Is the cardinality of the set of rational numbers, $\mathbb{Q}$ odd?

Question

Let $\mathbb{Q}$ be the set of unique rational numbers of the form $m/n$ where $m\in\mathbb{Z}$ and $n\in\mathbb{N}$.

Define the following two sets:

$$\begin{align} \mathcal{A}&=\{m/n\ \vert\ m/n \in \mathbb{Q},\ |m|>n\}\\ \mathcal{B}&=\{m/n\ \vert\ m/n \in \mathbb{Q},\ |m|<n,\ m\neq 0\} \end{align}$$

Now, $f:\ \mathcal{A}\to\mathcal{B},\ f(x)=1/x$ is bijective and so, the cardinality of $\mathcal{A}$ and $\mathcal{B}$ are the same. Denote $\kappa=\#\mathcal{A}$, where $\#$ denotes cardinality of the set.

Now we can write $\mathbb{Q}=\mathcal{A}\cup\mathcal{B}\cup\{0,1,-1\}$. Hence,

$$\begin{align} \#\mathbb{Q}&=\#\mathcal{A}+\#\mathcal{B}+\#\{0,1,-1\}\\ &=\kappa+\kappa+3\\ &=2(\kappa+1)+1 \end{align}$$

which is odd. Is this correct?

I do not know if this is well known or if my reasoning is correct at all. I apologize if this is a trivial question, but I'm not a mathematician and this has been nagging me for a while.

Answer 1

The problem is that the notions of "odd" and "even" don't make sense for infinite sets. Or rather, if you try to define "even" for cardinals as "equal to twice a cardinal", and define "odd" as "equal to twice a cardinal plus 1", then every infinite cardinal is both even an odd, so the notions become useless.

Explicitly, assuming the Axiom of Choice, for every infinite cardinal $\kappa$, $\kappa+1=\kappa$ (if we define cardinals to be ordinals that cannot be bijected with any smaller ordinal, then this follows from the fact that every infinite ordinal is bijectable with its successor).

And for every infinite cardinal $\kappa$, $2\kappa=\kappa+\kappa = \kappa$. So every infinite cardinal is twice an infinite cardinal (hence even), and also $$\kappa = 2\kappa = 2\kappa+1$$ hence also odd.

So, yes, you can express $\mathbb{Q}$ as the disjoint union of two infinite sets of equal cardinality and a finite set of odd cardinality. But you can also express $\mathbb{Q}$ as a disjoint union of two sets of even cardinality (first well-order the positive rationals, using any of the explicit bijections of $\mathbb{Q}$ with $\mathbb{N}$; then push everything forward by one by adding $0$ at the beginning; now biject with the negative rationals in the obvious way). So then you can just as well call it an "even cardinal".

So the notions become useless once you get to $\aleph_0$.

(Assuming the Axiom of Choice, if $\kappa$ and $\lambda$ are cardinals, at least one of them infinite, none of them $0$, then $\kappa+\lambda = \kappa\lambda = \max\{\kappa,\lambda\}$.)

Answer 2

One way to define an even cardinality is the cardinality of a set $X$ which is the union of two disjoint subsets $A, B$ of the same cardinality. Then every countably infinite set has even cardinality (and probably every infinite set does as well). This isn't completely pointless; it naturally occurs as a definition when describing square roots of infinite permutations.

Answer 3

This question doesn't mean anything. We know that $\mathbb{Q}$ is countably infinite, i.e. can be put into bijection with the natural numbers. Let $S=\{a_1,a_2,\ldots\}$ be any countably infinite set. There are obvious bijections between the sets $$\{a_{2i}\mid i\geq 1\}\hskip0.3in \{a_{2i+1}\mid i\geq 1\} \hskip0.3in \{a_{2i+1}\mid i\geq 0\}.$$ Thus

$$|S|=|\{a_{2i}\mid i\geq 1\}|+|\{a_{2i+1}\mid i\geq 1\}|+|\{a_1\}|=\text{''even''} + 1 = \text{''odd''}$$ $$|S|=|\{a_{2i}\mid i\geq 1\}|+|\{a_{2i+1}\mid i\geq 0\}|=\text{''even''} $$