The time has come (actually, it had come before 2002) to shred this page into a sub-directory and merge it with a page on linearity. It is presently a mess of fragments from different attacks on the subject matter, that doesn't validate, and uses an old notation (except for the initial part from 2002/Jan, using a newer one). Here I'll add dereferencing links if I remove any named anchors from this page:

additive domain characterized by V = (: (V: v+x←x :) ←v :) – or
maybe x←v+x, i.e. subtraction – so (:V|) are the vectors while
(|V:) are the translations – or displacements.

scalings of V commute with all other maps that respect V, but there's
more to it than that (or respect is a sophisticated notion),

mappings from V which respect V and commute with all scalings are linear
(the equivalent truths for mappings to V are implied by the point-wise combination laws)

there's a continuum-ness to be asked of the scalars, at some point, somehow

the scalars imply the positives, effectively as (equivalence classes of) least
upper bounds of increasing sequences of positive rationals

the scalars (as a linear space) support a quadratic form, scalar x*y for scalar
x, y, for which x*x is always positive, unless x is an additive identity, in
which case x*x is x.

one can build simplices out of {({positives}:h:n): natural n, sum(h) = one},
a.k.a. the universal simplex (for given positives)
[Using one for the unit in {positives} to distinguish from 1 = {{}} in {naturals}.]
the canonical simplices are left values of
 psimplex = (: {({positives}:h:N): sum(h) = one} ←N :{naturals})
 has vertices ({one}::{n}) for n in N
 the n-simplex, with 1+n vertices, is psimplex(1+n),
 hence the psilent prefix on the mappings name
 psimplex(0), i.e. the −1 simplex, is {}
 psimplex(1), the 0-simplex or point, is {[one]} 
 psimplex(2), the 1-simplex or chord, is
	{[s,t]: s+t=one, s, t positive}&unite;{({one}:|{i}): i in 2}}
 so unit = (| h(0)←h :psimplex(2)), which quietly ignores h if 0 isn't in (:h|),
	 = (| transpose(psimplex(2),0) :)
	 = unite(:(|h:)←h:psimplex(2)) is all the positives up to one

i in n implies psimplex(n) subsumes psimplex(i)
 r = ({mappings ({positives}::{naturals})}: sum(:h:N)=1; (:h:N)←N :{finite})
{r(n)} = psimplex(n)

these enable constant up to order n notions for natural n
constant up to order 0 constitutes continuity
constant up to order 1 amounts to piece-wise differentiable
differing by a linear map from constant at order 1 is differentiable
constant up to order 2 implies differentiable (with derivative zero)
constant up to order n may fairly be described as
 [continuous, piece-wise linear, quadratic, cubic, quartic, quintic, …](n)

simplices – and shrink – thus deliver our topology,
i.e. continuity – or regenerate it, if the topology was part of the
sophistication of scalars – along with the gradients of chords when
computing derivatives;

the ability to map a simplex (for pertinent positives)
non-degenerately (which more-or-less means monic) into a linear domain (fpp)
tells us that the simplex doesn't have too many vertices for our dimension

We can also examine unit and grow = {positives not in unit},

is q > p ?
if rationally commensurate, trivial.
so try: is q/p > 1 ? and promptly come back to:
is p + N > 0 ? looks scary for irrational p, natural N

hmm: values smaller than 1 should show themselves via powers tending to 0

for natural n:
      1<x: 0< power(n,x) < power(1+n,x)
1 is the identity, repeat(once)
0<x<1: 0< power(1+n,x) < power(n,x)
when available:
0 is an additive identity, easy to spot
−1 is a square root and additive inverse of 1, easy to spot
x<−1: power(2.n+3,x)<power(2.n+1,x)<−1<0<power(2.n,x)<power(2.n+2,x)

a<b means a+p=b for some positive p
initially the only positives we know of are rationals.
however, if


early Jan 2002

Scalar Domains

In many mathematical contexts, particularly those used to model physical reality, it is usual to involve some form of scalar. This is usually either the real numbers or the complex numbers. The predominant properties of the scalar domain are its additive, multiplicative, integral and differential structures. It may also have an ordering or a conjugation. For the purposes of my pages, I shall define a very basic scalar domain and point to two particular common cases as covering what I need of my scalars.

