... center-of-gravity).¹

Copyright 2002. May be reproduced non-commercial purposes.

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... explanation.²

I had hoped the intuitive argument had worked, but I guess not. Let $S$ be the subset of numbers smaller than the average, and $L$ be the subset larger. Let $E$ be the subset of all numbers equal to the average. $E$, $S$ and $L$ are mutually exclusive (a number cannot be both greater, less than, or equal to the average). Call the average $a_x$, and let $S_i$ be an element in $S$ (similarly for $L$ and $E$). The facts are

• $\sum{(S_i - a_x)} < 0$, true because each $S_i < a_x$ by definition
• $\sum{(L_i - a_x)} > 0$, true for similar reasons
• $\sum{(E_i - a_x)} = 0$, true because each $E_i = a_x$ by definition
• $\sum{a_i} = E + L + S$, since the sets are mutually exclusive and each number $a_i$ must fall into exactly one category
• $\sum{a_i - a_x} = 0$, shown earlier

Therefore,

$$\sum{a_i} - \sum{a_x} = E + L + S - \sum{a_x}$$

$$= \sum{(S_i - a_x)} + \sum{(L_i - a_x)} + \sum{(E_i - a_x)}$$ 0

$$= \sum{(S_i - a_x)} + \sum{(L_i - a_x)}$$ 0

$$-\sum{(S_i - a_x)} = \sum{(L_i - a_x)}$$

which is the result we sought to prove. Phew! Sorry for the roundabout proof, but you asked for it!

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