Notes on the Foundations
of Mathematics and Analysis
Eduardo Dueñez
Lucio Tavernini
March 13, 2014
Contents
0 Background
0.1
Introduction . . . . . . . . . . . . .
0.2
Proof and Paradox . . . . . . . . .
0.3
Essential Linguistic Concepts . . .
0.4
A Dog’s World . . . . . . . . . . . .
0.5
Peano’s Postulates, Induction . . .
0.6
Formal Grammars . . . . . . . . . .
0.7
Warning About the Notation Used
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1
1
3
6
7
11
13
16
1 The Propositional Calculus
1.1
Introduction . . . . . . . . . . . . . . . . . .
1.2
The Language of the Propositional Calculus
1.3
Substitution . . . . . . . . . . . . . . . . . .
1.4
Interpretations . . . . . . . . . . . . . . . . .
1.5
Tautologies and Contradictions . . . . . . .
1.6
Examples . . . . . . . . . . . . . . . . . . .
1.7
Some Useful Tautologies . . . . . . . . . . .
1.8
Proofs . . . . . . . . . . . . . . . . . . . . .
1.9
Appendix A: More Examples . . . . . . . . .
1.10 Appendix B: The Greek Alphabet . . . . . .
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1-1
1-1
1-2
1-5
1-6
1-8
1-11
1-16
1-18
1-21
1-24
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2-1
2-1
2-2
2-4
2-8
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3-1
3-1
3-3
3-8
3-9
2 The Predicate Calculus
2.1
Introduction . . . . . . . . . . . . .
2.2
Quantification of Predicates . . . .
2.3
Formulas of the Predicate Calculus
2.4
Interpretations Revisited . . . . . .
3 Sets
3.1
3.2
3.3
3.4
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Introduction . . . . . . . . . . . . . .
Axioms of Extension and Separation
Intersections and Differences . . . . .
Unions . . . . . . . . . . . . . . . . .
ii
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iii
CONTENTS
3.5
3.6
Powers, Products and Ordered Pairs . . . . . . . . . . . . . . . . . 3-12
Appendix A: The Axiom Schema of Separation . . . . . . . . . . . 3-16
4 The Natural Numbers
4.1
Introduction . . . . . . . . .
4.2
The Set of Natural Numbers
4.3
The Peano Postulates . . . .
4.4
Trichotomy . . . . . . . . .
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4-1
4-1
4-2
4-3
4-8
5 Functions and Relations
5.1
Functions, Forward and Inverse Images . . . .
5.2
Compositions and Identities . . . . . . . . . .
5.3
Injections, Surjections, Bijections and Inverses
5.4
Properties of Functions . . . . . . . . . . . . .
5.5
Indexed Sets . . . . . . . . . . . . . . . . . . .
5.6
Equinumerous Sets . . . . . . . . . . . . . . .
5.7
Relations . . . . . . . . . . . . . . . . . . . . .
5.8
The Recursion Theorem . . . . . . . . . . . .
5.9
Cantor and Schröder-Bernstein Theorems . . .
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5-1
5-1
5-5
5-7
5-11
5-15
5-17
5-21
5-26
5-30
6 Numbers and Arithmetic
6.1
The Arithmetic of the Natural
6.2
Order . . . . . . . . . . . . . .
6.3
The Integers . . . . . . . . . .
6.4
The Rationals . . . . . . . . .
6.5
Algebraic Structures . . . . .
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6-1
6-1
6-12
6-20
6-24
6-29
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7-1
7-1
7-2
7-6
7-8
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8-1
8-1
8-2
8-6
8-12
8-15
8-21
8-27
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Numbers
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7 The Axiom of Choice
7.1
Introduction . . . . . . . . . . . . .
7.2
Products and the Axiom of Choice
7.3
One-Sided Inverses . . . . . . . . .
7.4
Countable and Uncountable Sets . .
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8 The Real Numbers
8.1
Introduction . . . . . . . . . . . . . . . . . .
8.2
The Reals . . . . . . . . . . . . . . . . . . .
8.3
Addition . . . . . . . . . . . . . . . . . . . .
8.4
Multiplication . . . . . . . . . . . . . . . . .
8.5
The Real Field, Density . . . . . . . . . . .
8.6
Monotonic Functions. No-Gaps Condition. .
8.7
Powers, Roots, Exponentials and Logarithms
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iv
CONTENTS
8.8
Sequences and Numerals . . . . . . . . . . . . . . . . . . . . . . . 8-37
9 Important Subsets of the Reals
9.1
Preliminaries . . . . . . . . . . . . .
9.2
Countable and Uncountable Sets . .
9.3
Open and Closed Sets . . . . . . . .
9.4
Properties of Open and Closed Sets
9.5
Compactness . . . . . . . . . . . . .
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9-1
9-1
9-3
9-14
9-20
9-26
10 Sequences and Series
10.1 The Extended Reals . . . . . . . . . . .
10.2 The Topology of the Extended Reals .
10.3 Sequences and their Limits . . . . . . .
10.4 Subsequences and Subsequential Limits
10.5 Cauchy Sequences . . . . . . . . . . . .
10.6 Some Convergence Results . . . . . . .
10.7 Contractive Sequences . . . . . . . . .
10.8 The Binomial Theorem . . . . . . . . .
10.9 Monotonicity . . . . . . . . . . . . . .
10.10 Series . . . . . . . . . . . . . . . . . . .
10.11 Euler’s Number e . . . . . . . . . . . .
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10-1
10-1
10-5
10-9
10-15
10-19
10-21
10-24
10-26
10-33
10-42
10-58
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11 Continuity
11-1
11.1 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 11-1
11.2 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . 11-10
12 Answers to the Exercises
12.1 Note 1 . . . . . . . . .
12.2 Note 2 . . . . . . . . .
12.3 Note 3 . . . . . . . . .
12.4 Note 4 . . . . . . . . .
12.5 Note 5 . . . . . . . . .
12.6 Note 6 . . . . . . . . .
12.7 Note 8 . . . . . . . . .
12.8 Note 9 . . . . . . . . .
12.9 Note 10 . . . . . . . .
12.10 Note 11 . . . . . . . .
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12-1
12-1
12-10
12-12
12-24
12-28
12-41
12-58
12-69
12-76
12-87
Foundations Note 8
The Real Numbers
Eduardo Dueñez and Lucio Tavernini
March 13, 2014
Contents
8.1
8.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 8-1
8.2
The Reals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8-2
8.3
Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8-6
8.4
Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . 8-12
8.5
The Real Field, Density . . . . . . . . . . . . . . . . . . . . . 8-15
8.6
Monotonic Functions. No-Gaps Condition. . . . . . . . . . . 8-21
8.7
Powers, Roots, Exponentials and Logarithms . . . . . . . . . 8-27
8.8
Sequences and Numerals . . . . . . . . . . . . . . . . . . . . 8-37
Introduction
Note: In this note the symbol ⇒ is used to denote the sentential connective →
and also to denote a tautological implication (logical consequence). The symbol
⇔ will be used analogously (for esthetic symmetry). The purpose is to release →
from performing triple duty. Thus, → will be used to denote limits (e.g.: 1/x → 0
as x → +∞) and as a function symbol (e.g.: f : A → B).
Recall (see Note 6 ) that an ordered field (A, +, · , 0, 1, <) is a field (A, +, · , 0, 1)
together with a strict ordering∗ < of A such that for all a, b, c ∈ A we have
∗
Per Note 6, an ordering is a relation that is asymmetric and transitive. Do not confuse
“asymmetric” with “antisymmetric.” To say that < is asymmetric means that a < b ⇒ ¬(b < a).
8-1
Note 8: The Real Numbers – March 13, 2014
8-2
• a < b ⇔ a + c < b + c; and
• a < b ⇔ a · c < b · c, whenever 0 < c.
Further recall (see Note 6 ) that an ordered field (A, +, ·, 0, 1, <)
• is called dense if for all a and b in A such that a < b there is a c ∈ A such
that a < c < b; and
• is called complete if every nonempty subset of A that is bounded above has
a least upper bound in A. (Alternatively, if every nonempty subset of A that
is bounded below has a greatest lower bound in A.)
In Note 6 we proved that the rationals Q are dense, but not complete. That
Q is dense is easy to see: Given a, b ∈ Q with a < b, we have
a=
a+a
a+b
b+b
<
<
= b.
2
2
2
Therefore, by choosing c = (a + b)/2, we have a < c < b with c ∈ Q. To show that
Q fails to be complete is a little more subtle, but quite elementary: We define A
to be the set of all positive rationals a such that a2 < 2 and define B to be the
set of all positive rationals b such that 2 < b2. As shown in Remark 6.5.31, neither
sup A nor inf B exists in Q.
That Q fails to be complete means that not every subset of Q that is bounded
above has a least upper bound and that not every subset of Q that is bounded
below has a greatest lower bound. The reals are constructed to be complete.
8.2
The Reals
We write <, +, ·, 0, 1, −1, . . . instead of <Q , +Q, ·Q, 0Q , 1Q , −1Q , . . .. In other words,
by default, our numbers are the rationals Q and we think of Z as a subset of Q.
We write Z ⊂ Q even though Z is only isomorphic to a subset of Q. Likewise, we
write N ⊂ Z. As noted in Remarks 6.3.4, the identification of Z (and, likewise,
of N) with a subset of Q is unambiguous. As usual, we may omit the dot denoting
multiplication.
8.2.1 Definition. Initial and Final Segments: Let @ be a linear ordering of
a set A and let B ⊂ A.
B is an initial segment of A if
x∈B∧y@x⇒y∈B
8-3
Note 8: The Real Numbers – March 13, 2014
for all x, y ∈ A. In other words, B ⊂ A is an initial segment of A if whenever x is
in B so is everything in A that is to the left (less than) x.
B is a final segment of A if
x∈B∧x@y⇒y∈B
Note that some of the consequences of B being, say, an initial segment, are
x∈B∧y∈
/B⇒x@y
and x 6∈ B ∧ x @ y ⇒ y ∈
/ B.
8.2.2 Definitions. Let (A, @) be a linearly ordered set. For x ∈ A define
A<x
A≤x
A>x
A≥x
(8.1)
= {y
= {y
= {y
= {y
∈A|y
∈A|y
∈A|y
∈A|y
< x}
≤ x}
> x}
≥ x} .
Clearly, A<x and A≤x are initial segments of A whereas A>x and A≥x are final
segments.
8.2.3 Remarks. We constructed the integers Z from the natural numbers N by
using two natural numbers to name an integer. Then, we constructed the rationals
Q from the integers Z by using two integers to name a rational number. The
construction of the reals from the rationals is more complicated: infinitely many
rationals are needed to name a real.
8.2.4 Definition. Dedekind Cuts: A Dedekind cut is a nonempty proper
initial segment of Q without a largest element. That is, a Dedekind cut is a subset
x of Q such that
(1) x 6= ∅ and x 6= Q (x is nonempty proper subset of Q);
(2) x has no largest element, i.e.: for every a ∈ x there is a b ∈ x such that a < b;
(3) x is an initial segment.
8.2.5 Definition.
as
The Real Numbers: The set R of real numbers is defined
R = {x ∈ P(Q) | x is a Dedekind cut}.
8.2.6 Definition (Ordering of R). Define the ordering <R on R by
<R = {(x, y) ∈ R × R | x ⊂ y ∧ x 6= y}.
8-4
Note 8: The Real Numbers – March 13, 2014
That is, for all x and y in R we have
x <R y ⇔ x ⊂ y ∧ x 6= y.
We write x ≤R y if x = y or x <R y, i.e.:
x ≤R y ⇔ x ⊂ y.
We also write x >R y for y <R x and x ≥R y for y ≤R x.
8.2.7 Proposition (Trichotomy for the Real Numbers:). The relation <R is
a linear ordering on R.
Proof. The transitivity of <R follows immediately from the transitivity of set inclusion.
Let x and y be real numbers. We must show that exactly one of the alternatives
x <R y,
x = y,
y <R x,
holds. Clearly, at most one of them holds. We finish the proof by showing that
at least one of them holds. To do this, suppose that x <R y and x = y are both
false. That is, suppose that x 6⊂ y. We must show that y ⊂ x. Since x 6⊂ y, there
is some rational r in x \ y. Choose any s ∈ y. If r ≤ s, then, since y is an initial
segment, we have r ∈ y, contradicting r ∈ x \ y. Therefore, we have s < r. Then,
since r ∈ x and x is an initial segment, we have s ∈ x.
