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All numbers are real, WLOG positive.

$A + B + ... + N = T$ and $A' + B' + ... + N' = T$

I'm trying to figure out some function, f, such that if

$f(A,B,... ,N) > f(A',B',...,N')$

then,

$max(A,B,..,N) > max(A',B',...,N')$

I've been going at this for a couple weeks and starting to think no such function exists. Thanks in advance for help!

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  • $\begingroup$ What exactly is your goal with this? The class of functions is very broad (and includes the max function, as is pointed out below), so without some restriction, we have a trivial answer. Also, do you want "if and only if" instead of "if", since as you've stated it $f(A,B,\ldots,N)=0$ is a satisfactory answer. $\endgroup$ Feb 5 '16 at 3:13
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Such a function exists. It is defined by: $$f(A, B, ..., N) = \begin{cases} A & \text{if } A \ge B, A \ge C, ... , A \ge N\\ B & \text{if } B \ge A, B \ge C, ... , B \ge N\\ \vdots\\ N & \text{if } N \ge A, N \ge B, ... , N \ge M\end{cases}$$ Like cosine and sine, this function even has its own special three-letter symbol: $\max$.

Sarcasm aside, the point is, a function is just a means of selecting a unique output value based on input values. Any way that you can accomplish this defines a function. So your question is whether a function exists that satisfies the same inequalities as the maximum value function.

Perhaps you meant a function defined by algebraic formula? In that case, note that $$\max(x, y) = \frac{x + y + |x - y|}2$$ and that $$\max(a, b, c) = \max(a, \max(b, c)).$$

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  • $\begingroup$ This is awesome. $\endgroup$ Feb 5 '16 at 4:41

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