A scalar domain consists of a topological space, D, and a pair of epic continuous associative binary operators, + and . in {(D×D|:D)}, known as addition and multiplication respectively, with:

multiplication distributing over addition:
that is, for any a, b, c in D, (a+b).c = (a.c) + (b.c) and c.(a+b) = (c.a)+(c.b);
addition cancellable:
that is, a+b = a+c implies b=c; and
a multiplicative identity,
written 1: for any a in D, a.1 = a = 1.a.

Further Reading

Scalar domains contain just enough to be the foundation for discussions of linearity and, consequently, of both differentiation and vectors.

Variations on the theme.

Just this minimal one, for now. I must re-shuffle so that the familiar ones come and join in.


Any scalar domain must contain a multiplicative identity, 1. It does, in fact, suffice to have just this one element. There is precisely one binary operator ({1}×{1}|:{1}), which maps (1,1)->1; this is the trivial non-empty group. Furthermore, this binary operator distributes over itself: if we write it as both . and + (to separate out the roles in distributivity), we see 1.(1+1)=1.1=1, 1.1+1.1=1+1=1, (1+1).1=1.1=1. Thus, if we take both multiplication and addition to be this one binary operator, we obtain a scalar domain {1}. I'll call this the trivial scalar domain. I shall almost exclusively be interested in non-trivial scalar domains !

The rest of this page begins with a discussion of the algebraic and analytic consequences of this definition and then goes on to examine, in further detail, some of the more important types of scalar domain.


Particular scalar domains may, of course, possess further algebraic properties. Multiplication may also be cancellable; either or both of the binary operators may be complete or commutative (Abelian): addition may also have an identity and either may inverses. An additive identity, if present, is called a zero and written 0. If a scalar, d, has an additive inverse, we call this −d; if it has a multiplicative inverse, we call this d−1 and we write its product with any scalar, c, as c/d.

As I explain in discussing linearity, cancellability of multiplication is incompatible with completeness of addition, even going so far as to preclude any sum being equal to either of its summands (i.e. any r=r+t). In particular, a scalar domain can have a zero or be multiplicatively cancellable, but not both. By comparison, allowing the presence of a multiplicative identity, 1, presents no problems for the additive structure (as, indeed, additive cancellability was harmless without completeness). I'll describe a scalar domain which admits of no solution to a=a+e as non-cyclic: multiplicative completeness is only possible in non-cyclic scalar domains.

So, what's implied for a non-cyclic scalar domain ? What does that definition give us ?

[Huge removal to within the new sub-directory.]

I'll say that a scalar domain is fieldish if each of its members either is an additive identity or has a multiplicative inverse. An additively complete fieldish scalar domain is called a field. [Its addition is complete and cancellable, so forms a group: its multiplication, when zero is elided, has identity and inverses, so forms a group: thus this definition coincides with the conventional one.]

Since both addition and multiplication of scalars are associative binary operators, we can employ the standard bulk action construction to obtain products and sums of functions from finite ordered sets to any scalar domain: we can drop the ordering condition if our addition and multiplication are Abelian.

In non-commutative scalar domains, we should really distinguish between left- and right-zeros, -completeness and -inverses. However, we shall always deal with additively commutative scalar domains and nearly always with multiplicatively commutative domains. It is easy to show that, in a scalar domain with both a left-zero and a right-zero, the two zeros are equal: likewise, to show that the zero (if any) in a scalar domain is unique (simply consider the sum of two zeros). If both left-inverses and right-inverses are present (for either addition or multiplication) they must coincide (consider a−l . a . a−r; bracketed one way it delivers one inverse, the other the other; but associativity says the two bracketings give equal answers).

For a scalar domain D, I'll describe a function (D|f:D) as a conjugation precisely if: for any d in D, +0(f(d)) = f o +1(d); likewise for · in place of +: and f(1)=1. The identity is always a conjugation, in this sense: I call it trivial, if only for this reason.

