Tao Analysis: Exercise 2.2.2
The existence of at most one predecessor can be proven by contradiction. Say there existed two natural numbers, b and c, such that b++=a, c++=a, and b≠c. By substitution, this means that b++=c++ and b≠c, which violates Axiom 2.4.
The existence of at least one predecessor can be proven by induction. Let P(a) be the fact that if a is positive, then there exists at least one natural b such that b++=a. When a=0, P(a) is vacuously true because 0 is not positive, by definition of positive. Now let's assume that P(a) applies for some natural number a. By Lemma E.-1, a++ must be positive. We also know that there exists at least one predecessor to a++, that being a. Therefore, the implication holds that if a++ is positive, then it has at least one predecessor. Therefore, if P(a), P(a++), which closes the induction.
Because there exists a minimum and maximum of one predecessor to all positive numbers, we now know that there always exists exactly one.
Comments
Post a Comment