Download Chapter 3 Real numbers

Chapter 3
Real numbers
The notion of real number was introduced in section 1.3 where the axiomatic denition of
the set of all real numbers was done and some basic properties of the set of all real numbers
were shown. This chapter is dedicated to the properties of the set R and its subsets. These
properties follows from the Dedekind cut D(R) and from the axioms of ordering.
3.1 Basic properties of real numbers - continuation
3.1.1 Inmum and supremum of a set of real numbers
Having in mind we will work with the sets of real numbers, we will give a specic characterization of the inmum and supremum, respectively, in R.
THEOREM 3.1.1. Let B ⊂ R be nonempty. Then:
1. inf B = b ∈ R i
(a) ∀x ∈ B : b ≤ x;
(b) ∀² > 0 ∃x0 ∈ B : x0 < b + ².
2. sup B = b ∈ R i
(α) ∀x ∈ B : x ≤ b;
(β ) ∀² > 0 ∃x0 ∈ B : b − ² < x0 .
Proof. We will prove the rst part of the theorem. The second part can be proven in fully
analogical way.
Let inf B = b ∈ R. The (a) property coincides with the rs property of an inmum mentioned
in Theorem 1.5.3. The same theorem implies from b < b + ² (∈ R) that x0 < b + ² and this
veries the (b) property.
Now, let the properties (a) and (b) hold true. For y ∈ R such that b < y we put ² = y −b > 0.
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3.1 Basic properties of real numbers - continuation
6
Then the property (b) implies there exists x0 in B such that x0 < b + (y − b) = b. This
completes the proof of the second property of inmum from Theorem 1.5.3.
We have shown in the example 1.5.5 that the set B = {x ∈ Q; x2 ≤ 3, x ≥ 0} is bounded
from above (in Q) but has no supremum in Q. The Dedekind cut for real numbers implies
that the subset in R behave essentially dierently from the subsets in Q with respect to
supremum.
THEOREM 3.1.2. Any nonempty set M ⊂ R that is bounded from above has a supremum
in R.
Proof. Let us dene two sets A and B as follows:
B := {y ∈ R; ∀x ∈ M : x < y},
A := R \ B.
The following relations hold true:
∅ 6= A ⊂ R,
∅ 6= B ⊂ R,
A ∪ B = R.
We will prove that for any element a ∈ A and any element b ∈ B we have: a < b. Let us
suppose there are two numbers a0 ∈ A, b0 ∈ B such that a0 ≥ b0 , then for all x ∈ M we
have x < b0 ≤ a0 . Hence a0 must be in B and this is the contradiction with the fact that
A ∩ B = ∅. We have shown that the pair of sets A, B is a Dedekind cut (see section 1.3) and
therefore there is a real number c such that a ≤ c ≤ b for all a ∈ A and b ∈ B . By using
Theorem 3.1.1 we will show that c = sup M . The inclusion M ⊂ A veries the (α) property
for c. We have to prove that the (β ) property also holds true. Let ² > 0. Then
c − ² < c ≤ b, ∀b ∈ B,
and therefore c − ² ∈
/ B . If for any x ∈ M would be x < c − ² then it would be also c − ² ∈ B
and this would be in contradiction with the statement that c−² ∈
/ B . Thus, the (β ) property
holds true and this means:
∀² > 0 ∃x0 ∈ M : c − ² < x0 .
There is, naturally, an analogical theorem about inmum:
THEOREM 3.1.3. Any nonempty set M ⊂ R that is bounded from below has an inmum
in R.
Proof. Let k ∈ R is a lower bound of a set M ⊂ R, then −k is a upper bound of the set
M1 = {y ∈ R; y = −x, x ∈ M }.
Theorem 3.1.2 guarantees existence of sup M1 = c1 ∈ R. It is obvious that inf M = −c1 .
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3.1 Basic properties of real numbers - continuation
7
Example 3.1.1. Let ∅ 6= M1 ⊂ M ⊂ R and let M be bounded from above. The sup M1 ≤
sup M in R.
Solution: The existence of the following two real numbers
sup M =: b,
sup M1 =: b1
is guaranteed by Theorem 3.1.2. Using the denition of supremum (Denition 1.5.4) we
obtain
∀x ∈ M1 ⊂ M : x ≤ b and b1 ≤ b.