Order and locality

Reminders: A scalar domain is non-cyclic ⇔ it contains no solution to a+e=a. A relation on a set S is a subset of S×S, and you may wish to remind yourself of the notation I introduce for relations, extending the meaning of my basic bracket notation. A relation, f, is a partial order precisely if it is transitive and contains no member of form (a,a).

For any non-cyclic scalar sub-domain, R, of a scalar domain, S (which may, of course, be R), I define two relations on S

={(r+s,s): r in R, s in S}

I'll write > as a short-hand for this and t>s for (t,s) is in R-order(S). We then have, for s in S: (|>:{s}) = (R: +1(s) |) denoting everything greater than s; and ({s}:>|) denoting everything than which s is greater. I'll write R-between(s,t) for the intersection of (|>:{s}) and ({t}:>|), which is empty unless t>s.

={(s,s+r): r in R, s in S}

I'll write < as a short-hand for this and s<t for (s,t) is in order-R(S). Again, for s in S: ({s}:<|) = (R: +0(s) |) denoting everything less than s and (|<:{s}) denotes everything than which s is less. I'll write between-R(s,t) for the intersection of ({s}:<|) and (|<:{t}), which is empty unless s<t.

Because R is non-cyclic, and addition is associative and cancellable, both of these are partial orders. If addition is commutative, they are transpose to one another: otherwise, we have to distinguish between s<t and t>s, thus between everything than which s is less and everything greater than s. I'll develop the discussion in terms of < and leave you to work out the analogue for >. These R-orderings are, of course, only of any real interest when R is enough of S: e.g. when S is the reals, R needs to be the positive reals.

Both of these partial orderings are respected by R-multiplication and S-addition: that is, for a<b: r in R implies r.a<r.b (as a+t=b, with t in R, gives r.a+r.t=r.b with r.t in R); and s in S implies s+a<s+b (as s+a+t=s+b). Consequently: for a<b and s<t in S, s+a<t+b; and for a<b and r<t in R, r.a<t.b.

In a scalar domain S, define the locality of a in S to be locality(a) = {s in S: no r in S has a+r=s or s+r=a} and describe any s therein as local to a. Note that the relation is local to is symmetric: any a local to s implies s local to a. This is only of any real interest for a non-cyclic scalar domain, of course. In a non-cyclic scalar domain, is local to is also reflexive: every member of S is local to itself.

This definition lets us decompose any non-cyclic scalar domain, in terms of any element, s, into locality(s), ({s}:<|) and (|<{s}).

So now let's look at what faces locality can wear. In the case of the positive integers, rationals or reals, on which < and > are full orderings, we have locality(s) ={s}. This is not, however, the only possibility – nor, indeed, the only one in which locality is transitive (thus an equivalence relation). An important detail is that my definition of a scalar domain insisted that both + and × be epic. This means that 1=a+b for some a and b, which are thus less than 1.

In our non-cyclic scalar domain, from any value, a, less than 1, we can take progressive powers: each of which (as multiplication respects <) is less than 1 and those before it. Indeed, generally, for any s in S, a.s<s. Thus for any s<t, take u for which s+u=t, to obtain a.u<u whence s<s+a.u<s+u=t, giving us a value strictly between s and t. Thus each value less than 1 allows us to find a value between any ordered pair of values, whence (be applying this to the intervals between the end-points and this mid-point and repeating ad nauseam) there are infinitely many values between any ordered pair. Formally, for any positive integer n and any s<t in a non-cyclic scalar domain S, there is some monotonically increasing function (n|f:S) with each r in (f|) between s and t: s<r<t.

I suspect an important property to ask for is < = <&on;< – that is, whenever a<c, there is some b for which a<b<c.

Applying this to the interval between a and 1, with a+u=1: a.(1+u) = a+a.u <1, so we have some r, namely 1+u, for which a.r<1<r. This then implies a.a<a.a.r<a<a.r and so on; also, any s between r and 1 will do in place of r.