We have shown that s ∈ y ⇒ s ∈ x. Since x 6= y and s is arbitrary, we have
y ⊂ x.
8.2.8 Proposition. Least Upper Bound Property: Every nonempty subset
of R that is bounded above has a unique least upper bound.
Proof. Suppose that A is a nonempty subset of R that is bounded above. We prove
that ∪A is a least upper bound. By trichotomy, it will be unique.
First, we prove that ∪A is a Dedekind cut.
We prove that ∪A 6= ∅. Since A 6= ∅, we have a ∈ A for some a ∈ R. Note that
a 6= ∅ because the empty set is not a Dedekind cut. Since ∅ =
6 a ⊂ ∪A, we have
∪A 6= ∅.
We prove that ∪A 6= Q. Let b∗ be an upper bound of A. Then b∗ 6= Q. Since
x ∈ A ⇒ x ⊂ b∗ we have ∪A ⊂ b∗. Therefore ∪A 6= Q.
We prove that ∪A has no largest element. For every x ∈ ∪A we have x ∈ a for
some a ∈ A. Since a is a Dedekind cut, it has no largest element. Hence, x < y
for some y ∈ a. This means that y ∈ ∪A, showing that for all x ∈ ∪A there is a
y ∈ ∪A with x < y.
Note 8: The Real Numbers – March 13, 2014
8-5
We prove that ∪A is an initial segment. Choose any x ∈ ∪A. Then, x ∈ a for
some a ∈ A. If y < x then y ∈ a also, because a is an initial segment. Hence,
y ∈ ∪A, showing that ∪A is also an initial segment.
We have shown that ∪A is a Dedekind cut. We show that it is also the least
upper bound of A. Clearly, ∪A is an upper bound of A, i.e.: a ∈ A ⇒ a ≤R ∪A.
Consider any b <R ∪A. This means that there is an x ∈ ∪A with x ∈
/ b. Since
x ∈ ∪A, we have x ∈ a for some a ∈ A. This says that b <R a, so that b cannot be
a bound for A.
We have shown that ∪A = sup A.
8.2.9 Proposition. Every nonempty subset of R that is bounded below has a
unique greatest lower bound.
Proof. See the Exercise below.
8.1 Exercise. Prove Proposition 8.2.9.
Henceforth we will denote the order relation <R of R simply by <.
8.2.10 Definition. Intervals The following subsets of the real line R are called
intervals. For real a and b define
(a, b) = {x ∈ R | a < x < b},
[a, b) = {x ∈ R | a ≤ x < b},
(a, b] = {x ∈ R | a < x ≤ b},
[a, b] = {x ∈ R | a ≤ x ≤ b},
(a, +∞) = R>a = {x ∈ R | a < x},
(−∞, b) = R<b = {x ∈ R | x < b},
[a, +∞) = R≥a = {x ∈ R | a ≤ x},
(−∞, b] = R≤b = {x ∈ R | x ≤ b},
(−∞, +∞) = R.
Note that for a ≥ b we have (a, b) = (a, b] = [a, b) = ∅, the empty set. For a > b
we have [a, b] = ∅; on the other hand, [a, a] = {a}, the set with the single point a.
The notation (a, b) is ambiguous: it can denote an open interval or an ordered pair.
Context tells which.
8.2.11 Proposition. Intervals of R can be characterized as the subsets I ⊂ R
having the following intermediacy property:
(∀x, y, z ∈ R)(x, z ∈ I ∧ x < y < z ⇒ y ∈ I).
Note 8: The Real Numbers – March 13, 2014
8-6
Proof. See the Exercise below.
8.2 Exercise. Prove Proposition 8.2.11.
8.2.12 Corollary. An initial (or final) segment of an interval I ⊂ R is an interval.
Proof. It is obvious from the definitions that an initial or final segment satisfies
the intermediacy property, hence is an interval by Proposition 8.2.11.
8.3
Addition
We begin by defining the additive identity of R.
8.3.1 Definition. Real Zero: Define 0R to be the set of all negative rationals:
0R = {a ∈ Q | a < 0}.
8.3.2 Definition.
Real Addition: Addition +R of two reals is defined by
x +R y = {a + b ∈ Q | a ∈ x ∧ b ∈ y}.
8.3.3 Proposition.
x +R y is a Dedekind cut for all x, y ∈ R.
Proof. First, we prove that x +R y 6= ∅ and x +R y 6= Q. Since x and y are cuts,
neither is empty. Therefore, x + y 6= ∅. Further, x 6= Q and y 6= Q. Take a ∈ x
and b ∈ y. If a0 ∈
/ x and b0 ∈
/ y then a0 > a and b0 > b. Hence, a0 + b0 > a + b, giving
a 0 + b0 ∈
/ x +R y. This shows that x +R y 6= Q.
Next, we prove that x +R y has no largest element. Choose c ∈ x +R y. Then
c = a + b with a ∈ x and b ∈ y. Take r ∈ x with r > a. Then c < r + b and
r + b ∈ x +R y. This shows that x +R y has no largest element.
We prove that x +R y is an initial segment. Choose c ∈ x +R y. Then c = a + b
for some a ∈ x and b ∈ y. If d < c then d − b < a, giving d − b ∈ x and
d = (d − b) + b ∈ x +R y. This shows that x +R y is an initial segment.
8.3.4 Definition. The Negative of a Real Number For every real x we
want to define the real −x in such a way that x +R (−x) = 0R .
Guided by Figure 1, we may be tempted to define −x so that
a ∈ (−x) ⇔ −a ∈
/x
for all a ∈ Q, i.e.: −x = {a ∈ Q | −a ∈
/ x}.
8-7
Note 8: The Real Numbers – March 13, 2014
−x
z
...
|
}|
{
q
a
{z
x
q
q
0
−a
...
Q →
}
Figure 1: A Visualization of the Negative of a Cut.
Unfortunately, −x defined in this way may not be a cut because it may contain
a largest element. For example, we would have
−0R = {a ∈ Q | −a ∈
/ 0R }
= {a ∈ Q | −a ≥ 0}
= {a ∈ Q | a ≤ 0},
which is not a cut, since 0 is the largest element of −0R and 0 ∈ (−0R ). Therefore
we modify things a little and define
−x = {a ∈ Q | (∃b > a)(−b ∈
/ x)},
which does the trick quite nicely, as we prove in Proposition 8.3.6 below.
8.3.5 Proposition. For any Dedekind cut x ∈ R
a ∈ x ∧ b ∈ (−x) ⇒ a < −b;
a ∈ x ∧ a ≥ −b ⇒ b ∈
/ (−x);
b ∈ (−x) ∧ a ≥ −b ⇒ a ∈
/ x.
Before starting the proof of this easy proposition, we remark that it almost
characterizes (−x). In fact, the set of b ∈ Q for which −b > a for all a ∈ x is
exactly our early (not quite correct) definition {b ∈ Q | −b ∈
/ x} of (−x). The
proposition is useful because it is often all that is needed to prove of one half of an
identity involving (−x). The proof will not use the fact that (−x) is a Dedekind
cut; this is established in Proposition 8.3.6 below.
Proof. It suffices to prove the first statement since it is tautologically equivalent to
the second and third (in view of ¬(a < −b) ⇔ a ≥ b, by trichotomy).
Let a ∈ x and b ∈ (−x). By definition, there exists c > b with −c ∈
/ x. Since
c < b we have −b > −c; since x is an initial segment and −c ∈
/ x we have −b ∈
/x
and −b > a.
8.3.6 Proposition.
Proof.
If x ∈ R then −x ∈ R.
We prove that if x is a cut then −x is also a cut.
Note 8: The Real Numbers – March 13, 2014
8-8
• We prove that −x 6= ∅. There is some rational b that is not in x (because
x 6= Q). Define a = −b − 1. Then a ∈ x, since a < −b and −(−b) = b ∈
/ x.
• We prove that −x 6= Q. Consider any a ∈ x. Since a ∈ A and a ≥ a =
−(−a), we have −a ∈
/ x, by Proposition 8.3.5. Hence, −x 6= Q.
• We prove that −x is an initial segment. Suppose that a0 < a ∈ (−x). Then
there is some b > a with −b ∈
/ x. Hence, for any such b, we have b > a0 with
−b ∈
/ x. Therefore, a0 ∈ (−x).
• We prove that −x has no largest element. Consider any a ∈ (−x). There
is some b > a with −b ∈
/ x. Because the rationals are dense, there is some
rational a0 such that b > a0 > a. We have a0 > a with a0 ∈ (−x), as was to
be shown.
8.3.7 Proposition. For every positive rational c and every real x there is some
a ∈ x such that a + c ∈
/ x.
Proof. Let c and x be fixed and given as in the statement.
Fix a rational number number b ∈ x. Let
S = {n ∈ N | b + nc ∈
/ x}.
First, we show that S is a nonempty, proper subset of N.
Since a cut is a proper subset of Q, there exists d ∈
/ x. Then b < d (since x is an
initial segment). Since c is positive, the Archimedean property of Q implies that
b + mc > d for some positive integer m. Since d ∈
/ x and x is an initial segment
we must have b + mc ∈
/ x, hence m ∈ S and S is nonempty. Since b ∈ x we have
0∈
/ S so S is a proper subset of N.
Since N is well-ordered there exists a least element n0 in S; moreover n0 ≥ 1 and
n0 − 1 ∈ N \ S. Letting a = b + (n0 − 1)c, we have a ∈ x and a + c = b + n0c ∈
/ x.
8.3.8 Corollary. For every rational c > 0 and any real x there exist a ∈ x and
b∈
/ x with b − a < c.
(Intuitively speaking, there can be no gap of any positive length between a cut
x and its complement Q \ x.)
Proof. Let c and x be as in the statement. By Proposition 8.3.7 there exists a0 ∈ x
such that a0 + c ∈
/ x. Let a ∈ x such that a > a0 (x has no maximum) and let
b = a0 + c. Then b ∈
/ x and b − a < b − a0 = c.
An alternate (and perhaps simpler) proof is as follows. Choose d ∈ Q such that
0 < d < c and apply Proposition 8.3.7 with d replacing c. We obtain a ∈ x and
b∈
/ x such that b − a = d < c.
Note 8: The Real Numbers – March 13, 2014
8.3.9 Proposition.
8-9
Real addition +R satisfies the axioms of addition in a field.
(1) Real addition is associative: (x +R y) +R z = x +R (y +R z) for all x, y, z ∈ R.
(2) Real addition is commutative: x +R y = y +R x for all x, y ∈ R.
(3) There exists an additive identity 0R : x +R 0R = x for all x in R.
(4) Every real number has an additive inverse for all x in R there exists y ∈ R
with x +R y = 0R . In fact y = −x is one such additive inverse.
(It can be shown that 0R is the only additive identity and that −x is the only
additive inverse of x. See Exercise 8.3 below.)
Proof. We prove (1): Each sum (x +R y) +R z is the set of rational sums
(a + b) + c ∈ Q, where a ∈ x, b ∈ y, and c ∈ z. Thanks to the associativity of
rational addition, i.e.: (a + b) + c = a + (b + c), x +R (y +R z) yields the same set.
We prove (2): Each sum x +R y is the set of rational sums a + b ∈ Q,
where a ∈ x and b ∈ y. Thanks to the commutativity of rational addition, i.e.:
a + b = b + a, y +R x yields the same set.
We prove (3): Choose any x ∈ R. If a ∈ x and b ∈ 0R then a + b < a (since
b < 0). This gives a + b ∈ x, i.e.: x +R 0R ⊂ x. Now, for every a ∈ x there is some
b ∈ x such that a < b. We have a − b ∈ 0R , hence, a = b + (a − b) ∈ x +R 0R ,
giving x ⊂ x +R 0R . We have shown that x = x +R 0R .
We prove (4): We prove that −x is the additive inverse of x. The definition
of +R gives
(1)
x +R (−x) = {a + b | a ∈ x ∧ (∃c > b)(−c ∈
/ x)}.
• We prove that x +R (−x) ⊂ 0R . For all b ∈ (−x) we have b < c for some c
such that −c ∈
/ x. Since x is an initial segment, we have a < −c for all a ∈ x;
thus a + b < a + c < 0. This shows that a + b ∈ 0R , proving x +R (−x) ⊂ 0R .
• We prove that 0R ⊂ x +R (−x). Choose any c ∈ 0R . Then c < 0 and −c > 0.