Example 3.1.2. Show that the set of all irrational numbers R \ Q is nonempty.
Solution: Example 1.5.5 tells us that the set
B = {x ∈ Q; x2 < 3, x ≥ 0} ⊂ R
is bounded from above in Q, and therefore it is bounded from above in R also. However,
sup B ∈
/ Q. Thus, with respect to Theorem 3.1.2 we obtain sup B ∈ R and subsequently
sup B ∈ R \ Q.
3.1.2 Archimedean property
By using Theorem 3.1.2 we will show that the Archimedean property holds true for the real
numbers. (Compare this with the Archimedean axiom IVQ from Denition 1.3.1).
THEOREM 3.1.4. It holds true:
∀x ∈ R ∃n ∈ N : x < n.
Proof. By contradiction. Let
∃y ∈ R ∀n ∈ N : n ≤ y.
This means that the set of all natural numbers N is bounded from above and therefore
(Theorem 3.1.2) it exists
sup N = b ∈ R.
With respect to the (β ) property from Theorem 3.1.1 implies for ² = 1 that there exists n0
in N such that b − 1 < n0 . Subsequently, we have b < n0 + 1 ∈ N and this is in contradiction
with that b = sup N.
3.1.3 On density of the rational numbers and the irrational numbers in R
The set R, as shown in Example 3.1.2, is the extension of the set of all rational numbers Q
and these two sets diers by nonempty set of all irrational numbers. We will show in the
following theorem that both sets Q and R \ Q are "dense" in R. We will start with the
supply lemma:
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3.1 Basic properties of real numbers - continuation
8
Lemma 3.1.1. The statement:
∀x ∈ R ∃n ∈ Z : n ≤ x < n + 1.
The integer n is dened by x uniquely.
Proof. Let x ∈ R. By Archimedean property we have
∃n1 ∈ N, n2 ∈ N : −x < n1 , x < n2 .
Therefore
−n1 < x < n2 .
Let us construct the integers −n1 , −n1 + 1, . . . , n2 − 1, n2 and denote by n the greatest of
them such that n ≤ x. Such an integer is unique and it obeys
n ≤ x < n + 1.
Denition 3.1.1. Let x ∈ R. The integer n such that n ≤ x < n + 1 is called the integral
part of the real number x and it is denoted by [x].
For example
2
= 0,
3
8
− = −2.
5
THEOREM 3.1.5. Let a < b be a pair of real numbers. Then there exist such a rational
number x and such an irrational number y that
and
a < x < b,
a < y < b.
Proof. By the Archimedean property we know there exists such a natural number n that
1
< b − a,
n
b − a > 0.
Let k = [na]. Then we have k ≤ na < k + 1 and also
k
k+1
≤a<
.
n
n
The inequality k/n ≤ a implies
1
k+1
≤ a + < a + (b − 1) = b.
n
n
So we have shown that the rational number
x=
8
k+1
n
3.1 Basic properties of real numbers - continuation
9
obeys
a < x < b.
We know that the number sup B from Example 3.1.2 is irrational. Let us choose x in such
a way that
a − sup B < x < b − sup B.
(The existence of such a rational x is guaranteed). Then y = x + sup B is an irrational
number for which the requested inequality holds true.
Example 3.1.3. Show that for any x ∈ R we have:
x = sup{u ∈ Q; u < x}.
Solution: The (α) property of supremum from Theorem 3.1.1 is obvious. Let ² > 0. Theorem 3.1.5 implies that
∃r0 ∈ Q : x − ² < r0 < x.
Thus, r0 ∈ {u ∈ Q; u < x} and the (β ) property of Theorem 3.1.1 holds true.
3.1.4 Root of a real number
The following theorem guarantees the existence and the uniqueness of the nth root of a
positive real number.
THEOREM 3.1.6. It holds true that
∀a > 0 ∀n ∈ N ∃y > 0 : y n = a.
The number y is dened by the numbers a and n uniquely.
Proof. Let B := {x > 0; xn ≤ a}. The set B is nonempty since
b :=
a
∈ B,
a+1
in fact: 0 < b < 1 and bn < b < a.
