# The so-called Real numbers

As concerns numbers, one starts with the natural or counting numbers, […, 3, 2, 1, 0] and their susceptibility to addition and multiplication. There are various ways of building up formal structures providing numbers to fill the gaps between natural numbers and handle negative values; these aim to preserve what was true for the naturals,

• a.(b+c) = a.b + a.c,
• a+b = a+c implies b = c and
• a.b = a.c implies b = c unless a is 0,

but to provide extra values which fill in gaps that come to light when the last two, cancellation, lead us to ask, given the sum a+b or product a.b, and knowledge of either a or b, what is the value of the other ? The cancellation rules tell us that any value which works is the only value which works, but it doesn't tell us that we'll necessarily find a value which works, unless we have some clear prior knowledge about the numbers involved. Fortunately, this is easy to bypass, effectively by inventing imaginary numbers like 2/3 or −1 and then showing how practical and useful they are, and how well they model real phenomena.

So far so good: we've now got the rational numbers, each of which can be written as n/m for some integer n and positive integer m. At this point, missing numbers show up from two directions at once,

polynomials

We can build the rationals by asking for solutions to given x+y and x, tell me y and similar for x.y; so why not ask for solutions to other equations, too ? For example: given x.x, what is x ? or: given 4.x.x.x +3.x.x +2.x + 1, what is x ? The polynomial equations involved are all expressed in terms of addition, multiplication and natural numbers, so it surely can't be all that much harder than solving for one unknown summand or factor given the sum, or product.

Well, OK, it's harder, but human ingenuity won out and gave us the algebraic numbers, among which all polynomial equations are soluble, not just the easy ones that gave us the rationals, but also the famous simple cases like x.x = 2 and x.x = −1, along with many more. The algebraic numbers are, to the complex numbers, what the rationals are to the real line – once we've filled in the other kind of gap.

Here's a fore-taste of the gaps of smoothness. The rationals have a certain grain to them, albeit fractal; and the algebraics aren't the only holes in that grainy-ness, either. Take a look at the unit circle in two dimensions – that is, the set of pairs, [x, y], of numbers for which x.x +y.y =1. That's a polynomial equation and it produces a nice easy curve which is the epitome of smooth – the circle. So surely it can't be all that hard to measure lengths along such a well-behaved curve ?

convergence

It's possible to construct sequences of rationals for which, given any positive tolerance ε, there is some point beyond which any two numbers in the sequence lie within ε of one another. Formally, the sequence is a mapping ({rationals}: s(i) ←i |{naturals}) and, given ε, there is some N for which, whenever i > N and j > N, s(j) < s(i)+ε and s(i) < s(j)+ε. Such sequences are described as convergent.

An easy example is 1/(1+i) ←i. It's generally accepted that, if ε > 0, then there is some N for which N.ε > 1: for any i, j > N, we have 1/i < 1/N < 1/N +1/j < ε + 1/j and likewise with i, j swapped; so the sequence is convergent. Furthermore, along the way, we've seen that, for every tolerance there's a point in the sequence beyond which all its values are within that tolerance of 0. The sequence 1/(1+i) ←i is said to converge on 0.

A convergent sequence s is said to converge to and a value, S, precisely if: for any tolerance ε, there is some natural N for which s(i) is within ε of S whenever i > N. This worked fine for 1/(1+i) ←i, but it proves embarrassingly easy to construct convergent sequences of rationals which don't converge to any algebraic number. So we invent numbers to which our convergent sequences converge and show how they fit in with the rationals and the algebraic numbers.

which makes for a good moment to pause,

So we built all our original numbers by asking for solutions to equations expressible entirely in terms of addition and multiplication. Just as we started with the natural numbers, we can now start with the polynomial functions, on which we can do addition, multiplication and differentiation. So, naturally enough, we ask what equations we can invent, with the available operations on functions; and we try to solve them. Ask for a function which is equal to the square of its derivative and I will give you x.x/4 ←x, a.k.a. power(2)/4. But ask for a function which is simply equal to its derivative and you shall be hard pushed to find a suitable polynomial – though the sequence

• sum({polynomials}: power(i) / factorial(i) ←i |n) ←n

looks promising [factorial and power are defined, on natural numbers, by: factorial(0) is 1, power(0) is 1 ←x; and, for every natural i, factorial(i+1) is (i+1).factorial(i) and power(i+1) is x.power(i,x) ←x].

given some numbers n and m, for a number i for which n+i = m or n.i = m. Filling those holes gives us what are known as the rational numbers or proper fractions – each can be written as n/m for some integer n and positive integer m.

One might as readily have observed that, among the naturals, a.a = b.b implies a=b and begged the question: given a.a's value, what's a ?

The way I was introduced to the reals started from the naturals; equipped it with additive inverses and, for the positive naturals, multiplicative inverses, thus obtaining the rationals; defined a notion of convergence for sequences of rationals and constructed the reals as the collection of all convergent sequences of rationals subject to an equivalence which amounts to they converge together. This applies additive and multiplicative completion followed by topological completion. For ease of expression of the last, I'm going to delay additive completion to last: but here's a quick overview of the whole construction:

### The Classical Reals

Start with the positive (i.e. non-empty) naturals, on which we have addition and multiplication, both of which are cancellable, symmetric and associative. To perform multiplicative completion, we represent a ratio as an atomic relation, using n←m to encode the ratio between n and m. We can define multiplication, addition and an order on such pairs; and an equivalence which all three respect.

