KindiakMath

Is 0 a Natural Number?

Many mathematicians insist that 0 is not a natural number, motivated by the philosophical something-or-nothing dilemma. Computer scientists, on the other hand, mostly assert that 0 is a natural number, motivated by zero-indexing in for-loops.

Most otherwise sane academics try to harmonise both views by saying that 0 being a natural number is context-specific, and is included or excluded for purposes amounting to notational convenience.

I attempt to formalise this perspective by asserting that one thing is for certain: 0 must be a nonnegative integer. More precisely, we can define \mathbb N to be exactly one of the following sets: \mathbb N^+ := \mathbb N \backslash \{0\} or \mathbb N_0 := \mathbb N \cup \{0\}.

Theorem. Let K be any set and c denote any object. Then K is exactly one of K_1 := K \backslash \{c\} or K_2 := K \cup \{c\}. Particularising to K = \mathbb N and c = 0 yields the desired result.

Proof. We observe that

K_1 = K \backslash \{c\} \subseteq K \subseteq K \cup \{c\} = K_2.

The claim now is that K is exactly one of K_1 or K_2. Now, exactly one of K = K_1 or K \neq K_1 must hold. In the former, we are done.

Suppose the latter, that is, K \neq K_1. We aim to prove that K = K_2. To that end, since K \neq K_1, there exists x \in K \subseteq K_2 such that x \notin K_1.

If x \neq c, then x \in K \backslash \{c\} = K_1, a contradiction. Thus, c = x \in K. Hence,

K_2 = K \cup \{c\} \subseteq K \cup K = K \subseteq K_2,

as required.

To conclude: Using this formulation, mathematicians use the convention \mathbb N = \mathbb N^+, while computer scientists use the convention \mathbb N = \mathbb N_0. This allows both of them in their contexts to write one less symbol (+ for mathematicians, 0 for computer scientists).

Now end your foolish controversies and needless debates.

—Joel Kindiak, 3 Nov 24, 1957H

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