Thanks to Corollary 8.3.8 there exist a ∈ x and d ∈
/ x such that d − a < −c.
Let b = c − a. Then b = c − a < −d with −(−d) = d ∈
/ x, so b ∈ (−x) by
definition. Moreover, c = a + b ∈ x +R (−x).
8.3 Exercise. Show that 0R is the unique additive identity of R, and that −x is
the only additive inverse of x.
8.3.10 Proposition. Cancellation Law The Cancellation Law
x +R y = x +R z ⇒ y = z,
holds for all x, y and z in R.
Note 8: The Real Numbers – March 13, 2014
Proof.
8-10
Let x +R y = x +R z. Add (−x) to both sides:
(−x) + (x + y) = (−x) + (x + z)
((−x) + x) + y = ((−x) + x) + z
0R + y = 0R + z
y = z.
8.3.11 Proposition. Addition Preserves Order For all x, y and z in R we
have
x <R y ⇔ x +R z <R y +R z.
Proof.
We prove the biconditional in each direction.
• We prove that
(1)
x < R y ⇒ x +R z <R y +R z
for all x, y and z in R. We begin by proving
(2)
x ≤R y ⇒ x +R z ≤R y +R z.
Indeed, x ≤R y says that x ⊂ y. This, together with the definition of +R ,
x +R z = {a + b ∈ Q | a ∈ x ∧ b ∈ z},
gives
x +R z = {a + b ∈ Q | a ∈ x ∧ b ∈ z} ⊂ {a + b ∈ Q | a ∈ y ∧ b ∈ z} = y +R z,
which says that x +R z ≤R y +R z, which establishes (2). From Proposition 8.3.10, we obtain
(3)
x 6= y ⇒ x +R z 6= y +R z
Now, (1) follows from (2) and (3).
• We prove that x +R z <R y +R z ⇒ x <R y for all x, y and z in R. Suppose
that x +R z <R y +R z. Then, (x +R z ⊂ y +R z) ∧ (x +R z 6= y +R z). We
cannot have x = y, since this would imply x +R z = y +R z. Therefore, by
trichotomy, we only have two possibilities: either x <R y or y <R x.
If y <R x, then y ⊂ x, and we have
y +R z = {a + b ∈ Q | a ∈ y ∧ b ∈ z} ⊂ {a + b ∈ Q | a ∈ x ∧ b ∈ z} = x +R z,
which gives y +R z ≤R x +R z. Since x 6= y, we have y +R z <R x +R z, a
contradiction (thanks to trichotomy).
The only possibility left is x <R y.
8-11
Note 8: The Real Numbers – March 13, 2014
8.3.12 Definition (Subtraction of Real Numbers).
For x, y ∈ R define
x −R y = x +R (−y).
Subtraction is merely a cosmetic rewriting of the sum of the additive inverse.
8.4 Exercise. For all x ∈ R show
0R <R x
⇔
−x <R 0R .
8.5 Exercise. For x, y ∈ R show
x <R y
⇔
−y <R −x
⇔
0R < y −R x.
8.3.13 Definition. Real Absolute Value For x ∈ R we define its absolute
value |x| by |x| = x ∪ −x. Note that |x| is the larger of x and −x. Since x and −x
are both initial segments, |x| is the larger of the two, giving x ⊂ |x| and −x ⊂ |x|.
That is to say: x ≤R |x| and −x ≤R |x|. Note that when we shall finish making
the reals into a field, this definition will match the definition given in Note 6.
8.3.14 Proposition.
For all x ∈ R we have 0R ≤R |x|.
Proof. If 0R ≤R x then 0R ⊂ x ⊂ x ∪ −x = |x|. Hence, 0R ≤R |x|.
If x <R 0R then x ⊂ 0R . Hence, 0R ⊂ −x ⊂ x ∪ −x = |x|, which gives
0R ≤R |x|.
8.6 Exercise. Let A be a nonempty subset of R. Define
−A = {−x | x ∈ A} .
Prove the following:
1. A is bounded above if and only if −A is bounded below, in which case
inf(−A) = − sup A.
2. A is bounded below if and only if −A is bounded above, in which case
sup(−A) = − inf A.
8.7 Exercise. Let A be a nonempty subset of R and let c ∈ R. Define
A + c = {x +R c | x ∈ A}
Prove the following:
1. A is bounded above if and only if A + c is bounded above, in which case
sup(A + c) = c +R sup A.
Note 8: The Real Numbers – March 13, 2014
8-12
2. A is bounded below if and only if A + c is bounded below, in which case
inf(A + c) = c +R inf A.
8.8 Exercise. Let A, B be nonempty subsets of R, both bounded above. Define
A + B = {z ∈ R | z = x +R y for some x ∈ A, y ∈ B}.
Prove that A + B is nonempty and bounded above, and sup(A + B) = sup A +R
sup B.
We remark that the same relationship holds for the infima, as long as the
assumption on upper- is replaced by lower-boundedness.
8.4
Multiplication
We begin by defining the multiplicative identity of R.
8.4.1 Definition. Real One Define 1R = {a ∈ Q | a < 1}.
8.4.2 Definition. Real Multiplication It may be tempting to define multiplication ·R by
x ·R y = {ab | a ∈ x ∧ b ∈ y}.
Unfortunately, the above does not work: Since x and y are both initial segments,
they both contain negative rationals of arbitrarily large magnitude. This results in
x ·R y containing arbitrarily large rationals. Therefore we proceed as follows.
We define 0R ·R x = x ·R 0R = 0R . For 0R <R x and 0R <R y, we define
x ·R y = Q≤0 ∪ {ab ∈ Q | a ∈ x ∧ 0 < a ∧ b ∈ y ∧ 0 < b}.
More generally, we define

 (−x) ·R (−y) if x <R 0R ∧ y <R 0R ,
x ·R y =
−[(−x) ·R y] if x <R 0R ∧ 0R <R y,

−[x ·R (−y)] if 0R <R x ∧ y <R 0R .
8.4.3 Proposition.
Proof.
For all x, y ∈ R we have x ·R y ∈ R.
The proof is left as an exercise.
8.9 Exercise. Prove Proposition 8.4.3.
Note 8: The Real Numbers – March 13, 2014
8-13
8.4.4 Definition. The Multiplicative Inverse of a Real Number For every
real x we want to define its multiplicative inverse x−1 in such a way that x ·R x−1 =
1R .
A combination of the ideas involved in defining the negative −x and the product
x ·R y suggests the following definition: For x >R 0R , let
x−1 = Q≤0 ∪ {a ∈ Q>0 | (∃b > a)(b−1 ∈
/ x)}.
For x <R 0R , define x−1 = − (−x)−1 . (Naturally, 0−1
R is undefined.)
8.4.5 Proposition. If x ∈ R then x−1 ∈ R. Moreover, 0R <R x ⇒ 0R <R x−1 ,
and 0R >R x ⇒ 0R >R x−1.
Proof. The proof is left as an exercise (see below).
8.10 Exercise. Prove Proposition 8.4.5.
8.4.6 Proposition.
field.
The operation ·R satisfies the axioms of multiplication in a
(1) Real multiplication is associative: (x·R y)·R z = x·R (y ·R z) for all x, y, z ∈ R.
(2) Real multiplication is commutative: x ·R y = y ·R x for all x, y ∈ R.
(3) Real multiplication distributes over +R : x ·R (y +R z) = (x ·R y) +Z (x ·R z)
for all x, y, z ∈ R.
(4) There is a multiplicative identity 1R , with 1R 6= 0R : x·R 1R = x for all x ∈ R.
(5) There are multiplicative inverses: For every x 6= 0R there exists y 6= 0R such
that x ·R y = 1R . (In fact, y is x−1 as defined above.)
(6) There are no zero divisors: x ·R y = 0R ⇒ x = 0R ∨ y = 0R for all x, y ∈ R.
(7) (−1R ) ·R x = −x for all x ∈ R.
Proof.
The proof is left as an exercise.
8.11 Exercise. Prove the following consequences of Proposition 8.4.6:
1. 0R ·R x = 0R for all x ∈ R.
2. (−x) ·R y = −(x ·R y) for all x, y ∈ R.
8-14
Note 8: The Real Numbers – March 13, 2014
8.4.7 Definition (Division of Real Numbers). For real numbers x, y with
y 6= 0R define the division (or quotient) of x by y by
x/y =
x
= x ·R y −1 .
y
Division is just an alias for multiplication by the reciprocal.
8.4.8 Proposition.
For all x, y in R we have
(1) Cancellation Law for multiplication:
for all z 6= 0R , x ·R z = y ·R z ⇒ x = y;
(2) Monotonicity of multiplication by positive numbers:
x <R y ⇒ x ·R z <R y ·R z.
Proof.
for all z >R 0R ,
(1) Multiply both sides of x ·R z = y ·R z by z −1 .
(2) The particular case w >R 0R ∧ z >R 0R ⇒ w ·R z >R 0R follows immediately
from the definitions of multiplication and of the ordering. Apply the above
with w = y +R (−x) >R 0R to obtain y ·R z + (−x) ·R z = (y +R (−x)) ·R z >R
0R . Adding x ·R z to both sides and using Exercise 8.11:
y ·R z = y ·R z +R 0R
= y ·R z +R (−x) ·R z + x ·R z
>R 0R +R x ·R z
= x ·R z.
8.12 Exercise. Let A be a nonempty subset of (0, ∞). Define
A−1 = x−1 | x ∈ A .
Prove the following:
1. A−1 is bounded above if and only if A is bounded below by a positive number,
in which case sup(A−1) = (inf A)−1 .
2. A−1 is bounded below by a positive number if and only if A is bounded above,
in which case inf(A−1 ) = (sup A)−1 .
Try to generalize the problem to subsets of (−∞, 0).
8.13 Exercise. Let A be a nonempty subset of R. For any real number c define
cA = {c ·R x | x ∈ A} .
Prove the following
Note 8: The Real Numbers – March 13, 2014
8-15
1. If c > 0 then cA is bounded above if and only if A is bounded above, in which
case sup(cA) = c ·R sup A.
2. If c > 0 then cA is bounded below if and only if A is bounded below, in which
case inf(cA) = c ·R inf A.
3. If c < 0 then cA is bounded above if and only if A is bounded below, in which
case sup(cA) = c ·R inf A.
4. If c < 0 then cA is bounded below if and only if A is bounded above, in which
case inf(cA) = c ·R sup A.
8.14 Exercise. Let A, B be nonempty subsets of R+ , both bounded above. Define
AB = {z ∈ R | z = x ·R y for some x ∈ A, y ∈ B}.
Then AB is nonempty and bounded above, and sup(AB) = sup A · sup B.
8.5
The Real Field, Density
We have proved that the reals form a complete ordered field. Next, we have to
justify the customary notion that the rationals are a subset of the reals. They
are a proper subset (irrational numbers exist) and also dense (there are rational
numbers arbitrarily close to any given real number).
8.5.1 Proposition. Define Φ : Q → R by Φ(a) = Q<a = {b ∈ Q | b < a}.
Then, Φ is injective and for all a, b ∈ Q we have
(1) Φ(a + b) = Φ(a) +R Φ(b),
(2) Φ(ab) = Φ(a) ·R Φ(b),
(3) Φ(0) = 0R ,
(4) Φ(1) = 1R ,
(5) a < b ⇔ Φ(a) <R Φ(b).
Proof.
The proof is left as an exercise.
8.5.2 Definitions.
For every a ∈ Q define aR = Φ(a) and
Q∗ = {aR ∈ R | a ∈ Q} = range of Φ.
Then (Q∗ , +R , ·R , 0R , 1R , <R ) is an Archimedean field with Q∗ ⊂ R. Clearly,
Q 6= Q∗, but Q is isomorphic to the bona fide subset (subfield) Q∗ of R. We
Note 8: The Real Numbers – March 13, 2014
8-16
identify Q with Q∗ to simply write Q ⊂ R, with the understanding that it is not
Q, but the subset Q∗ of R, that we call “the rationals.” This means that we do
not distinguish between, for example, 3 ∈ Q and 3R = {a ∈ Q | a < 3} ∈ R.
We shall write <, +, ·, 0, 1, −1, . . . instead of <R , +R , ·R , 0R , 1R , −1R , . . .. In
other words, by default, our numbers are the reals R and we think of Q as a subset
of R. We now have
N ⊂ Z ⊂ Q ⊂ R.
The irrationals are the elements of R that are not rational, i.e.: the elements
of R \ Q. (There is no standard notation for the set R \ Q of irrational numbers;
some author use I.)
