5

Starting from the axiom of choice and passing the lemma of Zorn we arrive at the well-ordering theorem which states that for every set $X$ there exists a well-ordering with domain $X$.

I am aware that possibly nobody will ever be able to construct an explicit well-ordering of $\mathbb R$. There are in fact enough questions to be found here on math.stackexchange.com

But what about the "next best thing": an explicit (=constructed) well-ordering of $\mathbb Q$?

Martin Sleziak
  • 53,687
Achilles
  • 251
  • 2
    Let $\varphi\colon \mathbb Q\to \mathbb N$ be a bijection and define $q_1\leq_{\mathbb Q}q_2\iff \varphi(q_1)\leq_{\mathbb N}\varphi(q_2)$. – Git Gud Mar 08 '15 at 12:23
  • @Git Gud: Interesting idea! What about cases like $\frac{1}{2}$ and $\frac{2}{4}$? – Achilles Mar 08 '15 at 12:25
  • 1
    $\frac{1}{2}=\frac{2}{4}$ – Damian Reding Mar 08 '15 at 12:33
  • There shouldn't be a problem because $\left[\frac 1 2\right]=\left[\frac 1 4\right]$ and the rationals numbers are $\left[\frac 1 2\right]$ and $\left[\frac 1 4\right]$, not $\frac 1 2$ and $\frac 1 4$, see this. The real problem is finding an explicit bijection between $\mathbb Q$ and $\mathbb N$ with an explicit inverse. – Git Gud Mar 08 '15 at 12:34
  • @GitGud: The rational numbers are $\frac{1}{2}$ and $\frac{1}{4}$. They are not $(1,2)$ or $(1,4)$. – tomasz Mar 08 '15 at 12:43
  • @tomasz You're right. My comment still stands with the appropriate correction. – Git Gud Mar 08 '15 at 12:44
  • @Git Gud: Do you have an idea about the explicit bijection between $\mathbb Q$ and $\mathbb N$? – Moritz Mar 08 '15 at 12:51
  • I suspect it is hard to be more explicit than what Tomasz did. – Git Gud Mar 08 '15 at 13:37

1 Answers1

5

As suggested in the comments, all you need is an injective function from rationals into naturals.

For example, the function $f(p/q)=2^{p/d}3^{q/d}$ where $d$ is the greatest common denominator of $p$ and $q$ is quite explicit, I would say (as explicit as one can hope for, given ambiguity of representations of rational numbers!).

Then for any two rational numbers $q_1,q_2$, declare $q_1\prec q_2\iff f(q_1)<f(q_2)$.

tomasz
  • 35,474
  • If we use the proposed $f$, then $\frac{1}{2} = 2^{0-1}\cdot 3^{1-1} = f(0/1) = f(0) = f(0/2) = 2^{0-2}\cdot 3^{2-2} = \frac{1}{4}$ for $\gcd(0, s) = |s|$ with $s \in \mathbb Z$. Moreover, $f(\frac{-1}{2}) = 2^{-1-1}\cdot 3^{2-1} = \frac{3}{4} \notin \mathbb N$. Can this be remedied? – Moritz Mar 09 '15 at 16:40
  • @Moritz: It's my mistake, it was supposed to be $p/d,q/d$. Then there's no problem. – tomasz Mar 09 '15 at 17:24
  • Thank you for the edit. I think still $f(-\frac{1}{2}) \notin \mathbb N$. Do I make a mistake? – Moritz Mar 11 '15 at 07:42
  • @Moritz: No, you're right again, I was thinking of positive rationals. Still, this can be easily fixed by interleaving positive and negative rationals or putting all negatives after the positives (like you do when well-ordering the integers). But I will refrain from editing it, since this question is closed anyway, so I'd rather not bump it with a mostly trivial edit. – tomasz Mar 11 '15 at 08:13
  • Thank you very much for your effort! – Moritz Mar 11 '15 at 12:14