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I am not a mathematician so I might not be entirely accurate in my mathematical depictions, but I will correct them if I can.

My problem is like that: I have $f_n:\mathbb R ^+\to\mathbb R ^+$ with $f_{n}(x)=x_{m_{n}}$ For the moment I will explain the function using a $n=10$ as $n$ is the basis in which the number is written and "processed" by my function.

So given the normal decimal basis I can write the function as: $f(\sum c_i*10^i)=\sum c_i*10^{-1-i}$. I think the sum goes from $-\infty$ to $\infty$ for any given number as the digits that are not written are zeros.

So what my function does (or at least if think it does) is basically a mirroring of digits over the decimal point (ex.: $f(23.45)=54.32$).

I should be easy to prove that the function is bijective as $f\circ f=1_{\mathbb R ^+}$.

Now given the established function my problem comes when I restrict the function to the interval $(0;1)$ as i believe that $ Im_f(0;1)=\mathbb N ^* $ which seems to indicate that ${\mathbb R}$ could be countable.

EDIT: Since this was a commonly raised problem about my question I will add as many forms as I have encountered it in with what I hope are satisfying answers:

  • $\mathbb N$ has "finitely many digits": Note the "finite number of digits" with $k$. Take $10^{k+1}$.

  • $\mathbb R^+$ can not have numbers with infinite digits (to the left of the decimal point) then $lim_{x\to\infty}\int log_{10}x\ dx=?$

  • any other proof of something is finite using induction is also flawed from a simple perspective: $\mathbb N$ is not finite to begin with

  • the argument about the lack of a successor and a predecessor holds no ground. If i can't add or subtract 1 from a decimal number knowing all the digits involved (at least in theory) I would have to go and start school again from the first grade.

piciaxel
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    What is the value of your function at $x=\frac13$? – Steven Stadnicki Jan 27 '18 at 01:34
  • Or, at $x = \pi - 3 = 0.14159\ldots$? – xyzzyz Jan 27 '18 at 01:35
  • (Also, you probably mean $f(\sum_ic_i\times10^i)=\sum_ic_i\times10^{-1-i}$; that actually flips around the decimal point as you suggest, since otherwise you would have $f(1)=1\times 10^{1-0} = 10$, rather than $f(1)=.1$) – Steven Stadnicki Jan 27 '18 at 01:36
  • Since for $x=1/3$ that would be a infinity of $3$ to the right of the point, then i guess the value of my function would be a infinity of $3$ to the left of the point. Also thanks for noticing that mistake. – piciaxel Jan 27 '18 at 01:36
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    @piciaxel The problem is, that sum - $\sum_{i\geq0} 3\times 10^i$ - doesn't converge, and so it doesn't define a member of $\mathbb{R}^+$. – Steven Stadnicki Jan 27 '18 at 01:39
  • @Steven Stadnicki Yeah, but what kind of number is $\sum_{1 \ge 0} 3 \times 10 ^i$ – piciaxel Jan 27 '18 at 01:47
  • It is not a number at all. It does not exist. – fleablood Jan 27 '18 at 02:01
  • @piciaxel Note that you can sometimes give a meaningful definition to such a sum - for instance, they may converge in the p-adic numbers. But it won't be a member of $\mathbb{N}$, or even $\mathbb{R}$, so your purported bijection doesn't work at all. – Steven Stadnicki Jan 27 '18 at 02:11
  • @Steven Stadnicki I'm sorry for being a bit insistent about this subject, but I still feel that a number like $\sum_{i \ge 0} 3 \times 10^i$ would be a Natural number because it is positive and it does not have a fractional part. If $\mathbb N$ is truly infinite I expect it to contain all positive, non-fractional numbers, not just some of them. – piciaxel Jan 27 '18 at 02:29
  • @piciaxel $\mathbb{N}$ has a well-defined meaning, and the 'number' produced by applying your procedure to $0.33\bar{3}$ can't be a member of $\mathbb{N}$, essentially because it's not finite. For instance, it can be proven by induction that every whole number has a finite number of digits: see https://math.stackexchange.com/questions/1729362 and https://math.stackexchange.com/questions/1088877 which talk about essentially the same issue. – Steven Stadnicki Jan 27 '18 at 03:09
  • (In fact, this question is essentially a duplicate of the first one there) – Steven Stadnicki Jan 27 '18 at 03:10
  • If by $\Bbb N^$ you mean the set of "hyper-naturals" in an ordered-field extension $\Bbb R^$ of $\Bbb R,$ note that in general $\Bbb N^*$ need not be a countable set. – DanielWainfleet Jan 27 '18 at 03:46
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    It never helps to bring up things —p-adic numbers, hyperreal numbers, and so in— to deal with people that disprove the uncountability of the real numbers. By far, the most useful things one can do is to emphasize that the "argument" makes no sense, because there is something basic (here, that a simbol that looks like a sum does not denote anything) is broken. – Mariano Suárez-Álvarez Jan 27 '18 at 04:43
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    Here the problem is that the OP wrote something (that looks like a sum) that does not mean anything. Bringing up the fact that when the number is 1/3 then the sum has a meaning provided we consider the 3-adic valuation is of no help; for there is also 1/7, and so on. – Mariano Suárez-Álvarez Jan 27 '18 at 04:45
  • The problem is that bringing in technology is only satisfying to the one bringing it in, and not to be OP — how could the fact that weird sums converge p-adically help if one does not know that that series in the question does not. – Mariano Suárez-Álvarez Jan 27 '18 at 04:47
  • Being positive and having no fractional part is not the definition of a natural number. A natural number must have a value. And it must have successor and other than the first be a successor to another. Describing an infinite sum does not mean it's resolvable. – fleablood Jan 27 '18 at 05:09
  • If somehow you give meaning to this infinite sums and make them distinct and give them meaning, it still doesn't matter. The set of infinite sums is not countable. – fleablood Jan 27 '18 at 06:12
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    That "sum" is neither positive not negative: it is nothing, as it makes no sense. For something to be a natural number, it has to be something, and string of symbols you wrote does not denote anything. – Mariano Suárez-Álvarez Jan 27 '18 at 06:15
  • @piciaxel Regarding the latest edit N has "finitely many digits" ... - no offense, but that makes no sense whatsoever, and I don't think you grasped what the objections that were raised were about. – dxiv Jan 27 '18 at 06:40
  • @dxiv That is exactly the answer to the question that is associated with my post. It has a weird demonstration that overall does not explain anything. The main reason for which i pose the question was to get a real in-depth explanation which i can't find. I only find people who snap at me that what i wrote makes no sense and people who direct me something that has either a flawed or incomplete demonstration/explanation. – piciaxel Jan 27 '18 at 07:39
  • @piciaxel No, what was said is that each $n \in \mathbb{N}$ has finitely many digits. It doesn't make sense to say that $\mathbb{N}$ itself has finitely many digits. You need to be more careful with the math language, and that's largely how you managed to confuse yourself with the question itself, too. – dxiv Jan 27 '18 at 16:53

