= Solution
There are $2^{\aleph_0}$ distinct irrational cuts of the named rational order. Each defines a <consistent first-order theory>[consistent] complete one-type as above, already over the empty additional parameter <set>, since the rational constants are part of the language. Different cuts cannot be realized by the same element: a named rational between two cuts gives contradictory required inequalities.
An $\aleph_0$-saturated <first-order model>[model] must realize every one of these types, so it has at least $2^{\aleph_0}$ <Elements>. Consequently \b[there is no <countable> $\aleph_0$-saturated <first-order model>[model] of $T$]. This does not contradict existence of a <saturated model> from the preceding construction, whose size was unrestricted, or existence of the <countable> <prime model>, whose requirement is different. No continuum-hypothesis assumption is involved.
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