Furthermore, the set B is bounded from above by the number c := a + 1. In fact, if
some y0 ∈ B would obey y0 > a + 1 then it would be ynn > (a + 1)n , and this would
be in contradiction with our assumption that y0 belongs to B . At the moment Theorem
3.1.2 guarantees existence of R+ 3 y := sup B . This number y must obey just one of the
followings:
y n < a, y n > a, y n = a.
The rst two cases can be excluded:
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3.1 Basic properties of real numbers - continuation
10
(1) let us suppose y n < a. We can choose a positive real number h < 1 such that
h<
a − yn
,
(y + 1)n − y n
(a − y n > 0).
By making use of binomial theorem we obtain
‚
n
n
–‚
Œ
n
1
(y + h) = y +
Œ
‚
y
n−1
n
y +h
y n−1 + · · · +
1
y n + (a − y n ) = a.
h + ··· +
‚
n
n
n
Ϊ
n
n
Œ
hn ≤
= y n + h [(y + 1)n − y n ] <
Thus we have shown that y + h ∈ B and this is the contradiction since y = sup B .
(2) One can prove that y n > a cannot hold true by analogy.
(3) Finally, the relation
yn = a
must hold true.
The uniqueness can be shown indirectly. Let y1 , y2 ∈ R+ be two numbers such that
y1n = a,
y2n = a,
y 1 < y2 .
Then we obtain the following contradiction
a = y1n < y2n = a.
This completes the proof.
Theorem 3.1.6 allows us to write the following denition
Denition 3.1.2.
1. Let a > 0 and n ∈ N. The number y > 0 for which y n = n is called the nth root of the
√
number a. It us denoted by the symbol n a or by a1/n .
√
2. For a = 0 we put n 0 := 0 .
3. For a < 0 and an odd n ∈ N we dene
√
n
√
a = − n −a.
It follows from the proof of Theorem 3.1.6 and from Examples 3.1.2 and 3.1.3 that
√
3 = sup{x ∈ R+ ; x2 < 3} = sup{x ∈ Q; x2 < 3, x ≥ 0}
is an irrational number.
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3.1 Basic properties of real numbers - continuation
11
3.1.5 Extended set of real numbers
For the further purposes in Chapter 6 we will need some extension R̄ of the set of all real
numbers R. The set R̄ will be closed under operation of taking a limit, i.e. a mapping with
values in R ⊂ R̄ will have a limit in R̄. Such a set can be dened with help of relation of
natural ordering in R:
Denition 3.1.3.
R̄ := R ∪ {−∞, +∞},
where for all x ∈ R: −∞ < x < +∞.
For our purposes we need not to dene any operation with improper elements ±∞, it
will be sucient to know the topological structure of the set R̄ only (see Theorem 5.2.5).
By denition 3.1.3 and Theorems 3.1.2 and 3.1.3 we obtain:
THEOREM 3.1.7. Any nonempty set M ⊂ R̄ has a supremum as well as an inmum in
R̄.
For example,
sup Z = +∞,
inf Z = −∞
(in R̄),
where sup Z and inf Z do not exist in real numbers.
Problems
1. Prove in details the second part of Theorem 3.1.1.
2. Prove that Theorem 3.1.2 as well as Theorem 3.1.3 are equivalent with the Dedekind
cut mentioned in the third section of the rst Chapter.
3. Let ∅ 6= M1 ⊂ M ⊂ R and let M1 be bounded from below. Prove that inf M1 ≥ inf M
in R.
4. Show that for any real number x we have
x = inf{u ∈ Q : x < u} = sup{u ∈ Q : u < x}.
5. Show that
n−1
≤ b < an .
∀a ∈ R, a > 1 ∀b ∈ R+
0 ∃n ∈ Z : a
(Hint: you can use the Archimedean property.)
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3.2 Notion of interval, intervals over the real axis
12
6. Let ∅ 6= M ⊂ R and M1 = {y ∈ R; y = −x, x ∈ M }. Show that
sup M1 = − inf M,
inf M1 = − sup M.
7. Find supremum as well as inmum in R and in R̄ of the following mappings
x
, x ∈ R \ {1},
1+x
c1) f1 : x 7→ [x], x ∈ {t ∈ R; 0 < t < 2},
a) f : x 7→
€
n+1
nŠ
=: (an )n∈N , n(−1) n∈N ,
n
n∈N
c2) f2 : x 7→ [x], x ∈ {t ∈ R; 0 ≤ t ≤ 2},
b)
c3) f3 : x 7→ x − [x], x ∈ {t ∈ R; 0 ≤ t ≤ 1}.