The set of positive natural numbers is denoted by N+ . Thus,
N+ = {n ∈ N | 0 < n} = {n ∈ Z | 0 < n}.
In Note 6 we proved that < orders N well. This means that every nonempty subset
of N has a least element, i.e.: for every nonempty subset S of N there is an n0 ∈ S
such that n < n0 ⇒ n ∈
/ S.
8.5.3 Remark. Henceforth we will never again use the set-theoretical construction
of R as a set of Dedekind cuts. All we shall ever use in subsequent proofs is the
fact that R is a complete ordered field possessing an isomorphically embedded copy
Q∗ of Q. Analysis, in the classical sense, is the study of the reals as a structure
possessing the above properties, rather than the study of any specific set possessing
that structure. It is, of course, far from obvious that such an ordered complete field
should exist at all, and it is precisely to show the existence of any such structure
at all that we have provided a particular example constructed via Dedekind cuts.
The reader is henceforth urged to rely only on the structural rather than settheoretical properties of R in subsequent proofs. Of course any new results proved
in this fashion may also be used subsequently. As a first example, the Archimedean
property of R is proved in Proposition 8.5.9 below, using only the ordering and
completeness of the field R.
8.5.4 Proposition. The set R \ Q of irrational numbers is nonempty. In fact,
there exists a unique real number z such that z > 0√ and z 2 = 2, and we have
z ∈ R \ Q. (The number z is conventionally denoted 2.)
Proof. Strictly speaking, we already know that Q must be a proper subset of R
since R is complete but Q is not (Remark 6.5.31). Since this fact is so important,
we presently choose to reproduce the argument in a slightly different form, working
directly in R.
Let
A = {x ∈ (0, +∞) | x2 < 2}.
Note 8: The Real Numbers – March 13, 2014
8-17
Then A is nonempty (since 1 ∈ A) and bounded above by 2 because 22 > 2 and
the function x 7→ x2 is strictly increasing† in (0, +∞). Let z = sup A. We claim
that z 2 = 2. Certainly z is positive. Define
(8.2)
y=z−
z2 − 2
2z + 2
=
.
z+2
z+2
The above gives
(8.3)
y2 − 2 =
2(z 2 − 2)
(2z + 2)2
−
2
=
.
(z + 2)2
(z + 2)2
If z 2 < 2 then (8.2) gives y > z > 0 and (8.3) gives y 2 < 2, so y ∈ A,
contradicting the upper bound property of z. If z 2 > 2 then (8.2) gives 0 < y < z
and (8.3) gives y 2 > 2, showing (by strict monotonicity) that y is an upper bound
for A and contradicting the minimality of z.
Thus z 2 = 2. Recalling that the latter equation has no solution z in√Q (Remark 6.5.31) we conclude that
√ z is irrational. Conventionally one writes 2 for z
and the equation z 2 = 2 as ( 2)2 = 2: z is the (positive) square root
√ of 2. Note
2
that the strict monotonicity of x 7→ x on [0, +∞) implies that 2 is uniquely
defined.
8.15 Exercise. Prove that the set of irrational numbers is closed under neither
addition nor multiplication. (Hence R \ Q is not a ring.)
8.5.5 Proposition. For all real nonnegative x there exists a positive integer n
such that n − 1 ≤ x < n. In particular, N+ is unbounded above (and so is N).
Proof. First we prove that N+ is unbounded above. Assume, for the sake of
seeking a contradiction, that N+ is bounded above. Then z = sup N+ exists by
the completeness property of R, Proposition 8.2.8 (which is they key ingredient of
the present proof). Since z is least, there is n ∈ N+ such that z − 1 < n. Then
z < n + 1, and n + 1 ∈ N+ , contradicting that z is an upper bound for N+ . The
contradiction shows that N+ is bounded above (and so is N ⊃ N+ ).
Now fix x nonnegative. Since N+ is unbounded above, there exist positive
integers numbers n > x. Since N is well-ordered, there exists a least n ∈ N
satisfying x < n. Let n0 be such least n. By assumption n0 6= 0 since x ≥ 0, so
n0 − 1 ∈ N and, by minimality of n0 , n0 − 1 ≤ x. Then n0 ≤ x < n0 .
8.5.6 Corollary. For all real numbers x there exists a unique integer n such
that n ≤ x < n + 1. In particular, Z is unbounded above and below.
†
See Definition 8.6.1 below. The strict monotonicity is shown as follows: 0 < x < x0 ⇒ x2 =
xx < xx0 < x0 x0 = (x0 )2 .
Note 8: The Real Numbers – March 13, 2014
8-18
Proof. If x ≥ 0 there exists m ∈ Z+ with m − 1 ≤ x < m, so let n = m − 1. If
x ∈ Z− let n = x. If x < 0 and x is not an integer then m − 1 ≤ −x < m, hence
−m < x ≤ −m − 1 for some integer m. Moreover x 6= −m − 1 since x ∈
/ Z, so let
n = −m.
To show n is unique let n0 be another such. If n 6= n0 then, without loss of
generality, we may assume n < n0 , so x < n + 1 ≤ n0 ≤ x, a contradiction. Hence
n = n0 .
8.5.7 Notation. For each x ∈ R, the unique integer n such that n ≤ x < n + 1
is denoted by bxc and called the largest or greatest integer in x or sometimes the
integer part of x.
The real number x − bxc ∈ [0, 1) is called the fractional part of x and denoted
*x+ in these Notes. Observe that the identity x = bxc + *x+ uniquely determines
the numbers bxc and *x+ subject to the conditions bxc ∈ Z and *x+ ∈ [0, 1).
8.5.8 Corollary.
inf{1/n | n ∈ N+ } = 0.
Proof. Define S = {1/n | n ∈ N+ }. Then, S is nonempty and is bounded below
by 0. Hence, i = inf S ≥ 0. We claim that i > 0 is impossible. If i were positive,
then 1/i < n for some n ∈ N+ , by Proposition 8.5.5, whence i > 1/n would follow,
contradicting the fact that i is a lower bound for S. We conclude that i = 0.
8.5.9 Corollary (Archimedean Property of the Reals). The field R of real
numbers is Archimedean: For every x, y ∈ R with x > 0 there is some positive
integer n such that y < nx.
Proof. Let x, y ∈ R with x > 0. Since N+ is unbounded above there exists a
positive integer n such that x−1 y < n. Multiplying by x > 0 the inequality is
preserved and we obtain y < nx.
8.5.10 Theorem.
field.
Proof.
(R, +R , ·R , 0R , 1R , <R ) is an ordered complete Archimedean
This sums up the statements of 8.2.7, 8.2.8, 8.3.9, 8.4.6, 8.4.8 and 8.5.9.
8.5.11 Remarks. We leave it as an exercise to prove that the notion of Archimedean
field used above is equivalent to the notion used in Note 6.
Next, we show that Q is also dense in a more general sense.
8.5.12 Proposition (Q is dense in R).
there is a rational a such that x < a < y.
Proof.
For every two reals x and y with x < y
Let reals x and y with x < y be given. We consider the following cases.
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8-19
(i ) If 0 < x < y then y −x > 0 and the Archimedean property gives the existence
of a positive integer n such that 1 < n(y − x). Hence,
(1)
nx + 1 < ny.
Since nx > 0, there is a positive integer m such that
(2)
m − 1 ≤ nx < m.
From (1) and (2) we obtain
(3)
m ≤ nx + 1 < ny.
From (2) and (3) we obtain nx < m < ny. The desired rational is a = m/n.
(ii ) If x ≤ 0 and y > 0, take a = 1/n for sufficiently large n. Then, x < a < y.
(iii ) If x < y < 0 then −x > −y > 0. Thanks to (i ), with x replaced by −y and
y replaced by −x, we obtain a rational a with −y < a < −x, i.e.: such that
x < −a < y.
(iv) If x < 0 and y ≥ 0, take a = −1/n for sufficiently large n. Then, x < a < y.
8.5.13 Corollary (The irrationals are dense in R). For every two real
numbers a and b with a < b there is an irrational x such that a < x < b.
Proof. Let reals a and b with a < b be given. By Proposition 8.5.12, there exists
c ∈ Q with
a
b
√ <c< √ .
2
2
√
Hence, a < c 2 < b.
We remark that we can always choose c 6= 0 without loss of generality. For, if
a < 0 < b, we can take c = 1/n or c = −1/n
for sufficiently large n ∈ N+ .
√
Assuming thus c 6= 0 we take x =√c 2 noting that x must be irrational, since
if x were rational so would be x/c = 2, a contradiction.
√
n
8.16 Exercise. For any integers n, k with n ≥ 2, prove that k is √
either an
n
integer m (this means that k = mn is a perfect n-th power) or else k is an
irrational number. Your proof may rely on the Fundamental Theorem of Arithmetic
(existence of uniqueness of factorization of positive integers into primes) as well as
the existence and uniqueness of n-roots (a fact proved below in Section 8.7.)
Note 8: The Real Numbers – March 13, 2014
8-20
8.17 Exercise. Call two nonzero real numbers x, y commensurable if y/x is rational; otherwise x, y are incommensurable.
1. Show that the relation of commensurability is an equivalence relation in the
set of nonzero real numbers. The equivalence class of a nonzero real is called
its commensurability class. (For completeness, we may call {0R } the commensurability class of 0R .)
2. Show that the commensurability class of any nonzero real is dense in R.
8.18 Exercise. For every integer n ≥ 2, n given nonzero real numbers x1, x2 , . . . , xn
are called pairwise incommensurable if, for all i, j ≤ n with i 6= j, xi and xj are
incommensurable.
Prove that n pairwise incommensurable numbers exist for each n ≥ 2. (Note
that the special case n = 2 is equivalent to the existence of an irrational number.)
8.19 Exercise. Let any two real numbers a, b be given with a < b.
For each n ≥ 2, prove that n pairwise mutually incommensurable numbers exist
in (a, b).
8.20 Exercise. For any n ∈ N call n real numbers u1, u2 , . . . , un additively independent if, for all integers c1, c2 , . . . , cn ,
c1 u1 + c2u2 + · · · + cn un = 0 ⇒ c1 = c2 = · · · = cn = 0.
(For n = 0 the left-hand side of the equation above is the identity 1 = 1, since an
empty product is 1 by definition, hence ∅ is a set of numbers additively independent.)
An arbitrary subset S ⊂ R is called additively independent if every n (distinct)
elements of S are additively independent.
Show the following:
1. Any subset of a set of additively independent numbers is additively independent.
2. For n = 1, x1 is additively independent if and only if x1 is nonzero.
3. For n = 2, x1, x2 are additively independent if and only if they are nonzero
incommensurable numbers.
4. For any n ≥ 2, if x1, x2 , . . . , xn are additively independent, then they are
pairwise incommensurable.
5. When n = 3, find a counterexample to the converse of the statement above.
Note 8: The Real Numbers – March 13, 2014
8-21
8.21 Exercise. For any n ∈ N call n nonzero real numbers u1, u2 , . . . , un multiplicatively independent if, for all integers e1, e2 , . . . , en ,
ue11 ue22 · . . . · uenn = 1 ⇒ e1 = e2 = · · · = en = 0.
(For n = 0 the left-hand side of the equation above is the identity 1 = 1, since
an empty product is 1 by definition, hence ∅ is a set of numbers multiplicatively
independent.)
An arbitrary subset S ⊂ R \ {0} is called multiplicatively independent if every
n (distinct) elements of S are multiplicatively independent.
Assuming the results from section 8.7 (specifically, the Existence and Uniqueness of Roots and the algebraic laws of powers, logarithms and exponents), prove
the following:
1. If S is an additively independent set, then T = {2x | x ∈ S} is a multiplicatively independent set.
2. If T is a multiplicatively independent subset of R+ , then S = {log2 t | t ∈ T }
is an additively independent subset of R.
(Note that there is nothing special about the base 2 above. Any other base b 6= 1
would work just fine.)
8.5.14 Remark. The reader may ask whether additively/multiplicatively independent sets exist of large size (hitherto we only know this up to size n = 2, being
a restatement of the existence of irrational numbers). These exist, not only for any
finite size n, but even infinite. The integral powers of any transcendental number
(such as π or Euler’s number e) provide an explicit example of an additively independent countable subset of R. Much more is true: there exist subsets S of R
that are both uncountable and algebraically independent, in the sense that only
trivial algebraic relations (i.e., only trivial equations involving addition, subtraction, multiplication, division, hence integral powers) hold between elements of S.
