Usually the set of numbers are introduced starting from integers, from wich the rational numbers are defined using equivalence classes of couples of integer numbers. Than, from these rational numbers, we can construct the real numbers via Dedekind cuts or Cauchy sequences.

But real numbers can be defined in purely abstract way as a set that is a field with a total order that is Dedekind- complete (we can refer to Tarski's axiomatization).

Is it possible, starting from this abstract definition of reals, define the subfield of rationals (and the subset of integers)?

In other words can we define the usual sets of numbers starting from the real numbers instead of from integers?

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    $\begingroup$ The integers are the least subring of $\mathbb{R}$; the rationals are the least subfield of $\mathbb{R}$. $\endgroup$ – egreg Jan 11 at 14:01
  • $\begingroup$ You should clarify what kind of definitions you allow. egreg's comment gives you an algebraic definition, but, for example, we can show there is no definition of $\mathbb Q$ in $\mathbb R$ in the first-order language of rings. $\endgroup$ – Wojowu Jan 11 at 14:04
  • $\begingroup$ The last paragraph in the page you refer to seems to imply that the integers (and the rationals) are used for defining the multiplication. The integers are the subgroup generated by $1$. $\endgroup$ – egreg Jan 11 at 14:06

Yes, of course. Because $\mathbb R$ is a field, it has a unit, $1$. Then you can immediately construct the integers by considering the additive subgroup generated by $1$: you have $0$, $1+2$, $2+1=3$, etc., and $-1$, $-1+(-1)=-2$, etc.

And now you can consider the subfield of $\mathbb R$ generated by $\mathbb Z$: that requires you to have the multiplicative inverses of the nonzero integers: $1/n$, for all $n\in\mathbb Z$, and $m/n=1/n+\cdots+1/n$ for all $m\in\mathbb N$.

  • $\begingroup$ Doing it this way, we start with the integers which is not what the author wants. $\endgroup$ – Peter Jan 11 at 14:09
  • $\begingroup$ Thank you @Martin. This construction obviously works, but the difficult (as noted in the comment) is to find rationals from reals. What I don't see is how we can separate the rational from the irrational elements of the field of reals, giving a constructive definition (maybe that my question is not so clear about this). $\endgroup$ – Emilio Novati Jan 11 at 14:24
  • $\begingroup$ @EmilioNovati If you agree that this construction of $\mathbb Z$ works (and is satisfactory to you), then you can take quotients of elements of $\mathbb Z$ and this will give you $\mathbb Q$. $\endgroup$ – Wojowu Jan 11 at 14:53
  • $\begingroup$ @Emilio: I have to admit that I fail to see where you are going with this. In what construction would it be possible to separate rationals from irrationals other than by definition or some concrete cases? In the usual setting, it is not trivial to tell if a Dedekind cut is rational or not. $\endgroup$ – Martin Argerami Jan 11 at 16:38

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