We can construct the positive rationals from ratios of positive integers (and thus from pairs of finite ordinals).

Can we analogously construct the reals from pairs of countable ordinals?

  • $\begingroup$ I think that [fraction] was a bit off, and [set-theory] is somewhat fitting here. Perhaps [elementary-set-theory], but I'm not 100% sure about that. $\endgroup$
    – Asaf Karagila
    May 20, 2012 at 12:29

2 Answers 2


The answer is no from cardinality considerations.

There are only $\aleph_1$ many countable ordinals, and there can only be $\aleph_1$ many pairs of countable ordinals. If we could construct the real numbers as a fraction field of some sort, its cardinality would not exceed $\aleph_1$.

However we know that ZFC does not decide the cardinality of the continuum. It could be $\aleph_1$ but it could be larger, much larger.

Furthermore, while $\mathbb R$ with its natural order embeds every countable ordinal it cannot order-embed $\omega_1$ (that is, there is a function $f_\alpha\colon\alpha\to\mathbb R$ which is order preserving if and only if $\alpha$ is countable). If we take a field of fraction over all countable ordinals we either embed $\omega_1$ into it, or we end up using only countably many ordinals to begin with.

Therefore in ZF this fails due to both the order type of the real numbers and their cardinality.

Of course there is what Mark wrote, the ordinal operations are not commutative; and of course fractions would result with infinitesimals. One could ask whether an extended field in which infinitesimals are allowed may be constructed that way, but the above shows that this fails as well.

Slightly related:

  1. The Aleph numbers and infinity in calculus.
  2. Comparing infinite numbers
  • $\begingroup$ I would like to see a proof that there is an order-preserving $f_\alpha:\alpha\to\Bbb R$ for every countable ordinal $\alpha$. If you could provide a reference, I would be grateful. $\endgroup$
    – MJD
    May 22, 2012 at 17:59
  • 1
    $\begingroup$ @Mark: There are several arguments here, the best I can think of is that $\mathbb Q$ is universal, and embeds every countable linear order (without the need for choice because everything is already well-ordered...); you can look up the usual construction of an Aronszajn tree which also uses this (it requires choice, but for a different reason, we require a sequence of functions to exist, not just existence for every ordinal); Also this and that on this very site. Let me know if you want more. $\endgroup$
    – Asaf Karagila
    May 22, 2012 at 18:49

I haven't thought about this as much as it might deserve, so I don't know if you get anything interesting; you might. But your question was if you can get the reals this way, and that I can answer: whatever you do get, it is definitely not going to be the reals, because $\frac1\omega$ is an infinitesimal, and the reals do not contain infinitesimals. By this I mean that for any finite integer $n$, you will have $0 < \frac1\omega < \frac1n$, and there is no such real number.

Or to take a more obvious problem I should have mentioned before, addition and multiplication of countable ordinals are not commutative, so the resulting operations on quotients will be noncommutative as well. For example, $\omega+1\ne1+\omega$, so we will have $\frac\omega1+\frac11\ne \frac11+\frac\omega1$, and that is not how the reals work.

That said, I think you're going to get into trouble even trying to apply this construction, which only applies to integral domains, and $\omega_1$ is not an example. The usual construction of the rationals says that $\frac ab=\frac cd$ exactly when $ad=bd$. This question, Operations on ordinal numbers, points out that $(\omega\cdot2)\cdot\omega = \omega\cdot\omega$. So by the construction, we have $\frac11\ne\frac{\omega\cdot2}\omega = \frac\omega\omega = \frac11$, and now we are in trouble.

  • $\begingroup$ At the very end you probably want to write $2 = \frac{ \omega \cdot 2}{\omega} = \dots$, since if $\omega \cdot 2 = \omega$ you actually get $\frac{\omega \cdot 2}{\omega} = 1$. Very nice answer btw! $\endgroup$ May 20, 2012 at 5:41
  • $\begingroup$ Thanks! It was an interesting question. I think $\frac11\ne\frac11$ is worse than $2=1$. Also, your suggestion depends on the fussy details of how we interpret the $ad=bc$ law in the absence of commutativity, whereas mine doesn't. $\endgroup$
    – MJD
    May 20, 2012 at 5:44
  • $\begingroup$ Ok! : ) ${}{}{}$ $\endgroup$ May 20, 2012 at 5:44
  • $\begingroup$ In the last paragraph I believe that you want $\frac{a}b=\frac{c}d$ exactly when $ad=bc$, as in your comment to Matt. The example still depends on having $bc$ rather than $cb$: $\omega\cdot(\omega\cdot 2)=(\omega\cdot\omega)\cdot 2\ne\omega\cdot\omega$. $\endgroup$ May 20, 2012 at 7:43

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