This article was inspired by the question: is there a definable equivalence relation on the field of complex numbers, each of whose equivalence classes has exactly two elements? The answer turned out to be no, as we now explain in greater detail.Let Κ be an algebraically closed field and let E be a definable equivalence relation on Κ. [Note: By “definable” we will always mean “definable with parameters”.] Either E has one cofinite class, or all classes are finite and there is a number d such that all but a finite set of classes have cardinality d. In the latter case let B be the finite set of elements of Κ which are not in a class of size d. We prove the following result.Theorem 1. a) If char(Κ) = 0 or char(Κ) = p > d, then ∣B∣ ≡ 1 (mod d).b) If char(Κ) = 2 and d = 2, then ∣B∣ ≡ 0 (mod 2).c) If char(Κ) = p > 2 and d = p + s, where 1 ≤ s ≤ p/2, then ∣B∣ ≡ p + 1 (mod d).Furthermore, a)−c) are the only restrictions on ≡B≡.If one is in the right mood, one can view this theorem as saying that the “algebraic cardinality” of the complex numbers is congruent to 1 (mod n) for every n.§1 contains a reduction of the problem to the special case where E is induced by a rational function in one variable. §2 contains the main calculations and the proofs of a)−c). §3 contains eight families of examples showing that all else is possible. In §4 we prove an analogous result for real closed fields.