Given such an r, suppose we have some s, t in S local to one another. Then we have s<s+a.r<s+1<s+r and no u for which s+u=t or t+u=s. Consequently, we have no u for which s+r+u=t+r or t+r+u=s+r (I suspect I just required + to be commutative), so s+r is local to t+r: likewise, s+a.r is local to t+a.r and s+1 is local to t+1. For t to be local to s+r, there would have to be no u for which t+u=s+r, so we'd need (| +0(t) :{s+r}) empty, along with (| +0(s) :{t+r}). For any r, s in S and t local to s, we have s < s+r and no u for which s+u=t or t+u=s: consequently, no u for which (t+r)+u=s+r or (s+r)+u=t+r, so t+r is local to s+r. Likewise, t+a.r is local to s+a.r, which is greater than s but less than s+r (as a<1). Consider the rationals and all rational multiples of the square root of some prime: call this base irrational simply root. These form a non-cyclic scalar domain, with cancellable multiplication. Let q be root.root−1, which is a positive integer: consequently, we can divide by q or 1+q. In particular, dividing root by 1+q, we obtain a multiplicative inverse for root, as root.root/(1+q) =1. From root.root = 1+q, we obtain root.(1+root)/q = (root + 1+q)/q = 1 + (1+root)/q, making 1.(1+root)/q less than root.(1+root)/q: we also obtain root.(1+root)/q >1. Consider 1 and root: divide each by q and multiply by 1+root, to obtain (1+root)/q and (root+q+1)/q = 1+(1+root)/q, whence we have 1.(1+root)/q less than root.(1+root)/q (by 1). Before we can cancel the (1+root)/q from these, we need to express 1 as a sum of multiples of (1+root)/q. We know (1+root).(1+1/root)=2+root+1/root and the square of (root + 1/root) is just 2 + 1+q + 1/(1+q) root.root = 1+q, whence Is 1 less than root ? Is multiplication complete – in particular, does 1+root have a multiplicative inverse ? (If so, the positive integer root.root−1 multiplied by this inverse serves as root−1: thus 1 is less than root. Likewise, if 1 is less than root, we can divide root−1 by root.root−1 to obtain a workable reciprocal for 1+root and, I'm willing to believe, for anything else.) These give us arbitrarily small values and imply that localities are infinitesimal, in the sense that everything in the locality of some s must be less than the sum of s with any of these arbitrarily small values. I suspect this makes locality transitive (and, thus, an equivalence relation). In the familiar non-cyclic scalar domains of positive rationals and reals, all localities are singleton sets, which makes locality trivially transitive Of course, if localities are all singleton sets (whence, somewhat dully, locality is an equivalence relation) we obtain a full ordering. When is locality transitive ?

Polynomials Equations

On a scalar domain, D, we can define addition and multiplication on functions (S|:D), for any set S, via (S|f:D).(S|g:D) = (S| x->f(x).g(x) :D) with, likewise, (f+g)(x)=f(x)+g(x). These naturally inherit distributivity, associativity, additive cancellability and a multiplicative identity, 1= (S| x->1 :D). Additive or multiplicative completeness will be inherited if D has it Consequently, any additively and multiplicatively closed subset of {(S|:D)} forms a scalar domain, so long as it contains 1. This scalar domain is incidentally a linear space over D – as witnessed by D's natural embedding (D| d-> (S| x->d :D) :) in {(S|:D)}.

We can also define a function ({naturals}| power :{(D|:D)}), defined by: power(0)= constant(1)= (D| x->1 :D); for any natural number n, power(1+n) = (D| x-> x.power(n, x) :D), which is its product with power(1) (derived, by this formula, from power(0)). We find (:power|) is a multiplicatively complete subset of (D|:D), but not an additively complete one: however, its members are linearly independent in {(D|:D)}, a linear space over D as above (S=D). Their span in {(D|:D)} is both additively and multiplicatively complete: the functions in it are known as polynomials; they are sums of scaled powers. The finite span of the powers, in {(D|:D)}, is known as the polynomials on D.

