Real Numbers
Lawrence Tsang
01/10/2004
1
Axioms
Definition 1.1. A truth value is either true or false.
Definition 1.2. A proposition is a statement whose truth value is to be determined.
Definition 1.3. An axiom is a proposition that is assumed to be true.
Definition 1.4. An theorem is a proposition whose truth value is unequivocally
determined by logical deductions originating from the axioms.
Definition 1.5. A proof is a demonstration of a logical deduction leading from
the axioms, which unequivocally determines the truth value of a proposition.
Definition 1.6. A conjecture is a proposition whose truth value seems to be
true, but a proof has not been obtained yet.
2
Mathematical Notations
The following notations are ubiquitous throughout modern mathematics:
x∈S
x 6∈ S
S = {n | P }
S⊆T
T =S
S⊂T
T \S
3
x is in set S.
x is not in set S.
S is the set of all n such that proposition P is true.
S is a subset of set T , or x ∈ T ⇒ x ∈ S.
if T ⊆ S and S ⊆ T.
S is a proper subset of set T , or S ⊆ T but S 6= T.
set difference, or {x | x ∈ T and x 6∈ S}.
Types of Real Numbers
1. [natural] N = {1, 2, 3, 4, . . .}.
2. [integer] Z = {. . . , −2, −1, 0, 1, 2, . . .}.
3. [rational] Q = m
n | m, n ∈ Z and n 6= 0 .
4. [irrational] H = R \ Q.
1
(1)
(2)
(3)
(4)
(5)
(6)
(7)
4
4
AXIOMS FOR REAL NUMBERS
Axioms for Real Numbers
Let R be the set of real numbers with two binary operations: addition x + y and
multiplication x · y or just xy. In a written expression involving both additions
and multiplications, multiplications take precedence over addition.
Furthermore, R obeys the following sets of axioms.
4.1
Field Axioms
The Field Axioms describe the algebraic properties of real numbers.
Axiom 4.1 (Closure). For every a, b ∈ R,
a + b ∈ R,
ab ∈ R.
(8)
(9)
Axiom 4.2 (Commutative). For every a, b ∈ R,
a + b = b + a,
ab = ba.
(10)
(11)
Axiom 4.3 (Associative). For every a, b, c ∈ R,
(a + b) + c = a + (b + c),
(ab)c = a(bc).
(12)
(13)
Axiom 4.4 (Distributive). For every a, b, c ∈ R,
(a + b)c = ac + bc.
(14)
Axiom 4.5 (Identity). For every a ∈ R, there exists 0, 1 ∈ R such that
a+0=0+a=a
a1 = 1a = a
Additive Identity,
Multiplicative Identity.
(15)
(16)
Axiom 4.6 (Inverse). For every a ∈ R, there exists b, c ∈ R such that
a+b=b+a=0
ac = ca = 1
Additive Inverse,
Multiplicative Inverse.
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(18)
Axiom 4.7 (Nontrivial field). 0 6= 1.
Some consequences of the field axioms about real numbers:
Theorem 4.8 (Uniqueness of Identity). Additive and multiplicative identities are unique.
Proof. Suppose 0̂ is another additive identity. Then,
0 = 0 + 0̂
= 0̂
by identity axiom on 0̂
(19)
by identity axiom on 0.
(20)
The argument for the multiplicative identity is similar.
2
4.1
Field Axioms
4
AXIOMS FOR REAL NUMBERS
Theorem 4.9 (Uniqueness of inverse). Additive and multiplicative inverses
are unique.
Proof. Suppose ȧ and ä are additive inverses of a. Then,
ȧ = 0 + ȧ
= (ä + a) + ȧ
= ä + (a + ȧ)
= ä + 0
= ä
by
by
by
by
by
identity axiom
inverse axiom
associative axiom
inverse axiom
identity axiom.
(21)
(22)
(23)
(24)
(25)
The argument for the multiplicative inverse is similar.
Because of the uniqueness of inverses, we will denote −a as the additive
inverse of a, and a−1 as the multiplicative inverse of a. This notation allows us
to define substraction and division as followed.
