Supposing that numbers are reducible to sets, how would one go about reducing a complex number to one?

Question

I'm aware of the Benacerraf's identification problem, but suppose that numbers are reducible to sets such that the empty set is identified with zero, the power set of the empty set is identifiable with one, etc. Following this way of identification, how would one go about reducing a complex number (i.e. of the form a + bi) to a set?

Answer 1

The most standard set theoretic way to define the various types of numbers is probably as follows:

• Natural numbers are defined as sets via $0:=\emptyset$ and $n+1:=\{0,\ldots\,n\}$ for $n\geq 0$
• Integers are defined as equivalence classes of ordered pairs of natural numbers, under the relation $(a,b)\sim(c,d)\leftrightarrow a+d=b+c$, via $n:=[(n,0)]$ and $-n:=[(0,n)]$ for $n\ge 0$.
• Rational numbers are defined an equivalence classes of ordered pairs from $\Bbb{Z}\times(\Bbb Z\setminus\{0\})$ under the equivalence relation $(a,b)\sim(c,d)\leftrightarrow ad=bc$, with addition and multiplication defined appropriately, and with the integer $n$ being identified with the class $[(n,1)]$.
• Real numbers are defined as equivalence relations of Cauchy sequences of rational numbers in the usual way, where the rational number $q$ is identified with the equivalence class of the constant sequence $a_n=q$.
• Complex numbers are defined as ordered pairs of real numbers, with addition and multiplication being defined appropriately, and with the real number $x$ being identified with the pair $(x,0)$.

This certainly isn't the only way to do it, but it works.