Is there an easy to describe (using a formula or something more-or-less constructive, no zig-zag argument) an order-preserving bijection between the negative rational numbers and the rational numbers less than $\pi$?

Let $P$ be the set of all irrational numbers. Let $G$ be the set of all negative irrational numbers, let $H$ be the set of all irrational numbers greater than or equal to $\pi$, and let $Y=G\cup H$. Then $P$ and $Y$ are order-isomorphic, i.e. there is an order-preserving bijection between them.

One way to see this is to first construct an order-preserving bijection between the negative rational numbers and the rational numbers less than $\pi$, using a back-and-forth (zig-zag) argument (like the usual proof that every countably infinite dense linear order with no first or last element is order-isomorphic to the rationals). (Alternatively use that the negative rationals are a countably infinite dense linear order with no first or last element, and so are the rationals less than $\pi.$) Once we have this bijection we could (uniquely, e.g. using Dedekind cuts) extend it to an order-preserving bijection between $(-\infty,0)$ and $(-\infty,\pi)$ (and in fact between $(-\infty,0]$ and $(-\infty,\pi]$, but I prefer to only look at the order-preserving bijection between $(-\infty,0)$ and $(-\infty,\pi)$.) Note that the latter is an order-preserving bijection that sends rationals to rationals, and irrationals to irrationals. Taking it, together with the identity on $[\pi,\infty)$ (and restricting to the irrationals only) we get the required order-preserving bijection between $P$ and $Y$.

My question is whether there is an easier way to describe an order-preserving bijection between $P$ and $Y$. Or more "constructive" (in one way or another), or using some easy formula. E.g. one easy order-preserving bijection between $(-\infty,0)$ and $(-\infty,\pi)$ is translation by $\pi$, except it sends rationals to irrationals (and sends some irrationals to rationals, though it sends most irrationals to irrationals).

I feel there should be no easy way, since any easy way would send $0$ to a rational number (because that is in the nature of "easy" bijections, I would think), but $0$ must go to $\pi$ (if we extend, and get an order-preserving bijection between $(-\infty,0]$ and $(-\infty,\pi]$). (Or $0$ may go to some other irrational, even when we don't necessarily use the identity on $[\pi,\infty)$, but at any rate $\pi$ must go to an irrational, and $0$ must "simultaneously" go to a rational and to that irrational to which $\pi$ goes, which clearly is impossible.)

So, my related question is: Does anybody know what I am talking about ... and could you please give me directions, references that would confirm my guess that there is no "easy" order-preserving bijection between the negative irrationals and the irrationals strictly less than $\pi$? (Equivalently, that there is no "easy" order-preserving bijection between the negative rationals and the rationals less than $\pi$.) (Or, perhaps there is such an order-preserving bijection that might qualify as "easy?")

The answer would need to include a mathematically precise definition of "easy" (with the risk that I would not be able to understand it, but that is another matter, just try your best to come up with what you believe is an appropriate answer). What area of mathematics is involved (what books do I need to read)?

Edit. The answer posted by Brian M. Scott shows we could get a piecewise linear order-preserving bijection (which one might argue is as good as it gets). It is clear (or easy to see) that we could not remove "piecewise." I was asking for an "easy" bijection, and now I am starting to think that I was conflating "easy" with "nice," though without specific definitions one can interpret my question any way that might be reasonable. So here is a version of the problem where at least the term "nice" is given some possible interpretations. I guess I am hoping for a negative answer (for whatever reason) so I am putting some extra conditions on the bijection which may turn out to be impossible.

Is there an order-preserving bijection between $(-\infty,0)$ and $(-\infty,\pi)$ that sends rationals to rationals, sends irrationals to irrationals, and such that: (a) it is differentiable, or (b) it is $C^{\infty}$, or (c) it is analytic, or (d) it is something even nicer, e.g. given by an algebraic expression, or (somehow) given using the solutions (roots) of a polynomial? Is there any irrational number $p$ and order-preserving bijection between $(-\infty,0)$ and $(-\infty,p)$ that sends rationals to rationals, sends irrationals to irrationals, and such that any of (a) through (d) above might hold? I am thinking that perhaps (a) or (b) could have positive answer, while there is no way that (c) or (d) have positive answers, but I do not know, and I don't even know what area of math might turn out to be relevant. Any ideas or references (if something like this was already considered) would be appreciated. I don't really have any applications but this question seems interesting to me. (My guess is that these variations may just make the question difficult to handle.)