(a←b)<(c←d)
iff a.d < b.c
(a←b).(c←d)
= (a.c ← b.d)
(a←b)+(c←d)
= (a.d+b.c ← b.d)
Positive Rational
relates (a←b) to (c←d) iff a.d = b.c

Given pairs (a←b) and (c←d), we can compute the naturals a.b and b.c; either these are equal or one of them is < the other; so either Positive Rational relates (a←b) to (c←d) or one of them is < the other. To perform multiplicative completion, construct an equivalence on pairs (using the atomic relation n←m, for any given n and m, as representation of the pair) of naturals as follows:

Positive Rational
relates (a←b) to (c←d) iff: a.d = b.c with a, b, c and d all positive naturals

That Positive Rational is symmetric and reflexive follows directly from the symmetry of natural multiplication. If it relates (a←b) to (c←d) and the latter to (e←f) then we have a.d = b.c and c.f = d.e, so a.d.f = b.c.f = b.d.e whence, cancelling d from the first and last, a.f = b.e; so it also relates (a←b) to (e←f); so Positive Rational is transitive, hence an equivalence.

We can define multiplication and division between pairs of positive naturals by:

• (a←b).(c←d) = (a.c ← b.d)
• (a←b)/(c←d) = (a.d ← b.c) = (a←b).(d←c)

Since p/q = p.reverse(q) we can naturally interpret reverse as multiplicative inversion. First note that if (c←d) = (a←b).(n←n) = (a.n ← b.n) we obtain a.d = a.b.n = a.n.b = c.b so Positive Rational deems (n.a ← n.b) equivalent to (a←b) and (n←n) is a multiplicative identity for all n. Clearly n.m = n.m implies Positive Rational relates (n←n) to (m←m) for all positive naturals n, m, so all such multiplicative identities are equivalent; and (c←d).reverse(c←d) = (c←d).(d←c) = (c.d ← c.d) so q.reverse(q) is always a multiplicative identity.

Now suppose Positive Rational deems (a←b) equivalent to (n←m) and (c←d) equivalent to (h←k); so a.m=b.n and c.k=d.h; let (e←f) = (a←b).(c←d) = (a.c←b.d) and (i←j) = (n←m).(h←k) = (n.h←m.k); observe that e.j = a.c.m.k = b.n.d.h = f.i, so Positive Rational deems (a←b).(c←d) and (n←m).(h←k), i.e. our multiplication on pairs of positive naturals respects the equivalence, so serves as a multiplication on the equivalence classes of Positive Rational.

At this stage we can read (a←b) as a/b or as b/a by embedding the positive naturals in the pairs thereof as (: (a←1) ← a :) or (: (1←b) ←b :), respectively. Settling on the former, we can define addition on pairs by:

• (a←b)+(c←d) = (a.d+c.b ← b.d)

Again, consider (n←m) and (h←k) equivalent to (a←b) and (c←d) respectively: write (e←f) = (a.d+c.b ← b.d) and (i←j) = (n.k+h.m ← m.k) and observe:

e.j
= (a.d + c.b).m.k
= a.m.d.k + c.m.b.k
= b.n.d.k + d.h.m.b
= (n.k + h.m).b.d
= i.f

so addition also respects the equivalence.

## An alternative construction

In (at least a modern rendition of) Cantor's demonstration that there are more reals than naturals, the reals between 0 and 1 are represented in binary and, thereby, as collections of naturals – label the digits after the binary point, numbering from the left as 0, 1, 2, 3, …, n, … and identify any sequence of binary digits as the collection of labels for which the corresponding digit is a 1. This is a remarkably convenient representation for the relevant real numbers, provided we allow infinite sequences of digits. One has to take care about the equivalence of a sequence with finitely many 1 digits and the result of replacing the last 1 with a zero and having a 1 in every subsequent position; but one can define addition on subsets of the naturals as a model of the (limits of) diadic rationals.

To avoid trouble with infinite regress, it'd be nice to define addition using just intersection, union and the mutual complement consisting of those members of either set which aren't members of the other.

```S, c, n = {}, ?, infinity # ? in {0, 1}
while n-- > 0:
if n in U: c = 1 + c
if n in V: c = 1 + c
assert c in {0, 1, 2, 3}
c, m = divmod(c, 2)
if m: S.insert(n)
assert c in {0, 1}

```

S now represents the sum of the numbers represented by U and V. If c is 1 we're discarding the carried bit.

• if 1+n is in both U and V, we definitely carry to n
• if we carried to 1+n and 1+n is in either U or V, we carry to n
• if we carry to n and n is either both in U and in V, or in neither, n is in S
• if we don't carry to n and n is in just one of U, V, but not in the other, n is in S

C = {n: we carried to n} Approach via a sequence Q defined by: Q(0) = (|successor:U∩V) Q(1+n) = Q(n)∪(|successor:Q(n)∩(U∪V)) then use unite(|Q:{naturals}) as C. Is a given natural's membership of C always decidable from this ? Plausibly not, for suitable U and V. What if membership of U and V is decidable, then ?

In any case, we obtain a model of the non-negative real numbers which are less than two. There is a natural ordering; given U, V, consider {n in U: n not in V} and {n in V: n not in U}; these are manifestly disjoint so, if they have any members, we can identify the smallest of each and know the two are different, so imply an order – if the former's is smaller than the latter's, V is less than U; and vice versa – and if either is empty, one of U, V subsumes the other. If U and V subsume one another, they're the same collection so equal; otherwise, if one subsumes the other, the former has a member the latter lacks so the latter is less than the former.

Well, anyhow, the model should tell us how to build an arithmetic which can be used as a model of a continuum, albeit a cyclic one. We can cut the cycle at either 0 = {} or 1 = {0}; the former reads the numbers as ranging from 0 up to 2; the latter reads the numbers as ranging from −1 to 1 and, effectively, reads 0 as contributing −1 to the sum (rather than 1 as described above), so reads {0} as −1 (while {0,1} is −1+half = −half).

Written by Eddy.