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or counterexample.
Exercise 1.5.10. It is worth noting explicitly that ordered ¬elds may satisfy
the axiom of Archimedes but not the fundamental axiom. Show in particular
that the rationals satisfy the axiom of Archimedes. (This is genuinely easy
so do not worry if your answer is brief.)
Exercise 1.5.11. The reader may be interested to see an ordered ¬eld con-
taining Z which does not satisfy the axiom of Archimedes. We start by con-
sidering polynomials P (X) = N an X n with real coe¬cients an and form
n=0
the set K of rational functions P (X)/Q(X) where P and Q are polynomials
and Q is not the zero polynomial (that is Q(X) = M bm X m with bM = 0
m=0
for some M ). Convince yourself that, if we use the usual standard formal
algebraic rules for manipulating rational functions, then K is a ¬eld (that is,
it satis¬es conditions (A1) to (D) as set out in the axioms on page 379).
To produce an order on K we de¬ne the set P to consist of all quotients
of the form
N n
n=0 an X
M m
m=0 bm X
13
Please send corrections however trivial to twk@dpmms.cam.ac.uk

with aN , bM = 0 and aN bM > 0. Convince yourself that this is a consistent
de¬nition (remember that the same quotient will have many di¬erent repre-
sentations; P (X)/Q(X) = R(X)P (X)/R(X)Q(X) whenever R(X) is a non-
zero polynomial) and that P satis¬es conditions (P1) to (P3). If we de¬ne
P1 (X)/Q1 (X) > P2 (X)/Q2 (X) whenever P1 (X)/Q1 (X)’P2 (X)/Q2 (X) ∈ P
condition (P4) is automatically satis¬ed and we have indeed got an ordered
¬eld.
We note that the elements of K of the form a/1 with a ∈ R can be
identi¬ed in a natural way with R. If we make this natural identi¬cation, K
contains Z.
To see that the axiom of Archimedes fails, observe that 1/n > 1/X > 0
for all n ∈ Z, n ≥ 1.
Of course, since the axiom of Archimedes fails, the fundamental axiom
fails. By examining the proof of Theorem 1.5.1, show that the 1/n form a
decreasing sequence bounded below but not tending to any limit.
If the reader knows some modern algebra she will see that our presentation
can be sharpened in various ways. (It would be better to de¬ne K using equiv-
alence classes. We should take greater care over checking consistency. The
words ˜identi¬ed in a natural way™ should be replaced by ˜there is an isomor-
phism of (R, +, —, >) with a sub¬eld of K™.) Such readers should undertake
the sharpening as an instructive exercise.
Exercise 1.5.12. We shall not make any essential use of the decimal ex-
pansion of the real numbers but it is interesting to see how it can be obtained.
Let us write
D = {n ∈ Z : 9 ≥ n ≥ 0}.
(i) If xj ∈ D show that N xj 10’j ¤ 1.
j=1
N
(ii) If xj ∈ D show that ’j
j=1 xj 10 converges to a limit x, say, as
N ’ ∞. Show that 0 ¤ x ¤ 1 and that x = 1 if and only if xj = 9 for all j.
(iii) If y ∈ [0, 1] show that there exist yj ∈ D such that
N
y ’ 10’N < yj 10’j ¤ y
j=1

and that N yj 10’j ’ y as N ’ ∞.
j=1
(iv) Identify explicitly the use of the axiom of Archimedes in the proof of
the last sentence of (ii) and in the proof of (iii).
(v) Suppose that aj , bj ∈ D, aj = bj for j < M and aM > bM . If
N
’ a and N bj 10’j ’ b as N ’ ∞ show that a ≥ b. Give
’j
j=1 aj 10 j=1
the precise necessary and su¬cient condition for equality and prove it.
14 A COMPANION TO ANALYSIS