Given any rationals $p<q$ and $r<s$ there is a linear (or perhaps one should say affine) map $y=mx+b$ that sends $[p,q]$ to $[r,s]$ (as given in the answer by Brian M. Scott, $x\mapsto r+\frac{s-r}{q-p}(x-p),$ where $m=\frac{s-r}{q-p}$ and $b=r-\frac{s-r}{q-p}p$, both $m$ and $b$ rational). I just seem to be unable to take it that you couldn't do the same if the endpoints are irrational. Say $[\frac{-\pi}2,\frac{-\pi}3]$ and $[\pi-\frac{\pi}2,\pi-\frac{\pi}3]$ are clearly two intervals of the same length and with all endpoints involved being irrational. One is tempted to expect that in this case one could translate one interval to the other, sending rationals to rationals and irrationals to irrationals, but this is clearly not the case as the translation $x\mapsto \pi+x$ send rationals to irrationals. (All this is soo irrational... :)
I guess all this just illustrates that the irrationals are much "thicker" than the rationals, and the set of translations that sends rationals to rationals is "thin" in the set of all possible translations. (The translation $x\mapsto q+x$ "works" only if $q$ is rational.) I am tempted to visualize this using the integers instead: The intervals $(n,n+1)$ in between the integers (for integer $n$) are "thick" so the only translations $x\mapsto t+x$ that send integers to integers are when $t$ is an integer. So I may want to come up with a question about "nice" order-preserving functions from the integers into the integers (aiming at making the existence of such functions to be "difficult",: I just don't know how to come up with a condition in this case that is analogous to asking that $(-\infty,0)$ maps onto $(-\infty,\pi)$ (when we consider rationals and irrationals as in the original version of my question).
(End of edit.)

Regarding the extension of an order-preserving bijection from the rationals to the reals (if someone needs to see the details), see Extension of order-preserving bijection from rationals to reals.

A closely-related question seems to be: Order preserving bijection from ${\mathbb Q}\times{\mathbb Q}$ to $\mathbb Q$, where one asks for "wondering if something simpler could be given, more analytic." There are two answers there, the accepted one starts with "surreal numbers up to generation $\omega$," which is not something familiar to me. I would need to take a more careful look at those answers, but I feel that my version of this question is "different" because $\pi$ is an endpoint (and the irrationals rather than rationals are involved), and there are no endpoints in ${\mathbb Q}\times{\mathbb Q}.$

Regarding how I came up with this, I don't remember anymore, but I was thinking of subsets of the reals that are not $F_\sigma$ but are reverse order-isomorphic to themselves (the set $P$ being one example with $p\mapsto-p$), and at some point realized that there seemed to be no "easy" order-isomorphism between certain sets, and felt curious to find out more about this, as asked above.

(I am puzzled there is no tag "linear-orders." There is "order-theory," but that is too general, seems mostly about partial orders(?), and there is "well-orders," but this is too narrow.)

  • $\begingroup$ This is not an answer, but I find the question quite interesting. How come we cannot take the map $x \mapsto -1/(x-\pi)$? Edit: Oops, I did something backwards. $\endgroup$
    – marcelgoh
    Mar 28, 2021 at 2:34
  • 1
    $\begingroup$ @marcelgoh because I want my map to send rationals to rationals (and irrationals to irrationals, depending on which version of the problem you read), but if $x$ is rational then $-1/(x-\pi)$ is irrational. (That is, maynly because of that, in addition to your correction that you did something "backwards" ... I don't see which intervals it is supposed to send to which.) $\endgroup$
    – Mirko
    Mar 28, 2021 at 2:39
  • 2
    $\begingroup$ I mean, the "zig-zag argument" is actually perfectly constructive in this case; it's even computable. $\endgroup$ Mar 28, 2021 at 2:40
  • 3
    $\begingroup$ You can write a computer program that, when you input a rational number less than 0 (written as a fraction of integers), outputs a rational number less than $\pi$, such that the function it gives is an order-preserving bijection (and is in fact the bijection given by a certain implementation of the "zig-zag argument"). $\endgroup$ Mar 28, 2021 at 2:46
  • 1
    $\begingroup$ @D According to uio.no/studier/emner/matnat/ifi/nedlagte-emner/INF5170/v14/… (Any two countable densely ordered sets without endpoints are isomorphic a formal proof with KIV, Martin Giese Arno Schonegge), "Georg Cantor in [1] gave the first "informal" proof for the fact that any two countable densely ordered sets without endpoints are isomorphic." G. Cantor, Gesammelte Abhandlungen ch.9,p.303 Springer 1932. Also mathoverflow.net/questions/101996/cantor-theorem-on-orders addresses this question (answer unclear) $\endgroup$
    – Mirko
    Mar 28, 2021 at 20:24

1 Answer 1


It’s not precisely easy, but here is a piecewise linear order-preserving bijection.

First, suppose that $p,q,r,s\in\Bbb Q$, $p<q$, and $r<s$. Let

$$h_{p,q,r,s}:(p,q)\to(r,s)\setminus\Bbb Q:x\mapsto r+\frac{s-r}{q-p}(x-p)\,$$

$h_{p,q,r,s}$ is an order-preserving bijection that takes rationals to rationals and irrationals to irrationals, so its restriction to $(p,q)\setminus\Bbb Q$ is a bijection to $(r,s)\setminus\Bbb Q$.

Now let $a_n=-2^{-n}$ for $n\in\Bbb N$, and let $\langle b_n:n\in\Bbb N\rangle$ be a strictly increasing sequence in $\Bbb Q$ converging to $\pi$ such that $b_0=-1=a_0$. Let $f_0$ be the identity map on $(\leftarrow,-1)\setminus\Bbb Q$, and for $n\in\Bbb Z^+$ let $f_n=h_{a_{n-1},a_n,b_{n-1},b_n}$. Then

$$f=\bigcup_{n\in\Bbb N}f_n:(\leftarrow,0)\setminus\Bbb Q\to(\leftarrow,\pi)\setminus\Bbb Q$$

is an order-preserving bijection.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .