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philosopher Kant who proposed the following interesting answer. According
to Kant, there exist certain a priori principles such that our minds are bound
to interpret the world in terms of these a priori principles. The constraints
imposed by the axioms of Euclidean geometry provided an example of such
a priori principles.
Some mathematicians felt that the axioms stated by Euclid were not
yet in the best form. In particular, they believed that the so called parallel
axiom2 was unsatisfactory and they sought to prove it from the other axioms.
Between 1750 and 1840 a number of independent workers saw that the
questions.
(A) Why is the parallel axiom true?
2
This is now usually stated in the following form. Given a line l and a point P not on
the line, there exists one and only one line l through P which does not intersect l.
365
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(B) Why is Euclid™s geometry true?
should be replaced by the new questions
(A ) Do there exist mathematical systems obeying all the axioms of Euclid
except the parallel axiom and disobeying the parallel axiom?
(B ) Does Euclid™s geometry describe the world?
Although Gauss did not publish his views, he asked both of these new
questions and answered yes to the ¬rst and no to the second. He even tried
to check the validity of Euclidean geometry by physical experiment3 . In a
private letter he stated that geometry should not be classed ˜with arithmetic,
which is purely a priori, but with mechanics™.
Gauss, Bolyai, Lobachevsky and Riemann developed the theory of various
non-Euclidean geometries but did not show that they were consistent. Thus
although they believed strongly that the answer to question (A ) was no,
they did not prove it. (Certainly they did not prove it explicitly.) However,
in the second half of the 19th century mathematicians (notably Beltrami)
showed how to construct models for various non-Euclidean geometries using
Euclidean geometry. In modern terms, they showed that, if Euclid™s axioms
(including the parallel axiom) form a consistent system, then the same set
of axioms with the parallel axiom replaced by some appropriate alternative
axiom remains consistent.
Finally, in 1917, Einstein™s theory of general relativity con¬rmed the sus-
picion voiced from time to time by various farsighted mathematicians that
question (B ) should be replaced by
(B ) Is Euclid™s geometry the most convenient description of the world?
and that the answer to the new question was no.
In 1800, Euclid™s geometry was the only rigorously developed part of
mathematics. Initially, those, like Cauchy and Gauss, who sought to rigorise
the calculus, took the notion of magnitude (or as we would say real number)
as implicit (or, as Kant would say, a priori) but it gradually became clear that
the rigorous development of the calculus depended on (or was identical with)
a close study of the real numbers. However, as I tried to show in Chapter 2,
the properties of the real line that we want are by no means evident.
Faced with this problem, the natural way forward is to show that the
complicated system constituted by the real numbers can be constructed from
the simpler system of the rationals. Thus, if we believe that the rationals
exist, we must believe that the reals exist. (Gauss and others had already
shown how to construct C from R showing that, if the real numbers are
3
Euclid uses the parallel axiom to prove that the sum of the angles of a triangle add up
to 180 degrees. Gauss used surveying instruments to measure the angles of a triangle of
mountain peaks. To within the limits of experimental error, the results con¬rmed Euclid™s
prediction.
366 A COMPANION TO ANALYSIS

acceptable, then so are the complex numbers.) Simpler constructions enabled
mathematicians to obtain the rationals from the integers and the integers
from the positive integers.
In the next sequence of exercises we outline these simpler constructions.
The full veri¬cation of the statements made is long and tedious and the
reader should do as much or little as she wishes. Nothing in what follows
depends on these exercises. I have felt free to use notions like equivalence
relations, integral domains and so on which I have avoided in the main text.
Exercise 14.3.1. (Obtaining Z from N.) Here N = {0, 1, 2, . . . }. We
take (N, +, —) with its various properties as given. Show that the relation

(r, s) ∼ (r , s ) if r + s = r + s

is an equivalence relation on N2 .
We write Z = N2 / ∼ for the set of equivalence classes

[(r, s)] = {(r , s ) ∈ N2 : (r, s) ∼ (r , s )}.

Show that the following give well de¬ned operations

[(r, s)] + [(u, v)] = [(r + u, s + v)], [(r, s)] — [(u, v)] = [(ru + sv, su + rv)].