A polynomial equation is a pair of polynomials, subject to the natural equivalence relation induced by additive cancellability: namely, for any pair (P,Q) of polynomials and any polynomial R, (P,Q)~(P+R,Q+R). The order of a polynomial equation is, for any (P,Q) equivalent to that equation, the largest natural number n for which: for every natural number m>n, the coefficients of power(m) that appear in P and Q, respectively, when expressed as sums of scaled (by the given coefficients) powers, are equal. That is, if P has a term in xm, then so has Q and their coefficients are equal (i.e., when all such terms on each side are summed, they cancel). If either P or Q has no terms of form scalar.power(m), then the other, likewise, has none and the sums are both empty (so we can deem them equal, even if they are undefined). The equivalence induced by cancellability respects this definition of the order of an equation.

Single-variable polynomial equations in D have additive completeness even if D does not, in the sense that (so long as addition is commutative) adding P(x)=Q(x) to Q(x)=P(x) gives a sum which can necessarily be cancelled to a=a for any scalar a: this polynomial equation serves as an additive identity, even if D lacks one – and we have just seen that the additive inverse of (P,Q) is (Q,P). The additive identity of polynomial equations is described as degenerate.

For a polynomial equation, (P,Q), we define roots(P,Q)= {x in D: P(x)=Q(x)}. Because D is additively cancellable, some equations (such as x+a=x with a not an additive identity in D) have no solutions: I shall describe these, also, as degenerate. While these have no solutions, the other degenerate polynomial equation, a=a, has more than I care to count. If D isn't additively complete, equations of form x+a=b aren't guaranteed solutions: likewise, without multiplicative completeness, a.x=b needn't have solutions. However, regardless of these completions, some equations aren't guaranteed any roots: for example, x.x+2=1. We can ask for all non-degenerate equations to have roots: and the interesting upshots of this arise when the number of roots is (almost always) the order of the equation.

A topology on D induces one on any {(I|:D)} (via each (I|U:closed(D)) implies a subset, {(I|f:D): for each i in I, f(i) is in U(i)}, of {(I|:D)}, which is taken to be closed in {(I|:D)}; and any intersection of closed sets is closed, as is any finite union of closed sets). My chosen escape route from the detail of repeated roots is to say that almost any small perturbation of a polynomial equation will yield one with as many roots as its order. The small perturbations that matter are the ones which preserve order. The topology on {polynomials (D|:D)} thus induced from that on {(D|:D)} implies one on the collection of polynomial equations on D (which is worth thinking about: at the very least, notice that any polynomial equation of given order is a limit (member of the boundary a collection) of polynomials of arbitrary higher order: but not a limit of any collection of polynomials of lower order – the connectivity of the polynomials over a scalar domain is quite fascinating!).

Consider polynomial equations for which the number of elements of roots() is the order of the equation. Some closed sets of polynomial equations contain all these polynomial equations. Take the intersection of these. An arbitrary intersection of closed sets is closed. So this intersection is closed: and it also contains all the polynomial equations whose order is their number of roots.

The following definition of algebraic completeness simply says that the closure of the set of polynomial equations subsumes the set of non-degenerate polynomial equations.

Algebraic completeness

I'll describe a scalar domain, D, as

algebraically complete
precisely if, for every single-variable polynomial equation in D, P(x) = Q(x): there is some neighbourhood of the pair (P,Q) [in the topology on pairs of single-variable polynomials in D – i.e. single-variable equations] within which the set {equations whose order is no more than the number of solutions it has} is dense.

That is, if a polynomial equation has fewer roots than its order, the difference is made up for by multiple roots and almost all small perturbations of the polynomials will remove these degeneracies.

minimally so
precisely if it has no minimal proper scalar sub-domain which is also algebraically complete.

I need to prove that the intersection of two algebraically complete scalar domains is algebraically complete, at least when given that they have some common scalar sub-domain. The algebraic completion of a scalar domain can then be defined as the intersection of all scalar domains of which the given scalar domain is a scalar sub-domain.

