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Peano's axioms for Natural numbers.

3rd axiom from Introduction to topology by Bert Mendelson -

There is one and only one object in $\mathbb{N}$ denoted by $1$, which is not the successor of an object in $\mathbb{N}$, i.e, $1 \neq s(x)$ for each $x \in \mathbb{N}$.

How can $1$ not be a successor, isn't $0$ a natural number? Or Please explain what this statement says.

Note: I am not a maths student. Just reading that book out of curiosity.

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    $\begingroup$ Different authors have different opinions about whether to classify 0 as a natural number or not. What's really important about the definition is that the natural numbers have a least element $\endgroup$ – eepperly16 Feb 3 '16 at 18:05
  • $\begingroup$ I feel like many programmers like to consider $0$ a natural number, but usually, in mathematics, $1$ is the least natural number. The notation $\mathbb{N}_0$ is often used for $\mathbb{N}\cup \{0\}$. $\endgroup$ – vrugtehagel Feb 3 '16 at 18:14
  • $\begingroup$ If your personal $\Bbb N$ contains an element $0$ that is not the successor of anything else, then what the author denotes with the symbol $1$ is your $0$. (You probably also use the symbol $1$, but for the diffferent object $s(0)$; in this case confusion between the notations is due to arise; moreover, if the author will define two operations $+$ and $\cdot $ soon, rest assured that those are not your ordinary addition and multiplication). The theory is the same, but it may be easier for you to follow the author if you remove $0$ from $\Bbb N$. ;) $\endgroup$ – Hagen von Eitzen Feb 3 '16 at 18:15

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