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Math Help - Real Analysis

  1. #1
    Junior Member
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    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)]
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by MKLyon View Post
    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
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