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.
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.