Such S may even be taken to be a subset of an arbitrarily small nonempty open
interval (a, b). Any such sets S is both additively and multiplicatively independent, and much more. However, it is not possible to describe any single such set S
explicitly; nonconstructive proofs of the existence of S rely on indirect arguments.
(A relatively simple counting argument works, based on the fact that there are only
countably many algebraic relations between any countable set of numbers whereas
R is uncountable. See Proposition 9.2.16.)
8.6
Monotonic Functions. No-Gaps Condition.
8.6.1 Definition. Monotonic Function. Let f : I → R be a function defined
on a subset I ⊂ R. We call f
Note 8: The Real Numbers – March 13, 2014
8-22
• (monotonically) increasing if x ≤ y ⇒ f(x) ≤ f(y)
• strictly (monotonically) increasing if x < y ⇒ f(x) < f(y)
• (monotonically) decreasing if x ≤ y ⇒ f(x) ≥ f(y)
• strictly (monotonically) decreasing if x < y ⇒ f(x) > f(y)
for all x, y ∈ I. (The word “monotonically” is optional in the nomenclature above
and only used for emphasis.)
In either of the cases above, f is called monotonic. In either of the strict cases
above, f is called strictly monotonic.
8.6.2 Proposition. Let f, g be monotonic functions such that the composition
g ◦ f is defined. Then g ◦ f is monotonic. If both f, g are strictly monotonic, so is
g ◦ f. Specifically,
• If f, g are both (strictly) increasing or both (strictly) decreasing, then g ◦ f
is (strictly) increasing.
• If one of f, g is (strictly) increasing and the other is (strictly) decreasing, then
g ◦ f is (strictly) decreasing.
Proof. We provide the details only in the case when f is strictly increasing and g
strictly decreasing. The remaining cases are all cosmetic variations of this one.
Let x, y ∈ Dom(g ◦ f), x < y. Then f(x) < f(y) by the strict monotonicity of
f, and
(g ◦ f)(x) = g(f(x))
by definition of composition,
> g(f(y))
since f(x) < f(y) and g is strictly decreasing,
= (g ◦ f)(y).
This shows that g ◦ f is strictly decreasing.
8.6.3 Proposition. A strictly monotonic function is injective. In particular, a
strictly monotonic surjective function f : I → J ⊂ R is bijective.
(Note that no assumption is made about J other than it is the range of f. In
particular, J need not be an interval. However, see Proposition 8.6.13 below.)
Proof. The second assertion follows immediately from the first, which we now
prove.
First, assume f is strictly increasing. For any two real numbers x 6= y in the
domain of f we have either x < y or y < x; by renaming them we may assume the
8-23
Note 8: The Real Numbers – March 13, 2014
first inequality holds. By the strict monotonicity then f(x) < f(y), so f(x) 6= f(y).
Therefore f is injective.
The case of f strictly decreasing is deduced from the above applied to the
function g defined by g(x) = −f(x) and two obvious facts:
• g is (strictly) increasing if and only f is (strictly) decreasing, and
• g is injective if and only if f is injective.
8.6.4 Proposition. An injective monotonic function is strictly monotonic. In
particular, an invertible monotonic function is strictly monotonic.
Proof. Let f be increasing and injective. Then x < y implies f(x) ≤ f(y); however, by injectivity f(x) 6= f(y). Therefore f(x) < f(y). Since invertible implies
injective (and surjective), the second claim is obvious. The proof for f decreasing
is essentially identical and we omit it (or just apply the present proof to −f).
8.6.5 Definition. Let S ⊂ R. For any c ∈ R we define the initial segment
S<c = {x ∈ S | x < c} and the final segment S>c = {x ∈ S | x > c}.
(Recall Equation (8.1).)
8.6.6 Proposition. Let f : S → R be a monotonic function defined on a dense
subset S of a nonempty open interval I ⊂ R. For each c ∈ I,
• If f is increasing:
sup f(x) ≤ inf f(x).
x∈S<c
• If f is decreasing:
x∈S>c
sup f(x) ≤ inf f(x).
x∈S>c
x∈S<c
• If f is increasing and f(c) is defined:
sup f(x) ≤ f(c) ≤ inf f(x).
x∈S<c
x∈S>c
• If f is decreasing and f(c) is defined:
sup f(x) ≤ f(c) ≤ inf f(x).
x∈S>c
x∈S<c
We stress that the conclusion holds for all c ∈ I.
Note 8: The Real Numbers – March 13, 2014
8-24
Proof. Since c ∈ I, I is an open interval, and S is dense in I, both sets S<c and
S>c are nonempty.
For all x ∈ S<c and y ∈ S>c , if f is increasing (resp., for all y ∈ S<c and
x ∈ S>c , if f is decreasing) we have f(x) ≤ f(y). The existence of the relevant
suprema/infima follows by completeness in all cases; moreover, the first two statements are obvious. If additionally f(c) is defined, then f(x) ≤ f(c) ≤ f(y). The
last two assertions thus follow as well.
8.6.7 Proposition. Let I ⊂ R be a nonempty open interval, S be a dense subset
of I, and f : S → R be a monotonic function on S. For each dense subset T ⊂ S
and each c ∈ I,
• supx∈T<c f(x) = supx∈S<c f(x) and inf x∈T>c f(x) = inf x∈S>c f(x), if f is increasing, and
• supx∈T>c f(x) = supx∈S>c f(x) and inf x∈T<c f(x) = inf x∈S<c f(x), if f is decreasing.
Note that the conclusions hold for all c ∈ I, not merely for all c ∈ S.
Proof. Since T is dense in S, it is also dense in I, so T>c and T<c are also nonempty.
Let f be increasing (resp., decreasing). Since T ⊂ S, T>c ⊂ S>c and T<c ⊂ S<c ,
then supx∈T<c ≤ supx∈S<c and inf x∈T>c ≥ inf x∈S>c (resp., supx∈T>c ≤ supx∈S>c and
inf x∈T<c ≥ inf x∈S<c ). The reverse inequalities follow from the monotonicity of f
noting that for all x, y ∈ S such that x < c < y, the density of T implies that there
exist x0, y 0 ∈ T with x < x0 < c < y 0 < y.
8.6.8 Definition. No-Gaps Condition. Let I ⊂ R be a nonempty open interval, S a dense subset of I, and f : S → R a monotonic function on S. We say that
f satisfies the No-Gaps condition on I if, for every c ∈ I:
• If f is increasing:
sup f(x) = inf f(x).
x∈S<c
• If f is decreasing:
x∈S>c
sup f(x) = inf f(x).
x∈S>c
x∈S<c
(Note that, in view of Proposition 8.6.6, the equalities above can both be relaxed
to inequalities ≥.)
8.6.9 Remarks.
• By Proposition 8.6.6, if f is defined at c then the common
value of the supremum and infimum in Definition 8.6.8 is f(c).
Note 8: The Real Numbers – March 13, 2014
8-25
• By Proposition 8.6.6, the suprema/infima in Proposition 8.6.8 can be taken
over segments of S or of any dense subset T ⊂ S without changing their
value.
• The denomination “No-Gaps” is ad hoc terminology to be used in these Notes.
The No-Gaps condition for monotonic functions whose domain is an open
interval is equivalent to continuity, which is a more general concept applicable
to functions not necessarily monotonic and the main topic of Note 11. The
No-Gaps condition can be see as a prelude to continuity.
8.6.10 Proposition. Let p, q be real numbers. Let f : R → R be defined by
f(t) = pt + q (this is called a linear function or, more properly, an affine function
(or affine mapping) from R to R). Then f is monotonic (increasing if p ≥ 0,
decreasing if p ≤ 0) and satisfies the No-Gaps condition on R. For p 6= 0, the
monotonicity is strict and f is bijective.
Proof. The statement is obvious if p = 0, for then f is the constant function
f(t) = q. For p 6= 0 it is very easy to check that f −1 (t) = p−1 t − p−1q (incidentally,
also a linear function), so f is bijective.
Henceforth we will assume that p > 0, leaving p < 0 to the reader.
f is increasing, since x < y implies 0 ≤ p(y − x) = (py + q) − (px + q) =
f(y) − f(x). If p > 0 the inequality is strict, so f is strictly increasing.
Let c ∈ R and > 0. Choose x < c < y such that y − x < /p. Then
f(y) − f(x) = (py + q) − (px + q) = p(y − x) < . Since this is true for all > 0 we
conclude that inf x>c f(x) − supx<c f(x) ≤ 0, proving that f satisfies the No-Gaps
condition.
8.6.11 Proposition. Let f be a monotonic function defined on a dense subset
S of a nonempty open interval I. Assume that f has an inverse f −1 defined on
a dense subset T of a nonempty open interval J . Then f satisfies the No-Gaps
condition on I.
Proof. Let c ∈ I. We asume that f is increasing, leaving the trivial modifications
needed to handle the decreasing case to the reader. Let T+ = f∗ (S>c ) and T− =
f∗(S<c ). By bijectivity and monotonicity
it follows immediately that (i) x ∈ T− and y ∈ T + implies x < y, and (ii) T− ∪T+
is equal to T (if c ∈
/ dom(f)) or T \ {f(c)} (if c ∈ domf).
Let α = sup T− , β = inf T+ and T ∗ = T− ∪ T+ . Clearly α, β ∈ J . From (i),
α ≤ β. From (ii), T ∗ is dense in J (since T is). By the density of T ∗, the interval
(α, β) ⊂ J must be empty (since it contains no points of T ∗). Therefore α = β,
concluding the proof.
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8-26
8.6.12 Proposition (Intermediate Value Theorem: Monotone Case). Let
f be a monotonic function defined on a nonempty open interval I. Assume that f
satisfies the No-Gaps condition.
Let α = inf x∈I f(x) (or α = −∞ if f is not bounded below on I) and β =
supx∈I f(x) (or β = +∞ if f is not bounded above on I). Then the range of f is
an interval with endpoints α, β.
(Note that α, β themselves may or may not be in the range.)
Proof. We will present the details of the proof only in the case when f is increasing,
the decreasing case being analogous (or obtained upon replacing f by −f). Let
J = Im(f).
By definition, α ≤ f(x) (if α 6= −∞) and f(x) ≤ β (if β 6= ∞) for all x ∈ I. If
both α, β are real, we must have α ≤ β. The result is trivial if α = β so henceforth
we will assume that either α < β, or α = −∞, or β = ∞. We will show below that
J ⊃ (α, β), whence the conclusion that J is an interval with endpoints α, β will
follow immediately (note that α, β themselves may or may not be in J , but that
does not change the conclusion that the latter set is an interval).
Let w ∈ (α, β) and
S = {x ∈ I | f(x) ≤ w},
T = {x ∈ I | f(x) > w}.
The monotonicity of f implies that S is an initial segment, and T a final segment
of I, and also that elements of S are lower bounds for T , and elements of T are upper
bounds for S. S is nonempty, because w > α implies that w is not a lower bound
for f in I (and no w is a lower bound if α = −∞). Similarly, T 6= ∅. Let c = sup S,
d = inf T . Clearly c ≤ d and sup f(S) ≤ w ≤ inf f(T ). By Proposition 8.2.12, S
is an interval ending at c and T an interval starting at d (note that c may or may
not belong to S, and d may or may not belong to T ). However, S ∪ T = I is clear
from the definitions, so we must have c = d. Now, in the notation of the proof of
Proposition 8.6.6, S contains the initial segment I<c and T the final segment I>c
of I, so sup f∗ (I<c ) ≤ sup f∗ (S) and inf f∗ (T ) ≤ inf f∗ (I>c ). By Proposition 8.6.6
and the present hypothesis we get sup f∗ (I<c ) = f(c) = inf f∗ (I>c ).
Putting everything together we get:
w ≤ inf f∗ (T ) ≤ inf f∗ (I>c ) = f(c) = sup f∗ (I<c ) ≤ sup f∗ (S) ≤ w.
Therefore all of the above are equalities and w = f(c) showing that w ∈ J . Since
w ∈ (α, β) was arbitrary, this shows that (α, β) ⊂ J and concludes the proof.
8.6.13 Proposition. Let f be a strictly monotonic function defined on a nonempty
open interval I. Assume that f satisfies the No-Gaps condition. Then:
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8-27
• The range of f is an open interval J (with endpoints given in the statement
of Proposition 8.6.12);
• f has a strictly monotonic inverse f −1 : J → I (increasing or decreasing in
agreement with f); and
• f −1 satisfies the No-Gaps condition.