8. Prove Theorem 3.1.7.
Answers
7
a) There are no min f, max f, inf f, sup f in R and min f = inf f = −∞ and
max f = sup f = +∞ in R̄.
b) In the set R:
min an , min bn , max bn , sup bn
n∈N
n∈N
n∈N
n∈N
do not exist and
inf an = 1, max an = sup an = 2, inf bn = 0.
In the set R̄:
max bn = sup bn = +∞.
Further relations coincides with those valid in R.
c) In the set R min f1 , and max f1 do not exist, inf f1 = 0, sup f1 = 1; min f2 =
inf f2 = 0, max f2 = sup f2 = 2; min f3 = inf f3 = 0, max f3 does not exist and
sup f3 = 1.
3.2 Notion of interval, intervals over the real axis
Studying continuity, dierentiability and integrability of a mapping (function) we will see
that these properties do depend not only on a relation that denes the mapping but they
do depend also on denition domain of the mapping. This is the reason why we will be
interested in more details in most frequently used subsets of the real axis - intervals.
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3.2 Notion of interval, intervals over the real axis
13
3.2.1 Intervals
Denition 3.2.1.
1 A set I ⊂ R̄ is called an interval (from R̄) i it has the following properties:
a) The set I contains at least two distinct points,
b) If x1 ∈ I, x2 ∈ I then (∀x ∈ R̄ : x1 < x < x2 ) ⇒ x ∈ I .
2 A cartesian product of m intervals (from R̄) is called an m-dimensional interval (from
R̄).
It is easy to verify that if a ∈ R̄, b ∈ R̄ and if a < b then any of the following sets is an
interval from R̄:
• I1 = {x ∈ R̄ : a ≤ x ≤ b} =: [a, b] (closed interval)
• I2 = {x ∈ R̄ : a ≤ x < b} =: [a, b) (half-closed interval or half-open interval, open
from the right etc.)
• I3 = {x ∈ R̄ : a < x ≤ b} =: (a, b] (half-closed interval or half-open interval, open
from the left etc.)
• I1 = {x ∈ R̄ : a < x < b} =: (a, b) (open interval)
Let us remark that any of the sets Ii , i = 1, 2, 3, 4 represents four intervals from R̄ with
respect to whether a, b ∈ R or a = −∞ or b = +∞. For example, for a, b ∈ R̄ the interval
I1 can behave as
[a, b], [a, +∞], [−∞, b], [−∞, +∞] = R̄.
Analogically for other intervals.
A natural question arises whether any interval can be identied with one of the types
Ii , i = 1, 2, 3, 4. The answer to this question together with full characterization of the set of
all intervals is given in the following theorem.
THEOREM 3.2.1. A set I ⊂ R̄ is an interval from R̄ if and only if there exists an index i
in {1, 2, 3, 4} such that I = Ii .
Proof. 1. It is obvious from the text above this theorem that if I = Ii , i = 1, 2, 3, 4 then I
is an interval (in R̄).
2. Let know I be an interval. Then (Theorem 3.1.7) there exist a := inf I and b := sup I in
R̄. Since I contains at least two distinct points, we have a < b. The denitions of inmum
and supremum of the set I imply:
∀x ∈ I : a ≤ x ≤ b.
13
(?)
3.2 Notion of interval, intervals over the real axis
14
Four distinct cases can appear:
a) a ∈ I, b ∈ I,
b) a ∈ I, b ∈
/ I,
c) a ∈
/ I, b ∈ I,
a) a ∈
/ I, b ∈
/ I.
Let us consider case a). (In other cases the proof is similar.) Eq. (?). implies that I ⊂ [a, b].
Let x ∈ [a, b]. If x = a or x = b then x ∈ I . Let now x ∈ (a, b). Since a = inf I it exists an
x1 ∈ I such that x1 < x (see Theorem 3.1.5). Similarly, b = sup I implies that it exists an
x2 ∈ I such that x < x2 . So we have that x1 < x < x2 . The (b) property of the interval we
have x ∈ I , thus [a, b] ⊂ I . We have shown that I = [a, b].
