Construction using hyperreal numbers

 

Construction using hyperreal numbers[edit]

As in the hyperreal numbers, one constructs the hyperrationals ^*Q from the rational numbers by means of an ultrafilter.^[10][11] Here a hyperrational is by definition a ratio of two hyperintegers. Consider the ring B of all limited (i.e. finite) elements in ^*Q. Then B has a unique maximal ideal I, the infinitesimal numbers. The quotient ring B/I gives the field R of real numbers^{[citation needed]}. Note that B is not an internal set in ^*Q. Note that this construction uses a non-principal ultrafilter over the set of natural numbers, the existence of which is guaranteed by the axiom of choice.

It turns out that the maximal ideal respects the order on ^*Q. Hence the resulting field is an ordered field. Completeness can be proved in a similar way to the construction from the Cauchy sequences.

Construction from surreal numbers[edit]

Every ordered field can be embedded in the surreal numbers. The real numbers form a maximal subfield that is Archimedean (meaning that no real number is infinitely large or infinitely small). This embedding is not unique, though it can be chosen in a canonical way.

Construction from integers (Eudoxus reals)[edit]

A relatively less known construction allows to define real numbers using only the additive group of integers ${\displaystyle \mathbb{�}}$ with different versions.^[12][13][14] The construction has been formally verified by the IsarMathLib project.^[15] Shenitzer (1987) and Arthan (2004) refer to this construction as the Eudoxus reals, named after an ancient Greek astronomer and mathematician Eudoxus of Cnidus.

Let an almost homomorphism be a map ${\displaystyle �:\mathbb{�}\to \mathbb{�}}$ such that the set ${\displaystyle \{�(�+�)-�(�)-�(�):�,�\in \mathbb{�}\}}$ is finite. (Note that ${\displaystyle �(�)=\lfloor ��\rfloor }$ is an almost homomorphism for every ${\displaystyle �\in \mathbb{�}}$.) Almost homomorphisms form an abelian group under pointwise addition. We say that two almost homomorphisms ${\displaystyle �,�}$ are almost equal if the set ${\displaystyle \{�(�)-�(�):�\in \mathbb{�}\}}$ is finite. This defines an equivalence relation on the set of almost homomorphisms. Real numbers are defined as the equivalence classes of this relation. Alternatively, the almost homomorphisms taking only finitely many values form a subgroup, and the underlying additive group of the real number is the quotient group. To add real numbers defined this way we add the almost homomorphisms that represent them. Multiplication of real numbers corresponds to functional composition of almost homomorphisms. If ${\displaystyle [�]}$ denotes the real number represented by an almost homomorphism ${\displaystyle �}$ we say that ${\displaystyle 0\le [�]}$ if ${\displaystyle �}$ is bounded or ${\displaystyle �}$ takes an infinite number of positive values on ${\displaystyle {\mathbb{�}}^{+}}$. This defines the linear order relation on the set of real numbers constructed this way.

Other constructions[edit]

Faltin et al. (1975) write: "Few mathematical structures have undergone as many revisions or have been presented in as many guises as the real numbers. Every generation reexamines the reals in the light of its values and mathematical objectives."^[16]

A number of other constructions have been given, by:

• de Bruijn (1976), de Bruijn (1977)
• Rieger (1982)
• Knopfmacher & Knopfmacher (1987), Knopfmacher & Knopfmacher (1988)

For an overview, see Weiss (2015).

As a reviewer of one noted: "The details are all included, but as usual they are tedious and not too instructive."^[17]

1.1 Motivation for the Least Upper Bound Axiom Let us here consider the limitations of the rational numbers Q for conducting measurement. Theorem 1.2 (Q is Densely Ordered) Between any two distinct rational numbers p and q there exists a rational number r, and hence infinitely many rational numbers. Proof : If p 6= q are rational numbers, then since Q is totally ordered have two cases p < q and p > q Case 1: p < q: In this case, the rational number 1 2 (p + q) = 1 2 p + 1 2 q lies between p and q, as you can see directly: p < q =⇒ 2p < p + q < 2q =⇒ p < 1 2 (p + q) < q (Note: both implications seem to rely on Axiom 3 of R above, but looking at properties (10)-(11) of Theorem 3.12 in Numbers, and the subsequent constructions of Z and Q, it would seem these are in fact derived from 3 Theorem 3.12 in the case of Q: exercise!) Since this is true for each p and q, in particular it is true for p and 1 2 (p + q), so that p < 1 4 (p + q) = 1 2  1 2 (p + q)  < 1 2 (p + q) < q Repeating this process for all n ∈ N we have a sequence of rational numbers 1 2n (p + q)
n∈N lying between p and q. Case 2: p > q: reverse the roles of p and q in the proof of Case 1.  Theorem 1.3 (Archimedean Property of Q) For all p, q ∈ Q with p > 0, there exists a natural number n ∈ N such that pn > q. Remark 1.4 If we imagine p and q as masses on a see-saw, or balance, then supposing p a smaller mass than q, e.g. Archimedes the man versus a mountain, p < q, then by multiplying the power of Archimedes n-fold times, e.g. by increasing the length of p’s arm of the lever n-fold, the result is that Archimedes will lift the mountain. Our assumptions about N, Z, and Q make this a necessary consequence.  Proof : If q ≤ 0, then ∃1 ∈ N such that 1p = p > q. Thus, assume that q > 0. Let p = a b and q = c d , where a, b, c, d ∈ N. Let us consider the options: (1) p = q (2) p > q > 0 (3) 0 < p < q The first two cases are included only out of logical pedantry. In the first case ∃2 ∈ N such that q = p < 2p. In the second case, p > q > 0, and ∃1 ∈ N such that 1p = p > q. We move on to the third and only interesting case, when 0 < p < q. We need to find some n ∈ N such that q < np. I claim that n = b(c + 1) will work if we take p = a b and q = c d as above. To see this, note first that 0 < p < q means 0 < a b < c d , and therefore 0 < ad < bc (Theorem 3.12, Numbers). Since 0 < p < q, all a, b, c, d > 0, and in particular this implies a, b, c, d ≥ 1. In particular, d ≥ 1 implies dq ≥ 1q ≥ 1, which we insert here: c + 1 > c = d · c d = dq ≥ q citing again Theorem 3.12 in Numbers for the relevant maneouvers (that c + 1 > c and d ≥ 1 implying dq ≥ 1q ≥ 1). What do we gain from this? This: q < c + 1 ≤ a(c + 1) = b(c + 1)a b = np where n = b(c + 1) ∈ N.