SCHIZOPHRENIA IN CONTEMPORARY hL4THEMATICS 11 middle - - "A or not A. Constructively, this principle would mean that we had a method which, in finitely many purely routine steps, would lead to a proof of disproof of an arbitrary mathematical assertion A. Of course we have no such method, and nobody has the least hope that we ever shall. It is the principle of the excluded middle that accounts for almost all of the impor- tant unconstructivities of classical mathematics. Another incorrect principle is "(not not A) implies A. " In other words, a demonstration of the impos- sibility of the impossibility of a certain construction, for instance, does not constitute a method for carrying out that construction. I could proceed to list a more or less complete set of constructively valid rules of inference involving the connectives just introduced. This would be superfluous. Now that their meanings have been established, the rest is common sense. As an exercise, show that the statement "(A -+ 0 = 1) t--+ not A" is constructively valid. The classical concept of a set as a collection of objects from some pre- existent universe is clearly inappropriate constructively. Constructive rnathe- matics does not postulate a pre-existent universe, with objects lying around waiting to be collected and grouped into sets, like shells on a beach. The entities of constructive mathematics are called into being by the constructing intelligence. From this point of view, the very question "What is a set? is suspect. Rather we should ask the question, "What must one do to construct a set? " When the question is posed this way, the answer is not hard to find. Definition. To construct a set, one must specify what must be done to construct an arbitrary element of the set, and what must be. done to prove two arbitrary elements of the set are equal. Equality so defined must be shown to be an equivalence relation. As an example, let us construct the set of rational numbers. To con- struct a rational number, define integers p and q and prove that q # 0. To prove that the rational numbers p/q and p /q are equal, prove pql = plq. 1 1 While we are on the subject, we might as well define a function f: A + B. It is a rule that to each element x of A associates an element f(x) of B, equal elements of B being associated to equal elements of A.

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