2 Answers2

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What would happen if you mirrored an irrational number over the decimal point? f is not bijective since f isn't even defined for any number with an infinite decimal expansion. However, you could prove that the set of all numbers with finite decimal expansions is countable. Not what you wanted, but it's something I guess.

Noel Lundström
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Let's say that $n = \sum_{i=1}^{\infty} c_i 10^{-i}$ is an infinite nonterminating decimal. (It could be a repeating rational or an irrational).

If your function is well defined then $f(n) = \sum_{i=1}^{\infty} c_i 10^{i-1}$.

But this is an infinite sum of increasingly large powers of $10$. Such a sum is infinite and has no numeric value. The function does not return any real number for such non-terminating decimals.

This function is not a function and is not well defined unless we limit its domain only to terminating decimals.

fleablood
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  • assuming the given $n$, in which set would $f(n)$ fit? – piciaxel Jan 27 '18 at 02:05
  • None. It is garbage. It doesn't belong in any set. – fleablood Jan 27 '18 at 02:06
  • The same set that 5 sided squares, round triangles, and the integer solutions to $2^k = 13k$, and tiny smooth white elephants belong to. Just because you can write something doesn't mean it exists. There is no such thing. It does not exist. – fleablood Jan 27 '18 at 02:11
  • If it is not a number, then what is it? Help me understand. I don't believe that there is nothing there. Even garbage can be sorted so why is this type of whatever this is so ... unknown. Those examples u gave contradict definition clearly... What does this contradict? – piciaxel Jan 27 '18 at 02:12
  • What's wrong with letting $f_n : \mathbb{R} \to \mathbb{R}\cup{\infty}$ ? The function loses its bijectivity, but it still would be well defined. – Sudix Jan 27 '18 at 02:21
  • @Sudix And then what would the $Im_f(0;1)$ be? That infinity tag-along seems a bit weird... How would I define the function so it only takes values in $\mathbb R$, and why does it lose it's bijectivity even if it has a $f^{-1}$ being itself (or at least seems to). – piciaxel Jan 27 '18 at 02:51
  • It contradicts that numbers have finite distinct values. – fleablood Jan 27 '18 at 03:50
  • Infinity is in the extended real numbers but then this would map to the union of the naturals and the infinity. It would certainly not be bijective. – fleablood Jan 27 '18 at 03:53
  • @piciaxel The image would be all numbers with finite decimal representation as well as infinity: ${n\in\mathbb{Q}\mid \exists i,j\in\mathbb{N}: 2^i\cdot 5^j\cdot n\in\mathbb{N}}\cup{\infty}$ – Sudix Jan 27 '18 at 05:13
  • You can define a function pretty much however you want, the only necessities are that both domain and codomain are sets, and that every element of the domain gets mapped to exactly one point in the codomain. In the case of my definition, $f_n$ has no bijectivity because e.g. every irrational number maps to $\infty$, meaning it's not injective – Sudix Jan 27 '18 at 05:20
  • Sudix. I think the image is simply $\mathbb N \cap{\infty}$. It has no non natural rationals I think. If the infinite sums are infinity then irrationals an all repeating rationals all map to the same $\infty$ so the function isn't bejective. – fleablood Jan 27 '18 at 06:10