Example 3.2.1. Write down some examples of two dimensional intervals from R̄2 and nd
out what is their number.
Solution: By Theorem 3.2.1 and the second part of Denition 3.2.1 the following sets are
intervals in R̄2 (a, b, c, d ∈ R, a < b, c < d):
[a, b] × [c, d],
[a, b] × [c, +∞],
[a, b] × [−∞, d],
[a, b] × [−∞, ∞],
[a, +∞] × [c, d],
[a, +∞] × [c, +∞],
[a, +∞] × [−∞, d],
[a, +∞] × [−∞, +∞],
[−∞, b] × [c, d],
[−∞, b] × [c, +∞],
[−∞, b] × [−∞, d],
[−∞, b] × [−∞, +∞],
[−∞, +∞] × [c, d],
[−∞, +∞] × [c, +∞],
[−∞, +∞] × [−∞, d],
[−∞, +∞] × [−∞, +∞].
Other two dimensional intervals can be derived from Ii ×Ij , i, j = 1, 2, 3, 4 such that i+j 6= 2.
This means we have 44 = 256 distinct two dimensional intervals in R̄2 .
3.2.2 Embedded intervals
Denition 3.2.2. Let the sequences of real numbers (an )n∈N and (bn )n∈N obey:
a1 ≤ a2 ≤ · · · ≤ an ≤ · · · ≤ bn ≤ · · · ≤ b2 ≤ b1 .
Then the sequence of closed intervals in R
([an , bn ]n∈N )
is called the system of embedded intervals in R.
For any system of embedded intervals we have: [an+1 , bn+1 ] ⊂ [an , bn ], (n ∈ N).
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3.2 Notion of interval, intervals over the real axis
15
THEOREM 3.2.2. Let ([an , bn ]n∈N ) be a system of embedded intervals in R. Then
\
[an , bn ] 6= ∅,
n∈N
i.e. ∃c ∈ R ∀n ∈ N : c ∈ [an , bn ].
Proof. The sequence (an )n∈N is bounded from above by any of the numbers bn and therefore
(see Theorem 3.1.2) we know that it exists the number sup(an )n∈N =: a ∈ R and a ≤ bn .
Boundedness from below of the sequence (bn )n∈N by a implies that there exists inf(bn )n∈N =:
b ∈ R (see Theorem 3.1.3) and furthermore
for all n ∈ N.
an ≤ a ≤ b ≤ bn ,
Thus we have shown that at least a belongs to any interval of the system ([an , bn ]n∈N ).
The intersection
\
1 1
− ,
n n
n∈N
equals to one-point set {0}. The intersection of the embedded system of open intervals
\
n∈N
1
0,
n
is empty (but this fact is not in contradiction with statement of Theorem 3.2.2).
3.2.3 An example of uncountable set
THEOREM 3.2.3. The interval [0, 1] is uncountable set.
Proof. The proof will be done indirectly. Let us suppose the interval in question [0, 1] is an
countable set. Then it follows from Theorem 2.3.2 that this interval is the set of values of
an injective sequence of the elements of R. Thus, there exists a sequence (cn )n∈N , cn ∈ R
such that its values coincides with [0, 1]. Let us divide the interval [0, 1] into three parts:
1
0, ,
3
1 2
, ,
3 3
2
,1 .
3
Then c1 ∈ [0, 1] does not belong to at least one of these parts of the interval [0, 1]. Let
us denote it by [a1 , b1 ], so c1 ∈
/ [a1 , b1 ]. This interval can be also divided into three parts
analogically as we have done with [0, 1] one step before, i.e.:
–
™
b1 − a1
a1 ,
,
3
–
™
b1 − a1 2(b1 − a1 )
,
,
3
3
15
–
™
2(b1 − a1 )
, b1 .
3
3.2 Notion of interval, intervals over the real axis
16
The one of the above written intervals which is such that c2 does not belong to it will be
denoted as [a2 , b2 ], obviously [a2 , b2 ] ⊂ [a1 , b1 ]. By induction we can construct the sequence
of embedded closed intervals in R with property:
∀n ∈ N : cn ∈
/ [an , bn ].
By making use of Theorem 3.2.2 we know that
∃c ∈ R ∀n ∈ N : c ∈ [an , bn ].