Proof. By Proposition 8.6.12 the range of f is an interval J . If J has one or two
real endpoints we need to show they do not belong to J . We write the proof for
strictly increasing f only.
Say J has a left endpoint α = inf x∈I f(x). If α ∈ J then α = f(a) for some
a ∈ I. Since I is open, there exists b ∈ I with b < a. By strict monotonicity,
f(b) < f(a). Since α is the left endpoint of J then f(b) ≥ α = f(a). Contradiction.
The proof that J does not contain its right endpoint is completely analogous.
Therefore J is an open interval.
By Proposition 8.6.3, f is both injective and surjective from I to J , hence has
an inverse f −1 : J → I.
Let u, v ∈ J with u < v. Then u = f(x) and v = f(y) for some x, y ∈ J (namely,
for x = f −1 (u) and y = f −1 (v)). If y ≤ x then the monotonicity of f implies that
v = f(y) ≤ f(x) = u contradicting u < v. Hence f −1 (u) = x < y = f −1 (v) and
f −1 is strictly increasing.
Finally, let γ ∈ J and c = f −1 (γ). Since both f and f −1 are strictly increasing, I<c = (f −1 )∗(J<γ ) and I>c = (f −1 )∗ (J>γ ). Therefore inf y>γ f(y) =
inf(f −1 )∗ (J>γ ) = inf I>c = c = sup I<c = sup(f −1 )∗(J<γ ) = supy<γ f(y), so f −1
satisfies the No-Gaps condition.
8.7
Powers, Roots, Exponentials and Logarithms
8.7.1 Proposition. Existence and Uniqueness of Roots For each positive
integer n, the function f : R+ → R+ defined by f(x) = xn has an inverse f −1 :
R+ → R+ .
The function f −1 is strictly increasing and satisfies the No-Gaps condition.
8.7.2 Remark. For any y ∈ R+ the unique solution to xn = y, namely x = f −1 (y),
√
is denoted by x = y 1/n or x = n y.
The proof of Proposition 8.7.1 requires several preliminary steps.
8.7.3 Proposition. For any real b > 1 and any n ∈ N+ :
bn > 1.
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Note 8: The Real Numbers – March 13, 2014
Proof. This is an easy induction on n, the n = 1 case being the tautology b > 1 ⇒
b > 1. Assume bn > 1 for some n ∈ N+ . Multiplying both sides by the positive
real number b we obtain bn+1 > b. By the hypothesis b > 1 and transitivity we
obtain bn+1 > 1.
8.7.4 Proposition. The function f : R+ → R+ defined by f(x) = x−1 is strictly
decreasing.
Proof. Multiply both sides of x < y by y −1 x−1 to get y −1 < x−1.
8.7.5 Proposition. For any positive (resp., negative) integer n the n-th power
function f : R+ → R+ defined by f(x) = xn is strictly increasing (resp., strictly
decreasing).
Proof. Let 0 < x < y. Then x−1y > 1. If n > 0, from Proposition 8.7.3, (x−1 y)n >
1. From the ordinary laws of integer exponents valid in any Abelian group (such
as the Abelian multiplicative group R+ ) we have (x−1 y)n = (x−1 )n y n = (xn )−1 y n ;
hence (xn )−1 y n > 1. Multiplying both sides of this inequality by the positive
number xn we obtain xn < y n as desired.
For n < 0 the result follows from Proposition 8.6.2 applied to f and the function
g : x → x−1 , which is strictly decreasing by Proposition 8.7.4.
8.7.6 Proposition. For any real b > 1 (respectively, 0 < b < 1) the function f :
Z → R defined by f(k) = bk is strictly increasing (respectively, strictly decreasing).
Proof. Let m, n be integers with m < n.
First assume b > 1. By Proposition 8.7.3 with k = n−m, bn−m > 1. Multiplying
both sides by the positive number bm and using the ordinary laws of exponents we
get bn > bm as desired.
Now assume 0 < b < 1. Then bm−n = (b−1 )n−m > 1 by proposition 8.7.3
applied to b−1 > 1. Multiplying by the positive number bn we obtain bm > bn as
desired.
8.7.7 Proposition. Let a, b be commuting elements in a ring (that is, ab = ba).
For any n ∈ N+ :
bn − an = (b − a)(bn−1 + bn−2 a + · · · + an−1 ).
In Sigma notation the above reads
n
n
b − a = (b − a)
(Note that
P1
k=1
n
X
bn−k ak−1 .
k=1
b1−k ak−1 = 1, the unity element of the ring.)
Note 8: The Real Numbers – March 13, 2014
8-29
Proof. Easy exercise on induction and the laws of exponents safely left to the
reader.
Proof. We are ready to prove Proposition 8.7.1. We show that the function
f : R+ → R+ , x 7→ xn , is strictly increasing, surjective, and satisfies the No-Gaps
condition. By Proposition 8.6.13 and the definition of inverse function, it will follow
that xn = f(x) = y has the unique solution x = f −1 (y).
f is strictly increasing by Proposition 8.7.5.
We claim that α = inf x>0 f(x) = 0. Clearly α ≥ 0. For any > 0 take
x < min{1, }. By strict monotonicity of f we have f(x) < f(1) = 1, and by
Proposition 8.7.6 we have α ≤ f(x) = xn ≤ x < . Since this holds for all > 0 we
have α = 0.
Now we claim that f is unbounded above. For any y > 0 take x > max{1, y}.
By strict monotonicity of f we have f(x) > f(1) = 1 and by Proposition 8.7.6 we
have f(x) = xn ≥ x > y. Since this holds for all y > 0, f is unbounded above, so
in the notation of Proposition 8.6.12, β = +∞. By the same proposition, Ran(f)
is the open interval (0, ∞), so f is surjective.
Finally, we show that f satisfies the No-Gaps condition.
Let > 0 be arbitrary. Choose x, y such that x < c < y < c + 1 and
y−x<
.
n(c + 1)n−1
Then
n
n
0 < y − x = (y − x)
≤ (y − x)
= (y − x)
n
X
k=1
n
X
y n−k xk−1
by Proposition 8.7.7,
y n−k y k−1
since y > x,
y n−1
laws of exponents,
k=1
n
X
k=1
≤ (y − x) · n(c + 1)n−1
<
since y < c + 1, by Proposition 8.7.5,
by the assumed upper bound on y − x.
Then inf (c,∞) f − sup(0,c) f ≤ f(y) − f(x) = y n − xn < . Since > 0 was arbitrary,
we conclude that inf (c,∞) f ≤ sup(0,c) f. This concludes the proof that f satisfies
the No-Gaps condition.
8.7.8 Corollary.
For positive real numbers a and b and positive integer n,
(ab)1/n = a1/n b1/n.
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Note 8: The Real Numbers – March 13, 2014
Proof. (The reader is invited to provide a formal proof by induction based on the
following informal argument.)
Define c = a1/n and d = b1/n. We have, by commutativity,
· · d} = (cd) · · · (cd) = (cd)n .
ab = cn dn = |c ·{z
· · }c |d ·{z
|
{z
}
n
n
n
Since the nth root is unique,
(ab)1/n = cd = a1/nb1/n .
8.7.9 Corollary.
For positive real a and positive integers m and n, we have
(am )1/n = (a1/n)m .
Proof. (The reader is invited to provide a formal proof by induction based on the
following informal argument.)
Thanks to Corollary 8.7.8, we have
1/n
(am )1/n = (a
· · a})1/n = (a
· · a1/n}) = (a1/n)m .
| ·{z
| ·{z
m
m
8.7.10 Remark. For positive real b and rational r = m/n > 0 it is tempting to
define br = (bm )1/n . Before we do so we must verify that for m/n = p/q we have
(bm )1/n = (bp )1/q , i.e.: the definition of br is independent of the representative we
use for r. This we do below.
8.7.11 Corollary.
m/n = p/q; then
If b is a positive real; m, n, p, q are positive integers; and
(bm )1/n = (bp)1/q .
Proof. If m/n = p/q then mq = pn. Hence, bmq = bpn , giving
(bm )q = (bp )n .
(8.4)
The functions defined by b 7→ bn and b 7→ b1/n are inverses of each other, i.e.: for
real positive a
(8.5)
(ak )1/k = (a1/k )k = a.
From (8.4) and (8.5) we obtain
((bm )1/n )n )q = (((bp )1/q )q )n .
Hence,
((bm )1/n )nq = (((bp )1/q )qn .
Therefore, by the uniqueness of roots, the above yields (bm )1/n = (bp)1/q .
Note 8: The Real Numbers – March 13, 2014
8-31
8.7.12 Definition. In view of the above result, for any x > 0 and for m, n > 0 we
define xm/n = (xm )1/n, noting that this definition does not depend on the choice of
m and n, but only on the value r = m/n. This defines xr for all rationals r > 0.
For rationals r < 0, we define xr = (x−r )−1 = 1/x−r . Finally, we define x0 = 1 for
all x > 0.
8.7.13 Proposition. For each rational number r > 0 (resp., r < 0) the rth -power
function f : R+ → R+ , x 7→ xr , is strictly increasing (resp., strictly decreasing).
Proof. Write r = m/n for some nonzero integers m, n with n > 0. Then, by
definition, f = h ◦ g where g(x) = xm and h(y) = y 1/n. By Proposition 8.7.1, h is
strictly increasing, whereas by Proposition 8.7.5, g is strictly increasing if m > 0
(resp., strictly decreasing if m < 0). The assertion follows from Proposition 8.6.2.
8.7.14 Proposition. For positive real a, b and rational r and s, ar br = (ab)r ,
br+s = br bs and brs = (br )s .
Proof. Write r = m/n and s = p/q with integers m, n, p, q and n, q positive. We
have
(br+s )nq =
=
=
=
=
=
=
(bm/n+p/q )nq
(b(mq+np)/(nq))nq
bmq+np
bmq bnp
(bm/n )nq (bp/q )nq
(bm/n bp/q )nq
(br bs )nq .
From the above and the uniqueness of roots, we obtain br+s = br bs .
Note 8: The Real Numbers – March 13, 2014
8-32
Similarly,
(brs )nq = (bm/n·p/q )nq
= (b(mp)/(nq))nq
= bmp
= (bm )p
= ((bm/n )n )p
= (bm/n )pn
= ((bm/n )p )n
= (((bm/n )p/q )q )n
= ((bm/n )p/q )nq
= ((br )s )nq .
From the uniqueness of roots, we obtain br+s = (br )s .
Finally,
(ar br )n = (ar )n (br )n
= anr bnr
= a m bm
= (ab)m
= ((ab)m/n )n
= ((ab)r )n .
From the uniqueness of roots, we obtain ar br = (ab)r .
We wish to extend the definition of the exponential functions x 7→ bx to arbitrary real exponents x. This is accomplished by proving monotonicity and the
No-Gaps condition together with a general result extending functions satisfying the
No-Gaps condition from a dense subset of an interval to the whole interval. First
we prove strict monotonicity.‡
8.7.15 Proposition. Let b > 1 (resp., 0 < b < 1). Let f : Q → R+ be defined by
f(x) = bx . Then f is strictly increasing (resp., strictly decreasing).
Proof. We first assume b > 1. Let x, y ∈ Q, x < y. Write x = m/n, y = p/q for
integers m, n, p, q with n, q > 0. Then mq < np. By Proposition 8.7.6, bmq < bnp .
1
By the strict monotonicity of t 7→ t nq (Proposition 8.7.1) and the laws of exponents
mq
np
1
1
(Proposition 8.7.14): bx = bm/n = b nq = (bmq ) nq < (bnp ) nq = b nq = bp/q = by .
‡
Note that Proposition 8.7.15 generalizes 8.7.6.
Note 8: The Real Numbers – March 13, 2014
8-33
The result for 0 < b < 1 follows similarly, or can be derived from the above
using the identity bx = (b−1 )−x .
As a first step towards proving that f satisfies the No-Gaps condition, we prove
a very useful inequality.
8.7.16 Proposition.
(8.6)
Bernoulli’s Inequality: If t ≥ −1 and n ∈ N then
(1 + t)n ≥ 1 + nt.
(For t = −1 and n = 0 we interpret 00 as 1.)
Equality holds if and only if t = 0 or n = 0, 1.