Explain why you think we chose these de¬nitions. Show that (Z, +, —) obeys
the algebraic rules that it should (in more formal terms, that (Z, +, —) is an
integral domain). If you just wish to prove a selection of results, you might
choose
(a) There exists a unique 0 ∈ Z such that x + 0 = x for all x ∈ Z.
(b) If x ∈ Z, then we can ¬nd a y ∈ Z such that x + y = 0 (we write
y = ’x).
(c) If x, y, z ∈ Z, then x — (y + z) = x — y + x — z.
(d) There exists a unique 1 ∈ Z such that x — 1 = x for all x ∈ Z.
(e) If z = 0 and x — z = y — z, then x = y.
Show that the mapping θ : N ’ Z given by θ(n) = [(n, 0)] is injective
and preserves addition and multiplication (that is θ(n + m) = θ(n) + θ(m)
and θ(n — m) = θ(n) — θ(m)). Explain brie¬‚y why this means that we may
consider N as a subset of Z.
Show that, if P = θ(N), then P obeys the appropriate version of axioms
(P1) to (P3) in Appendix A. We write x > y if x + (’y) ∈ P.
Exercise 14.3.2. (Obtaining Q from Z.) We take (Z, +, —) with its var-
ious properties as given. Show that the relation

(r, s) ∼ (r , s ) if r — s = s — r
367
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is an equivalence relation on Z — (Z \ {0}).
We write Q = Z — (Z \ {0})/ ∼ for the set of equivalence classes

[(r, s)] = {(r , s ) ∈ Z — (Z \ {0}) : (r, s) ∼ (r , s )}.

Show that the following give well de¬ned operations

[(r, s)] + [(u, v)] = [(rv + su, sv)], [(r, s)] — [(u, v)] = [(ru, sv)].

Explain why you think we chose these de¬nitions. Show that (Q, +, —) obeys
the algebraic rules (A1) to (A4), (M1) to (M4) and (D) set out in Appendix A
(in algebraists™ terms (Q, +, —) is a ¬eld). If you just wish to prove a
selection of results you might choose
(a) There exists a unique 0 ∈ Q such that x + 0 = x for all x ∈ Q.
(b) If x ∈ Q, then we can ¬nd a y ∈ Q such that x + y = 0 (we write
y = ’x).
(c) If x, y, z ∈ Q, then x — (y + z) = x — y + x — z.
(d) There exists a unique 1 ∈ Z such that x — 1 = x for all x ∈ Q.
(e) If x = 0, then we can ¬nd a y ∈ Q such that xy = 1 .
De¬ne an appropriate P which obeys the appropriate version of axioms
(P1) to (P3) in Appendix A. [If you cannot do this, you do not understand
what is going on.] We write x > y if x + (’y) ∈ P. In algebraists™ terms,
(Q, +, —, >) is an ordered ¬eld.
De¬ne an injective map θ : Z ’ Q which preserves addition, multiplica-
tion and order (thus n > m implies θ(n) > θ(m)) and show that it has the
stated properties. Explain brie¬‚y why this means that we may consider Z as
a subset (more strictly as a sub-ordered integral domain) of Q.
Exercise 14.3.3. (Obtaining C from R.) This is somewhat easier than
the previous two exercises, since it does not involve equivalence classes and
it is obvious that the de¬nitions actually work. We take (R, +, —) with its
various properties as given.
We write C = R2 and de¬ne the following operations on C.

(r, s) + (u, v) = (r + u, s + v), (r, s) — (u, v) = (ru ’ sv, rv + su).

Explain why you think we chose these de¬nitions. Show that (C, +, —)
obeys the algebraic rules (A1) to (A4), (M1) to (M4) and (D) set out in
Appendix A (in algebraists™ terms (C, +, —) is a ¬eld). If you just wish to
prove a selection of results you might choose
(a) There exists a unique 0 ∈ C such that z + 0 = z for all z ∈ C.
(b) If z ∈ C, then we can ¬nd a w ∈ C such that z + w = 0 (we write
w = ’z).
368 A COMPANION TO ANALYSIS

(c) If z, w, u ∈ C, then z — (w + u) = z — w + z — u.
(d) There exists a unique 1 ∈ C such that z — 1 = z for all z ∈ C.
(e) If z = 0, then we can ¬nd a w ∈ C such that zw = 1 .
De¬ne an appropriate injective map θ : R ’ C which preserves addition
and multiplication (thus θ is a ¬eld morphism) and show that it has the
stated properties. Explain brie¬‚y why this means that we can consider R as
a sub¬eld of C.
Show that (0, 1) — (0, 1) = θ(’1). Show that, if we adopt the usual con-
vention of considering R as a sub¬eld of C and writing i = (0, 1), then every
element z of C can be written as z = a + bi with a, b ∈ R in exactly one way.
Show also that (0, ’1) — (0, ’1) = θ(’1). [This corresponds to the ambi-
guity introduced by referring to i as ˜the square root of ’1™. In C there are
two square roots of ’1 and there is no reason to prefer one square root rather
than the other.]

