Algebraic Problems And Exercises For High School

The Educational Publisher
Columbus, 2015
ALGEBRAIC PROBLEMS
AND EXERCISES
FOR HIGH SCHOOL
Ion Goian ■ Raisa Grigor ■ Vasile Marin ■ Florentin Smarandache

Algebraic problems and exercises for high school
1
ALGEBRAIC PROBLEMS AND EXERCISES FOR HIGH SCHOOL
Sets, sets operations ■ Relations, functions ■ Aspects of combinatorics

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First imprint: Algebra în exerciții și probleme pentru liceu (in Romanian),
Cartier Publishing House, Kishinev, Moldova, 2000
Translation to English by Ana Maria Buzoianu
(AdSumus Cultural and Scientific Society)
Zip Publishing
1313 Chesapeake Ave.
Columbus, Ohio 43212, USA
Email: info@edupublisher.com
AdSumus Cultural and Scientific Society
Editura Ferestre
13 Cantemir str.
410473 Oradea, Romania
Email: ad.sumus@laposte.net
ISBN 978-1-59973-342-5

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ALGEBRAIC PROBLEMS
AND EXERCISES
FOR HIGH SCHOOL
■ Sets, sets operations
■ Relations, functions
■ Aspects of combinatorics
Columbus, 2015

4
© Ion Goian, Raisa Grigor, Vasile Marin, Florentin Smarandache,
and the Publisher, 2015.

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Foreword
In this book, you will find algebraic exercises and problems,
grouped by chapters, intended for higher grades in high schools or
middle schools of general education. Its purpose is to facilitate
training in mathematics for students in all high school categories,
but can be equally helpful in a standalone work. The book can also
be used as an extracurricular source, as the reader shall find
enclosed important theorems and formulas, standard definitions
and notions that are not always included in school textbooks.

6
Contents
Foreword / 5
Notations / 7
1. Sets. Operations with sets / 9
1.1. Definitions and notations / 9
1.2. Solved exercises / 16
1.3. Suggested exercises / 32
2. Relations, functions / 43
2.1. Definitions and notations / 43
2.2. Solved exercises / 59
3. Elements of combinatorics / 101
3.1. Permutations. Arrangements. Combinations.
Newton’s binomial / 101
3.2. Solved problems / 105
Answers / 137
References / 144

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Notations
= equality;
≠ inequality;
belongs to;
doesn’t belong to;
subset;
superset;
union;
intersection;
∅ empty set;
(or) disjunction;
(and) conjunction;
⇔
ℕ = { , , , , , … }
ℤ = {… , − ,− , , , , … }
p equivalent with ;
natural numbers
integer numbers set;
ℚ = { | , ℤ, ≠ } rational numbers set;
ℝ real numbers set;
ℂ = { + �| , , � = − } complex numbers set;
+ = { | > }, {ℤ, ℚ, ℝ};
− = { | < }, {ℤ, ℚ, ℝ};
= { | ≠ } = { }, {ℕ, ℤ, ℚ, ℝ, ℂ};
| |
[ ]
{ }
absolute value of ℝ;
integer part of x ∈ R;
fractional part of ℝ,
{ } < ;
, pair with the first element
and the second element
(also called "ordered pair");

8
, , triplet with corresponding elements
, , ;
× = { , | , } Cartesian product of set
and set ;
× × = { , , | , , }
Cartesian product of sets , , ;
universal set;
= { | } set of parts (subsets) of set ;
= ⇔ ⇔
equality of sets and ;
⇔ ⇔
is included in ;
= { | } union of sets and ;
= { | } intersection of sets and ;
= { | } difference between sets and ;
∆ = symmetrical difference;
� = ̅ = complement of set relative to ;
× relation defined on sets and ;
: → function (application) defined on
with values in ;
domain of definition of function :
domain of values of function .

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1. Sets. Operations with sets
1.1. Definitions and notations
It is difficult to give an account of the axiomatic theory of sets at
an elementary level, which is why, intuitively, we shall define a set as a
collection of objects, named elements or points of the set. A set is
considered defined if its elements are given or if a property shared by
all of its elements is given, a property that distinguishes them from the
elements of another set. Henceforth, we shall assign capital letters to
designate sets: , , , … , , , , and small letters for elements in sets:
, , , … , , , etc.
If is an element of the set , we will write and we will
read " belongs to " or " is an element of ". To express that is not
an element of the set , we will write and we will read " does
not belong to ".
Among sets, we allow the existence of a particular set, noted as
∅, called the empty set and containing no element.
The set that contains a sole element will be noted with { }.
More generally, the set that doesn’t contain other elements except the
elements , , … , will be noted by { , , … , }.
If is a set, and all of its elements have the quality , then we
will write = { | verifies } or = { | } and we will read: "
consists of only those elements that display the property (for which
the predicate is true)."
We shall use the following notations:
ℕ = { , , , , … } – the natural numbers set;
ℕ = { , , , , … } – the natural numbers set without zero;
ℤ = {… , − ,− , , , , … } – the integer numbers set;
ℤ = {± ,± , ± … } – the integer numbers set without zero;
ℚ = { | , ℤ, ℕ } – the rational numbers set;

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ℚ = the rational numbers set without zero;
ℝ = the real numbers set;
ℝ = the real numbers set without zero;
ℝ+ = { ℝ| }; ℝ+ = { ℝ| > };
ℂ = { + �| , , � = − } = the complex numbers set;
ℂ = the complex numbers set without zero;
{ , , … , } = ,
̅̅̅̅̅;
= { ℤ | } = the set of all integer divisors of number
ℤ;
= | | = the number of the elements of finite set A.
Note. We will consider that the reader is accustomed to the
symbols of Logic: conjunction (…and…), disjunction (…or… ,
implication , existential quantification (∃) and universal quantification
( ).
Let and be two sets. If all the elements of the set are also
elements of the set , we then say that is included in , or that is
a part of , or that is a subset of the set and we write . So
⇔ .
The properties of inclusion
a. , (reflexivity);
b. (transitivity);
c. , ∅ .
If is not part of the set , then we write , that is,
⇔ ∃ .
We will say that the set is equal to the set , in short = ,
if they have exactly the same elements, that is
= .
The properties of equality
Irrespective of what the sets , and may be, we have:
a. = (reflexivity);
b. = = (symmetry);