Proof. We prove the inequality by induction on n, for any fixed t > −1. For
n = 0 we have (1 + t)0 = 1 = 1 + 0t. Suppose that (8.6) holds for some n ∈ N. We
have
(1 + t)n+1 = (1 + t)n (1 + t)
≥ (1 + nt)(1 + t)
= 1 + (n + 1)t + nt2
≥ 1 + (n + 1)t,
which is (8.6) with n replaced by n + 1. If both n ≥ 1 and t 6= 0, the last inequality
is strict. Hence, we have shown that (8.6) holds for all n ∈ N, with strict inequality
if t 6= 0 and n > 1. Since equality clearly holds when t = 0 or n = 0, 1, the proof
is complete.
8.7.17 Proposition. Let b be a positive real number, b 6= 1. Let f : Q → R+ be
defined by f(x) = bx . Then f satisfies the No-Gaps condition on R.
Proof. It follows from the identity bx = (b−1 )−x that the statement for 0 < b < 1
follows from that for b > 1, so we assume henceforth the latter holds.
From Bernoulli’s Inequality (8.6) with t = (b − 1)/n, (1 + b−1
)n ≥ b, so by the
n
strict monotonicity of n-th roots:
(8.7)
1+
b−1
≥ b1/n .
n
Let c ∈ R be arbitrary. For any > 0, choose n ∈ N such that n > −1 bc+1 .
Now choose rational numbers x, y such that x < c < y and y − x < 1/n. Then
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Note 8: The Real Numbers – March 13, 2014
by−x < b1/n by strict monotonicity (Proposition 8.7.15). Now,
by − bx = bx (by−x − 1)
< bc (b1/n − 1)
b−1
c
≤b
n
c+1
b
<
n
< .
by (8.7)
Since > 0 was arbitrary, this shows that inf y>c by − supx<c bx ≤ 0, so f satisfies
the No-Gaps condition.
We are ready to extend the definition of bx to irrational exponents x via the
following general result about extension of monotonic functions satisfying the NoGaps condition.
8.7.18 Proposition. Let I ⊂ R be a nonempty open interval, S a dense subset
of I, and f : S → R a monotonic function satisfying the No-Gaps condition (on I).
Then the function f¯ : I → R defined by
(
if f is increasing,
¯ = supx<c f(x) = inf x>c f(x)
f(c)
inf x<c f(x) = supx>c f(x)
if f is decreasing
extends f to a monotonic function satisfying the No-Gaps condition. Moreover, f¯
is increasing (resp., decreasing) if f is (and strictly so if f is).
8.7.19 Remark. In Proposition 8.7.18 above, is essential that I be an open interval. The function x 7→ 1/x is monotonically decreasing on S = (0, +∞), but does
not extend to I = [0, +∞) although S is dense in I.
Proof. We will assume that f is increasing, leaving the formal modifications needed
to treat the case of decreasing f to the reader.
The function f¯ is well defined and extends f by Propositions 8.6.6 and 8.6.8.
For any x, z ∈ I with x < z the density of S allows choosing y ∈ S such that
x < y < z. Then monotonicity of f implies that f¯(x) = inf f∗ (Sx+ ) ≤ f(y) ≤
¯
sup f∗ (Sz− ) = f(z)
(with inequalities strict if f is strictly monotonic), so f¯ is
monotonically increasing (and strictly so if f is).
To show that f satisfies the No-Gaps condition, it suffices to observe that,
for each c ∈ I, since f¯ extends f and S<c ⊂ I<c , sup f∗ (S<c ) ≤ sup f¯∗ (I<c );
similarly inf f∗ (S>c ) ≥ inf f¯∗ (I>c ). By the No-Gaps condition for f it follows that
¯
sup f¯∗ (I<c ) ≥ inf f¯∗ (I>c ), proving the No-Gaps condition for f.
Note 8: The Real Numbers – March 13, 2014
8-35
8.7.20 Definition. Combining Propositions 8.7.15, 8.7.18, and 8.6.13 we immediately obtain, for each fixed positive real number b 6= 1 (called the base), functions
expb : R → R+ (the exponential function base b) and logb : R+ → R (the logarithm
function base b) satisfying the following properties:
• expb and logb are mutually inverse, strictly monotonic functions (increasing,
for b > 1; decreasing, for 0 < b < 1) satisfying the No-Gaps condition, and
• expb (x) = bx for all x ∈ Q.
Henceforth we use the notation bx as an alternative to expb (x). (By Proposition 8.7.18, expb (x) = bx for all x ∈ Q, so no confusion should arise.)
We also define, 1x = 1 for all x ∈ R.
Note that, since logb is inverse to expb , the only solution x to the equation
y = bx is x = logb (y), and conversely.
8.7.21 Proposition (Laws of real exponents).
real x, y we have
For positive real a, b and all
bx+y = bx by ,
(bx )y = bxy , and
(ab)x = ax bx.
Proof. All of the above statements are obvious if a = 1, b = 1, x = 0, or y = 0, so
we assume henceforth a, b 6= 1 and x, y 6= 0.
We prove the first identity above for b > 1, 0 < b < 1 following similarly.
For now, let S = Q. Let also x, y ∈ R be arbitrary. By definition of the
exponential function and monotonicity (Proposition 8.6.7), bx = sup f(Sx− ) and
by = sup f(Sy− ). By Proposition 8.7.14, for all u ∈ Sx− , v ∈ Sy− we have u + v ∈
S(x+y)− and bu bv = bu+v . By Exercise 8.8, the supremum of all such u + v is x + y.
By Exercise 8.14, the supremum of all bu bv is bx by . Therefore, bxby = bx+y .
(This is one of the few instances we will meet when an identity is proved in
one step; however, note that we needed two auxiliary propositions: one for sums,
another one for products. With the help of just one of those propositions, we would
have proved an inequality in one direction, and the reverse direction could have
been obtained using infima instead of suprema. We invite the reader to ponder
that, in the absence of the auxiliary propositions of Exercises 8.8 and 8.14, no
inequality could be proved.)
Now we prove the second identity assuming b > 1 and x, y > 0. Let S = Q+ .
Let x, y ∈ R+ , u ∈ Sx− and v ∈ Sy− . By the laws of exponents 8.7.14, buv = (bu )v .
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Note 8: The Real Numbers – March 13, 2014
We clearly have supv∈Sy− supu∈Sx− (uv) = supv∈Sy− (xv) = xy. Since the set
of such products uv (resp., products xv) is clearly dense in (0, xv) (resp., dense
in (0, xy)), the monotonicity of expb (Proposition 8.6.7) implies that
sup
sup buv = sup bxv = bxy .
v∈Sy −
u∈Sx−
v∈Sy −
On the other hand, by the definition of the exponential function expb we have
supu∈Sx− bu = bx , so by the monotonicity of the v th-power function (Propositions 8.7.13), we have supu∈Sx− (bu )v ≤ (bx )v , by Proposition 8.6.6. By the same
Proposition and the obvious observation bx > bu > 1, the monotonicity of expbx
implies that supv∈Sy− (bx )v = (bx )y . Altogether we obtain
sup sup (bu )v ≤ sup (bx)v = (bx )y .
v∈Sy − u∈Sx−
v∈Sy −
Combining the results of the two paragraphs above, we obtain bxy ≤ (bx)y .
The reverse inequality follows by working with infima instead. This completes the
proof of the second identity in the case b > 1 and x, y > 0. The case 0 < b < 1 is
completely analogous.
At this point we note that the special case y = −1 of the second identity,
namely the identity b−x = (bx )−1 , actually follows from the first, already proved
identity by putting y = −x. We also have the special case (b−1 )y = b−y , which
easily follows taking suprema (if b < 1) or infima (if b > 1) over all rationals v < y
in the identity (b−1 )v = b−v . The second identity now follows in full generality (for
x, y 6= 0) by writing x = ±s and y = ±t with s, t > 0 and combining the two
special cases above with the identity (already proved) applied to s, t.
We now prove the third identity assuming a, b > 1 for every x ∈ R. Recall
that (ab)u = au bu for every rational u. Since ab > 1, by definition of expab we have
supu∈Qx− (ab)u = (ab)x. From the definition of supremum we also have au ≤ ax
and bu ≤ bx for all u ∈ Q− , so au bu ≤ axbx and supu∈Qx− (au bu ) ≤ ax bx . We
conclude that (ab)x ≤ axbx and the reverse inequality follows by working with
infima taken over Qx+ , so the third inequality is established for a, b > 1. The cases
when a < 1 or b < 1 are obtained by first applying the identity with a−1 replacing
a or b−1 replacing b (respectively, which we can do by the case already proved) and
subsequently using the special cases x = −1, y = −1 of the second identity.
8.7.22 Proposition (Laws of logarithms).
b 6= 1, all a ∈ R and all x, y > 0 we have
For any fixed positive real base
logb (xy) = logb (x) + logb (y),
logb (xa) = a logb (x).
and
Note 8: The Real Numbers – March 13, 2014
8-37
Proof. By the laws of real exponents and the mutually inverse relation between
expb and logb ,
expb (logb (x) + logb (y)) = expb (logb x) expb (logb y) = xy.
Applying logb to both sides above we obtain the first identity claimed in the statement.
Similarly,
expb (a logb x) = ba logb x = (blogb x )a = xa .
The second identity now follows applying logb to both sides.
8.7.23 Definition. Fix a real number κ 6= 0. Define the κth -power function pow κ :
R+ → R+ by:
pow κ (x) = expx (κ) = xκ .
(Recall that 1κ = 1 for all κ ∈ R.) Then:
8.7.24 Proposition.
• powκ is a strictly monotonic function (increasing for
κ > 0, decreasing for κ < 0);
• powκ is bijective, with inverse pow1/κ ; and
• powκ satisfies the No-Gaps condition.
Of course the usual notation for pow κ (x) is simply xκ .
Proof. We assume κ > 0, leaving the other case to the reader. First we prove powκ
is monotonic. We assume κ > 0 leaving κ < 0 to the reader. Let now 0 < x < y
and z = y/x > 1. By the (strictly increasing) monotonicity of the v th-power map
for rational v > 0 we have z v > 1v = 1, so z κ > 1 follows. Multiplying both
sides of the inequality by the positive number xκ and using the laws of exponents
(Proposition 8.7.21) we get xκ < y κ , proving that expκ is strictly increasing for
κ > 0.
By the laws of exponents, (x1/κ )κ = x = (xκ )1/κ , so pow κ is bijective, with
inverse pow1/κ .
Finally, pow κ satisfies the No-Gaps condition by Proposition 8.6.11.
8.8
Sequences and Numerals
8.8.1 Definition. Sequences A sequence is simply an element of a (finite or
infinite) Cartesian product. Below are the details.
Note 8: The Real Numbers – March 13, 2014
8-38
For any set S and k ∈ N a function f : k → S is called a finite sequence in S.
If we define an = f(n) for every§ n ∈ k = {0, 1, . . . , k − 1}, we denote the finite
k−1
sequence by {an }n=0
, or {a0, a1, . . . ak−1 }. The values a0, a1, . . . ak−1 , which need
not be distinct, are called the terms of the finite sequence.
For any set S a function f : N → S is called an infinite sequence in S. In other
words, S N is the collection of all infinite sequences in S. If we define an = f(n), we
denote the infinite sequence by {an }∞
n=0 , or {an }, or {a0 , a1 , a2 , . . .}. The values a0 ,
a1, a2 ,. . . , which need not be distinct, are called the terms of the infinite sequence.
The term sequence is used to refer to a finite or infinite sequence. The notation
k−1
is ambiguous: {an }n=0
, {a0 , a1, . . . ak−1 }, {an }∞
n=0 , {an }, {a0 , a1 , a2 , . . .} can denote
a sequence or the range of a sequence. Context tells which is meant. This notation
k−1
is used for historical reasons. Some use the notation (an )n=0
, (a0, a1, . . . ak−1 ),
∞
(an )n=0 , (an ), (a0, a1, a2, . . .) instead.
8.8.2 Proposition. Division Algorithm Given integers m and n, with n ≥ 1,
there exist unique integers q (the quotient) and r (the remainder ), with 0 ≤ r < n,
such that m = nq + r.
Proof. Define the set S = {m − kn ∈ N | k ∈ Z}. Since n ≥ 1, it follows that
|m| n ≥ |m|. Hence,
m − (− |m|)n = m + |m| n ≥ m + |m| ≥ 0.
Therefore, by choosing k = − |m| we obtain m − kn ≥ 0, i.e.: m − kn ∈ S, showing
that S is not empty.
Thus, S is a nonempty subset of N (all the members of S are nonnegative).