Exercise 1.5.13. It seems, at ¬rst sight, that decimal expansion gives a nat-
ural way of treating real numbers. It is not impossible to do things in this
way, but there are problems. Here is one of them. Let 0 < a, b < 1 and
c = ab. If N aj 10’j ’ a, N bj 10’j ’ b, and N cj 10’j ’ c ¬nd cj
j=1 j=1 j=1
in terms of the various ak and bk . (The reader is invited to re¬‚ect on this
problem rather than solve it. Indeed, one question is ˜what would constitute
a nice solution?™)
Exercise 1.5.14. Here is a neat application of decimal expansion.
(i) De¬ne f : [0, 1] ’ [0, 1] as follows. Each x ∈ [0, 1] has a unique
non-terminating decimal expansion

xj 10’j
x=
j=1

with the xj integers such that 0 ¤ xj ¤ 9. If there exists an integer N ≥ 2
such that x2j = 1 for all j ≥ N but x2N ’2 = 1 we set

x2(j+N )+1 10’j .
f (x) =
j=1

Otherwise we set f (x) = 0. Show that given any y ∈ [0, 1], any > 0 and
any t ∈ [0, 1] we can ¬nd an x ∈ [0, 1] with |x ’ y| < such that f (x) = t.
In other words f takes every value in [0, 1] arbitrarily close to every point.
(ii) Show that if 0 ¤ a < b ¤ 1 then, given any t lying between f (a)
and f (b) (that is to say, t with f (a) ¤ t ¤ f (b) if f (a) ¤ f (b) or with
f (b) ¤ t ¤ f (a) if f (b) ¤ f (a)), there exists a c ∈ [a, b] such that f (c) = t.
(Thus the fact that a function satis¬es the conclusion of the intermediate
value theorem (Exercise 1.4.2) does not show that it is well behaved.)
(iii) Find a g : R ’ [0, 1] such that, given any y ∈ R, any > 0 and any
t ∈ [0, 1], we can ¬nd an x with |x ’ y| < with f (x) = t.
(iv) (This may require a little thought.) Find a g : R ’ R such that given
any y ∈ R, any > 0 and any t ∈ R we can ¬nd an x with |x ’ y| < such
that f (x) = t.
Although decimal expansion is a very useful way of representing numbers
it is not the only one. In Exercises K.13 and K.14 we discuss representation
by continued fractions.


1.6 Lion hunting
Having dealt with the axiom of Archimedes, we can go on at once to prove
the intermediate value theorem.
15
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Theorem 1.6.1. (The intermediate value theorem.) We work in R. If
f : [a, b] ’ R is continuous and f (a) ≥ 0 ≥ f (b), then there exists a c ∈ [a, b]
such that f (c) = 0.
Proof. Since the method of proof is important to us, I shall label its three
main parts.
Part A Set a0 = a and b0 = b. We observe that f (a0 ) ≥ 0 ≥ f (b0 ). Now set
c0 = (a0 + b0 )/2. There are two possibilities. Either f (c0 ) ≥ 0, in which case
we set a1 = c0 and b1 = b0 , or f (c0 ) < 0, in which case we set a1 = a0 and
b1 = c0 . In either case, we have

f (a1 ) ≥ 0 ≥ f (b1 ),
a0 ¤ a 1 ¤ b 1 ¤ b 0 ,
and b1 ’ a1 = (b0 ’ a0 )/2.

Continuing inductively we can ¬nd a sequence of pairs of points an and bn
such that

f (an ) ≥ 0 ≥ f (bn ),
an’1 ¤ an ¤ bn ¤ bn’1 ,
and bn ’ an = (bn’1 ’ an’1 )/2,

for all n ≥ 1.
Part B We have a0 ¤ a1 ¤ · · · ¤ an ¤ b0 so that the an form an increasing
sequence bounded above. By the fundamental axiom there is real number c,
say, such that an ’ c as n ’ ∞. Since a = a0 ¤ an ¤ b0 we have a ¤ c ¤ b.
We note also that bn ’ an = 2’n (b0 ’ a0 ) so, by the axiom of Archimedes,
bn ’ an ’ 0 and thus