Since c ∈ [0, 1] we have:
∃n0 ∈ N : c = cn0 ∈
/ [an0 , bn0 ],
and this contradicts the previous statement.
3.2.4 Real line
In order to be able to interpret geometrically various fact and properties of mappings we will
give a description of relation between the set of all real numbers R and the set of points of
a line. This relation will be described using the real line axiom. We have already used this
axiom in an intuitive way.
Let us denote by ρ(A, B) the length of the line segment with end-points A and B . (We
suppose we can measure the lengths of line segments.) Let us have two distinct points O
and J on the line o. Let us suppose ρ(O, J) = 1. Then we can dene a mapping:
f : R 7→ o
in the following way:
(α) For x1 ∈ R+ we dene f (x1 ) =: Px ∈ o is such a point on a half-line OJ that ρ(O, Px ) =
x1
(β) For x2 ∈ R− we dene f (x1 ) =: Qx ∈ o is such a point on a half-line OJ that
ρ(O, Qx ) = −x2
(γ) For x = 0 we dene f (x) = O.
This situation is shown on Figure 3.1.
Real line axiom 3.2.4. The mapping f : R 7→ o with properties (α), (β), (γ) is bijective.
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3.2 Notion of interval, intervals over the real axis
17
x2
0
1
x1
Qx
O
J
Px
o
Figure 3.1:
The line o which is with respect to Real line axiom the image of the set R under the
mapping f is called real line, its point O is called the beginning and the line segment OJ is
called the unity.
One can dene on the real line an inner binary relation of addition of two points A, B ∈ o
(we denote it by the symbol ⊕) and an external operation of multiplication of a point A ∈ o
by a number s ∈ R (we denote it by the symbol ¯) as follows:
A ⊕ B =: E ∈ o,
(† )
s ¯ A =: F ∈ o,
where the point E is obtained as the end point of the addition of oriented segments OA and
OB (see Figure 3.2). The point O is the beginning of the real line o and ρ(O, B) = ρ(O0 , B 0 ).
o
O
A = O0
J
B
B0
Figure 3.2:
The point F can be constructed by homothety as shown in Figure 3.3. The point S ∈ o
is an image f (s) (f is the mapping from the real line axiom) and p 6= o is a line crossing
through the beginning O of the real line o.
F0
p
S0
O
J
S
A
F
o
Figure 3.3:
The following theorem holds true.
THEOREM 3.2.5. The vector space of the points of real line over R (o, R; ⊕, ¯, =) is
isomorphic with the vector space of the real numbers over R (R, R; +, ·, =).
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3.2 Notion of interval, intervals over the real axis
18
Proof. It is easy to nd out that (o, R; ⊕, ¯, =) is in fact the vector space. Let us consider the
bijective mapping f dened in the real line axiom. For a, b ∈ R we put f (a) =: A, f (b) =: B .
Then we obtain by the properties (α), (β), (γ) and (†) that
f (a + b) = A ⊕ B = f (a) ⊕ f (b),
f (sa) = s ¯ (A) = s ¯ f (a).
This theorem allows for considering the set of real numbers R and the real line o as
equivalent from the algebraic point of view. Furthermore, we can write a = A(= f (A)) for
any a ∈ R and we can use the real line o as a geometrical tool for representation of the real
numbers.
Problems
1. Let I1 , I2 ⊂ R be intervals. Decide when the sets
I1 ∪ I2 ,
I1 ∩ I2
are intervals.
2. Write down few patterns of two dimensional intervals from R2 and nd out their
number.
3. Prove the following: A set I ⊂ R is an interval if and only if it can be written as one
of the next nine possibilities (a, b, c, d ∈ R, a < b):
[a, b], [a, b), (a, b], (a, b), [c, +∞), (c, +∞), (−∞, d], (−∞, d), (−∞, +∞).
(Hint: see the proof of Theorem 3.2.1)
4. Prove the following: for any a, b ∈ R, a < b the intervals [a, b], (a, b) and R are
uncountable sets. (Hint: the function
f : [0, 1] → [a, b]
dened by
f : x 7→ (b − a)x + a
is bijective. To prove that the set R is uncountable you can use the function
g : (0, 1) → R,
)
5. Prove Theorem 3.2.5 in details.
18
g(x) =
1
1
+
.
x x−1
3.3 Further extensions of the real numbers
19
Answers
1 The set I1 ∪ I2 is an interval i I1 ∩ I2 6= ∅. The set I1 ∩ I2 is an interval i it contains
at least two distinct points.
