1
$\begingroup$

This question already has an answer here:

According to the book "An introduction to the analysis of algorithms (written by Michael Soltys)", the author says in chapter 1 as follows.

Let $\mathbb N = \{0, 1, 2,...\}$ be the set of natural numbers. Suppose that $S$ is a subset of $N $ with the following the two properties: first $0 \in S$, and second, whenever $n\in S$, then $n+1\in S$ as well. Then, invoking the Induction Principle(IP) we can conclude that $S=\mathbb N$.

I am wondering if it is fine to define the set of natural numbers like the author. Why the author put the number $0$ into the set of natural number $\mathbb N$?

$\endgroup$

marked as duplicate by Brian M. Scott, user61527, Dominic Michaelis, William, TZakrevskiy Sep 1 '13 at 8:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ If you’re asking whether it’s correct to include $0\in\Bbb N$, the answer is yes. It’s also correct to omit it: there are two different conventions. Most of us with a set-theoretic background automatically include it. See this question for more information; I think that it answers your question, so I’m voting to close yours as a duplicate. $\endgroup$ – Brian M. Scott Sep 1 '13 at 6:31
  • $\begingroup$ " Regrettably, there seems to be no general agreement about whether to include 0 in the set of natural number":mathworld.wolfram.com/NaturalNumber.html and en.wikipedia.org/wiki/… $\endgroup$ – lab bhattacharjee Sep 1 '13 at 6:41
1
$\begingroup$

Whether or not $0$ is considered a "natural" number is a matter of convention. I think including it is the better convention, since it is nice to have an additive identity even if inverses don't exist. To avoid any doubt, use the terms "non-negative" integer and "positive" integer.

$\endgroup$
  • $\begingroup$ One convention I've seen is to call the set $\omega$ when 0 is meant to be included and $\mathbb{N}$ when it is not. $\endgroup$ – G Tony Jacobs Sep 1 '13 at 6:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.