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c. = = = (transitivity).
With we will note the set of all parts of set �, in short
.
Obviously, ∅, .
The universal set, the set that contains all the sets examined
further, which has the containing elements’ nature one and the same,
will be denoted by .
Operations with sets
Let and be two sets, , .
1. Intersection.
= { | },
i.e.
, (1)
, ’
2. Union.
= { | },
i.e.
, (2)
, ’
3. Difference.
= { | },
i.e.
, (3)
, ’
4. The complement of a set. Let . The difference
is a subset of , denoted � and called the complement of
relative to , that is
� = = { | }.
In other words,
� , (4)
� . ’

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Properties of operations with sets
= , = (idempotent laws)
= , = (commutative laws)
= ;
= (associativity laws)
= ;
= (distributive laws)
= ;
= (absorption laws)
� = � � ;
� = � � (Morgan’s laws)
Two "privileged" sets of are ∅ and .
For any , we have:
∅ ,
∅ = , ∅ = ∅, � ∅ = ,
= , = , � = ∅,
� = , � = ∅,
�( � ) = (principle of reciprocity).
Subsequently, we will use the notation � = ̅.
5. Symmetric difference.
∆ = .
Properties. Irrespective of what the sets , and are, we
have:
a. ∆ = ∅;
b. ∆ = ∆ (commutativity);
c. ∆∅ = ∅∆ = ;
d. ∆ ∆ = ;
e. ∆ ∆ = ∆ ∆ (associativity);
f. ∆ = ∆ ;
g. ∆ = .

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6. Cartesian product. Let and be two objects. The set
{{ }, { , }}, whose elements are the sets { } and { , }, is called an
ordered pair (or an ordered couple) with the first component and
the second component and is denoted as , . Having three objects
, and , we write , , = , , and we name it an ordered
triplet.
Generally, having objects , , … , we denote
, , … , = (…( , ) … )
and we name it an ordered system of elements (or a cortege of
length ).
We have
, , … , = , , … ,
= = … = .
Let , . The set
× = { , | }
is called a Cartesian product of sets and . Obviously, we can define
× × = { , , | }.
More generally, the Cartesian product of the sets , ,… ,
× × … × = { , , … , | � �, � = ,
̅̅̅̅̅}.
For = = = = = . . . = , we have
× ≝ , × × ≝ , × × … ×
⏟ ≝ .
ori
For example, ℝ = { , , | , , ℝ}. The subset
Δ = { , | }
is called the diagonal of set .
Examples.
1. Let = { , } and = { , , }. Then
× = { , , , , , , , , , , , }
and
× = { , , , , , , , , , , , }.
We notice that × ≠ × .

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2. The Cartesian product ℝ = ℝ × ℝ can be geometrically
represented as the set of all the points from a plane to wich a
rectangular systems of coordinates has been assigned,
associating to each element , ℝ a point , from the plane
of the abscissa and the ordinate .
Let = [ ; ] and = [ ; ], , ℝ . Then × can be
geometrically represented as the hatched rectangle (see Figure
1.1), where , , , , , , , .
Figure 1.1.
The following properties are easily verifiable:
a. × × ;
b. × = × × ,
× = × × ;
c. × = ∅ = ∅ = ∅ ,
× ≠ ∅ ≠ ∅ ≠ ∅ .

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7. The intersection and the union of a family of sets. A family of
sets is a set { �|� } = { �}� � whose elements are the sets �, �
, � . We say that �,� , � is a family of sets
indexed with the set .
Let there be a family of sets { �|� }. Its union (or the union of
the sets �, � ) is the set
. �
The intersection of the given family (or the intersection of the
sets �, � ) is the set
In the case of = { , ,… , }, we write
8. Euler-Wenn Diagrams. We call Euler Diagrams (in USA –
Wenn’s diagrams) the figures that are used to interpret sets (circles,
squares, rectangles etc.) and visually illustrate some properties of
operations with sets. We will use the Euler circles.
Example. Using the Euler diagrams, prove Morgan’s law.
Solution.
In fig. 1.2.a, the hatched part is ; the uncrossed one
(except ) represents � .

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Figure 2.2.a Figure 3.2.b
In fig. 1.2b the side of the square hatched with is equal
to � and the one hatched with / / / / is equal to � . All the
hatched side forms � � (the uncrossed side is exactly
).
From these two figures it can be seen that � (the
uncrossed side of the square in fig. 1.2.a coincides with �
� (the hatched side of fig. 1.2.b), meaning
� = � � .
1.2. Solved exercises
1. For any two sets and , we have
=
Solution. Using the definitions from the operations with sets, we
gradually obtain:
From this succession of equivalences it follows that:
which proves the required equality.

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Remark. The equality can also be proven using Euler’s diagrams.
So = .
2. Whatever , are, the following equality takes place:
Solution. The analytical method. Using the definitions from
the operations with sets, we obtain:
This succession of equivalences proves that the equality from
the enunciation is true.
The graphical method. Using Euler’s circles, we have

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Figure 4.3.a Figure 5.3.b
In fig. 1.3.a, we have ( ) ( ), which represents the
hatched side of the square. From fig. 1.3.b it can be seen that
( ) ( ) = ( ).
3. For every two sets , , the equivalence stands true:
Solution. Let = . We assume that ≠ . Then there
is with or with .
In the first case we obtain and , which
contradicts the equality = . In the second case, we obtain
the same contradiction.
So, if = = .
Reciprocally, obviously.
4. Sets = { , , , , , , , , , }, = { , , , , } and =
{ , , } are given. Verify the following equalities:
a) = ;
b) = .
Solution. a) We have = { , , , , , , }, =
{ , , }, = { , , , , }, = { , , , , , , }, =
{ , , } = .