Mathematicians and philosophers have by now become so skilled in doubt
that, whatever their private beliefs, few of them would care to maintain in
public that the system of the positive integers (often called the ˜natural
numbers™) is given a priori4 . None the less, most mathematicians feel that,
in some sense, the system of the positive integers is much more ˜basic™ or
˜simple™ or ˜natural™ than, say, the system of the complex numbers. If we can
construct the complex numbers from the integers we feel that much more
con¬dent in the system of the complex numbers.
Much of mathematics consists in solving problems and proving theorems,
but part of its long term progress is marked by changing and extending the
nature of the problems posed. To a mathematician in 1800, the question ˜Can
you construct the reals from the rationals?™ would be incomprehensible. One
of the many obstacles to understanding the question would be the nature of
the object constructed. Let us look at our construction of Z from N in
Exercise 14.3.1. We say, in e¬ect, that ˜the integer 1 is the equivalence
class of ordered pairs (n, m) of natural numbers such that their di¬erence
n ’ m is the natural number 1™. The non-mathematician is entitled to object
that, whatever the integer 1 may be, it is certainly not that. To this, the
mathematician replies that the system constructed in Exercise 14.3.1 has
exactly the properties that we wish the integers to have and ˜if it looks like
4
It is all very well to maintain that one plus one must always make two but this fails
when the one plus one are rabbits of di¬erent sexes or rain drops which run together.
A stronger objection deals with statements like n + m = m + n. This statement might,
perhaps, be a priori true if n = 2 and m = 3 but can it really be a priori true if n and m
are integers so large that I could not write them down in my lifetime?
369
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a duck, swims like a duck and quacks like a duck then it is a duck5 ™. To this,
the philosopher objects that she can construct a clockwork toy that looks like
a duck, swims like a duck and quacks like a duck. The mathematician replies
that, if all that we want from a duck is that it should look like a duck, swim
like a duck and quack like a duck, then, for our purposes, the clockwork toy
is a duck.


How do we construct the reals? ™
14.4
As the reader no doubt expects, we model the construction of the reals from
the rationals on the completion arguments in Sections 14.1 and 14.2. As the
reader, no doubt, also expects, the construction will be long and relatively
easy with a great deal of routine veri¬cation left to her. However, it is impor-
tant not to underestimate the subtlety of the task. Our earlier completion
arguments used the existence and properties of the real numbers, our new
arguments cannot. The reader should picture a street mime juggling non-
existent balls. As the mime continues, the action of juggling slowly brings
the balls into existence, at ¬rst in dim outline and then into solid reality.
We take the ordered ¬eld (Q, +, —, >) as given. We recall that our
de¬nitions of convergence and Cauchy sequence apply in any ordered ¬eld.
If xn ∈ Q and x ∈ Q, we say that xn ’ x if, given any ∈ Q with > 0, we
can ¬nd an N ( ) > 0 such that

|xn ’ x| < for all n ≥ N ( ) > 0.

We say that a sequence xn in Q is Cauchy if, given any ∈ Q with > 0, we
can ¬nd an N ( ) > 0 such that

|xn ’ xm | < for all n, m ≥ N ( ) > 0.

We start by considering the space X of Cauchy sequences x = (xn )∞ in
n=1
Q. We write x ∼ y if x, y ∈ X and xn ’ yn ’ 0 as n ’ ∞. The reader
should verify the statements made in the next exercise.

Exercise 14.4.1. Suppose x, y, z ∈ X. Then
(i) x ∼ x.
(ii) If x ∼ y, then y ∼ x.
(ii) If x ∼ y and y ∼ z, then x ∼ z.
5
This should be contrasted with the position in medical diagnosis. ˜If it looks like a
duck, swims like a duck and barks like a dog, then it is a duck. Every case presents atypical
symptoms.™
370 A COMPANION TO ANALYSIS

We write [x] = {y : y ∼ x} and call [x] the equivalence class of x. We
take R to be the set of all such equivalence classes. The reader should verify
the statements made in the next exercise.

Exercise 14.4.2. Each x ∈ X belongs to exactly one equivalence class in R.