Definition 4.10 (Substraction). a − b = a + (−b).
Definition 4.11 (Division). a/b = a(b−1 ) for b 6= 0.
Theorem 4.12. For all a ∈ R, a0 = 0a = 0.
Proof.
a + 0a = 1a + 0a
= (1 + 0)a
= 1a
=a
by
by
by
by
multiplicative indentity
distributive axiom
additive identity
multiplicative identity.
(26)
(27)
(28)
(29)
Subtracting both sides by a gives 0a = 0. a0 = 0a is true because of commutative axiom.
Theorem 4.13. −a = (−1)a
Proof.
a + (−1)a = (1)a + (−1)a
= (1 − 1)a
= 0a
=0
by
by
by
by
identity axiom
distributive axiom
additive inverse
Theorem 4.12.
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(31)
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(33)
Subtracting both sides by a gives the theorem.
Theorem 4.14 (Zero Factor). If ab = 0, then a = 0 or b = 0.
Proof by Contradiction. Suppose both a and b are nonzero. Then say a must
have a multiplicative inverse a−1 . Then,
a−1 ab = a−1 0
b = 0.
(34)
(35)
But by assumption b 6= 0. Thus a contradiction is reached, and the assumption
is false. Thus both a and b cannot be nonzero, or at least one of them is zero.
3
4.2
4.2
Ordering Axiom
4
AXIOMS FOR REAL NUMBERS
Ordering Axiom
The Ordering Axiom describe the ordering properties of real numbers.
Axiom 4.15 (Ordering). Let P ⊂ R such that
1. [real] If a, b ∈ P, then a + b, ab ∈ P.
2. [trichotomy] If a ∈ R, then either a ∈ P, a = 0, or −a ∈ P is true
exclusively.
Definition 4.16 (Inequality). We say a < b if b − a ∈ P.
With this definition, we note that if a ∈ P, then a − 0 ∈ P and thus a > 0
for every a in P. For this reason, we call the set P the set of positive numbers.
Also, if a number is not zero and positive, we say the number is negative.
We will also denote a ≤ b to mean a < b or a = b. Also, we denote a > b to
be b < a.
Theorem 4.17. 0 < 1.
Proof. By the first part of the Ordering Axiom, there are three possibilities:
1 = 0, 1 ∈ P, or −1 ∈ P. The first case is impossible because of Axiom 4.7.
Now, we know −1 · −1 = 1. If −1 were positive, then 1 must also be positive
by the first part of the Ordering Axiom. But this is a contradiction because
both 1 and −1 would be positive and this would violate the trichotomy part of
the Ordering Axiom.
Since −1 cannot be positive and 1 cannot be zero, 1 must be positive.
Theorem 4.18. a2 ≥ 0. a2 = 0 if and only if a = 0.
Proof. We consider the three possibilities:
a = 0: Then, 0 · 0 = 0.
a > 0: Then, aa ∈ P or a2 > 0 by the first part of the Ordering Axiom.
a < 0: Then, −a is positive. Let b = −a, a positive number. Then,
aa = (−1)b(−1)b = (−1)(−1)bb = bb ∈ P.
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a2 > 0.
(37)
Thus,
Theorem 4.19. If a ≤ b and b ≤ a, then a = b.
Proof. The first condition a ≤ b means b − a ∈ P or a = b. The second condition
b ≤ a means a − b ∈ P or b = a. But a − b = −1(b − a) and by the Trichotomy
axiom only b − a or −1(b − a) can be in P. Thus, we rule out b − a ∈ P and
a − b ∈ P. The only other condition is a = b.
Theorem 4.20 (Inequality Transitivity). If a < b and b < c, then a < c.
4
4.2
Ordering Axiom
4
AXIOMS FOR REAL NUMBERS
Proof. a < b means b−a ∈ P, and b < c means c−b ∈ P. Thus, (b−a)+(c−b) ∈
P. But, (b − a) + (c − b) = c − a. Thus, c − a ∈ P or a < c.
Theorem 4.21. For all c ∈ R, if a < b, then a + c < b + c.
