# Prove that if m is the average of x1,x2, ,xk, then

## Homework Statement

http://img440.imageshack.us/img440/9242/captureyeq.jpg [Broken]

## Homework Equations

Given in problem statement.

## The Attempt at a Solution

Basically I've come up with the contrapositive, but I'm unsure of how to proceed with the proof. Can anyone help?

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So x_1, x_2, ..., x_k are all less than m. You're trying to show that m cannot be (x_1 + x_2 +...+ x_k)/k. What can you say about the sum x_1 + x_2 +... + x_k?

So x_1, x_2, ..., x_k are all less than m. You're trying to show that m cannot be (x_1 + x_2 +...+ x_k)/k. What can you say about the sum x_1 + x_2 +... + x_k?

(x_1 + x_2 +... + x_k) is greater than (x_1 + x_2 +...+ x_k)/k ?

thanks for the help btw :)

Here are some more hints:

Yes, (x_1 + x_2 +... + x_k) is greater than (x_1 + x_2 +...+ x_k)/k. But I think that's true in general, since basically whenever you divide something by a positive integer greater than 1 you make it smaller. But we want to somehow use the fact that x_1, x_2,...,x_k are all less than m. Is there anything you can say about the sum that involves m? Remember that we're trying to show that m cannot be the sum divided by k. Think it over for a while and see if you can come up with something. Also if you really get stuck, maybe play around with some examples, like k = 5 or something and then make something up for x_1 to x_k...I dunno. Anyway, good luck!

Those are exactly my thoughts. It makes sense intuitively, I just don't know how to put it down in "math speak" haha.

If anyone can help that'd be sweet...

Hi,

OK, one more hint until I'm off to do other things:
Each of the x's are less than m. And there are k of them. So when you add all the x's together, what can you say? I mean, you want to show that m cannot be the sum of the x's divided by k. That's the same as: m times k is not the sum of the x's.