Having trouble proving this. Any suggestions?
For all a, b (they are real numbers) show that:
max (a,b) = (1/2)[a+b+abs(a-b)]
min (a,b) = (1/2)[a+b-abs(a-b)]
Without loss of generality suppose a>=b, then
(1/2)[a+b+abs(a-b)]=(1/2)[a+b+a-b]=a.
If b were the larger we would observe that max(a,b)=max(b,a) then
we would have:
max (b,a) = (1/2)[a+b+abs(b-a)],
and the result would again follow.
This is similar.min (a,b) = (1/2)[a+b-abs(a-b)]
RonL