1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pigeonhole Principle and the Arithmetic Mean

  1. Feb 26, 2012 #1
    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 [itex]\lfloor[/itex][itex]\frac{m-1}{k}[/itex][itex]\rfloor[/itex] 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?
  2. jcsd
  3. Feb 26, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook