Lakhasly

The aim of this chapter is to introduce axiomatically the set of Real numbers.W. Rudin Principles of Mathematical Analysis.Z+ with b < a. In order to make this equation soluble we have to enlarge the set Z+ by introducing negative integers as unique solutions of the equations a + x = 0 (existence of additive inverse) 28 CHAPTER 2.N. This necessitates adding {0} to N, declaring 0 < 1, thereby obtaining the set of non-negative integers Z+.(Fundamental theorem of arithmetic) Every positive integer except 1 can be expressed uniquely as a product of primes.N. Then (a b

c d ) <==> ( ad > bc) . The following theorem provides a very important property of rationals. Theorem 2.1.2. Between any two rational numbers there is another (and, hence, infinitely many others).Z. In order to solve (2.1.1) (for a = 0) we have to enlarge our system of numbers again so that it includes fractions b/a (existence of multiplicative inverse in Z - {0}).Those of you familiar with basic concepts of algebra will find that axioms A.1 - A.11 characterize R as an algebraic field.N. Our extended system, which is denoted by Z, now contains all integers and can be arranged in order Z = {.Hence, appealing to the Fundamental Theorem of Arithmetic, p 2 is even, and hence p is even.Multiplying this equality by n 2 we obtain m3 n = mn + 7n 2 , which is impossible since the right-hand side is an integer and the left-hand side is not.2.2 The Field of Real Numbers In the previous sections we discussed the need to extend N to Z, and Z to Q. The rigorous construction of N can be found in a standard course on Set Theory.The last axiom links the operations of summation and multiplication. The set of rationals Q also forms an algebraic field (that is, the rational numbers satisfy axioms A.1 - A.11).(order) The first difficulty occurs when we try to come up with the additive analogue of a.1 = 1.a = a for a ?Indeed, since b, d and m are positive we have [a(b + md) > b(a + mc)] <==> [mad > mbc] <=> (ad > bc), and [d(a + mc) > c(b + md)] <=> (ad > bc).Suppose for a contradiction that the rational number p q (p ? Z, q ? N, in lowest terms) is such that ( p q ) 2 = 2.In this course we postulate the existence of the set of real numbers R as well as basic properties summarized in a collection of axioms.R) [(ab)c = a(bc)] (associativity of multiplication).We also mention at this point the Fundamental theorem of arithmetic.Z. The equation (2.1.1) ax = b need not have a solution x ?Here hcf(p, q) stands for the highest common factor of p and q, so when writing p/q for a rational we often assume that the numbers p and q have no common factor greater than 1.All the arithmetical operations in Q are straightforward.Q and consider the equation (2.1.2) x 2 = a. In general (2.1.2) does not have rational solutions.The last statement contradicts our assumption that p and q have no common factor.The last theorem provides an example of a number which is not rational.No rational x satisfies the equation x 3 = x + 7.First we show that there are no integers satisfying the equation x 3 = x + 7.For a contradiction suppose that there is. Then x(x + 1)(x - 1) = 7 from which it follows that x divides 7.Direct verification shows that these numbers do not satisfy the equation.Second, show that there are no fractions satisfying the equation x 3 = x+7.Theorem 2.1.1..1 = 1 ....{0} ?!!!!

The aim of this chapter is to introduce axiomatically the set of Real numbers.
2.1 Various Sorts of Numbers
2.1.1 Integers
We take for granted the system N of natural numbers
N = {1, 2, 3, 4 . . .}
stressing only that for N the following properties hold.
(∀ a ∈ N)(∀ b ∈ N)(∃ c ∈ N)(∃ d ∈ N)[(a + b = c) ∧ (ab = d)].
(closure under addition and multiplication)
(∀ a ∈ N)[a · 1 = 1 · a = a].
(existence of a multiplicative identity)
(∀ a ∈ N)(∀ b ∈ N)[(a = b) ∨ (a < b) ∨ (a > b)].
(order)
The first difficulty occurs when we try to come up with the additive analogue of a·1 = 1·a = a
for a ∈ N. Namely, there is no x ∈ N : a + x = a, for all a ∈ N. This necessitates adding {0}
to N, declaring 0 < 1, thereby obtaining the set of non-negative integers Z+. Still, we cannot
solve the equation
a + x = b
for x ∈ Z+ with b < a. In order to make this equation soluble we have to enlarge the set Z+
by introducing negative integers as unique solutions of the equations
a + x = 0 (existence of additive inverse)
28
CHAPTER 2. NUMBERS
for each a ∈ N. Our extended system, which is denoted by Z, now contains all integers and
can be arranged in order
Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . .} = N ∪ {0} ∪ {−a | a ∈ N}.
We also mention at this point the Fundamental theorem of arithmetic.
Theorem 2.1.1. (Fundamental theorem of arithmetic) Every positive integer except 1
can be expressed uniquely as a product of primes.
This is proved in the Unit Number Theory and Group Theory.
2.1.2 Rational Numbers
Let a ∈ Z, b ∈ Z. The equation
(2.1.1) ax = b
need not have a solution x ∈ Z. In order to solve (2.1.1) (for a ̸= 0) we have to enlarge our
system of numbers again so that it includes fractions b/a (existence of multiplicative inverse
in Z − {0}). This motivates the following definition.
Definition 2.1.1. The set of rational numbers (or rationals) Q is the set
Q =
{
r =
p
q
: p ∈ Z, q ∈ N, hcf(p, q) = 1}
.
Here hcf(p, q) stands for the highest common factor of p and q, so when writing p/q for a
rational we often assume that the numbers p and q have no common factor greater than 1.
All the arithmetical operations in Q are straightforward. Let us introduce a relation of order
for rationals.
Definition 2.1.2. Let b ∈ N, d ∈ N. Then
(a
b