bn = an + (bn ’ an ) ’ c + 0 = c

as n ’ ∞.
Part C Since f is continuous at c and an ’ c, it follows that f (an ) ’ f (c)
as n ’ ∞. Since f (an ) ≥ 0 for all n, it follows that f (c) ≥ 0. A similar
argument applied to the bn shows that f (c) ¤ 0. Since 0 ¤ f (c) ¤ 0, it
follows that f (c) = 0 and we are done.
Exercise 1.6.2. (i) Give the complete details in the inductive argument in
Part A of the proof of Theorem 1.6.1 above.
(ii) Give the details of the ˜similar argument applied to the bn ™ which
shows that f (c) ¤ 0.
(iii) We use various parts of Lemma 1.2.2 in our Theorem 1.6.1. Identify
the points where we use Lemma 1.2.2.
16 A COMPANION TO ANALYSIS

Exercise 1.6.3. (i) Think how the argument used to prove Theorem 1.6.1
applies to [a, b] = [0, 1], f (x) = 2 ’ 4x2 . (You are not asked to write anything
though you may well choose to draw a diagram.)
(ii) Think also how the argument used to prove Theorem 1.6.1 applies to
[a, b] = [0, 1], f (x) = (1 ’ 5x)(2 ’ 5x)(3 ’ 5x).

The method used to prove Theorem 1.6.1 is called ˜Lion hunting™4 . The
method is also called ˜successive bisection™, ˜bisection search™ or simply ˜bi-
section™.
Let us summarise the proof. In Part A we have evidence of a lion in the
interval [an’1 , bn’1 ]. We split the interval into two halves [an’1 , cn’1 ] and
[cn’1 , bn’1 ] and show that, since there is evidence of a lion in the interval
[an’1 , bn’1 ], either there is evidence of a lion in [an’1 , cn’1 ], in which case we
take [an , bn ] = [an’1 , cn’1 ], or, if there is no evidence of a lion in [an’1 , cn’1 ]
(this does not mean that there are no lions in [an’1 , cn’1 ], simply that we do
not have evidence of one), then it follows that there must be evidence of a
lion in [cn’1 , bn’1 ] and we take [an , bn ] = [cn’1 , bn’1 ].
In Part B we use the fundamental axiom of analysis to show that these
successive bisections ˜close in™ on a point c which we strongly suspect of being
a lion. Finally in Part C we examine the point c to make sure that it really
is a lion. (It might be a wolf or a left handed corkscrew.)
Let us see what goes wrong if we omit parts of the hypotheses of Theo-
rem 1.6.1. If we omit the condition f (a) ≥ 0 ≥ f (b), then we cannot even
start Part A of the argument. The example [a, b] = [0, 1], f (x) = 1 shows
that the conclusion may indeed be false.
If we have f (a) ≥ 0 ≥ f (b) but replace R by another ordered ¬eld for
which the fundamental axiom does not hold, then Part A goes through per-
fectly but Part B fails. Example 1.1.3 with which we started shows that the
conclusion may indeed be false. (Take [a, b] = [’2, 0].)
If we have f (a) ≥ 0 ≥ f (b) and we work over R but we do not demand
f continuous then Part C fails. Working over R we may take [a, b] = [0, 1]
and de¬ne f (x) = 1 for x ¤ 1/3 and f (x) = ’1 for x > 1/3. Parts A and B
work perfectly to produce c = 1/3 but there is no lion (that is, no zero of f )
at c.
Exercises 1.6.4 to 1.6.6 are applications of the intermediate value theorem.