One thing falls rather nicely out of this definition of algebraic completeness: it implicitly states that the topology is not discrete. A topology is described as discrete if every subset of the whole is open: this includes all sets containing exactly one member. For any a in an algebraically complete scalar domain, consider the polynomial equation a2+x2=2ax. If this is guaranteed to have precisely one root (x=a), the set whose one member is this polynomial equation cannot contain an open neighbourhood of the given equation: this, in turn, implies that the topology on our scalar domain cannot be discrete.

If a scalar domain's locality is an equivalence relation, one can count the number of localities which contain roots to a polynomial equation, instead of counting the roots, and thereby sustain a definition of algebraic completeness which is still meaningful in the presence of infinitesimal localities – such as arise when, for some a, there is more than one solution to x2+a2 = 2ax. This naturally reads as saying that the square of x−a is zero – which, in a non-cyclic scalar domain, would tell us that there's no r in S for which x+r=a or a+r=x – i.e. there's no x−a in S.


A scalar domain contains (deliberately) just enough to enable us to define differentiation on it (though not enough to guarantee that it supports differentiation – only enough that it can tell whether it does). The familiar definitions for a field assume additive and multiplicative inverses in various ways; it suffices to unwrap these a little to obtain definitions for a scalar domain. Likewise, a scalar domain contains enough to enable a definition of measure, on an arbitrary space, taking values in the scalar domain; and just enough to extend this to include a definition of integration of scalar and vector functions over the space.

Some Variants

Natural Numbers

The integers and the natural numbers (or non-negative integers) are scalar domains, with discrete topology and a zero (hence their multiplications are not cancellable).

Notice that the positive integers (or positive natural numbers) form a self-linear action but, since + is not epic (there is no pair whose sum is 1), they do not form a scalar domain: however, their multiplication is cancellable and epic. [Proof of epic: {1} is a subset of PN, the positive integers, whence {1}×PN is a subset of PN×PN so ({1}×PN|.|), which =PN, is a subset of (.|).]

Because every scalar domain has 1, we can induce a function from the positive integers to any scalar domain via: calling the multiplicative identity of the scalar domain unit to avoid confusion with 1, used as name for that of the positive integers, f=(positive integers| 1->unit, 1+n -> unit+f(n) :), which is so natural that we will often refer to each f(n) simply as n. In particular, it always allows us to refer to unit as 1 without confusion. This embedding trivially preserves the linear structure of the positive integers, in the sense f(n)+f(m)=f(n+m), f(n).f(m)=f(n.m), though it need not be monic – there may be positive integers n, m for which f(n)=f(n+m), in which case f(m) is an additive identity.

Combining this natural embedding, f, of the positive integers in an arbitrary scalar domain, with the multiplication on that scalar domain gives us a linear action of the positive integers on the scalar domain. This is the usual inductive 1.v=v, (1+n).v= v+n.v, as follows from the above, which leads us to treat the positive integers as though they were members of every scalar domain (so we are happy enough to talk about 7×3 even when working in the cyclic field 5, which doesn't have 7 as a member – f(7) is, in fact, 2, and what's strictly meant is f(7)×3, which is 2×3 = f(6) = 1).

Additive Completion

Note that, for any scalar domain S whose addition is Abelian, we can use the standard difference construction on binary operators to obtain an additive domain with zero and inverses. This enables us to construct, as follows, the additive completion of S: a scalar domain with additive identity and inverses, in which S may be embedded so as to preserve all its scalar structure.