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b) For the second equality, we have
= { } , = { , , , , , , , , , } ,
= { , , , , , , , , , } = .
5. Determine the sets A and B that simultaneously satisfy the
following the conditions:
1) = { , , , , };
2) = { , , };
3) ;
4) .
Solution. From 1) and 2) it follows that { , , } and
{ , , } . From 3) it follows that or . If ,
then from = { , , , , } it follows that . But, if ,
then , because, on the contrary it would follow that =
{ , , }. So and remain. Similarly, from 4) it follows that
and so . In other words, { , , } { , , , } and
{ , , } { , , , } with , and that is why
= { , , , }, and = { , , , }.
Answer: = { , , , }, = { , , , }.
6. The sets = { + │ ℤ}, = { │ ℤ} and =
{ + − │ ℤ}, are given. Prove that = .
Solution. To obtain the required equality, we will prove the truth
of the equivalence:
.
Let . Then and and that is why two
integer numbers , ℤ, exist, so that = + =
= − . In this equality, the right member is divisible by 4,
and 11, 4 are prime. So, from it follows that , giving =
for one ℤ. Then
= + = ∙ + = ∙ + − = + −

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which implies , in other words, we have proven the implication
⇒ . (1)
Similarly, let . Then ℤ exists with
= + − = ∙ + − = ∙ + = +
Taking = ℤ and + = ℤ, we obtain
= + = ,
which proves the truth of the implication
⇒ . (2)
From (1) and (2) the required equality follows.
7. The following sets are given
and = ℕ.
Prove that:
a) all sets with = { , , , , , , };
b) all = { ℤ│ }, so that = { }.
Solution. We determine the set :
Then = ; = { , , }.
a) All possible subsets of are
∅, { }{ }{ }{ , }{ , }{ , }{ , , } =
The required sets are such that = { , , , , , , } and
will thus be like = { , , , }, where , namely, the sets
required at point a) are:
{ , , , } , { , , , , } , { , , , , } , { , , , , } ,
{ , , , , , }, { , , , , , }, { , , , , , }, { , , , , , , }..
b) Because ℤ, then and vice versa. Considering
that = { , , , , , , }, we obtain { , }, namely
{− ,− , , } = . The parts of the set are:

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∅ , {− } , {− } , { } , { } , {− ,− } , {− , }, {− , } , {− , } ,
{− , }, { , }, {− , − , }, {− , − , }, {− , , }, {− , , }, .
From the condition = { } it follows that is one of the
sets
= { }, = {− , }, = {− , }, = { , }, = {− ,− , },
= {− , , }, = {− , , }, = = {− ,− , , }.
Answer: a) { , , , , , , , };
b) { , , , , , , , }.
8. Determine , , and ∆ , if
= { , , , , , } , ∆ = { , } , ∆ = { , } , =
= { , }.
Solution. From = = { , }, it follows that { , }
.
We know that
∆ = = ,
∆ = = .
Then:
∆ ⇔ (
).
The following cases are possible:
a) and ;
b) and
(case no. 3, and ⇒ = { , }, is impossible).
In the first case and , from ∆ = { , } it follows
that , because otherwise and ⇒
∆ = { , }. So, in this case we have = { , }which is
impossible, so it follows that , . Similarly, we obtain
and and , and , and , ,
.
In other words, we have obtained:

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= { , , , }, = { , , , }, = { , } and ∆ = { , , , }.
Answer: = { , , , }, = { , , , }, = { , } and
∆ = { , , , }.
9. The sets = { , }, = { , } are given. Determine the fol-
lowing sets:
a) × ; b) × ; c) ;
d) ; e) × × ; f) × ;
g) × × .
Solution. Using the definition for the Cartesian product with
two sets, we obtain:
a) × = { , , , , , , , };
b) × = { , , , , , , , };
c) = { , , , , , , , };
d) = { , , , , , , , };
e) × × = { , };
f) = { , , }; × = { , , , , ,
, , , , , };
g) × × = { , , , , , , , ,
, , , } = × .
10. The sets = { , , }, = { , , } are given. Determine
and , knowing that { , } × { , } × .
Solution. We form the sets × and { , } × { , } :
Because { , } × { , } × , we obtain
For = and = , we have , × .
Answer: = , = .

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11. If , then
Prove.
Solution. ⇒ = and that is why
(we have used the equality × = × × ).
12. How many elements does the following set have:
Solution. The set has as many elements as the fraction
+ ∕ + + has different values, when takes the values
, , , … , . We choose the values of for which the fraction takes
equal values.
Let , ℕ , < with
Then
But ℕ and consequently
For = , we obtain = , and for = , we have = .
Considering that < , we obtain = and = . So, only for =
and = , the same element of the set : = ∕ is obtained.

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Answer: the set has elements, namely = .
13. Determine the integer numbers , that make the following
statement true − ⋅ − = .
Solution. We write
As is a prime number, and , ℤ, the following cases are
possible:
meaning the proposition is true only in these situations:
14. Determine the set
Solution. Because:
it follows that that equality
is possible, if and only if we simultaneously have:
Then:
a) if at least two of the numbers , and are different, we
cannot have the equality:
namely in this case = ∅;
b) if = = , then = − and = {− }.
Answer: 1) for = = , we have = {− }; 2) if at least two of
the numbers , and are different, then = ∅.

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15. Determine the set
Solution. Possible situations:
+ − / or + > − / .
We examine each case:
In this case we have:
All integer numbers for which − ∕ applies, are:
− , , .
Then
All integer numbers for which ∕ < x applies, are:
, , , , , , , .
We thus obtain:
Answer: = {− , , , , , , , , , , }.
16. Determine the values of the real parameter for which the
set has:
a) one element;
b) two elements;
c) is null.
Solution. The set coincides with the set of the solutions of
the square equation
and the key to the problem is reduced to determining the values of the
parameter ℝ for which the equation has one solution, two
different solutions or no solution.

