All sets of rational numbers are bigger than the set containing infinite integers - or are they?

Question

Intro

This started with me learning the different types of infinity. I like to call them types instead of sizes due to the fact, that infinite is defined by being endless or "not-finite" - meaning not a size. (I do know the right definition is different sizes).

My Hypothesis

I started trying to match integers to rational numbers one-to-one (or bijection). I have found, in the top answer in this question, that:

Two sets $A$ and $B$ are said to have the "same size" if there is a some function $f:A\to B$ which is a bijection. Note that we do NOT require that ALL functions be bijections, just that there is SOME bijection.

The way I found it possible was by limiting the set of rational numbers to $[0,1)$. Now by reversing the order of the decimals I could map all rational numbers excluding fractions resulting in an endless repeating sequence of decimals, thus matching:

• 0.034 to 430
• 0.2331 to 1332
• ...

As said, this maps any rational numbers excluding fractions resulting in an endless repeating sequence of decimals. Now the method to mapping those.

($\overline{\text{Overline}}$ means repeated endlessly)

The workaround with this is made in two steps:

• $\frac47 = 0.571428\overline{571428}$

If we accept this as a number I assume that this following is acceptable as well:

• $824175\overline{824175}$

Question

Does this mean, that there are exactly as many rational numbers in the set $[0,1)$ as there are integers in the set $[0,\infty]$?

And furthermore that in the rational set $[0,1]$ contains one more number that the set of integers $[0,\infty]$?

Bonus Quention

Is this allowed: $10\times824157\overline{824157}$?

EDIT

Winther made it clear that $824157\overline{824157}$ not is a real number. Thanks for that. However ... (I don't give up that easy)

If every fraction NOT ending in an infinite repeating is saved as already stated except set as next free even integer (multiplying with 2) - Just as in Hilbert's paradox of the Grand Hotel - with infinite new guests. Thus making:

• 0.1 -> 1 -> 2
• 0.2 -> 2 -> 4
• ...
• 0.5 -> 5 -> 10
• ...
• 0.01 -> 10 -> 20

The fractions not already mentioned (the ones with infinite repeating) will then get the uneven integers. The way to list these will then be done with Cantor's Diagonal listing:

• 1/1 is not in [0;1[ so moving on
• 1/2 is already represented (0.5 -> 5 -> 10) see above
• 2/2 again not in [0;1[
• 1/3 is not represented yet. 1/3 -> 1
• 2/3 is not represented yet. 2/3 -> 3
• ...
• 1/6 is not represented yet. 1/6 -> 5
• ...
• 4/6 is not represented yet. 4/6 -> 7

New Question

Is this proof then?

Answer 1

Integers can be broken up into 3 subsets: $\{ 0 \}$, ${+,\mathbb{N}}$ and $\{-, \mathbb{N}\}$, i.e. 0 and two instances of the set of naturals, one for positive integers and the other for negatives. The integer $0 $ corresponds with the rational $0$. If we can construct a correspondence between the positive rationals and the naturals, then we can use it twice to map both of the aforementioned infinite subsets of the integers to the positive and negative rationals.

The Stern-Brocot tree is a very useful organization of the positive rationals. Each path in the tree corresponds to a distinct positive rational, and every positive rational is present in the tree.

Naturals and finite binary strings have an obvious one-to-one relation through the use of binary numerals, with the exception of the empty string. But you can accommodate it by just sticking it in at the beginning.

Finite binary strings and paths in the Stern-Brocot tree correspond by replacing '0' with "Left" and '1' with "Right".

Each finite path in the Stern-Brocot tree corresponds to a positive Rational. All paths lead to a rational, and every positive rational is present in the tree.

So the whole chain of equivalences would be:

$$\mathbb{Z} \Leftrightarrow \{0\} \cup \{+,\mathbb{N}\} \cup \{-,\mathbb{N}\} $$ $$\text{let} B = \left[\mathbb{N} \Leftrightarrow \text{finite binary strings} \Leftrightarrow\text{finite paths in Stern-Brocot tree} \Leftrightarrow \{ \mathbb{Q} > 0\}\right]$$

$$\{0\} \Leftrightarrow \frac01$$ $$\{+,\mathbb{N}\} \Leftrightarrow_B \{\mathbb{Q}>0\}$$ $$\{-,\mathbb{N}\} \Leftrightarrow_B \{\mathbb{Q}<0\}$$

Answer 2

Firstly, you should not say "infinite integers" if you mean "infinitely many integers". If one admits such a thing as an infinite integer, and $n$ is an infinite integer, then $n$ and $n+1$ are infinite integers, but they are not infinitely many, since there are only two of them. This is standard usage in mathematics, regardless of what usages may prevail in informal English used in contexts other than mathematics.

The sequence below has a first term, a second term, a third term, and so on, so there are just as many terms as there are positive integers $1,2,3,\ldots\,{}$. \begin{align} \frac 1 2, \underbrace{\frac 1 3, \frac 2 3},\underbrace{\frac 1 4,\frac 3 4},\underbrace{\frac 1 5,\frac25,\frac35,\frac45},\ \underbrace{\ldots\ldots\ldots\ldots}_\text{sixths},\ \underbrace{\ldots\ldots\ldots\ldots}_\text{sevenths},\ \ldots \end{align} And all rational numbers between $0$ and $1$ are in this list (you'll notice I skipped $2/4$ since it's not in lowest terms).

If you want to add one more rational number, or two more (e.g. $0$ and $1$) then just append them to the front of the list and you'll see that there are still just as many as there are integers.

This differs from your argument and doesn't directly answer your question, but I hope it clears up a couple of confusions that might get in the way of your thinking about this.

Answer 3

Your proof (after the edit, with the separate even and odd sequences of non-repeating and repeating decimals) seems correct, except for the claim (just before the edit) about one "extra" rational number in the interval $[0,1]$. (Remember how you can always book one more guest in the Hilbert Hotel.)

The part that you added to the proof to handle the repeating decimals is actually very similar to a more usual way of enumerating all the rationals (including repeating and non-repeating decimals) in $(0,1]$, so with just a little tweaking you could actually drop the first half of the proof and use only the second half.

As a side note, a slight variation in the order in which we try different numerators and denominators (including numerators that are larger than their denominators) gives an enumeration of all positive rationals.