3)? I mean if all the real numbers will be listed, 0.(3) should be there, mapped to a number with an infinite number of digits. This is what I was actually trying to capture in f's definition. – ctapus Jan 07 '17 at 17:42
  • @ctapus Yes, the problem is numbers like ${1\over 3}$ - they won't appear anywhere in your proposed list. Indeed, no matter how you define $f$, it will miss most of the reals; this is suggested by the experience you get playing around with a bunch of possible $f$s (and seeing how they each break), and proved by Cantor's argument. – Noah Schweber Jan 07 '17 at 17:50