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a) The equation has one (two equal) solution, if and only if
= for = − ≠ or if = − = .
We examine these cases:
2) For = , the equation becomes − + = ⇔ = . So,
the set consists of one element for
b) The equation has two different roots, if and only if > ,
namely
c) The equation doesn’t have roots < ⇔ −
− >
Answer:
17. Let the set = { , , } and = { , , }. Write the elements
of the sets and ( ) × .
Solution. a) For the first set we have:

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= { , , , , , , , , , , , , , , , , , },
= { , , , , , , , , , , , , , , , , , },
= { , , , , , , , }.
b) For the second set we have:
Then
Answer:
18. Given the sets = { , , , , , , } and = { , , }, solve
the equations ∆ = and ∆ ∆ = .
Solution. In order to solve the enunciated equations, we use the
properties of the symmetrical difference: ∆ ∆ = , its
associativity and commutativity .
But
Consequently,
We calculate
Thus
Answer: = , = ∅.

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19. The following sets are given
Determine the sets , , , , , ,
and � × .
Solution. 1) We determine the sets and .
The inequation is equivalent to the totality of the three
systems of inequations:
So
In other words,
2) We determine the required sets with the help of the graphical
representation

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20. 40 students have a mathematics test paper to write, that
contains one problem, one inequation and one identity. Three
students have successfully solved only the problem, 5 only the
inequation, 4 have proven only the identity, 7 have solved not only the
problem, 6 students – not only the inequation, and 5 students have
proven not only the identity. The other students have solved
everything successfully. How many students of this type are there?

30
Solution. Let be the set of students who have correctly solved
only the problem, – only the inequation, – that have proven only
the identity, – the set of students that have solved only the problem
and the inequation, – the set of students that have solved only the
problem and have proven the identity, – the set of students who
have solved only the inequation and have proven the identity, and –
the set of students that have successfully solved everything.
From the given conditions, it follows that = , = ,
= , = , = , = .
But, as each of the students who have written the test paper
have solved at least one point of the test correctly and, because the
sets , , , , , , have only elements of the null set in common,
the union of the sets , , , , , , is the set of students that have
written the paper.
Consequently,
So
Answer: 4 students out of all that have taken the test have
solved everything successfully.
21. (Mathematician Dodjson’s problem)
In a tense battle, 72 out of 100 pirates have lost one eye, 75 –
one ear, 80 – one hand and 85 – one leg. What is the minimum number
of pirates that have lost their eye, ear, hand and leg at the same time?
Solution. We note with the set of one-eyed pirates, with –
the set of one eared pirates, with – the set of pirates with one hand
and with – the set of the pirates with one leg.
The problems requires to appreciate the set .

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It is obvious that the universal set is composed from the set
and the number of pirates that have kept either both
eyes, or both ears, or both hands, or both legs.
Consequently,
It follows that the set is not smaller doesn’t have fewer
elements) than the sum of the elements of the sets ̅, ̅, ̅, ̅ and
(equality would have been the case only if the sets , ,
and , two by two, do not intersect).
But,
Substituting, we have = namely
+ + + + .
Consequently, − − − = .
Therefore, no less than 10 pirates have lost their eye, ear, hand
and leg at the same time
Answer: No less than 10 pirates.
22. Of a total of 100 students, 28 study English, 8 – English and
German, 10 - English and French, 5 – French and German, 3 students -
all three languages. How many study only one language? How many
not even one language?
Solution. Let be the set of students that study English, -
German, - French.
Then the set of students attending both English and German is
, English and French - , French and German - ,
English, German and French - , and the set of students
that study at least one language is .
From the conditions above, it follows that the students who only
study English form the set , only German -
, only French - .

32
But, because , we have =
− = − + −
= − − + .
Similarly, ( ) = − −
+ ; ( ) = −
− + .
Let be the set of students who only study one language, then
Consequently,
= + + − . − . − . + . = . = .
The number of students who don’t study any language is equal
to the difference between the total number of students and the
number of students that study at least one language, namely =
− , where is the set of students that don’t study
any language and the set of all 100 students.
From problem 20 we have
So = − = .
Answer: 63 students study only one language, 20 students don’t
study any language.

33
1.3. Suggested exercises
1. Which of the following statements are true and which are
false?
2. Which of the following statements are true and which are
false ( , and are arbitrary sets)?
3. Let = { ℚ│ − + = },
= { ℚ│ + + = }; = { ℚ│ + + = }
a) Determine the sets , and .
b) State if the numbers / , , , / belong to these sets or
not.

34
4. Determine the sets:
5. Let = { ℕ│x = n + m,n, m ℕ };
a) Write three elements belonging to set .
b) Determine if 26, 28, 33 belong to .
6. Indicate the characteristic properties of the elements
belonging to the sets: = { , , }, = { , , } , =
{ , , , , }, = { , , , , }.
7. How many elements do the following sets have?
8. Let there be the set = {− , − , − , , , }. Determine the
subsets of :
9. Determine the sets:

35
10. Compare , , =, , the sets and , if:
11. Let = { , , , , , , }, = { , , , , , }. Using the
symbols , , , (complementary), express (with the help of , and
ℕ the sets:
.

