Cardinality of rationals less than 1, and greater than 1

Question

Say you arrange the elements of $\mathbb{Q}$ in an $m \times n$ matrix, similar to other proofs involving $\mathbb{Q}$:

$$ \begin{matrix} & 1 & 2 & 3 & \ldots\\ 1 & \frac{1}{1} & \frac{2}{1} & \frac{3}{1} & \ldots\\ 2 & \frac{1}{2} & \frac{2}{2} & \frac{3}{2} & \ldots\\ 3 & \frac{1}{3} & \frac{2}{3} & \frac{3}{3} & \ldots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{matrix} $$

I've noticed that the diagonal entries $\frac{1}{1}, \frac{2}{2}, \frac{3}{3}, \ldots$ are all equivalent to $1$, and that the numbers above and below each diagonal entry are greater than and less than $1$ respectively. Extending the matrix infinitely in both direction makes a sort of square with a diagonal line of $1$'s through the middle, and triangles containing entries less than or greater than $1$ below and above it. I'm pretty sure you could write a bijective function relating the two.

So it seems that there are just as many rationals less than 1 as there are greater than 1. But this seems strange. I understand the proof of $|\mathbb{Q}| < |\mathbb{R}|$, and that there are more reals between $0$ and $1$ than there are natural numbers, but how can it be that there are as many rationals between $0$ and $1$ as there are $1$ and infinity?

I'm guessing that all intuitions about respective size go out the window with infinite sets, but is there more to the answer than that? Thanks.

Answer 1

They don't go out the window but they stand on the sill and wave to the passerbys in the street.

Basically the bijection you describe is $f:(0,1)\cap \mathbb Q \rightarrow \mathbb (0,\infty) \cap Q$ so $f (m/n) = n/m $. Or in other words $f (x)=1/x $.

It's a perfectly good bijection. As to seeming strange. It's a bit, but if you think about it, it's on par with "there are the same number of even integers as there are all integers". And neither are as strange as "there are the same number of integers as rational numbers".

Bottom line: unlike finite sets, infinite proper subsets of countable infinite sets can still (actually must) have the same cardinality.

Bottom line below: "same size" doesn't actually conflict with "can fit inside each other". It's the concept that "if something fits inside another it must be smaller" that goes out the window. Cardinality and size stretch and sit on the edge, but "fit inside means smaller" jumps out to its well deserved death.

Answer 2

You can explicitly write a bijection between $\mathbb Q\cap(0,1]$ and $\mathbb Q\cap [1,\infty)$ by $$ f(a)=\frac1q. $$ That is, you can explicitly see $[1,\infty)$ as a stretch of $(0,1]$.

Answer 3

You are correct that you have a bijection between the sets, so they must have the same cardinality. You can say more. As the rationals have cardinality $\aleph_0$, which is the smallest infinite set, any infinite subset also has cardinality $\aleph_0$. All infinite sets of rationals have the same cardinality.