Constructing the reals from fractions of ordinals 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?
 A: 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.
A: 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:


*

*The Aleph numbers and infinity in calculus.

*Comparing infinite numbers
