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Is there any algebraic definition or construction of real numbers ?

If not, why ?

halfpog
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    What does "algebraic" mean in this context? – Ulrik Jan 26 '14 at 19:19
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    Do dedekind cuts count? – Git Gud Jan 26 '14 at 19:19
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    You can construct them as the quotient ring $C/M$ where $C$ is the ring of Cauchy sequences in ${\mathbf Q}$ (with pointwise addition and multiplication) and $M$ is the maximal ideal of null sequences (sequences in $\mathbf Q$ that tend to $0$). – KCd Jan 26 '14 at 19:27
  • Without using Dedekind axiom, Cauchy sequences, upper bound axiom or equivalents. I don't know how to explain well. My algebra book says :<< the construction of the reals has not an algebraic nature, so refer to a calculus textbook >>. – halfpog Jan 26 '14 at 19:30
  • What "algebra book" are you reading? Construction of the real numbers would not be covered in most calculus textbooks either. – hardmath Jan 26 '14 at 21:11
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    Here's a suggestion I'm not sure about (feedback required): $\Bbb C/\Bbb Q$ is the algebraic closure. However, it isn't ordered. Could it be that $\Bbb R$ is a (the?) maximal ordered field $\mathbb{Q}\subset F\subset\mathbb{C}$? – Jonathan Y. Jan 26 '14 at 21:15
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    You might look up "real closed field" – Robert Israel Jan 26 '14 at 21:47
  • I mean Mathematical analysis book – halfpog Jan 27 '14 at 14:18

1 Answers1

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There is a strong sense in which the answer is no:

It's an informal concept, but we might reasonably say that an "algebraic" construction of the reals is one which only refers to algebraic operations - e.g., polynomials and their roots - and does not talk about more complicated things, such as arbitrary sets of rational numbers (e.g., Dedekind cuts and Cauchy sequences). In particular, we might ask for this construction to take place in first-order logic: https://en.wikipedia.org/wiki/First-order_logic.

This, we can prove, is impossible. Here's one way to say this:

There is a countable field $F$, with $\mathbb{Q}\subset F\subset\mathbb{R}$, such that $F$ satisfies every first-order sentence in the language of fields which is true in $\mathbb{R}$.

This is a direct consequence of the Lowenheim-Skolem Theorem, and it is very powerful and flexible; for instance, we can replace "the language of fields" with "the language of fields with exponentiation, $\sin$, $\cos$, and the Gamma function," and we could still find such an $F$. So no algebraic construction is guaranteed to get all of $\mathbb{R}$.

Noah Schweber
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