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)]

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- Sep 19th 2006, 11:25 AMMKLyonReal 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)] - Sep 19th 2006, 11:37 AMCaptainBlack
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.

Quote:

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

RonL