$\max\{a,b\} = \frac12(a+b+|a-b|)$ and $\min\{a,b\} = \frac12(a+b-|a-b|)$
how would you go about solving this?
I started with suppose $a \leq b$
Also, show min{a,b,c} = min{min{a,b},c}.
How would I go about showing that?
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There are proofs of the $\max$ case here: http://math.stackexchange.com/questions/429622/show-that-the-max-x-y-dfracxyx-y2?rq=1 – Clarinetist Oct 13 '14 at 02:56
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thank you! now how do i show the min case? – Anon123 Oct 13 '14 at 03:00
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Suppose that $a\le b$.
$$\frac{a+b+|a-b|}{2}=\frac{a+b+b-a}{2}=b=\max \{a,b\}$$
And
$$\frac{a+b-|a-b|}{2}=\frac{a+b-(b-a)}{2}=a=\min \{a,b\}$$
Paul
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You can use the fact that $$\min\{x,y\}=-\max\{-x,-y\}$$ to get the result about mins from the one about maxes.
Mariano Suárez-Álvarez
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