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How does one select some $i$ to prove $\exists i$ s.t. $s_i \leq A$ where $A$ is the average of real numbers $(s_1 + s_2 + . . . + s_n) / n$

It seems like a trivial proof, but writing a proof for such an existence conventionally does not seem obvious to me.

Perhaps I am reading it wrong, but I am sure it says something like: For any real number $A$ which is the average of a set, $S$, of real numbers, there exists a number $s \in S$ s.t. $s$ is less than or equal to $A$.

Is there a special way to choose $i$ to prove this existence? It seems I cannot trivially select a set of numbers and show there is a value that satisfies $s_i \leq A$ since it is some arbitrary set. How would you select $i$ for this proof?

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No, you don't know which $i$ it must be. You go about it as a proof by contradiction. – Calvin Lin Feb 5 '13 at 5:54
Take $s_i$ to be the smallest element in $S$ ($S$ is finite I suppose) – Ewan Delanoy Feb 5 '13 at 5:55

1 Answer 1

up vote 2 down vote accepted

If not, then $s_i> A $ for all $i$. So $$A=\frac{s_1+s_2+\cdots+s_n}{n}>\frac{\overbrace{A+A+\cdots+A}^{n\text{ times}}}{n}=A$$ So derive a contradiction.

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So how might I write this then. To yield a contradiction, assume for all $i$, $i > A$. This implies the average of such a series is actually equal to, and not greater than, $A$. And this is the contradiction? Very much as you have written? Do I need to declare $i$ be a natural or something like that? – Leonardo Feb 5 '13 at 6:02
@Leonardo Assume $s_i >A$ for all $i$, then this assumption implies $A>A$ and it implies $A\neq A$. – Hanul Jeon Feb 5 '13 at 6:06
Yes thank you, the more I stare at the answer the more obvious it seems, its like one of those hidden paintings or something. – Leonardo Feb 5 '13 at 6:11

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