c
d
)
⇐⇒ (
ad > bc)
.
The following theorem provides a very important property of rationals.
Theorem 2.1.2. Between any two rational numbers there is another (and, hence, infinitely
many others).
Proof. Let b ∈ N, d ∈ N, and
a
b

c
d
.
Notice that
(∀m ∈ N) [ a
b

a + mc
b + md >
c
d
] .
Indeed, since b, d and m are positive we have
[a(b + md) > b(a + mc)] ⇐⇒ [mad > mbc] ⇔ (ad > bc),
and
[d(a + mc) > c(b + md)] ⇔ (ad > bc).
29 October 6, 2013
2.2. THE FIELD OF REAL NUMBERS
2.1.3 Irrational Numbers
Suppose that a ∈ Q and consider the equation
(2.1.2) x
2 = a.
In general (2.1.2) does not have rational solutions. For example, the following theorem holds.
Theorem 2.1.3. No rational number has square 2.
Proof. Suppose for a contradiction that the rational number p
q
(p ∈ Z, q ∈ N, in lowest terms)
is such that ( p
q
)
2 = 2. Then p
2 = 2q
2
. Hence, appealing to the Fundamental Theorem of
Arithmetic, p
2
is even, and hence p is even. Thus (∃k ∈ Z) [ p = 2k ]. This implies that
2k
2 = q
2
and therefore q is also even. The last statement contradicts our assumption that p and q have
no common factor.
The last theorem provides an example of a number which is not rational. We call such
numbers irrational. Here are some other examples of irrational numbers.
Theorem 2.1.4. No rational x satisfies the equation x
3 = x + 7.
Proof. First we show that there are no integers satisfying the equation x
3 = x + 7. For a
contradiction suppose that there is. Then x(x + 1)(x − 1) = 7 from which it follows that x
divides 7. Hence x can be only ±1, ±7. Direct verification shows that these numbers do not
satisfy the equation.
Second, show that there are no fractions satisfying the equation x
3 = x+7. For a contradiction
suppose that there is. Let m
n with m ∈ Z, n ∈ Z+, n ̸= 0, 1, and m, n have no common factors
greater than 1, is such that ( m
n
)
3 =
m
n + 7. Multiplying this equality by n
2 we obtain
m3
n = mn + 7n
2
, which is impossible since the right-hand side is an integer and the left-hand
side is not.
We leave the following as an exercise.
Example 2.1.1. No rational satisfies the equation x
5 = x + 4.
2.2 The Field of Real Numbers
In the previous sections we discussed the need to extend N to Z, and Z to Q. The rigorous
construction of N can be found in a standard course on Set Theory. The rigorous construction
of the set of real numbers from the rationals is rather complicated and lengthy. An excellent
exposition can be found in [5]. W. Rudin Principles of Mathematical Analysis. McGraw Hill,
2006. In this course we postulate the existence of the set of real numbers R as well as basic
properties summarized in a collection of axioms.
Notation. We use the symbol ∃ ! to say that there exists a unique... So (∃ ! x ∈ R) is read:
’there exists a unique real number x’.
Those of you familiar with basic concepts of algebra will find that axioms A.1 - A.11 characterize R as an algebraic field.
30 October 6, 2013
CHAPTER 2. NUMBERS
A.1. (∀ a ∈ R)(∀ b ∈ R) [(a + b) ∈ R] (closed under addition).
A.2. (∀ a ∈ R)(∀ b ∈ R) [a + b = b + a] (commutativity of addition).
A.3. (∀ a ∈ R)(∀ b ∈ R)(∀ c ∈ R) [(a + b) + c = a + (b + c)] (associativity of addition).
A.4. (∃ 0 ∈ R)(∀ a ∈ R) [0 + a = a] (existence of additive identity).
A.5. (∀ a ∈ R)(∃ ! x ∈ R) [a + x = 0] (existence of additive inverse). We write x = −a.
Axioms A.6-A.10 are analogues of A.1-A.5 for the operation of multiplication.
A.6. (∀ a ∈ R)(∀ b ∈ R) [ab ∈ R] (closed under multiplication).
A.7. (∀ a ∈ R)(∀ b ∈ R) [ab = ba] (commutativity of multiplication).
A.8. (∀ a ∈ R)(∀ b ∈ R)(∀ c ∈ R) [(ab)c = a(bc)] (associativity of multiplication).
A.9. (∃ 1 ∈ R)(∀ a ∈ R) [1 · a = a] (existence of multiplicative identity).
A.10. (∀ a ∈ R − {0})(∃ ! y ∈ R) [ay = 1] (existence of multiplicative inverse). We write
y = 1/a.
The last axiom links the operations of summation and multiplication.
A.11. (∀ a ∈ R)(∀ b ∈ R)(∀ c ∈ R) [(a + b)c = ac + bc] (distributive law).
Familiar rules for the manipulation of real numbers can be deduced from the axioms above.
Here is an example.
Example 2.2.1. (∀ a ∈ R)[0a = 0].
Indeed, we have
a + 0a = 1a + 0a (by A.9)
= (1 + 0)a (by A.11)
= 1a (by A.2 and A.4)
= a (by A.9)
Now add −a to both sides.
−a + (a + 0a) = −a + a
⇒ (−a + a) + 0a = 0 (by A.3 and A.5)
⇒ 0 + 0a = 0 (by A.5)
⇒ 0a = 0 (by A.4).
Remark 2.2.1. The set of rationals Q also forms an algebraic field (that is, the rational
numbers satisfy axioms A.1 - A.11). But the integers Z do not form a field, as axiom A.10
does not hold.
Axioms A.1 - A.11 represent algebraic properties of real numbers. Now we add axioms of
order.
31 Oct