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

[blashmet]

Homework Statement

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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?

 

[murmillo]

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?

 

[blashmet]

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 :)

 

[murmillo]

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!

 

[blashmet]

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...

 

[murmillo]

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.