Thread: Real Analysis

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)]

Originally Posted by MKLyon

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)]

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.

min (a,b) = (1/2)[a+b-abs(a-b)]

This is similar.

RonL