Where am I making a mistake in showing that countability isn't a thing?

Question

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.

Answer 1

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.

Answer 2

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.