Lemma 14.4.3. (i) If x, y ∈ X, then, taking x+y to be the sequence whose
nth term is xn + yn , we have x + y ∈ X.
(ii) If x, y, x , y , ∈ R and x ∼ x , y ∼ y , then x + y ∼ x + y .
(iii) If x, y ∈ X, then, taking x — y to be the sequence whose nth term
is xn — yn , we have x — y ∈ X.
(iv) If x, y, x , y , ∈ R and x ∼ x , y ∼ y , then x — y ∼ x — y .

Proof. I leave parts (i) and (ii) to the reader.
(iii) Since x and y are Cauchy sequences they are bounded, that is we
can ¬nd a K such that |xn |, |yn | ¤ K for all n. Thus (writing ab = a — b)

|xn yn ’xm ym | ¤ |xn yn ’ xn ym | + |xn ym ’ xm ym |
= |xn ||yn ’ ym | + |ym ||xn ’ xm | ¤ K(|yn ’ ym | + |xn ’ xm |),

and so the xn yn form a Cauchy sequence.
(iv) As in (iii), we can ¬nd a K such that |xn |, |yn | ¤ K for all n, and so

|xn yn ’xn yn | ¤ |xn yn ’ xn yn | + |xn yn ’ xn yn |
= |xn ||yn ’ yn | + |yn ||xn ’ xn | ¤ K(|yn ’ yn | + |xn ’ xn |) ’ 0

as n ’ ∞, as required.

Lemma 14.4.3 tells us that we can de¬ne addition and multiplication of real
numbers by the formulae

[x] + [y] = [x + y], [x] — [y] = [x — y].

Lemma 14.4.4. The system (R, +, —) is a ¬eld (that is to say, it satis¬es
the rules (A1) to (A4), (M1) to (M4) and (D) set out in Appendix A.)

Proof. I claim, and the reader should check, that the veri¬cation is simple
except possibly in the case of rule (M4). (We take 0 = [a] with aj = 0 for all
j and 1 = [b] with bj = 1 for all j.)
Rule (M4) states that, if [x] ∈ R and [x] = [a] (where aj = 0 for all j),
then we can ¬nd a [y] ∈ R such that [x] — [y] = [b] (where bj = 1 for all j).
To prove this, observe that, if [x] = [a], then xn = xn ’ an 0 as n ’ 0.
Thus we can ¬nd an ∈ Q with > 0 such that, given any N , we can ¬nd
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an M ≥ N such that |xM | > . Since the sequence xn is Cauchy, we can ¬nd
an N (0) such that

|xn ’ xm | < /2 for all n, m ≥ N (0) > 0.

Choose N (1) ≥ N (0) such that |xN (1) | > . We then have

|xn | ≥ |xN (1) | ’ |xn ’ xN (1) | > /2

for all n ≥ N (0).
Set yj = 1 for j < N (0) and yj = x’1 for j ≥ N (0). If n, m ≥ N (0) we
j
have

|yn ’ ym | = |xn ’ xm ||xn |’1 |xm |’1 ¤ 4 ’2
|xn ’ xm |,

so the sequence yn is Cauchy. Since xj yj = 1 = bj for all j ≥ N (0), we have
[x] — [y] = [b], as required.

We now introduce an order on R. This is an essential step before we
can introduce distance and limits. We shall need the result of the following
exercise.

Exercise 14.4.5. (i) Let aj = 0 for all j. By adapting the argument of
the third paragraph of the proof of Lemma 14.4.4, show that, if [x] ∈ R and
[x] = [a], then either
(a) there exists an N > 0 and an ∈ Q with > 0 such that xn >
for all n ≥ N , or
(b) there exists an N > 0 and an ∈ Q with > 0 such that ’xn >
for all n ≥ N .
(ii) Suppose that [x] ∈ R and there exists an N > 0 and an ∈ Q with
> 0 such that xn > for all n ≥ N . Show that, if [y] = [x], then there
exists an N > 0 such that yn > /2 for all n ≥ N .

Lemma 14.4.6. The set P of [x] ∈ R such that there exists an N > 0 and
an ∈ Q with > 0 such that xn > for all n ≥ N is well de¬ned. It has
the following properties.
(P1) If [x], [y] ∈ P, then [x] + [y] ∈ P.
(P2) If [x], [y] ∈ P, then [x] — [y] ∈ P.
(P3) If [x] ∈ R, then one and only one of the following three statements
is true [x] ∈ P, ’[x] ∈ P or [x] = [a] where aj = 0 for all j.

