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I had an argument about the following statement, which I think is wrong:

One third of odd numbers (33.333%) will be divisible by 3.

  1. Let $\Bbb{N}$ refer to the positive integers.

  2. Let $\Bbb{O} := \{o | o = 2n-1; n \in \Bbb{N}\}$, the set of odd numbers.

  3. $2n-1$ is a (trivial?) bijective function to $\Bbb{N}$.

  4. Thus $\Bbb{O}$ is countably infinite.

  5. Thus $|\Bbb{O}| = |\Bbb{N}| = \aleph_{0}$.

  6. Let $\Bbb{M}(m) := \{o | o = 2n-1 \land o\mod m = 0; m,n \in \Bbb{N}\}$, the set of odd numbers which are divisible by $m$ without remainder.

  7. The additional condition $ o\mod m = 0$ does not remove the bijective property of the odd numbers, as I think we can do a kind of Hilbert Hotel here: For $\Bbb{O}$ we have $1 \to 1$, $2 \to 3$, $3 \to 5$, $4 \to 7$, $5 \to 9$, $6 \to 11$, $7 \to 13$, $8 \to 15$, $7 \to 17$, .. For e.g. $\Bbb{M}(3)$ we have $2 \to 3$, $5 \to 9$, $8 \to 15$, .. which can be renumbered as $1 \to 3$, $2 \to 9$, $3 \to 15$, ..

  8. Thus $\Bbb{M}(m)$ is countably infinite.

  9. Thus $|\Bbb{M}(m)| = |\Bbb{O}| = \aleph_{0}$.

  10. The quoted statement is now in those terms $\frac {|\Bbb{M}(3)|}{|\Bbb{O}|} = \frac {\aleph_{0}}{\aleph_{0}} = 1 \neq \frac {1}{3}$, which is a contradiction.

Where am I wrong? Or where is more precision in my argumentation needed?

ste5an
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    It's difficult to work with infinities this way. To get at the sort of thing you (probably) are looking for, take a limit. Show that for odd numbers with absolute value $<N$ roughly $\frac 13$ of them are divisible by $3$ and that this becomes exact in the limit. – lulu May 09 '23 at 18:32
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    Welcome to Mathematics Stack Exchange. For an infinite set, its cardinality can be the same as that of a subset – J. W. Tanner May 09 '23 at 18:52
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    You might find the concept of “natural density” interesting. https://en.wikipedia.org/wiki/Natural_density. – Joe May 09 '23 at 18:57
  • Okay, so 10) is out of question and I need to look into the limit/density argument. But to get my wording and math correct: 3) am I allowed to call the generator function trivial? Is the Hilbert Hotel approach in 7) the correct argument or do I need to go for something different here? Thus, is 9) correct? – ste5an May 09 '23 at 22:04
  • @ste5an everything except for 10 looks good to me. Your reasoning shows that you have to be careful doing arithmetic with cardinals and talking about densities of infinite subsets of a set – doetoe May 09 '23 at 22:28

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In step 6, allowing "$m,n\in\mathbb N$" is too permissive. Consider that $\mathbb M(2)$ is the empty set! It would be better to require that $m$ is an odd prime in the definition, or at least that $m$ is an odd number. Then you can reason about general $\mathbb M(m)$ by calling on the Chinese remainder theorem.

In step 7, you're right that we can "do a kind of Hilbert Hotel", but you might want to write down the explicit bijection $\mathbb N\to\mathbb M(m)$ instead of hand-waving it. It's a simple enough formula, and proving that it works will help to show what kind of condition you need on $m$.

In step 10, as the comments say, that sort of division isn't going to work. Cardinal arithmetic is tricky; for example, $3\aleph_0=\aleph_0$.

Chris Culter
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