2 81.
3.3 Further extensions of the real numbers
The extension of the set of all rational numbers to the set of real numbers was motivated by
practical purposes, e.g. to give sense to the solution to the algebraic equation:
x2 = 3.
Another question is related to the fact that we have not dened even root of a negative
number (see Denition 3.1.2). The equation
x2 = −3
has no solution in R. However, high school knowledge tells us that this equation can be
solved in the set of complex numbers C. And this set C is, in some sense, also an extension
of the set of all real numbers. In this section we will remind basic properties of the set of all
complex numbers.
3.3.1 Field of complex numbers
Since the set R is isomorphic with the set of all pairs R×{0} (an isomorphism is given by the
mapping: f : x 7→ (x, 0), x ∈ R), the following denition will be a reasonable generalization
of the set of all real numbers.
Denition 3.3.1. The set of all complex numbers C is the cartesian product R×R on which
we have dened an inner binary operation of addition and multiplication (the equivalence of
pairs is dened in the fourth section of the rst Chapter) in this way:
1 Let z1 = (x1 , y1 ) ∈ C, z2 = (x2 , y2 ) ∈ C. Then the addition of the complex numbers
z1 and z2 is the complex number:
z1 + z2 = (x1 + x2 , y1 + y2 ).
2 The product of z1 and z2 is the complex number:
z1 z2 = (x1 x2 − y1 y2 , x1 y2 + x2 y1 ).
19
3.3 Further extensions of the real numbers
20
The following theorem is a simple consequence of previous denition:
THEOREM 3.3.1. The set of all complex numbers taken as (C, +, ·, =) is a eld and taken
as (C, R, +, ·, =) is a vector space over R.
Proof. It suces to verify the properties of eld and vector space, respectively (they are
introduced in Denitions 1.6.4 and 1.6.5). The external operation - product "·" of a vector
z = (x, y) ∈ C with a scalar s ∈ R (the set {(x, 0); x ∈ R} is isomorphic with R) in the
vector space (C, R, +, ·, =) is dened by:
sz = (s, 0) · (x, y) = (sx, sy) ∈ C.
The neutral element with respect to addition is 0 := (0, 0) ∈ C. The neutral element
with respect to multiplication is (1, 0) ∈ C. The inverse element (with respect to addition)
to an element z = (x, y) ∈ C is −z = (−x, −y). The inverse element (with respect to
multiplication) to an element z = (x, y) is dened if z 6= 0 and is denoted by z −1 ≡ (x0 , y0 )
and is given by the condition
z −1 z = (1, ) ⇔ (xx0 − yy0 , xy0 + yx0 ) = (1, 0).
This system of equation has unique solution that reads
x0 =
x2
Thus,
x
,
+ y2
‚
z0 =
y0 =
−y
.
+ y2
x2
Œ
x
−y
, 2
.
2
2
x + y x + y2
If z = (x, y) ∈ C then we dene:
x =: Rez,
y =: Imz
and we call them the real part and the imaginary part of the complex number z .
If we denote complex number (0, 1) as i ((0, 1) =: i, the imaginary unit) then
i2 = (−1, 0) = −1,
and
z = (x, y) = (x, 0) + (0, y) = (x, 0) + (0, 1)(y, 0) =: x + iy.
Complex number
z̄ = (x, −y) = x − iy
20
3.3 Further extensions of the real numbers
21
is called the complex conjugate to the complex number
z = (x, y) = x + iy.
It is obvious that
z z̄ = x2 + y 2 ,
1
x = (z + z̄),
2
z + z̄ = 2x,
1
y = (z − z̄).
2i
3.3.2 Complex plane
The geometrical interpretation of complex numbers is given in the following theorem
THEOREM 3.3.2. Let o1 , o2 be two real axes with common beginning. Then the vector space of complex numbers (C, R; +, ·, =) is isomorphic with the vector space (o1 ×
o2 , R, ⊕, ¯, =). An inner binary operation ⊕ means the vector addition and the external
binary operation ¯ represents the multiplication of a vector by a real scalar.
