explicit sequence

**particlejohn** · July 30th 2008, 10:54 PM

Let $\bold{Z} \times (\bold{Z}- \{0 \}) = \{(a,b): a,b \in \bold{Z}, b \neq 0 \}$ . This is a countable set. Now let us suppose that we have a function $f: \bold{Z} \times (\bold{Z}- \{0 \}) \to \bold{Q}$ defined by $f(a,b) := a/b$ . Then $f(\bold{Z} \times (\bold{Z}- \{0 \})) = \bold{Q}$ which is at most countable. Since $\bold{Q}$ in infinite, then it must be countable. Then we can arrange the rational numbers in a sequence: $\bold{Q} = \{a_{0}, a_{1}, a_{2}, a_{3}, \ldots \}$ .

How would you come up with a sequence, such that every element of the sequence is different from every other element, and the elements of the sequence exhaust $\bold{Q}$ ?

**particlejohn** · July 31st 2008, 01:21 PM

Is there an explicit formula?

**ThePerfectHacker** · July 31st 2008, 02:18 PM

Originally Posted by particlejohn

Is there an explicit formula?

Yes. It is not nice looking, one such formula can take advantage of unique factorization of the numbers into primes.

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