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Counting set of rationals

  1. Sep 15, 2014 #1
    Let ##T = \{ \frac{n}{m}\in \mathbb{Q} \vert n, m \in \{ 1, 2, ..., 9 \} \}##

    No values can repeat (e.g. ##\frac{2}{2},\frac{3}{3},...##)

    How many elements does the set have. I could just go ahead and count the elements and eliminate the repeats, but I'm wondering if there is a simpler (and more elegant) way to do it?

    Thanks
     
    Last edited: Sep 15, 2014
  2. jcsd
  3. Sep 15, 2014 #2

    mathman

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    What is the question?
     
  4. Sep 15, 2014 #3
    Sorry, edited the mistake. The question would be; is there a simpler (and more elegant) way to count the number of elements in the set?
     
  5. Sep 16, 2014 #4

    mathman

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    I don't believe there is a more elegant way. At least I can't think of one.
     
  6. Sep 16, 2014 #5
    I think this way is more "elegant", but is probably a slower way of counting elements in ##T_9##. I think this does provide a faster way of counting ##T_k = \left\{\dfrac{n}{m}: \enspace n,m\in\{1,...,k\}\right\}## for larger ##k\in \mathbb N##.

    Let ##P=\{2,3,5,7\}##, the set of primes that divide ##9##. For each ##D\subseteq P##, let ##M_D## denote the set of numbers whose prime factorization has positive exponents for exactly the primes ##D##. Let ##\mathcal D = \{(D,E): \enspace D,E\subseteq P, \enspace D\cap E=\emptyset\}##.

    Then ##T_9 = \bigcup_{(D,E)\in \mathcal D} \left\{ \dfrac{n}{m}: \enspace n\in M_D, m\in M_E \right\}##, and the last expression has no repetition anywhere.
     
  7. Sep 17, 2014 #6
    I don't think there is as simpler way, but you never know....
     
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