Tao Analysis: Exercise 2.2.2

This is a proof that for any positive number $a$, there exists exactly one natural number $$b$$ , such that $b++=a$. To do this, we will first prove that there always exists at most one, then we will prove that there always exists at least one.

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 $b \neq c$. By substitution, this means that $b++=c++$ and $b \neq c$, 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.