36
12. Determine the set , in the case it is not indicated and its
parts , , simultaneously satisfy the conditions:
has more elements than ;
has more elements than ;

37
13. Determine the sets , , , , , , ∆ ,
, , , × , × , × , × , if:

38
14. Let
a) Determine the sets:
b) How many elements does the set = { │
⁄ } have?
15. Determine the sets , , if
16. Determine the set and its parts and , if
17. Compare the sets and , if

39
18. Determine the values of the real parameter for which the
set has one element, two elements or it is void, if:
19. Determine the number of elements of the set :

40
20. The sets , , are given. Determine .
21. Determine , if:
22. Prove the properties of the operations with sets.
23. Determine the equalities ( , , etc. are arbitrary sets):

41
24. Given the sets = { , , }, = { , , }, = { , , } and
= { , , }, write the elements of the following sets:
25. Given the sets , and , solve the equations ∆ =
, where { , , ∆}:
26. Determine the sets , , �, ,
̅ , ,
� , � , if:

42

43
2. Relations, functions
2.1. Definitions and notations
Relations, types. Relations formation
Let and be two non-empty sets, and × their Cartesian
product. Any subset × is called a relation between the
elements of and the elements of . When = , a relation like
× is called a binary relation on set .
If there exists a relation like × , then for an ordered pair
, × we can have , or , . In the first case,
we write and we read " is in relation with ", and in the second
case – , that reads " is not in relation with ". We hold that
, .
By the definition domain of relation A X B, we understand
the subset � consisting of only those elements of set that are in
relation with an element from , namely
By values domain of relation × we understand the
subset ρ consisting of only those elements of set that are in
relation with at least one element of , namely
The inverse relation. If we have a relation × , then by
the inverse relation of the relation we understand the relation
−
× as defined by the equivalence
namely

44
Example 1. Let us have = { , }, = { , , } and also the
relations
Determine the definition domain and the values domain of
these relations and their respective inverse relations.
Solution.
The formation of relations. Let , , be three sets and let us
consider the relations × , × . The relation ×
formed in accordance with the equivalence
namely
is called the formation of relation and .
Example 2. Let us have , , , , , from Example 1.
Determine
− −
and −
.
Solution. We point out that the compound of the relations
× with × exists if and only if = .
Because × , × , it follows that and
doesn’t exist.
We determine :

45
, and { , , , } ⇒ { , , , } ;
, , but in we don’t have a pair with the first
component 6. It follows that
c) × , × ⇒ exists. By repeating the
reasoning from b),
and −
= { , }.
and − −
= { , , , }.
The equality relation. Let be a set. The relation
is called an equality relation on . Namely,
Of a real use is:
1stTheorem. Let , , , be sets and × , × ,
× relations. Then,
1) = (associativity of relations formation)
The equivalence Relations. The binary relation is called:

46
a) reflexive, if whatever is;
b) symmetrical, if ⇒ , ⩝ , ;
c) transitive, if ⋀ ⇒ , (⩝ , , ;
d) anti-symmetrical, if ⋀ ⇒ = , (⩝ , , ;
e) equivalence relation on A, if it is reflexive, symmetrical and
transitive;
f) irreflexive, if whatever is.
Let be an equivalence relation on set . For each element
, the set
is called the equivalence class of modulo (or in relation to R), and
the set
is called a factor set (or quotient set) of through .
The properties of the equivalence classes. Let be an
equivalence relation on set and , . Then, the following
affirmations have effect:
Partitions on a set. Let be a non-empty set. A family of
subsets { �│� } of is called a partition on (or of ), if the
following conditions are met:
2nd
Theorem. For any equivalence relation on set , the factor
set / = { �│ } is a partition of .

47
3rd
Theorem. For any partition = { �│� } of , there exists
a sole equivalence relation on , hence
The relation is formed according to the rule
It is easily established that is an equivalence relation on and
the required equality.
Example 3. We define on set ℤ the binary relation according to
the equivalence ⇔ − , where ℕ , fixed, ⩝ ,
ℤ.
a) Prove that is an equivalence relation on ℤ.
b) Determine the structure of the classes of equivalence.
c) Form the factor set ℤ/ . Application: = .
Solution. a) Let , , ℤ. Then:
1) reflexivity − = ⇒ ;
2) symmetry
3) transitivity
From 1) - 3) it follows that is an equivalence relation on ℤ.
b) Let ℤ. Then
In accordance with the theorem of division with remainder for
integer numbers and , we obtain
Then

48
where = + ℤ, and therefore
where is the remainder of the division of through . But,
In other words, the equivalence class of ℤ coincides with the
class of the remainder of the division of through .
c) Because by the division to only the remainders , , , . . .,
− , can be obtained, from point b) it follows that we have the
exact different classes of equivalence:
The following notation is customarily used
. Then
where �̂ consists of only those integer numbers that divided by yield
the remainder
.
For n = 5, we obtain
with
Definition. The relation is called a congruency relation
modulo on ℤ, and class â = is called a remainder class modulo
and its elements are called the representatives of the class.
The usual notation:
⇔ − ⇔ ≡ (mode ) ( is congruent with
modulo ),and

49
is the set of all remainder classes modulo .
Example 4 (geometric). Let be a plane and the set of all the
straight lines in the plane. We define as the binary relation in
accordance with
a) Show that is an equivalence relation on .
b) Describe the equivalence classes modulo .
c) Indicate the factor set.
Solution. a) It is obvious that is an equivalence relation (each
straight line is parallel to itself; if ∥ ⇒ ∥ and
b) Let . Then the class
consists only of those straight lines from that are parallel with the
straight line .
c) / = { │ } is an infinite set, because there are an
infinity of directions in plane .
Example 5. The set = { , , , , , , , , }is given and its parts
a) Show that = { , , , } is a partition of .
b) Determine the equivalence relation on .
c) Is = { , , } a partition on ?
Solution. a) 1) We notice that
meaning the system of the subsets of the set A defines a partition on
.

