# Counting set of rationals

1. Sep 15, 2014

### Born

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. Sep 15, 2014

### mathman

What is the question?

3. Sep 15, 2014

### Born

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?

4. Sep 16, 2014

### mathman

I don't believe there is a more elegant way. At least I can't think of one.

5. Sep 16, 2014

### economicsnerd

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.

6. Sep 17, 2014

### Hornbein

I don't think there is as simpler way, but you never know....