Exercise 1.6.4. Show that any real polynomial of odd degree has at least
one root. Is the result true for polynomials of even degree? Give a proof or
counterexample.
4
The name probably comes from A Contribution to the Mathematical Theory of Big
Game Hunting by H. P´tard. This squib is reprinted in [8].
e
17
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 1.6.5. Suppose that g : [0, 1] ’ [0, 1] is a continuous function. By
considering f (x) = g(x) ’ x, or otherwise, show that there exists a c ∈ [0, 1]
with g(c) = c. (Thus every continuous map of [0, 1] into itself has a ¬xed
point.)
Give an example of a bijective continuous function k : (0, 1) ’ (0, 1) such
that k(x) = x for all x ∈ (0, 1).
Give an example of a bijective (but, necessarily, non-continuous) function
h : [0, 1] ’ [0, 1] such that h(x) = x for all x ∈ [0, 1].
[Hint: First ¬nd a function H : [0, 1] \ {0, 1, 1/2} ’ [0, 1] \ {0, 1, 1/2}
such that H(x) = x.]

Exercise 1.6.6. Every mid-summer day at six o™clock in the morning, the
youngest monk from the monastery of Damt starts to climb the narrow path
up Mount Dipmes. At six in the evening he reaches the small temple at the
peak where he spends the night in meditation. At six o™clock in the morning
on the following day he starts downwards, arriving back at the monastery at
six in the evening. Of course, he does not always walk at the same speed.
Show that, none the less, there will be some time of day when he will be at
the same place on the path on both his upward and downward journeys.

Finally we give an example of lion hunting based on trisecting the interval
rather than bisecting it.

Exercise 1.6.7. Suppose that we have a sequence x1 , x2 , . . . of real num-
bers. Let [a0 , b0 ] be any closed interval. Show that we can ¬nd a sequence of
pairs of points an and bn such that

either xn ≥ bn + (bn’1 ’ an’1 )/3 or xn ¤ an ’ (bn’1 ’ an’1 )/3,
an’1 ¤ an ¤ bn ¤ bn’1 ,
and bn ’ an = (bn’1 ’ an’1 )/3,

for all n ≥ 1.
Show that an and bn tend to some limit c ∈ [a0 , b0 ]. Show further that,
for each n ≥ 1, either xn ≥ c + (bn’1 ’ an’1 )/3 or xn ¤ c ’ (bn’1 ’ an’1 )/3
and so in particular xn = c.
Thus we cannot write the points of [a0 , b0 ] as a sequence. (We say that
[a0 , b0 ] is uncountable. The reader may know a proof of this via decimal
expansions.)

For more on countability and an important extension of this exercise see
Appendix B and Exercise B.7 within that appendix.
18 A COMPANION TO ANALYSIS

1.7 The mean value inequality
Having disposed of one of the three tigers with which we started, by proving
the intermediate value theorem, we now dispose of the other two by using
the ˜mean value inequality™.
Theorem 1.7.1. (The mean value inequality.) Let U be the open inter-
val (±, β) on the real line R. Suppose that K ≥ 0 and that a, b ∈ U with
b > a. If f : U ’ R is di¬erentiable with f (t) ¤ K for all t ∈ U then

f (b) ’ f (a) ¤ (b ’ a)K.