50
b) In accordance to the 3rd Theorem, we have
So:
= { , , , , , , , , , , , , , , , , , , , , , ,
, , , , , , , , , , , , , , , , , , , , , , , }.
which proves that system does not define a partition on .
Order relations. A binary relation on the set is called an
order relation on , if it is reflexive, anti-symmetrical and transitive.
If is an order relation on , then −
is also an order relation
on (verify!). Usually, the relation is denoted by " " and the
relation −
by " ", hence
With this notation, the conditions that " " is an order relation
on the set are written:
 reflexivity ⇒ ;
 asymmetry ⋀ ⇒ = ;
 transitivity ⋀ ⇒ .
Example 6. On the set ℕ we define the binary relation in
accordance with
Show that is an order relation on ℕ.
Solution. We verify the condition from the definition of order
relations.
1) Reflexivity

51
2) Asymmetry. Let , ℕ with and . It follows that
there are the natural numbers , ℕ with = . and = . .
Then
which implies that
3) Transitivity. Let , , ℕ with and . Then there
exists , ℕ with = and = hence,
Because , ℕ, the implication is true
which proves that is an order relation on the set ℕ .
Remark. The order relation is called a divisibility relation on ℕ
and it is written as , namely
( │ reads "b divides a", and "a is divisible through b").
Functional relations
Let and be two sets. A relation × is called an
application or a functional relation, if the following conditions are met:
1)⩝ ∃ , so as ;
An application (or function) is a triplet = , , , where
and are two sets and × is a functional relation.
If × is an application, then for each there exists,
according to the conditions 1) and 2) stated above, a single element
, so that ; we write this element with . Thus,
The element is called the image of element
through application , set is called the definition domain of noted
by = , and set is called the codomain of and we usually say

52
that is an application defined on with values in . The functional
relation is also called the graphic of the application (function) ,
denoted, later on, by . To show that is an application defined on
with values in , we write : ⟶ or → , and, instead of
describing what the graphic of (or ′ ) is, we indicate for each
its image . Then,
i.e.
The equality of applications. Two applications = , ,
and = , , are called equal if and only if they have the same
domain = , the same codomain = and the same graphic
= . If , : ⟶ , the equality = is equivalent to =
, ⩝ , that is to say
The identical application. Let be a set. The triplet , , is,
obviously, an application, denoted by the same symbol (or ) and it
is called the identical application of set .
We have
Consequently,
: ⟶ and = for ⩝ .
By , we will note the set of functions defined on with
values in . For = , we will use the notation instead of
, .
In the case of a finite set = { , , … , }, a function
�: ⟶ is sometimes given by means of a table as is:

53
where we write in the first line the elements , , … , of set , and
in the second line their corresponding images going through �, namely
� , � , … , .
When = { , ,… , }, in order to determine the application,
we will use �: ⟶ and the notation
more frequently used to write the bijective applications of the set in
itself. For example, if = { , }, then the elements of are:
If : ⟶ , , , then we introduce the notations:
- the image of subset through the application .
Particularly, � = � , the image of application � ;
is the preimage of subset through the application .
Particularly, for , instead of writing − { } , we simply
note −
, that is
- the set of all preimages of through the application , and
- the complete preimage of set through the application .
The formation of applications. We consider the applications
= , , and = , , , so as the codomain of to coincide
with the domain of . We form the triplet = , , .
Then is also an application, named the compound of the
application with the application , and the operation " " is called the
formation of applications. We have

54
namely
4th
Theorem. Given the applications
. It follows that a) (ℎ ) = ℎ ( )
(associativity of the formation of applications); b) = = .
Example 7. We consider the relations ℝ × ℝ and
[ , +∞ × [ , +∞ , ℝ × ℝ, defined as it follows: ⇔ = ,
⇔ = , ⇔ = + .
A. Determine which of the relations , −
, , −
, , −
are
functional relations.
B. Find the functions determined at point A.
C. Calculate , , ℎ, ℎ , ℎ ℎ−
, ℎ−
ℎ (f, g, h are
the functions from point B.)
D. Calculate ℎ − , ℎ−
ℎ / , ℎ / .
Solution. We simultaneously solve and .
a) We examine the relation .
ℝ ⇒ ℝ ⇒ ℝ, i.e.
1)⩝ ℝ ∃ = ℝ, so that ;
which means that is a functional relation. We write the application
determined by through , = ℝ, ℝ, .
b) We examine the relation −
.
which means that if ℝ = { ℝ│ < }, then there is no ℝ,
so that −
, which proves that −
is not a functional relation.

55
c) For the relation , we have: and −
are functional relations.
We denote by = [ , +∞ , [ ,+∞ , ,and −
= [ , +∞ , [ ,+∞ ,
−
, the functions (applications) defined by and −
, respectively.
d) We examine the relation :
1) ℝ ⇒ + ℝ and so
so that ;
which proves that is a functional relation.
e) For the relation −
, we obtain:
1) ⩝ ℝ ∃ = − ℝ, so that y −
;
meaning that −
is also a functional relation. We denote by ℎ =
ℝ,ℝ, and ℎ−
= ℝ,ℝ, −
.
C. From A and B, it follows that:
a) Then , and ℎ don’t exist, because the domains
and codomains do not coincide:
= ℝ, ℝ, , = [ , +∞ , [ ,+∞ , , ℎ = ℝ, ℝ, ).
b) We calculate ℎ, ℎ , ℎ ℎ−
and ℎ−
ℎ.
So ℎ ≠ ℎ and ℎ ℎ−
= ℎ−
ℎ = ℝ.
D. We calculate:

56
Injective, surjective and bijective applications. An application
: ⟶ is called:
1) injective, if
(equivalent: ≠ ⇒ ≠ ;
2) surjective, if
(any element from B has at least one preimage in );
3) bijective, if is injective and surjective.
Remark. To prove that a function : ⟶ is a surjection, the
equation = must be solvable in for any .
Of a real use are:
5th
Theorem. Given the applications → → , we have:
a) if and are injective, then is injective;
b) if and are surjective, then is surjective;
c) if and are bijective, then is bijective;
d) if is injective, then is injective;
e) if is surjective, then is surjective.
6th
Theorem. The application : ⟶ with the graphic =
is a bijective application, if and only if the inverted relation −
is a
functional relation (f−
is an application).
This theorem follows immediately from
Inverted application. Let : ⟶ be a bijective application
with the graphic = . From the 6th
Theorem, it follows that the
triplet −
= , , −
is an application (function). This function is
called the inverse of function .
We have
−
: ⟶ , and for ,

