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Let's say I have a number $q \in \mathbb{Q}$, i.e. $$\forall a \in \mathbb{Z}. \forall b \in \mathbb{N}^+. q = \frac{a}{b}$$

Now I look at $a$ as an endless sequence $(a_0, a_1, a_2, \ldots)$ where $a_i$ can be determined by the expression $a = \sum_{i=0} a_i \cdot 10^{i} $

and I look at $b$ as an endless sequence $(b_0, b_1, b_2, \ldots)$, where $b_i$ can be determined by the expression $b = \sum_{i=0} b_i \cdot 10^{i}$

Now I can combine these two sequences into one sequence, that looks like $(a_0, b_0, a_1, b_1, \ldots)$. Note that this mapping of a rational number to an endless sequence is a bijection. For every rational number, there is an endless sequence.

Now let's say I take all existing rational numbers and created endless sequences like this, and I would write them like this:

$$ \begin{align} a_{0_{q0}} b_{0_{q0}} a_{1_{q0}} b_{1_{q0}} \ldots \\ a_{0_{q1}} b_{0_{q1}} a_{1_{q1}} b_{1_{q1}} \ldots \\ a_{0_{q2}} b_{0_{q2}} a_{1_{q2}} b_{1_{q2}} \ldots \\ \vdots \end{align}$$

Could I use this structure to apply Cantor's diagonal theorem and prove rationals to be uncountable? If not, where is the contradiction?

hgiesel
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    Rationals being countable is kind of the point – hgiesel Jun 05 '17 at 01:49
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    Thing is, are you sure that those terms in the sequence are rationals? Actually, one step earlier, are $a$ and $b$ rationals as you defined? – AspiringMathematician Jun 05 '17 at 01:49
  • @AspiringMathematician Of course, I can map them back to the rationals, just like I did, this mapping I did is bijective – hgiesel Jun 05 '17 at 01:50
  • a and b are not rationals, and that is fine, q is the rational number – hgiesel Jun 05 '17 at 01:50
  • If I can't prove it, there must be some kind of contradiction along the way. If I can prove it, there is none – hgiesel Jun 05 '17 at 01:53
  • From what I read there, you assume that $a=\sum_{k=0}^n a_k\times 10^k$, so your sequence is finite. – AspiringMathematician Jun 05 '17 at 01:55
  • If you meant $a=\sum_{k=0}^\infty a_k\times 10^{-k}$, then the contradiction lies within the first statement, that $a \in \mathbb{Z}$ – AspiringMathematician Jun 05 '17 at 01:56
  • @AspiringMathematician That was just a typo, fixed it, I first thought that I had to deal with the case of being less than zero, but I came to the conclusion that I did not – hgiesel Jun 05 '17 at 02:00
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    Let's say we have $a = 17$ and $b=12$ then the sequence is: ${7,2,1,1,0,0,0,\ldots}$ so it only has finite non-zero elements. This is the case for all rationals so the number you construct which are not supposed to be in the list will have infinite non-zero elements and will not correspond to a rational number. – Winther Jun 05 '17 at 02:01
  • @hgiesel Your sequence $a_i$, as defined, is finite, because $a$ must have finite digits. If not, then $a=\infty \notin \mathbb{Z}$. – AspiringMathematician Jun 05 '17 at 02:04
  • @Winther Why do infinite non-zero elements not correspond to rational numbers? A rational number is a number $q = a \div b$, there is nothing about finite digits in that definition – hgiesel Jun 05 '17 at 02:09
  • @AspiringMathematician infinite digits does not incur inifinity. – hgiesel Jun 05 '17 at 02:12
  • Let's say $(a_0,b_0,a_1,b_1,\ldots) = (1,1,1,1,1,1,\ldots)$. Then $a = b = 1 + 10 + 100 + 1000 + \ldots$ right? This is not a number. In order for it to be a number the digits must eventually be just $0$'s. – Winther Jun 05 '17 at 02:12
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    Your encoding scheme is flawed (why not just choose the usual decimal representation of a rational?) but pointing that out doesn't really get to the heart of the question IMO. However, there are questions that address the fixed version, this one for example. Searching for "Cantor diagonal rational" returns more. – pjs36 Jun 05 '17 at 02:56
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    besides all other errors, the formula $\forall a \in \mathbb{Z}. \forall b \in \mathbb{N}^+. q = \frac{a}{b}$ is nonsense. – miracle173 Jun 05 '17 at 03:33
  • @miracle173 Then how do you formulate it correctly? $\forall a \in \mathbb{Z} \wedge \forall b \in \mathbb{N}^{+} . \exists q = \frac{a}{b}$ ? – hgiesel Jun 06 '17 at 11:39
  • $$ \exists a \in\mathbb{Z} ; \exists b , \in \mathbb {N^+} ;\frac{a}{b}=q$$ or $$\forall q \in \mathbb{Q} ; \exists a \in\mathbb{Z} ; \exists b , \in \mathbb {N^+} ;\frac{a}{b}=q$$ – miracle173 Jun 06 '17 at 11:55
  • @miracle173 Thank you very much – hgiesel Jun 06 '17 at 11:59
  • @infinitylord Your comments are based on a misconception that many people have. Just because you can prove something it doesn't mean that you can't disprove it. After all, mathematics could turn out to be inconsistent. – Jonathan Schilhan Jun 07 '20 at 11:37

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The error is here:

Now I look at $a$ as an endless sequence $(a_0,a_1,a_2, \dots)$ where $a_i$ can be determined by the expression $a=\sum_{i=0}a_i\cdot 10^i$ and I look at $b$ as an endless sequence $(b_0,b_1,b_2,\cdots)$, where $b_i$ can be determined by the expression $b=\sum_{i=0}b_i\cdot 10^i$.

While these "endless sequences" are, I suppose, technically "endless", after finitely many terms the sequences will become all zeros. This is because $a$ and $b$ are both integers, and therefore have only finitely-many digits in their decimal expansion.

Likewise, when you interlace the sequences of $a$s and $b$s, you get a sequence of $c$s that eventually becomes all zeros.

When you apply the Cantor diagonalization trick, you end up with a new sequence that, by construction, ends with infinitely-many nonzero entries. While it's true that that sequence of digits does not already exist in the list, it does not correspond to a rational number, because the way you have set up your correspondence, a rational number corresponds to a sequence that eventually becomes all zeros.

mweiss
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    So actually the error is "Note that this mapping of a rational number to an endless sequence is a bijection" – miracle173 Jun 05 '17 at 03:31
  • I guess the issue is, that while you can go indefinitely behind the decimal point, you can't do so in reverse – hgiesel Jun 06 '17 at 11:35