Before we can do this we must de¬ne di¬erentiability and the deriva-
tive. The reader will almost certainly be familiar with a de¬nition along the
following lines.
De¬nition 1.7.2. Let U be an open set in R. We say that a function f :
U ’ R is di¬erentiable at t ∈ U with derivative f (t) if, given > 0, we can
¬nd a δ(t, ) > 0 such that (t ’ δ(t, ), t + δ(t, )) ⊆ U and
f (t + h) ’ f (t)
’ f (t) <
h
whenever 0 < |h| < δ(t, ).
In Chapter 6 we shall de¬ne a more general notion of di¬erentiation and
derive many of its properties. For the moment all we need is De¬nition 1.7.2.
Exercise 1.7.3. Let U be an open interval in R and suppose functions f, g :
U ’ R are di¬erentiable at t ∈ U with derivatives f (t) and g (t). If », µ ∈
R, show that »f + µg is di¬erentiable at t with derivative »f (t) + µg (t).
To obtain Theorem 1.7.1 we prove an apparently weaker result.
Lemma 1.7.4. We use the notation and assumptions of Theorem 1.7.1. If
> 0, then f (b) ’ f (a) ¤ (K + )(b ’ a).
Proof of Theorem 1.7.1 from Lemma 1.7.4. Since f (b)’f (a) ¤ (K+ )(b’a)
for all > 0, it follows that f (b) ’ f (a) ¤ K(b ’ a).
Proof of Lemma 1.7.4. We suppose that f (b) ’ f (a) > (K + )(b ’ a) and
use lion-hunting to derive a contradiction. Set a0 = a, b0 = b. We observe
that f (b0 ) ’ f (a0 ) > (K + )(b0 ’ a0 ). Now set c0 = (a0 + b0 )/2. Since

f (c0 ) ’ f (a0 ) ’ (K + )(c0 ’ a0 ) + f (b0 ) ’ f (c0 ) ’ (K + )(b0 ’ c0 )
= f (b0 ) ’ f (a0 ) ’ (K + )(b0 ’ a0 ) > 0,
19
Please send corrections however trivial to twk@dpmms.cam.ac.uk

at least one of the expressions f (c0 ) ’ f (a0 ) ’ (K + )(c0 ’ a0 ) and (f (b0 ) ’
f (c0 ) ’ (K + )(b0 ’ c0 ) must be strictly positive. If f (b0 ) ’ f (c0 ) ’ (K +
)(b0 ’ c0 ) > 0, we set a1 = c0 and b1 = b0 . Otherwise, we set a1 = a0 and
b1 = c0 . In either case, we have
f (b1 ) ’ f (a1 ) > (K + )(b1 ’ a1 ),
a0 ¤ a 1 ¤ b 1 ¤ b 0 ,
and b1 ’ a1 = (b0 ’ a0 )/2.
Continuing inductively, we can ¬nd a sequence of pairs of points an and
bn such that
f (bn ) ’ f (an ) > (K + )(bn ’ an ),
an’1 ¤ an ¤ bn ¤ bn’1 ,
and bn ’ an = (bn’1 ’ an’1 )/2,
for all n ≥ 1.
We have a0 ¤ a1 ¤ · · · ¤ an ¤ b0 so that the an form an increasing
sequence bounded above. By the fundamental axiom there is real number c,
say, such that an ’ c as n ’ ∞. Since a = a0 ¤ an ¤ b0 we have a ¤ c ¤ b
and similarly aN ¤ c ¤ bN for all N . We note also that bn ’an = 2’n (b0 ’a0 )
so, by the axiom of Archimedes, bn ’ an ’ 0 and thus
bn = an + (bn ’ an ) ’ c + 0 = c
as n ’ ∞.
Since f is di¬erentiable at c, we can ¬nd a δ > 0 such that (c’δ, c+δ) ⊆ U
and
f (c + h) ’ f (c)
’ f (c) < /2
h
whenever 0 < |h| < δ. Thus
|f (c + h) ’ f (c) ’ f (c)h| ¤ |h|/2
whenever |h| < δ, and so, since f (c) ¤ K,
f (c + h) ’ f (c) ¤ (K + /2)h for 0 ¤ h < δ
f (c) ’ f (c + h) ¤ ’(K + /2)h for ’δ ¤ h ¤ 0
Since an ’ c and bn ’ c, we can ¬nd an N such that |aN ’ c| < δ and
|bN ’ c| < δ. It follows, ¬rst taking h = aN ’ c and then h = bN ’ c, that
f (c) ’ f (aN ) ¤ (K + /2)(c ’ aN )
and f (bN ) ’ f (c) ¤ (K + /2)(bN ’ c).
20 A COMPANION TO ANALYSIS

Thus

f (bN ) ’ f (aN ) = f (bN ) ’ f (c) + f (c) ’ f (aN )
¤ (K + /2)(bN ’ c) + (K + /2)(c ’ aN )
= (K + /2)(bN ’ aN ),

contradicting our earlier assumption that f (bn ) ’ f (an ) > (K + )(bn ’ an )
for all n.
Thus our initial assumption must be wrong and the theorem is proved.


