Editing Well-ordering principle (section)

===Integer summation===
Theorem: <math>1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2}</math> for all positive integers <math>n</math>.

''Proof''. Suppose for the sake of contradiction that the above theorem is false. Then, there exists a non-empty set of positive integers <math>C = \{n \in \mathbb N \mid 1 + 2 + 3 + ... + n \neq \frac{n(n + 1)}{2}\}</math>. By the well-ordering principle, <math>C</math> has a minimum element <math>c</math> such that when <math>n = c</math>, the equation is false, but true for all positive integers less than <math>c</math>. The equation is true for <math>n = 1</math>, so <math>c > 1</math>; <math>c - 1</math> is a positive integer less than <math>c</math>, so the equation holds for <math>c - 1</math> as it is not in <math>C</math>. Therefore, 

<math display=middle>\begin{align}
1 + 2 + 3 + ... + (c - 1) &= \frac{(c - 1)c}{2} \\
1 + 2 + 3 + ... + (c - 1) + c &= \frac{(c - 1)c}{2} + c\\
&= \frac{c^2 - c}{2} + \frac{2c}{2}\\
&= \frac{c^2 + c}{2}\\
&= \frac{c(c + 1)}{2}
\end{align}</math>,

which shows that the equation holds for <math>c</math>, a contradiction. So, the equation must hold for all positive integers.<ref name='mit' />