Proof. With help of bijective mapping f introduced in Axiom 3.2.4 we can dene an bijection
from R2 onto o1 × o2 :
g(x, y) = Z ∈ o1 × o2 ,
(x, y) ∈ R × R,
by Figure 3.4.
o2
Z
y =: B
O
x =: A
o1
Figure 3.4:
Denitions of the operations ⊕ and ¯ can be also easily formulated by Figures 3.5 and
3.6.
Let (x1 , y1 ), (x2 , y2 ) ∈ R2 and let f1 , f2 be a bijection from R onto o1 and o2 , respectively,
such that
f1 (xi ) = Ai , f2 (yi ) = Bi , i = 1, 2.
21
3.3 Further extensions of the real numbers
22
o2
S1
B1 ⊕ B2
Z2
y2 =: B2
y1 =: B1
O
Z1
x2 =: A2
A1 ⊕ A2
x1 =: A1
o1
Figure 3.5:
Let
g(x1 , y1 ) = Z1 ∈ o1 × o2 ,
g(x2 , y2 ) = Z2 ∈ o1 × o2 ,
s ∈ R,
then
Z1 ⊕ Z2 = S1 ∈ o1 × o2 ,
sZ1 = S2 ∈ o1 × o2 .
The mapping g is then the requested isomorphism.
If the axes o1 and o2 from Theorem 3.3.2 are perpendicular then their cartesian product
o1 × o2 is called the complex plane. The axis o1 is called real axis and the axis o2 is called
the imaginary axis.
It is exactly Theorem 3.3.2 that allows for considering the set of all complex numbers C
and the complex plane to be equal from the algebraic point of view. We will write
(x, y) = Z(= g(x, y)),
for any (x, y) ∈ C.
3.3.3 Polar form of a complex number
From the above mentioned geometrical interpretation of a complex number one easily obtains
the following polar form of a complex number.
THEOREM 3.3.3. For any z = (x, y) ∈ C, z 6= 0 there exists just one φ ∈ [0, 2π) - called
the argument or the phase of the complex number z - and just one r > 0 - called the absolute
22
3.3 Further extensions of the real numbers
23
o2
S2
sB1 =: E
S0
Z1
y1 =: B1
x1 =: A1
J
O
sA1 =: D
s =: S
o1
Figure 3.6:
value or the modulus of the complex number z - such that
z = (r cos(φ), r sin(φ)) = r(cos(φ) + i sin(φ)).
Proof. The proof is evident from Figure 3.7.
o2
Z
y
r
φ
x
Figure 3.7:
It holds true that the numbers:
r=
È
x2 + y 2 ∈ R+ ,
23
φ ∈ [0, 2π)
o1
3.3 Further extensions of the real numbers
24
are dened uniquely by equations:
x = r cos(φ),
y = r sin(φ).
The number r is often denoted by |z|. An exact denition of used goniometric functions
is given in section 4.4 of this book.
Problems
1. Prove in details Theorems 3.3.1 and 3.3.2.
2. Formulate and explain the algorithm of construction of the image of:
a) product of complex numbers
b) symmetric element z −1 to an element z = (x, y) ∈ C in the complex plane by
using the isomorphism g from Theorem 3.3.2
3. Show that for z1 = r1 (cos(φ1 ) + i sin(φ1 )) ∈ C and z2 = r2 (cos(φ2 ) + i sin(φ2 )) ∈ C, for
any n ∈ N we have:
a) z1 z2 = r1 r2 (cos(φ1 + φ2 ) + i sin(φ1 + φ2 )),
b) z1n = rn (cos(nφ) + i sin(nφ)) - Moivre theorem,
c)
√
n
¨
z1 = z ∈ C; z =
(let us remind that
√
n
√
n
–
‚
φ + 2kπ
r cos
n
Œ
‚
φ + 2kπ
+ i sin
n
Ϊ
«
, for k = 0, 1, 2, . . . , n − 1 ,
z1 is a set of all complex numbers z for which z n = z1 ),
d) |z1 | ≥ 0 and |z1 | = 0 ⇔ z1 = 0 ∈ C,
e) |z1 · z2 | = |z1 ||z2 |,
f) ||z1 | − |z2 || ≤ |z1 ± z2 | ≤ |z1 | + |z2 |.
24