57
i.e.
7th
Theorem. Application : ⟶ is bijective, if and only if
there exists an application : ⟶ with = and = .
In this case, we have = −
.
Real functions
The function : ⟶ is called a function of real variable, if
= ℝ. The function of real variable : ⟶ is called a real
function, if ℝ. In other words, the function : ⟶ is called a
real function, if A ℝ and B ℝ . The graphic of the real
function : ⟶ is the subset of ℝ , formed by all the
pairs , ℝ with and = , i.e.:
If the function is invertible, then
Traditionally, instead of −
= we write = −
. Then
the graphic of the inverted function −
is symmetrical to the graphic
function in connection to the bisector = .
Example 8. For the function
the inverted function is
The graphic of function is the branch of the parabola =
comprised in the quadrant I, and the graphic of function −
is the
branch of parabola = comprised in the quadrant I (see below Fig.
2.1).
Algebra operations with real functions
Let , : ⟶ ℝ be two real functions defined on the same set.
We consider the functions:

58
= + : ⟶ ℝ ,
defined through = + , ⩝ ;
= + g – sum-function.
= ∙ : ⟶ ℝ ,
defined through = · , ⩝ ;
= . – product-function.
= − : ⟶ ℝ,
defined through = − , ⩝ ;
= − – difference-function.
= : ⟶ ℝ,
defined through =
�
�
, ⩝
where = { │ ≠ };
= – quotient-function.
│ │: ⟶ ℝ,
defined through │ │ = │ = {
, � ,
− ,� < ,
}
for ⩝ ;
│ │- module-function.
Figure 2.1.

59
Example 9. Let , : [ ,+∞ ⟶ ℝ with = and =
√ . Determine ± , ∙ and / ·
Solution. As , share the same definition domain, the
functions ± , ∙ and / make sense and
Example 10. Let : ℝ ⟶ [ , +∞ , = and : [ , +∞ ⟶
ℝ, = √ . Determine a) and b) .
Solution. a) = � makes sense and we obtain the function
�: ℝ ⟶ ℝ with
b) = : [ , +∞ ⟶ [ ,+∞ , also makes sense, with
2.2. Solved exercises
1. Let = { , , , } and = { , , , }, × , =
{ , | } :
a) What is the graphic of the relation ?
b) Make the schema of the relation .
Solution.
a) Because , and , it follows that
So:
b) See Fig. 2.2 below.

60
Figure 2.2.
2. Let = { , , , } and = { , , , , }. Write the graphic of
the relation = { , │ + = } × .
Solution. We observe that + = ⇔ = −
⇒ is even and − . Considering that , and ,
we obtain: , , and , . The equality + = is
met only by = and = . So = { , }.
3. Let = { , , , } and = { , , , }. Write the relation
using letters , and , if the graphic of the relation is known.
Solution. , ⇔ .
4. On the natural numbers set ℕ we consider the
relations , , , ℕ , defined as it follows: , ℕ : =
{ , , , , , , , } ; ⇔ ; ⇔ − = ;
⇔ = .
a) Determine , , , , , , �, �.
b) What properties do each of the relations , , , have?
c) Determine the relations −
, −
, −
and −
.
d) Determine the relations , , − −
, −
,
and −
.

61
Solution:
(for any natural numbers, there are natural numbers that exceed it).
(for any natural numbers there is at least one natural number that
exceeds it).
(for example, equation = + doesn’t have solutions in ℕ).
(for any ℕ, we have = / .
b) 1) is symmetrical, transitive, but is not reflexive. For
example , ; because (3,3) :, it follows that isn’t
antireflexive either.
2) is - reflexive , ⩝ ℕ ,
- anti-symmetrical ⋀ ⇒
⋀ ⇒ = ,
- transitive ⋀ ⇒ .
So is an order relation on set ℕ.
3) is antireflexive , ⩝ ℕ , anti-symmetrical
( ⇒ − = and − = ⇒ − = − ⇒
= ), but it is not symmetrical ⇒ − = ⇒ − =
− ⇒ , and it is not transitive ⋀ ⇒ − =
⋀ − = ⇒ − = ⇒ .
4) is not antireflexive ≠ , ⩝) ℕ , = . , it is anti-
symmetrical ⋀ ⇒ = ⋀ = ⇒ = ⇒ =
= , and for ≠ ≠ , the equalities = and = cannot
take place), it is not reflexive (for example, ≠ . ⇒ ), it is not

62
transitive ⋀ ⇒ = ⋀ = ⇒ = ≠ ,
generally ⇒ .
So
i.e.
i.e.
i.e.
which implies
i.e.
In other words,

63
⇔ + = , which implies
which implies
5. We consider the binary relation defined on ℝ as it follows:
a) Prove that is an equivalence relation.
b) Determine the factor set ℝ/ .
Solution.
1) Because = , ⩝) ℝ, we have , ⩝) ℝ.
3) Let and , , , ℝ. Then
In other words, the binary relation is reflexive, symmetrical
and transitive, which proves that is an equivalence relation on ℝ.
b) Let ℝ. We determine the class of equivalence of in
connection to .
Then the set factor is
We observe that
In other words, for ≠ , we have ≠ − and therefore
= − , and the factor can also be written

64
6. Let = { , , , , } and = { , , , , }. Which of the
diagrams in Fig. 2.3 represents a function defined on with values in
and which doesn’t?
Solution. The diagrams a), c), e) reveal the functions defined on
with values in . Diagram b) does not represent a function defined
on with values in , because the image of 2 is not indicated.
Diagram d) also does not represent a function defined on with values
in , because has two corresponding elements in , namely , .
7. Let A = { , , } and = { , }. Write all the functions
defined on with values in , indicating the respective diagram.
Solution. There are eight functions defined on with values in .
Their diagrams are represented in Fig. 2.4.
Figure 2.3.

