Results 1 to 10 of 10

Math Help - Hilbert's Hotel is Troubling Me

  1. #1
    Newbie
    Joined
    Nov 2009
    From
    London
    Posts
    5

    Hilbert's Hotel is Troubling Me

    I am about to do an inspiring lesson on Hilbert's hotel to my 15 year old pupils (they are bright, don't worry) but I have an issue that needs resolving for my own benifit.

    The problem is when infinitely many coaches turn up with infinitely many people on them. The solution, as I'm led to believe, is this:

    1) Move all the current guests from their room m to 2m, freeing up all the odd numbered rooms.
    2) Label each coach q = 1, 2, 3, 4, ...
    3) Label each seat on each coach n = 1, 2, 3, 4, ...
    4) Place people from the 1st coach into rooms numbered 3^n, 2nd coach into 5^n, 3rd coach into 7^n and qth coach into the (p+1)^n where p = (q + 1)th prime number.

    Thus everyone is accomodated. My problem is that the hotel is now not full. In fact, every odd numbered room which has 2 or more distinct factors is empty.

    Please can someone tell me where I have got things wrong or whether the hotel is indeed now not full.

    Thanks,

    James (this is my first post so please be kind).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    Quote Originally Posted by jac1999 View Post
    The problem is when infinitely many coaches turn up with infinitely many people on them. The solution, as I'm led to believe, is this:
    1) Move all the current guests from their room m to 2m, freeing up all the odd numbered rooms.
    2) Label each coach q = 1, 2, 3, 4, ...
    3) Label each seat on each coach n = 1, 2, 3, 4, ...
    4) Place people from the 1st coach into rooms numbered 3^n, 2nd coach into 5^n, 3rd coach into 7^n and qth coach into the (p+1)^n where p = (q + 1)th prime number.
    Thus everyone is accomodated. My problem is that the hotel is now not full. In fact, every odd numbered room which has 2 or more distinct factors is empty.
    Well of course the hotel is not full.
    If you want it full each odd numbered room must be occupied.

    The mapping \Phi (j,k) = 2^{j - 1} \left( {2k - 1} \right) is a bijection \Phi :\mathbb{Z}^ +   \times \mathbb{Z}^ +   \leftrightarrow \mathbb{Z}^ +  .
    That is a fun fact to prove.

    Assign the k^{th} person on the j^{th} coach to room # 2\Phi (j,k)-1.
    So the first person on the first coach is assigned to room #1.

    That will fill the hotel.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    From
    London
    Posts
    5
    Thanks Plato. I appreciate the answer.

    I guess then that we have to be happy that the hotel can absorb an infinite number of people from an infinite number of coaches and, at the end of it, have an infinite number of spare rooms if we use the solution I laid out.

    My class will love this tomorrow (I think I'll leave the bijection out)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,501
    Thanks
    765
    There is a nice graphical representation of a bijection between \mathbb{N}\times\mathbb{N} and \mathbb{N}: see Wikipedia.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2009
    From
    London
    Posts
    5
    That is useful, thanks.
    Looks very similar to the method of showing the rationals are countably infinite. Why is the 'path' different in this one?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,501
    Thanks
    765
    I agree, though I don't know what picture exactly you have in mind for rationals. I think that showing that \mathbb{N}\times\mathbb{N} is equinumerous with \mathbb{N} is simpler than the proof for rationals because you don't need to identify certain pairs, exclude division by zero, etc.

    I am used to the picture that differs from the one in Wikipedia: from (0, 1) one goes to (0, 2) instead of (2, 0) and then descends, rather than ascends, from (0, 2) to (2, 0). I believe this bijection has a simple polynomial formula in one direction, different from the one above. However, this is not important at all.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2009
    From
    London
    Posts
    5
    The one I'm talking about looking like the one at this link
    Proof that rational numbers are countable - from Homeschool Math
    However I have issues with this as it doesn't have negative numbers or zero.
    Although I agree that NxN = N is easier to show than Q=N.

    By the way, how do people get maths symbols on here?
    [IMG]file:///C:/DOCUME%7E1/JAC/LOCALS%7E1/Temp/moz-screenshot-1.png[/IMG][IMG]file:///C:/DOCUME%7E1/JAC/LOCALS%7E1/Temp/moz-screenshot-2.png[/IMG][IMG]file:///C:/DOCUME%7E1/JAC/LOCALS%7E1/Temp/moz-screenshot-3.png[/IMG][IMG]file:///C:/DOCUME%7E1/JAC/LOCALS%7E1/Temp/moz-screenshot.png[/IMG]
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,501
    Thanks
    765
    Yes, this is the picture I was talking about.

    You can write LaTeX code inside the [tex] tag. Also, when you are writing a message, there is a Sigma on the toolbar that surrounds the selected text with [tex]. I also had to search for this. I think, how to write math should be more prominently shown in the forum help section.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    Quote Originally Posted by jac1999 View Post
    The one I'm talking about looking like the one at this link
    Proof that rational numbers are countable - from Homeschool Math
    However I have issues with this as it doesn't have negative numbers or zero.
    Although I agree that NxN = N is easier to show than Q=N.


    By the way, how do people get maths symbols on here?
    Why would you have any issues with that argument?
    The countable union of countiable sets is a countable set.

    You know that f(x)=-x is a bijection.
    The finite set \{0\} is countable.

    See this forum.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Nov 2009
    From
    London
    Posts
    5
    I don't have any issues with it mathematically for precisely the reasons you make. But I do have issues with it as a 'job done' pictoral representation.

    Only a small issue, though.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 3 guys go to a Hotel
    Posted in the Math Puzzles Forum
    Replies: 9
    Last Post: November 22nd 2011, 04:32 AM
  2. Foundations of mathematics - something is troubling me
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: July 5th 2011, 06:55 AM
  3. Hilberts Hotel
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 21st 2009, 10:29 AM
  4. A troubling limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 5th 2008, 01:51 PM
  5. A troubling for me...
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 14th 2007, 05:20 PM

Search Tags


/mathhelpforum @mathhelpforum