Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Intuitive way to explain |Q| = aleph zero?

  1. Jan 8, 2007 #1


    User Avatar
    Science Advisor
    Homework Helper

    What is an intuitive, calculus-grade way to explain that rationals have the same cardinality as N? (Same question for perfect squares.)
  2. jcsd
  3. Jan 8, 2007 #2


    User Avatar
    Gold Member

    I'm not trying to be difficult, but aren't the standard proofs pretty inuitive? Especially for the perfect squares. To every element of the set of perfect squares you can correspond exactly on element of the set of naturals: its square root. The same goes for the other way around. So you have a one to one correspondance between the squares and naturals, so they have the same cardinality. The proof is nearly identical for the rationals, except that the bijection is somewhat more complicated
  4. Jan 8, 2007 #3
  5. Jan 8, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    Heres a pictorial way to show it for the rationals. Start with the positive rationals and start listing them like so
    1/1 1/2 1/3 1/4 1/5......
    2/1 2/2 2/3 2/4 2/5 .......
    3/1 3/2 3/3 3/4 3/5 ....

    Now, to count, follow this sequence: 1/2, 1/2, 2/1, 3/1, 2/2, 1/3, 1/4, 2/3,.... (I hope you can see what I'm doing). Now, it should be clear that this list is countable. Now, insert the negative rationals just before their positive counterparts, i.e. -1/1, 1/1, -1/2,..., and just put zero at the top left. Now, using the same arguement, the full set of Q is countable.

    I'm not sure whether this was something like what you were after, but it seems quite intuitive!
  6. Jan 8, 2007 #5

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Ah, diagonal method - possibly because a proof that Q is countable is called diagonal, and that R isn't yet has the same adjective, people have wrong impressions.

    One way is a standard method using prime decomposition.

    Send a/b to 2^a*3^b (for a and b >0) or 2^a*3^b*5 for negative rationals. This is a bijection from Q onto a subset of N. This proof shows that any finite product of countable sets is countable, and shows an important difference between union and product (in the categorical sense, perhaps).
  7. Jan 9, 2007 #6


    User Avatar
    Science Advisor
    Homework Helper

    I thought both the "perfect squares" and the "rationals" demonstration can use the cross-tabulation square where row and column headings are the natural numbers. For the perfect squares, each cell holds the product "row # * column #"; all perfect squares are on the diagonal, which you can count linearly.

    For the rationals, each cell is row #/column # (or the inverse), but I wasn't sure how to count them. Now I know. Thanks.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Intuitive way to explain |Q| = aleph zero?
  1. Aleph null problem. (Replies: 2)

  2. Aleph null ! (Replies: 4)