Proof. If a < b, then b − a ∈ P. Because
b − a = b − a + c − c = b + c − a − c = (b + c) − (a + c),
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(b + c) − (a + c) ∈ P,
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a + c < b + c.
(40)
we have
or
Theorem 4.22. If c > 0 and a < b, then ac < bc. If d < 0 and a < b, then
ad > bd.
Proof. In both cases, a < b implies b − a ∈ P. If c > 0, then
(b − a)c ∈ P.
(41)
bc − ac ∈ P
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ac < bc.
(43)
(b − a)(−d) ∈ P.
(44)
−bd + ad = ad − bd ∈ P
(45)
bd < ad.
(46)
Thus,
or
If d < 0, then −d > 0 and
Thus,
or
Theorem 4.23. If n is positive, then n−1 is positive.
Proof. By Theorem 4.18, n−1 n−1 > 0. Thus,
nn−1 n−1 > 0
Corollary 4.24. If 0 < a < b, then
(47)
1n
−1
>0
(48)
n
−1
> 0.
(49)
1
b
< a1 .
5
4.3
Completeness Axiom
4
AXIOMS FOR REAL NUMBERS
Proof. By Theorem 4.23, a−1 and b−1 are also positive. Thus,
a<b
−1 −1
a(a
b
b
4.3
(50)
−1 −1
) < b(a
−1
b
)
−1
<a
1
1
<
b
a
(51)
(52)
(53)
Completeness Axiom
Definition 4.25. Let E ⊆ R. E is bounded above by u if u ≥ x for all x ∈ E.
u is called an upper bound of E.
Likewise, E is bounded below by l if l ≤ x for all x ∈ E. l is called a lower
bound of E.
Definition 4.26. If s is an upper bound of E and s ≤ u for every upper bound
u of E, then s is called the least upper bound denoted by
s = sup E.
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Likewise, if t is a lower bound of E and t ≤ l for every lower bound l of E, then
t is called the greatest lower bound denoted by
t = inf E.
(55)
Theorem 4.27. Least upper bound and greatest lower bound are unique.
Proof. Suppose s and s0 are two least upper bounds of a set. Then by definition
of the least upper bound,
s ≤ s0 and s0 ≤ s
(56)
But this implies s = s0 by Theorem 4.19.
Similarly, by definition of the greatest lower bound,
t ≥ t0 and t0 ≥ t
(57)
But this also implies t = t0 .
Axiom 4.28 (Completeness). Every subset of R that is bounded above has a
least upper bound.
Theorem 4.29. Every subset of R that is bounded below has a greatest lower
bound.
Proof. Suppose E is a set bounded below by m. Then let F = {x | −x ∈ E}.
F is bounded above by −m, because m being a lower bound of E means m ≤ x
for all x ∈ E, and thus −m ≥ −x for all −x ∈ F .
By the Completeness Axiom, F has a least upper bound s and s ≤ u for
every upper bound u of F . Thus, −s ≥ −u for every lower bound −u of E. −s
by definition then is the greatest lower bound of E.
6
6
DENSITY OF RATIONAL NUMBERS
Lemma 4.30 (Archimedean principle). For any real number a, there exists
an integer n such that a < n.
Proof. Suppose the lemma is false and a > n for all integer n. Then the set
of integers N is bounded above, and by the Completeness Axiom it has a least
upper bound M .
Since M is the least upper bound, M − 1 cannot be an upper bound. Thus,
there is an integer n such that M − 1 < n.
But this implies M < n + 1 where n + 1 is an integer greater than the upper
bound. This is a contradiction, thus the lemma cannot be false.
Corollary 4.31. For any real number > 0, there exists an integer n such that
0<
1
<
n
Proof. By the Archimedean Principle, there is an integer n such that 0 <
By Theorem 4.24, we have 0 < n1 < .
5
(58)
1
< n.
Well-Ordering Principle
Axiom 5.1 (Well Ordering Principle). Every nonempty subset of the set of
natural number has a least element. That is, if E ⊆ N, then there exists m ∈ E
such that m ≤ n for all n ∈ E.
