The Road to Reality

Exercise [16.07]
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Author:  robin [ 15 May 2010, 08:27 ]
Post subject:  Exercise [16.07]

We wish to define a way of ordering the fractions. We can do this by consider the fractions as the ordered pairs (a,b)\in\mathbb{Z}\to\mathbb{Z}, such that b\neq 0, and with the equivalence (a,b)\equiv(\lambda a,\lambda b) for any \lambda\in\mathbb{Z}\backslash\{0\}.

In Exercise 16.8 we found a 1-1 mapping f:\mathbb{N}\times\mathbb{N}\to\mathbb{N}. Thus there exists a well-defined inverse function g = f^{-1}:\mathbb{N}\to\times\mathbb{N}.

We first construct the sequence \{a_n\}, n\in\mathbb{N} via:
a_0=g(0), a_1=-g(0), a_2=g(1), a_3=-g(1), a_4=g(2), a_5=-g(2), \dots

Note that we needed to include the negative values because the fractions take both positive and negative values. Now we edit the sequence to remove duplicates and values which do not represent fractions. Namely we go through the sequence \{a_n\} term by term starting from \{a_0\}. If a_k=(p,0) or a_k\equiv a_r for some r<k then we delete the term. Otherwise we leave it in, identifying a_k=(p,q) with the fraction p/q.

After relabelling indices, this provides a new sequence \{b_n\}, n\in\mathbb{N} which defines an ordering for the fractions.

Author:  deant [ 03 Apr 2011, 05:40 ]
Post subject:  Re: Exercise [16.07]

An even easier way is just to order fractions first by (|numerator|+denominator) and then by numerator (assuming the denominator is always positive; i.e. the sequence that begins:

0/1; -1/1, 0/2, 1/1; -2/1, -1/2, 0/3, ....

You still need to then go through and form the final sequence by removing all fractions that aren't in their simplest form, though. (e.g. 5/10 -> 1/2, so 5/10 gets removed, etc.).

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