# Proof that $1$ is the least natural number using the Principle of Well Ordering

I'm doing a question from a PDF on Abstract Algebra.(The pdf can be found here http://abstract.pugetsound.edu/ 2012 edition). I have to show that the Principle of Well-Ordering implies that $1$ is the smallest natural number; this is only part of the whole question.(Question 14 of Chapter 2).

This is what I got so far.

Proof.

Let $S$ be an arbitrary subset of the Natural Number excluding one. By the Principle of Well-Ordering there exist a least element $X \in S.$ Now add $1$ to $S.$ $1$ is now the least element. Since there is no $X < 1$ in any such set $S,$ $1$ is the smallest element of any set $S$ and therefore the smallest natural number.

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I made a formatting edit to your question. I changed all instances of "one" into the numeral $1.$ Please make sure I did not change your question. –  user2468 Aug 13 '12 at 1:05
In the reasoning that you have so far, you are assuming that 1 is smaller than any other natural number by saying "there is no $X < 1$ in any such set"; so you have begged the question. What are the axioms you are starting from? –  Niel de Beaudrap Aug 13 '12 at 1:12
This is a weird question, because that is usually an axiom, or else part of the definition of the $<$ relation. I join Niel de Beaudrap in asking what the axioms are, or at least a link to the PDF, if it is online. –  MJD Aug 13 '12 at 2:25
One possibility that makes the question make sense is to define $\mathbb{Z}$ as the essentially unique ordered integral domain with well-ordered positive elements, and $\mathbb{N}$ as the set of positive elements. –  Chris Eagle Aug 13 '12 at 9:54

By the Well-Ordering principle, there is a least element $a$ in $\mathbb{N}$. Suppose, to the contrary, that $a<1$. Multiplying both sides of the inequality by $a$, we have $a^2<a$. Since $\mathbb{N}$ is closed under multiplication, $a^2\in\mathbb{N}$, which contradicts the assumption that $a$ is the least element in $\mathbb{N}$.
You may need to prove as a lemma that $a<b\implies ac<bc$. –  crf Aug 13 '12 at 1:46