Theorem 1.7.1 immediately proves Theorem 1.1.1.

Theorem 1.7.5. (The constant value theorem.) Let U be the open in-
terval (±, β) or the real line R. If f : U ’ R is di¬erentiable with f (t) = 0
for all t ∈ U , then f is constant.

Proof. Let b, a ∈ U with b ≥ a. By applying Theorem 1.7.1 to f with K = 0
we see that f (b) ’ f (a) ¤ 0. By applying Theorem 1.7.1 to ’f with K = 0,
we see that f (a) ’ f (b) ¤ 0. Thus f (a) = f (b). But a and b were arbitrary,
so f is constant.

In section 1.1 we noted the importance of the following simple corollary.

Theorem 1.7.6. Let U be the open interval (±, β) or the real line R. If the
functions f, g : U ’ R are di¬erentiable with f (t) = g (t) for all t ∈ U then
there exists a constant c such that f (t) = g(t) + c for all t ∈ U .

Proof. Apply Theorem 1.7.5 to f ’ g.

In A Tour of the Calculus [5], Berlinski greets this theorem with a burst
of rhetoric.

. . . functions agreeing in their derivates, the theorem states,
di¬er on an interval only by a constant. It is the derivative of
a real-valued function that like some pulsing light illuminates
again the behaviour of the function, enforcing among otherwise
anarchic and wayward mathematical objects a stern uniformity
of behaviour. Such is the proximate burden of the mean value
theorem, which is now revealed to play a transcendental role in
the scheme of things.
21
Please send corrections however trivial to twk@dpmms.cam.ac.uk

To which the present author, constrained by the conventions of textbook
writing from such active expressions of enthusiasm, can only murmur ˜Hear,
hear™.
It is fairly easy to see that Theorem 1.7.1 is equivalent to the following
result.

Theorem 1.7.7. Let U be the open interval (±, β) or the real line R. Sup-
pose that a, b ∈ U and b > a. If g : U ’ R is di¬erentiable with g (t) ≥ 0
for all t ∈ U then

g(b) ’ g(a) ≥ 0.

Exercise 1.7.8. (i) By taking f = ’g and K = 0, prove Theorem 1.7.7
from Theorem 1.7.1.
(ii) By taking g(t) = Kt’f (t), prove Theorem 1.7.1 from Theorem 1.7.7.

Thus a function with positive derivative is increasing.
The converse result is ˜purely algebraic™ in the sense that it does not
involve the fundamental axiom.

Lemma 1.7.9. If g : (a, b) ’ R is di¬erentiable and increasing on (a, b)
then g (t) ≥ 0 for all t ∈ (a, b).

Exercise 1.7.10. Use the de¬nition of the derivative to prove Lemma 1.7.9.
[Hint: Show ¬rst that given any > 0 we have g (t) > ’ .]

Readers who know the mean value theorem (given as Theorem 4.4.1 later)
may wish to extend Theorem 1.7.1 as follows.

Exercise 1.7.11. Suppose that a, b ∈ R and b > a. If f : [a, b] ’ R is
continuous and f is di¬erentiable on (a, b) with f (t) ¤ K for all t ∈ (a, b),
use Theorem 1.7.1 on intervals (an , bn ) with a < an < bn < b and continuity
to show that

f (b) ’ f (a) ¤ (b ’ a)K.

Experience shows that students do not fully realise the importance of
the mean value inequality. Readers should take note whenever they use
Theorem 1.7.6 or Theorem 1.7.7 since they are then using the mean value
inequality directly.

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