Proof. The fact that P is well de¬ned follows from part (ii) of Exercise 14.4.5.
If 1 ∈ Q, 1 > 0, xn ≥ 1 for n ≥ N1 and 2 ∈ Q, 2 > 0, yn ≥ 2 for n ≥ N2 ,
372 A COMPANION TO ANALYSIS

then xn + yn ≥ 1 + 2 > 0 and xn — yn ≥ 1 — 2 > 0 for all n ≥ max(N1 , N2 ),
so conclusions (P1) and (P2) follow.
By part (i) of Exercise 14.4.5, we know that, if [x] ∈ R, then at least
one of the three statements [x] ∈ P, ’[x] ∈ P or [x] = [a] is true. Since the
statements are exclusive, exactly one is true.

As usual, we write [x] > [y] if [x] ’ [y] ∈ P. If [x] ∈ R then, as we have just
shown, either [x] ∈ P and we de¬ne |[x]| = [x] or ’[x] ∈ P and we de¬ne
|[x]| = ’[x] or [x] = [a] (where aj = 0 for all j) and we set |[x]| = [a].
If w ∈ Q, we set wn = w for all n and write θ(w) = [w]. We leave it to
the reader to verify the next lemma.

Lemma 14.4.7. The map θ : Q ’ R is injective and preserves addition,
multiplication and order.

Note that θ(0) = [a] where aj = 0 for all j.
We have now shown that (R, +, —, >) is an ordered ¬eld and so our
de¬nitions of convergence and Cauchy sequence apply. If [x(n)] ∈ R and
[x] ∈ R, we say that [x(n)] ’ [x], if given any [ ] ∈ R with [ ] > θ(0), we
can ¬nd an N ([ ]) > 0 such that

|[x(n)] ’ [x]| < [ ] for all n ≥ N ([ ]).

We say that a sequence [x(n)] in R is Cauchy if, given any [ ] ∈ R, with
[ ] > θ(0) we can ¬nd an N ([ ]) > 0 such that

|[x(n)] ’ [x(m)]| < [ ] for all n, m ≥ N ([ ]).

Lemma 14.4.8. Given any [ ] ∈ R with [ ] > θ(0), we can ¬nd a y ∈ Q
with y > 0 such that [ ] > θ(y).

Proof. This follows from our de¬nition of > via the set P and part (ii) of
Exercise 14.4.5.

Lemma 14.4.9. Given any [x], [y] ∈ R with [x] > [y], we can ¬nd a z ∈ Q
such that [x] > θ(z) > [y].

Proof. By de¬nition, [x] ’ [y] > θ(0) so, by Lemma 14.4.8, we can ¬nd a
v ∈ Q with v > 0 such that [x] ’ [y] > θ(v) > θ(0). Setting u = v/5, we
obtain a u ∈ Q with u > 0 such that [x] ’ [y] > θ(5u) > θ(0). It follows that
we can ¬nd an M such that

xj ’ yj > 4u for all j ≥ M .
373
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Now choose an N ≥ M such that |yj ’ yk | < u and so, in particular,

yj ’ yk > ’u for all j, k ≥ N .

Set z = yN + 2u. By the displayed inequalities just obtained, we have

xk > 4u + yk = (yN + 2u) + (yk ’ yN ) + 2u > z + u

and

yk < y N + u ¤ z ’ u

for all k ≥ N . Thus [x] > θ(z) > [y] as required.

Up to now, our arguments have only used the fact that (Q, +, —, >) is
an ordered ¬eld. We now use the fact speci¬c to (Q, +, —, >) that every
rational x > 0 can be written as x = p/q with p and q strictly positive
integers. Since p/q ≥ 1/q, Lemma 14.4.8 implies the axiom of Archimedes.

Lemma 14.4.10. We have θ(1/n) ’ θ(0) as n ’ ∞.

Exercise 14.4.11. Give the details of the proof of Lemma 14.4.10.

We now show that every Cauchy sequence in (R, +, —, >) converges.
The proof runs along the same lines as the proof of Lemma 14.2.4

Lemma 14.4.12. If [y(1)], [y(2)], [y(3)], . . . is a Cauchy sequence in R,
then we can ¬nd an [x] ∈ R such that [y(j)] ’ [x] as j ’ ∞.

Proof. Since [y(j)] + θ(j ’1 ) > [y(j)], it follows from Lemma 14.4.9 that we
can ¬nd an xj ∈ Q such that

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