<s>[INST] How to solve a particular well ordering problem I want to prove that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}=\frac{n}{n+1}$$ with well ordering principle. 
I know that $$n \in N $$
because the numbers 1,2,3... are discrete and are positive. I was thinking of using a set of natural numbers that don't satisfy the equation and proving by contradiction  but I'm not sure that works. 
 [/INST] Yes, you can make it like that. It is not different than induction, though. 
Let $E$ be the set of $n\in\mathbb N$ such that your equality does not hold. Note that $1\not\in E$. Let $k\geq2$ be the minimum of $E$ (here is where one uses the principle). Then $k-1\not\in E$, so it satisfies the formula:
$$
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{(k-1)k}=\frac{k-1}{k}.
$$
But then
\begin{align}
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{k(k+1)}&=
\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{(k-1)k}+ \frac{1}{k(k+1)}\\ \ \\
&=\frac{k-1}k+\frac{1}{k(k+1)}=\frac{(k-1)(k+1)+1}{k(k+1)}\\ \ \\
&=\frac{k^2}{k(k+1)}=\frac{k}{k+1}.
\end{align}
Then $k$ would satisfy the equality a contradiction. That is, the minimum of $E$ cannot exist. So it has to be that $E$ is empty, and the formula holds for all $n\in\mathbb N$. 
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