Proof by Cases: $\operatorname{max}\{x,y\} + \operatorname{min}\{x,y\}=x+y$ So I'm told to "[u]se proof by cases to prove that $\operatorname{max}\{x,y\} + \operatorname{min}\{x,y\}=x+y$ for all real numbers $x$ and $y$." What does this mean?
 A: Given $x,y \in \mathbb{R}$, we have one of the following to be true:
\begin{align}x & < y\\ x & =y\\ x & >y \end{align}
Now consider each case and conclude what you want.
For instance, consider the case $x<y$. We then have $\max\{x,y\} = y$ and $\min\{x,y\}=x$. This gives us $$\max \{x,y\} + \min \{x,y\} = y + x  = x +y$$ Similarly, argue out the remaining two cases as well.
A: Well, there are three natural cases: $x\lt y$, $x=y$, and $x\gt y$.
But breaking up into cases is not really necessary. Because of the symmetry, we could say something like "Without loss of generality we may assume that $x\le y$."
If, for example, you examine the case $x\lt y$, and compute $\max(x,y)$ and $\min(x,y)$, you will quickly be able to verify that the equation holds in that case. Same with the others.
A: Note that $\min\{x,y\}=x$ when $x\le y$, but $\min\{x,y\}=y$ when $x>y$. Similarly, when $x\le y$ we have $\max\{x,y\}=y$, but when $x>y$ we have $\max\{x,y\}=x$. The suggestion is that you break the problem into these two cases: treat $x\le y$ as one case and $x>y$ as the other.

Case 1: If $x\le y$, then $\min\{x,y\}=x$ and $\max\{x,y\}=y$, so ... .

And similarly for the other case.
This isn’t the only possible division into cases, by the way: you could also use $x<y$ and $x\ge y$, or the three-way division into $x<y$, $x=y$, and $x>y$. 
The point is that $\max\{x,y\}$ and $\min\{x,y\}$ depend on the relationship between $x$ and $y$. If you split the proof up according to the possibilities for that relationship, you can arrange to make the argument uniform within each case, even though the argument in one case may have to be a bit different from the argument in another.
