Question

Prove for fun -

Answer 1

and what should we prove?

Answer 2

The maximum of two numbers x and y is denoted by max (x,y). The max (-1, 3) = max (3,3) = 3 and max (-1, -3) = max(-3, -1) = -1. The minimum of x and y is denoted b min (x,y). Prove that:
\[\max(x,y) = \frac{ x+y+|y-x| }{ 2 },\]
\[mix (x,y) = \frac{ x+y-|y-x| }{ 2 }.\]

Answer 3

typo, it's min not mix LOL

Answer 4

if y>x then |y-x| =(y-x) so max(x,y)=y you can easily prove that
if x>y then |y-x|=(x-y) max(x,y) = x you can prove that

This Sucks!

Answer 5

@nincompoop min is also same.

Answer 6

are you sure that's the way to prove it?

Answer 7

Is there a problem with this proof?

Answer 8

\[\frac{x+y+|y-x|}{2}=\cases{\frac{x+y+(y-x)}{2}\qquad, y>x\\\frac{x+y+(x-y)}{2}\qquad ,x>y}\\=\cases{y\qquad ,y>x\\x\qquad ,x>y}\]

Answer 9

max(x,y) = (x+y +|y| - |x|)/2 -> (x+y +y-x)/2 =x and (x+y-y+x)/2 = y

Answer 10

max x if x>y and max y if y>x

Answer 11

thanks guys

Answer 12

@nin I think |y-x| is not eual to |y|-|x|

Answer 13

it's the same. I read that recently LOL

Answer 14

spivak calculus 4th edition

Answer 15

|5-(-3)| is not |5|-|-3|

Answer 16

|5 -(-3)| = |5+3| = |5| + |3|