6
Density of Rational Numbers
Theorem 6.1 (Density of Rational Number). Given any two distinct numbers a and b, there is a rational number p such that
a < p < b.
(59)
Proof. Let d = b − a. We have two cases here.
1. Suppose d ≥ 1. By Archimedean Principle, there is an integer m such
that
a < m.
(60)
By the Well-Ordering Principle, there is a smallest integer n such that
a<n
(61)
n − 1 < a.
(62)
n < a + 1.
(63)
1 ≤ d,
(64)
a + 1 ≤ a + d = b.
(65)
and
Thus,
Now, since
we have
7
7
DENSITY OF IRRATIONAL NUMBERS
Combining (61), (63) and (65), we have
a<n<b
(66)
Thus, n is our desired rational number.
2. Suppose d < 1. By Corollary 4.31, there is an integer n such that
1
<b−a
n
1 < nb − na.
0<
(67)
(68)
Now, we can apply the result of the previous case to nb and na. Thus,
there is an integer m such that
na < m < nb
m
a<
< b.
n
7
(69)
(70)
Density of Irrational Numbers
Lemma 7.1.
√
2 is irrational.
√
Proof. Suppose 2 = m
n . Suppose also that we have reduced the fraction so
that not both m and n are even. Then,
√
2n = m,
(71)
2n2 = m2 .
(72)
Since the left side has a factor of 2, the right side must be even. Thus, m must
be even. Let, m = 2p.
2n2 = (2p)2 ,
2
(73)
2
n = 2p .
(74)
Since the right side has a factor of 2, the left side must be even. Thus, n must
be even. But this contradicts the fact that both m and n cannot be even.
Lemma 7.2. If h is irrational and p is rational, then hp is irrational.
Proof. Let p =
and n0 . So,
m
n.
Suppose hp is rational, then hp =
m0
m
= hp = h ,
n0
n
m0 n
m0 n
h= 0
= 0 .
n m
nm
m0
n0
for some integers m0
(75)
(76)
But this contradicts the fact that h cannot be written as a fraction of integers.
8
8
EXERCISE
Theorem 7.3 (Density of irrational number). Given any two distinct numbers a and a, there is an irrational number h such that
a < h < b.
(77)
a
b
√ <√ .
2
2
(78)
Proof. Since a < b, we have
By Theorem 6.1, there exists a rational number p such that
b
a
√ <p< √
2
2
√
a < p 2 < b.
√
(79)
(80)
√
By Lemma 7.2, p 2 is irrational. Let h = p 2.
8
Exercise
Exercise 8.1. Identify the type(s) of number for each of the following: (Note
that some number may be considered to be of more than one type. In that case,
list all the types the number can be associated with.)
a. −7
b. 0.7
√
c. 3
√
d. 9
e. 0
f.
g.
h.
21
7
√
√4
3
√
√12
3
Exercise 8.2. Specify the axiom or theorem used to justify each step of the
following procedure for solving the algebraic equation. Note that part a is the
given equation.
2
= 3x + 5
x
2 = x(3x + 5)
2 = x3x + x5
by a) Given.
(81)
by b)
by c)
(82)
(83)
2 = 3x2 + 5x
by d)
(84)
by e)
by f)
(85)
(86)
2
0 = 3x + 5x − 2
0 = (3x − 1)(x + 2)
9
8
3x − 1 = 0
3x = 1
1
x=
3
by g)
by h)
x+2=0
x = −2
EXERCISE
by j)
by k)
by i)
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(88)
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Exercise 8.3. Check the boxes in the table below for each number type obeying
the specified axioms.
Types of Number
N Z Q H R
Additive Closure
Multiplicative Closure
Additive Commutativity
Multiplicative Commutativity
Additive Associativity
Multiplicative Associativity
Distributive
Additive Identity
Multiplicative Identity
Additive Inverse
Multiplicative Inverse
Ordering Axiom
Completeness Axiom
Table 1: Axioms and Numbers
Exercise 8.4. Prove if M ≥ 0, then there is a real number a such that
a2 = M.
10
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