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

• blashmet
In summary: So what can you say about the sum of the x's?In summary, the conversation is about proving that m cannot be equal to the sum of x_1, x_2, ..., x_k divided by k. The person is struggling with how to proceed with the proof, but has come up with the contrapositive. They are asked to consider what they can say about the sum of the x's, given that they are all less than m. The conversation ends with a hint to think about what the sum of the x's can tell us about m times k.

## 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?

murmillo said:
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.

## What does it mean for m to be the average of x1, x2, ..., xk?

When we say that m is the average of a set of numbers (x1, x2, ..., xk), it means that m is the sum of all the numbers in the set divided by the total number of numbers in the set.

## How do I prove that if m is the average of x1, x2, ..., xk, then m = (x1 + x2 + ... + xk) / k?

To prove this statement, you can use the definition of average and manipulate the equation to show that both sides are equal. Alternatively, you can use mathematical induction to prove this statement for any number of values in the set.

## Can m be equal to one of the values in the set (x1, x2, ..., xk)?

Yes, it is possible for m to be equal to one of the values in the set. In fact, if all the values in the set are the same, then m will be equal to that value.

## What is the relationship between the average m and the median of a set of numbers (x1, x2, ..., xk)?

The average m and the median are two different ways of measuring the central tendency of a set of numbers. While the average is the sum of all the numbers divided by the total number of numbers, the median is the middle number when the numbers are arranged in ascending or descending order. In some cases, the average and the median may be the same, but in other cases, they may be different.

## Is it possible for m to be negative or zero?

Yes, it is possible for m to be negative or zero. This can happen if the sum of the numbers in the set is negative or zero, or if the total number of numbers in the set is zero. However, in most cases, the average will be a positive number.