65
Figure 2.4.
8. Let = { , , , }, = { , , }, = { , , , } and =
{ , }.
a) Draw the corresponding diagrams for two surjective functions
defined on with values in .
b) Draw the diagram of the injective functions defined on with
values in .
c) Draw the diagram of a function denied on with values in ,
that is not injective.
d) Draw the corresponding diagrams for two bijective functions
defined on with values in .

66
Solution. a) The diagrams of two surjective and two non-
surjective functions defined on with values in are represented in
Fig. 2.5 a) and b).
Figure 2.5.
b) The diagram of the injective functions defined on with
values in are represented in Fig. 2.6.
c) The diagram of a non-injective function defined on with
values in is represented in Fig. 2.7.
d) The diagrams of the two bijective functions defined on with
values in are represented in Fig. 2.8.

67
Figure 2.5.
Figure 2.6.
Figure 2.7.
Figure 2.8.

68
9. Let = { , , , }, = { , , , } and the function
: ⟶ as defined by the equality = − + , ⩝
.
a) Is the function surjective?
b) Is injective?
c) Is bijective?
Solution. a) We calculate = { , , , } =
{ , , , } = . So, is surjective.
b) The function is also injective, because = , = ,
= and = , i.e. ≠ ⇒ ≠
c) The function is bijective.
10. Using the graphic of the function : ⟶ , : ℝ ⟶ ℝ,
show that the function is injective, surjective or bijective.
a) = {
− , � −∞, ,
− , � [ , +∞ .
c) : [− ,+∞] ⟶ ℝ, = {
− , � − < ,
− , � < ,
− . + ,� > .
Solution. If any parallel to the axis of the abscises cuts the
graphic of the function in one point, at most (namely, it cuts it in one
point or not at all), then the function is injective. If there is a parallel to
the abscises’ axis that intersects the graphic of the function in two or
more points, the function in not injective. If is the values set of
the function and any parallel to the abscises’ axis, drawn through the
ordinates axis that are contained in , cuts the graphic of the
function in at least one point, the function is surjective. It follows that
a function is bijective if any parallel to the abscises’ axis, drawn
through the points of , intersects the graphic of in one point.
a) The graphic of the function is represented in Fig. 2.9.

69
We have = −∞,+∞ and any parallel = , ℝ to
the abscises’ axis intersects the graphic of the function in one point.
Consequently, is a bijectve function.
= {
− + , � − + ,
− + − , � − + <
=
= {
− + , if −∞, ] [ ,+∞ ,
− + − , if , .
Figure 2.9.
The graphic of the function is represented in Fig. 2.10.
We have = [ ; ] [− ; ]. Any parallel = ,
, intersects the graphic of the function in four points; the
parallel = intersects the graphic in three points; any parallel =
, ; ] intersects the graphic of the function in two points.
So the function is neither surjective, nor injective.
c) The graphic of the function is represented in Fig. 2.11. We
have

70
Any parallel = to the abscises’ axis cuts the graphic of the
function in one point, at most = − , = , = and that
is why is injective. The equation = ℝ has a solution
only for , hence is not a surjective function.
d) We explain the function :
= max + , − = {
+ , � − + ,
− , � + < −
=
= {
+ , � ,
− , � < .
Figure 2.10.
Figure 2.11.

71
The graphic of the function is represented in Fig. 2.12.
We have = ℝ, = [ , +∞ . Any parallel = ,
, +∞ intersects the graphic in two points, therefore is not
injective. Straight line = / [ ,+∞ doesn’t intersect the
graphic of the function , so is not surjective either.
Figure 2.12.
11. Determine which of the following relations are applications,
and which ones are injective, surjective, bijective?
For the bijective function, indicate its inverse.
Solution.
Because
the equation = + doesn’t have solutions in ℕ, it follows that �
is not a functional relation.
Then, because = = [− ; ], and

72
with √ / ≠ −√ / , and is not a functional relation.
c) We have: , ⇔ ⇔ = = . Then:
because , , +∞ ;
4) The equation = ℝ has solutions in [ , +∞ , if and
only if , which proves that is not a surjection (number -2, for
example, doesn’t have a preimage in [ ,+∞).
d) We have:
Repeating the reasoning from c), we obtain that is an
injection. Furthermore, the equation = [ , +∞ has the
solution = √ [ , +∞ , and that is why is a surjective
application. Then is a bijective application and, in accordance with
the 6th
Theorem, the relation −
is also an application (function).
In accordance with the same 6th
Theorem, we have
12. Let { = , , , , , , , , } and � given; using the
following table:
determine:
a)

73
b)
c) Calculate �−
; �−
; �−
.
Solution.
13. Let = { , , } and = { , , }. We examine the
relations
a) Determine , , , and , , .
b) Which of the relations , �� are applications? Indicate the
application type.
c) Determine the relations −
, −
and −
. Which one is a
function?
d) Determine the relations −
, −
, −
, −
,
−
and −
. Which ones are applications?
Solution.

74
b) is not an application, because element 3 has two images
and . is a bijective application; is an application, nor injective (the
elements ≠ have the same image c), nor surjective (b doesn’t have
a preimage).
We have:
−
is not an application, because the element has two
images 1 and 3; −
is a bijective application; −
is not an application,
because − ≠ (the element doesn’t have an image);
Only the relation −
is an application, neither injective, nor
surjective.
14. Given the functions: , : ℝ ⟶ ℝ,
= {
− , � ,
− , � > ,
= max − , − ,
show that = .
Solution. The functions and are defined on ℝ and have
values in ℝ. Let’s show that = , ⩝ ℝ.
We explain the function :
= {
− , � − − ,
− ,� − < − ,
= {
− , � ,
− , � > ,
=
no matter what ℝ is. So = .

75
15. The functions , : ℝ ⟶ ℝ,
= {
+ , if ,
�+
, if > ,
= {
− , if < ,
− , if
are given.
a) Show that and are bijective functions.
b) Represent graphically the functions and −
in the same
coordinate register.
c) