For $x, y \in \mathbb{R}$, prove that $\max(x, y) = \frac{x + y + |x - y|}{2},$ and $\min(x, y) = \frac{x + y - |x - y|}{2}$. Prove that for all real numbers $x$ and $y$, 
$$\max(x, y) = \dfrac{x + y + |x - y|}{2},$$
and 
$$\min(x, y) = \dfrac{x + y - |x - y|}{2}.$$
For any real number $x$, the absolute value of $x$, denoted $|x|$ is defined as follows:
\begin{equation}
|x| = \begin{cases} x; & \text{ if } x \geq 0 \\
-x; & \text{ if } x< 0
\end{cases}
\end{equation}
What I understand from this is that $|x| = x$ if $x \geq 0$ or $|x| = −x$ if $x<0$. Other than that I don't really know how to start this. 
 A: \begin{align}
   \max\{x,y\} + \min\{x,y\} &= x + y \\
   \max\{x,y\} - \min\{x,y\} &=  |x-y| \\ 
\text{Adding we get} \\
2\max\{x,y\} &= x + y + |x-y| \\
\text{Subtracting we get} \\
2\min\{x,y\} &= x + y - |x-y| \\
\end{align}
A: This is a good approach, but it might be useful to rewrite $|x-y|$ in terms of $x$ and $y$:
$$
|x-y|=\left\{\begin{array}{}
x-y&\text{if }x\ge y\\
y-x&\text{if }x\lt y
\end{array}\right.
$$
Using this in your formula should make things simpler.
A: $|x-y|$ means the distance between $x$ and $y$ in the number line. 
On the other hand, $\frac{x+y}{2}$ means the midpoint between $x$ and $y$. 
(you can check those two statements if you didn't know them!)
Therefore, $\frac{x+y+|x-y|}{2}=\frac{x+y}{2}+\frac{|x-y|}{2}$, which means adding half the distance between $x$ and $y$  to the midpoint of $x$ and $y$, is $\max(x, y)$.
Similarly, $\frac{x+y-|x-y|}{2}=\frac{x+y}{2}-\frac{|x-y|}{2}$, which means subtracting half the distance between $x$ and $y$  from the midpoint of $x$ and $y$, is $\min(x, y)$.
A: 
If $\displaystyle x>y$ then $x-y>0$:

$|x-y|=x-y$
$\max(x,y) = (x+y+x−y)/2 = x$
$\min(x,y) =  (x+y-(x−y))/2 = y$

If $\displaystyle x<y$ then $x-y<0$:

$|x-y|=y-x$
$\max(x,y) = (x+y+y-x)/2 = y$
$\min(x,y) =  (x+y-(y-x))/2 = x$

If $x=y$ then:

$|x-y|=0$
$\max(x,y) = (x+y)/2 = x=y$
$\min(x,y) =  (x+y)/2 =x=y$
