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I would appreciate if someone could demonstrate how to show $\sqrt{2}, e, \pi$ are real numbers in the axiomatic approach to defining $\mathbb{R}$ (without reference to a model).

The Wikipedia page for Real Numbers gives a summary of the different approaches to defining the reals. The article highlights two seperate ideas that apply to defining them. On one hand, there is an axiomatic definition that is stated something like "The Real Numbers are the unique, up to Isomorphism, Dedekind-complete ordered field". On the other hand, specific models of the reals can be used to define a set of objects which are then shown to satisfy these axioms. I am familiar with the Cauchy construction and it is straightforward to me how to show that e.g. $\sqrt{2}, e, \pi$ are real numbers.

What I don't understand is how you can do that without a model, i.e. just using the axiomatic definition. To make the issue worse, I don't understand what a general "number" is defined to be in the first place, so that we can say there is a subset of all "numbers", the reals, that satisfies the axioms. I'd be more confortable if they were just considered "objects" that satisfy these axioms. I am left with the impression that the axiomatic definition without a model is meaningless. I think a way to say that is it a characterization of the reals, but cannot serve as the definition.

Note: I take it as given that the rational numbers have been previously well-defined.

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    $\begingroup$ Well, how do you define those numbers in the first place? $\endgroup$ – Tobias Kildetoft Jun 27 '15 at 15:26
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    $\begingroup$ @TobiasKildetoft Yes, that is a very important question I am trying to answer. I added in a note that I accept rationals have been defined. By that I mean, the naturals using Peano, and then integers, and then ratios of integers. $\endgroup$ – muaddib Jun 27 '15 at 15:37
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    $\begingroup$ By "those numbers" I meant $e$, $\pi$ and $\sqrt{2}$. $\endgroup$ – Tobias Kildetoft Jun 27 '15 at 15:38
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    $\begingroup$ @TobiasKildetoft I personally use the Cauchy construction. The point of what I am asking is how could one use just the axiomatic definition to say they are real numbers. $\endgroup$ – muaddib Jun 27 '15 at 15:43
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    $\begingroup$ @muaddib The question is, how you define/characterise $\sqrt{2},e,\pi$. I can say $\sqrt{2} := 4, e := \frac{5}{6}, \pi := 0$. Then I have defined these symbols as real (rational even) numbers. But these are not what one would expect. On the other hand, I can define $$\begin{gather} x := \sup \{ r\in\mathbb{Q} : r^2 < 2\},\\ y := \sup \Biggl\{ \sum_{k = 0}^n \frac{1}{k!} : n \in \mathbb{N}\Biggr\},\\ z:= \sup \Biggl\{ \sum_{k = 0}^n \frac{8}{(4k+1)(4k+3)} : n \in \mathbb{N}\Biggr\}.\end{gather}$$ Then these are real numbers that have all the properties we expect from $\sqrt{2},e,\pi$. $\endgroup$ – Daniel Fischer Jun 27 '15 at 15:50
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Updating the answer according to Daniel's comment.

You can define $\sqrt 2$ as the least upper bound of the bounded set $S =\{x : x^2 <2\}$.

And defining $$e_n=\sum_{k=0}^n \frac{1}{n!}$$ for $n \ge 0$ integer, $e$ can be defined as the least upper bound of the bounded set $$U=\{x : \exists n (x < e_n) \}$$

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  • $\begingroup$ I need an answer without reference to a model. Dedekind cuts is one of the multiple models. Just using the axiomatic definition. $\endgroup$ – muaddib Jun 27 '15 at 15:35
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    $\begingroup$ The least upper bound only uses Dedekind-completeness, nothing in this answer actually uses Dedekind cuts. If you remove the reference to Dedekind cuts, you have a definition of a real number $x$ that happens to have the properties $x > 0$ and $x^2 = 2$. $\endgroup$ – Daniel Fischer Jun 27 '15 at 15:42

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