Thank to the well ordering principle, S has a least element r. Thanks to the
definition of S, we have r = m − qn for some q ∈ Z.
We show that r < n. Suppose the contrary. Then r ≥ n gives
m − (q + 1)n = (m − qn) − n = r − n ≥ 0,
which says that m − (q + 1)n ∈ S. Hence m − (q + 1)n = r − n < r contradicting
the minimality of r in S. We conclude that 0 ≤ r < n.
We prove the uniqueness of q and r. Suppose that
m = qn + r
and m = q 0n + r0 ,
with 0 ≤ r < n and 0 ≤ r0 < n. Then, we have r − r0 = n(q − q 0). Hence, recalling
that n ≥ 1,
(1)
§
|r − r0 | = n |q − q 0| .
Recall that zero in N is simply the empty set and k = (k − 1) ∪ {k − 1} for k ∈ N, k 6= 0.
Hence, we have 1 = ∅ ∪ {∅} = {∅} = {0}, 2 = 1 ∪ {1} = {0} ∪ {{0}} = {0, {0}} = {0, 1},
3 = 2 ∪ {2} = {0, 1} ∪ {{0, 1}} = {0, 1, {0, 1}} = {0, 1, 2} and so on.
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Note 8: The Real Numbers – March 13, 2014
We add the inequalities −n < −r ≤ 0 and 0 ≤ r0 < n to obtain −n < r0 − r < n,
which implies
|r0 − r| < n.
(2)
Thanks to (1) and (2) we have
n |q − q 0| = |r − r0 | < n,
which implies |q − q 0| < 1. Hence, since |q − q 0| ∈ N, we have q = q 0. Then, with
q = q 0 in (1) we obtain r = r0 .
8.8.3 Definition (Base Expansions for Integers). Let b be an integer greater
than 1 (the base). Let m ∈ Z. First consider the case m > 0. By repeated
applications of the division algorithm we have
m = m1b + n0 ,
m1 = m2b + n1 ,
..
.
mk−1 = mk b + nk ,
with mj = 0 for all j ≥ k; 0 ≤ n0 < b, . . . , 0 ≤ nk−1 < b; and 0 < nk < b. This
process yields
m = nk bk + · · · + n1 b + n0 .
The sequence {nk , . . . , n1 , n0 } is called the base-b representation of m, or base-b
numeral representing m, and we usually write
(8.8)
+(ñk . . . ñ1 ñ0 )b
to denote (or name) the number m =
k
X
nj bj
j=0
(the leading lexical symbol ‘+’ and the parentheses are optional in practice; the
suffix b may also be omitted if it is ten or otherwise clear from context), where ñi
stands for a lexical symbol (called a numeral in this context) representing ni , and
where the juxtaposition in (8.8) is purely lexical and does not indicate product.
For example, 12710 does not mean 1 · 2 · 7, but names the number 1 · 102 + 2 · 10 + 7,
so that 12710 , 1778 , 11111112 and 7F16 all name the same number, since
1 · 102 + 2 · 10 + 7 = 1 · 82 + 7 · 8 + 7
= 1 · 26 + 1 · 25 + 1 · 24 + 1 · 23 + 1 · 22 + 1 · 2 + 1
= 7 · 16 + 15.
Note 8: The Real Numbers – March 13, 2014
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(The numerals A, B, C, . . . , F, . . . denote ten, eleven, twelve, . . . , fifteen, . . . .) The
numbers n0 , n1 , . . . , nk are called the base-b digits of m. Note that any base-b
digit is an integer n such that 0 ≤ n < b. The subscript indicating the base is
frequently omitted when b is ten. Note that k is characterized by the inequalities
bk ≤ m < bk+1 . By the strict monotonocity of logb and Notation 8.5.7, it follows
that k = blogb mc.
The integer m = 0 is represented by the single numeral 0̃ = +(0̃)b . We may let
k = 0 in this case.
For m < 0, with the representation of |m| given in (8.8), the representation of
m is obtained replacing the symbol ‘+’ by ‘−’.
8.8.4 Definition. Let b be a base (an integer greater than 1). Let x be a nonnegative real. Let m = bxc be the largest m ∈ N such that m ≤ x. For this m, define
k and nk , nk−1 , . . . , n0 as in Definition 8.8.3 above, and

n

n−1 = largest n ∈ N such that m + ≤ x,


b


n−1
n



+ 2 ≤ x,
n−2 = largest n ∈ N such that m +


b
b

.
..
(8.9)



n−1
n−j+1
n


n−j = largest n ∈ N such that m +
+ · · · + −j+1 + −j ≤ x,



b
b
b



..
. .
Note that each n−j exists since the set of n ∈ N whose largest element is defined
to be n−j always contains n = 0 but no element n ≥ b. Therefore n−j is a base-b
digit for j = 1, 2, . . . .
We note that for arbitrary x > 0, the first (that is, largest) j such that nj 6= 0
is still characterized by the inequality bj ≤ x < bj+1 , hence its value is blogb xc.
Let S denote the collection of all such sums:
n−1
n−j
S = sj = m +
+ · · · + j ∈ R j ∈ N ∧ j > 0 ∧ (8.9) holds .
b
b
We have sj ≤ x < sj + b−j , so clearly x = sup S. It follows that x both determines
and is in turn uniquely determined by the sequence of its base-b digits
{nk , . . . , n1 , n0 , n−1 , n−2 , . . .}.
We usually write that sequence of digits as the base-b expansion of x
(8.10)
+(ñk . . . ñ1 ñ0 . ñ−1 ñ−2 . . .)b ,
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Note 8: The Real Numbers – March 13, 2014
which is a lexical string of numerals conventionally denoting x (where the leading
lexical symbol ‘+’ and the parentheses are optional in practice).¶ The case x < 0
is handled replacing ‘+’ by ‘−’ in the base-b expansion of |x|.
We call any lexical string E as in (8.10) (possibly prefixed by a minus symbol
instead) a (lax) base-b expansion.
By convention we shall call a base-b expansion nonpositive if it is prefixed
by the minus symbol, otherwise it is nonnegative. The positive expansions are the
nonnegative ones except +(0̃.0̃0̃0̃ . . . )b . The negative expansions are the nonpositive
ones except −(0̃.0̃0̃0̃ . . . )b .
Given a base-b expansion E, let
(8.11)
sj =
k
X
ni bi
for j = 0, 1, 2, . . . ,
i=−j
and let S = {sj | j ∈ Z≤k }—obviously a monotonically increasing sequence. The
base-b expansion is said to denote (or represent, or name, or have the value of ) the
real number
(
sup S
if the expansion E is nonnegative;
x=
− sup S if the expansion E is nonpositive.
While we have hitherto tacitly assumed k ≥ 0, we may as well take k as the largest
j such that nj 6= 0 (perhaps k < 0). For the base-b representations ±(0̃.0̃0̃0̃ . . . )b
we may regard k as being minus infinity and sj as the empty sum with value 0 for
all j.
A priori we do not know whether every base-b expansion denotes a real number x. Moreover, we have shown that every real number is denoted by some base-b
expansion, but we do not know if that expansion is unique. In fact, ±(0̃.0̃0̃0̃ . . . )b
are two (lexically) different representations of the real number 0, so expansions are
not always unique.
We call a base-b expansion strict if it is not −(0̃.0̃0̃0̃ . . . )b and its digits {nj |
j ∈ Z≤k } have the following property:
(∀j ∈ Z)(∃i < j)(ni 6= b − 1).
¶
In Note 10 we will study infinitely long sums (series) and we shall be able to interpret
(ñk . . . ñ1 ñ0 . ñ−1 ñ−2 . . .)b
as
k
X
nj b j .
j=−∞
For the time being the reader may just take sup S as the definition of the series above.
Note 8: The Real Numbers – March 13, 2014
8-42
In other words, an expansion is strict if not all its digits are b − 1 from any point
on. For instance, in base ten,
+(0.999 . . . )10
and
+ (1.000 . . . )10
both name the number 1. The second expansion is strict, but the first one is not.
8.8.5 Remark. It is easy to see that digits chosen according to (8.9) give a strict
expansion of the nonnegative real number x.
8.8.6 Theorem (Existence and uniqueness of base-b expansions). Fix an
integer base b > 1. The relation “E denotes x” is a bijection between strict base-b
expansions E and real numbers x.
A real number x is denoted by exactly two different base-b expansions if and
only if an integer l exists such that bl x is an integer. Otherwise x is denoted by
exactly one base-b expansion, necessarily strict.
Proof. We already proved the existence of a base-b expansion with digits chosen
according to (8.9).
We show that any base-b representation denotes certain real number x. It
suffices to check that the sequence {sj } with sj given by (8.11) is bounded above.
On the one hand, we clearly have
sj ≤
k
X
i=−j
(b − 1)bi = bk+1 − b−j ,
so the sequence {sj } is bounded above by bk+1 , hence its supremum x exists. In
fact, sincePthe sequence {sj } is monotonically increasing as j decreases and also
j−1
si ≤ sj + l=i (b − 1)bl = sj + bj − bi , it follows that sj + bj is an upper bound for
{si }, hence
(8.12)
sj ≤ x ≤ sj + b j
for all integers j. Equality holds on the left (resp., on the right) if and only if
ni = 0 (resp., ni = b − 1) for all i < j.
We claim that two base-b representations
E : +(ñk . . . ñ1ñ0 . ñ−1 ñ−2 . . .)b
and
E0 : +(ñ0k . . . ñ01ñ00 . ñ0−1 ñ0−2 . . .)b
agreeing down to but excluding the j-th position, that is, such that ni = n0i for all
i > j and (for the sake of definiteness) with nj > n0j , the inequalities
(8.13)
(nj − n0j − 1)bj ≤ x − y ≤ (nj − n0j + 1)bj
Note 8: The Real Numbers – March 13, 2014
8-43
hold, with equality on the left (resp., on the right) if and only if ni = 0 and
n0i = b − 1 (resp. ni = b − 1 and n0i = 0)
all i < j.
Pfor
Pk
k
In fact, for the partial sums sj = i=j ni bi of E and s0j = i=j n0i bi of E0, we
have, by (8.12),
x − y = (x − sj ) + (sj − s0j ) − (y − sj )
≥ 0 + (nj − n0j )bj − bj
= (nj − n0j − 1)bj ,
with equality if and only if y − s0j = bj and x = sj , that is, ni = 0 and n0i = b − 1 for
all i < j. This proves the left inequality in (8.13) and justifies the case of equality;
the analysis of the right inequality is similar and omitted.
Assuming x = y and E different from E0, both agreeing down to but excluding
the j-th position as above, it follows from (8.13) that nj − n0j − 1 ≤ 0; since also
n0j < nj by assumption, we conclude nj = n0j + 1. Therefore the left inequality
in (8.13) is an equality and we conclude that E is a strict base-b representation and
E0 a lax but not strict representation, both of x, and E0 is obtained decreasing the
rightmost nonzero numeral of E by one, and replacing all subsequent zero numerals
of E by the numeral denoting b − 1.
Clearly, positive expansions denote positive numbers, and negative expansions
denote negative numbers. As shown above, every positive number is denoted by a
strict positive expansion E, but some are also denoted by a non-strict one E0. This
also obviously applies to negative numbers and their expansions.
The number zero has two expansions differing only by their syntactic sign, as
noted in Definition 8.8.4. Note that 0 = 1b0 is an integer. Every nonzero real x
having two distinct expansions will have the same sign as either of them. Denote
by l the last nonzero position of the strict expansion E of x. Then 2−l x ∈ Z.
Conversely, if a real x satisfies 2l ∈ Z, we have either x = 0 (which has both a
proper and an improper expansion, or else x 6= 0. In the latter case n0i = 0 for
all digits with i < −l in the strict expansion of x, but nj 6= 0 for some j ≥ −l.
Let j the least such, so nj is the right-most nonzero digit of the expansion of x.
Then x also has the non-strict expansion E0 obtained from E as above. The proof
is complete.
8.8.7 Corollary. Every real number possessing a non-strict base-b expansion to
some base b is rational.
Proof. By Theorem 8.8.6, any such number is of the form b−l n for some integer n,
hence is rational.
8.22 Exercise. Prove that there exist countably many numbers in [0, 1] having a
nonstrict binary (base-2) expansion.
Note 8: The Real Numbers – March 13, 2014
(A significant strengthening of this Exercise is given in Corollary 9.2.12.)
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