Pigeonhole Principle and the Arithmetic Mean

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The discussion centers on the pigeonhole principle and its connection to the arithmetic mean. The principle states that if m objects are placed into k containers with m greater than k, at least one container will hold more than a specific number of objects. The corollary suggests that the arithmetic mean of a set of numbers lies between the smallest and largest values, which is considered obvious. The explanation provided interprets the mean as redistributing values into containers, demonstrating that the smallest number must be represented in each container. This reasoning, while clear, lacks details for cases where the mean is not an integer, yet still supports the conclusion that the mean is bounded by the smallest and largest values.
Dschumanji
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My discrete mathematics book gives the following definition for the pigeonhole principle:

If m objects are distributed into k containers where m > k, then one container must have more than \lfloor\frac{m-1}{k}\rfloor objects.

It then states as a corollary that the arithmetic mean of a set of numbers must be in between the smallest and largest numbers of the set. No proof is given, it pretty much just says "well it's just obvious that this is the case."

I think it is obvious that the arithmetic mean of a set of numbers is in between its smallest and largest values. What isn't obvious to me is how their definition of the pigeonhole principle leads to the corollary. Can anyone help me out?
 
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Hmmm... well I can interpret it so that the arithmetic mean is larger than the smallest number:

Taking the mean of numbers n1,...,nk is the same thing as having k containers with ni objects in the ith container, and then re-distributing them so that each container has an even amount. (the number of objects in each container is the mean) WLOG suppose that n1 is the smallest number (otherwise just re-arrange). There must be at least n1*k objects, and hence by the pigeonhole principle, one container (and hence all of them) must have more than the floor of (n1k-1)/k = n1-1objects. Hence they have at least n1 objects each.

There are missing details in the case that the mean is not exactly an integer, but then the containers each hold a number of objects that is smaller than the mean so it still works.
 
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