Show that the $\max{ \{ x,y \} }= \frac{x+y+|x-y|}{2}$.

Question

Show that the $\max{ \{ x,y \} }= \dfrac{x+y+|x-y|}{2}$.

I do not understand how to go about completing this problem or even where to start.

Answer 1

Here is another way of looking at it:

We have $|x| = \max(x,-x)$. Also, $\max(a,b)+c = \max(a+c,b+c)$, and if $c \geq 0$, then $c \max(a,b) = \max(ac,bc)$.

Hence \begin{eqnarray} \frac{1}{2}(x+y+|x-y|) &=& \frac{1}{2}(x+y+\max(x-y,y-x)) \\ &=& \frac{1}{2}(\max(x-y+x+y,y-x+x+y)) \\ &=& \frac{1}{2}(\max(2x,2y)) \\ &=& \max(x,y) \end{eqnarray}

Answer 2

This probably isn't as rigorous as it should be, but I think it's intuitive enough.

Hmm... We don't know which of $x$ or $y$ is bigger, but we do know one thing: their average. If we call the average $z$, then $z=\frac{x+y}{2}$. Now, the distance between $x$ and $y$ is $|x-y|$, so the distance from $z$ to both $x$ and $y$ is $\frac{|x-y|}{2}$.

So if we imagine a number line, the distance from $0$ to $z$ is $\frac{x+y}{2}$, and the distance from $z$ to max(x, y) is $\frac{|x-y|}{2}$. Thus, the total distance from $0$ to max(x, y) is $\frac{|x-y|}{2}$ + $\frac{x+y}{2}$, as desired.

Answer 3

Hint: If $x\ge y$ then $|x-y|=x-y$. If $x\lt y$ then $|x-y|=-(x-y)=y-x$.

We have used the fact that in general $|w|=w$ if $w\ge 0$ and $|w|=-w$ if $w\lt 0$.

Answer 4

$$ \max\{x,y\} =\frac{x+y+|x-y|}{2} $$ $$ => 2.\max\{x,y\} =x+y+|x-y| $$ there are two possible situation :
1. $ y>x $, i.e $\max\{x,y\}=y$ then $y-x=|x-y|$, this equation is true because we assume that
$ y>x$
2. $x>y$, or $\max\{x,y\}=x$, then $x-y=|x-y|$, which is true if $x>y$

Answer 5

For another approach we can do as follows:

$$max(x, y) - max(-x, -y) = x + y$$ and, $$max(x, y) + max(-x, -y) = |x - y|$$

Adding those together yeilds: $$2\times max(x, y) = x + y + |x - y|$$ Dividing by two, gives us what we want.

Answer 6

Conceptually, focus on $|x-y|$ as the absolute difference of the two numbers.

Without loss of generality, assume $x > y$. Then $y + |x - y| = x$. This can be understood as representing that if we add the difference between two numbers to the smaller, we get the larger.

The specific equation will naturally fall out with this observation.

Answer 7

Without loss of generality, let $y=x+k$ for some nonnegative number $k$. Then,

$$ \frac{x+(x+k)+|x-(x+k)|}{2} = \frac{2x+2k}{2} = x+k = y $$

which is equal to $\max(x,y)$ by the assumption.