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How can we define a well-ordering relation on set Q(set of all fractions) [We're looking for a total order on Q such that every subset of Q has a least element by applying that order. Clearly the conventional < relation we're familiar with doesn't do the job.]

Dana Mirafzal
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    The rationals are countable. So use any bijection with the natural numbers and transfer the order. – Ethan Bolker Jan 21 '22 at 16:55
  • @EthanBolker How can one find a bijection between natural numbers and rationals? – Dana Mirafzal Jan 21 '22 at 17:15
  • If you search for bijection between rationals and natural numbers you find many links, some right here on SE. https://math.stackexchange.com/questions/7643/produce-an-explicit-bijection-between-rationals-and-naturals – Ethan Bolker Jan 21 '22 at 17:58

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You just need to find a way to through them systematically. Let's ignore the negatives for a moment. List all of the integers, then those of the form $\frac{n}{2}$ (except those that you have already done), now those of the form $\frac{n}{3}$, etc.

Woops, that doesn't work since the first set does not end and you never start the second. However, we can fix the procedure.

List all of the form $\frac{m}{n}$ where $m + n = 1$, that's just $0$. Now those of the form $m + n = 2$ that's just $1$. Now $m + n = 3$, that's $2$ and $\frac{1}{2}$. Etc. This time, each set is finite and hence, it will end and you can go onto the next. Again, skip ones which have already appeared. Every positive rational will eventually show up.

What about the negatives? Just pop them in after the corresponding positive.

badjohn
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