We can combine the difference construction's addition, (S×S|+:S), on S with a multiplication on S×S: (h,k).(m,n) = (h.m+k.n,k.m+h.n). It is not hard to show that this distributes over the addition; with a little more effort, one may show that it respects the equivalence relation used in the difference construction – that is, if (h,k)~(m,n) then, for any (a,b): (a,b).(h,k) ~ (a,b).(m,n). Proof: (a,b).(h,k) = (a.h+b.k,a.h+b.h) and (a,b).(m,n) = (a.m+b.n,a.n+b.m): now, (h,k)~(m,n) means h+n=m+k: whence a.h+b.k + a.n+b.m = a.(n+h) + b.(k+m) = a.(k+m) + b.(n+h) = a.m+b.n + a.k+b.h; which is just the statement (a.h+b.k,a.k+b.h)~(a.m+b.n,a.n+b.m), as required. Consequently, this induces a multiplication on the equivalence classes and the resulting collection of equivalence classes forms a scalar domain: the difference construction's embedding of S in the collection of equivalence classes not only preserves the additive structure, but actually preserves all of the scalar domain structure.

Thus, if the addition in a scalar domain is Abelian, it is for all practical purposes invertible – in the sense that the above construction turns it into such a one. Consequently, where additive invertibility causes problems, avoiding it (if possible) presents no serious problems to the development of the theory.

A Positive Scalar Domain

As shown above, the presence of a zero precludes multiplicative cancellability. This leads us naturally to consider a positive scalar domain, which is a scalar domain whose multiplication is cancellable: it thus has no zero. We can, on this, define an ordering via: for a, b in D: a<b ⇔ b∈(D|d->a+d|).

That this is an ordering, in the usual sense, follows from the absence of additive inverses (despite additive cancellability), which in turn follows from that of the zero. It has some excellent advantages (especially when the multiplication is also complete, so that we have a multiplicative group). Its lack of a zero presents no real problem: we can always additively complete it. Alternatively, the positive scalar domain's natural embedding in its algebraic completion will do this job for us. I shall use a positive scalar domain as what amounts to the (strictly) positive real line.

The usual embedding of the positive integers in any positive scalar domain necessarily preserves order and is, as a result, monic: 1 < 1+1=2 < 1+2=3 < … are all distinct. Thus any positive scalar domain subsumes a copy of the positive integers: because it is multiplicatively complete, we can find a solution x to any equation of form p.x=q with p and q positive integers – thus any positive scalar domain also subsumes (a copy of) the positive rationals. We describe a scalar domain as rational precisely if it has a positive scalar sub-domain. Thus the usual real and complex numbers have the positive reals as a positive scalar sub-domain. Since the rationals are, themselves a positive scalar domain (the minimal non-trivial one at that)

ItsAlgebraic Completion

The other highly interesting scalar domain is the algebraic completion of our positive scalar domain – that is, the minimal algebraically complete scalar domain in which our positive scalar domain may be embedded consistently with the scalar structure on both scalar domains. This fills the rôle of the complex numbers.

An algebraically complete scalar domain necessarily has a unit, a zero and additive inverses (because of the implied roots of polynomial equations of order 1). It cannot, as a result, be a positive scalar domain. However, if it is the algebraic completion of a positive scalar domain, this implies that it subsumes an ordered field whose positive half is our positive scalar domain. We shall refer to this ordered field as the real line in the algebraic completion.

[I haven't yet worked out how to persuade myself that] On the algebraic completion, C, of a positive scalar domain, P, there is a self-inverse function (C|*:C), called conjugation, for which:

It follows from the last two that the sum of any value with its conjugate is real (since conjugation of this sum preserves it, term by term). Note that the taken in reverse order clauses are redundant (but harmless) in the Abelian case: they are there so that I can apply the same reasoning to (for instance) the Quaternions. Anyway, why assume commutativity when I don't need to ?

The real part of a value, c, is the unique real value, r, for which the square of (c−r) is a non-positive real value. The imaginary part of c is then (c−r) and the conjugate is r−(c−r) or 2r−c. However, proving that this definition of the real part, hence of conjugation, works in general is not clear. Yet.

I glibly suppose that C is a 2-dimensional vector space over P supporting a (scalar domain) faithful embedding of P in C (as the positive real half-line) and having members −1 and i of C not in P's image under this embedding (which I shall hereafter treat as synonymous with P) for which −1 + 1 = 0 and i2 = −1.

I haven't finished writing this page yet (and I'll probably shred it instead of trying to do so).

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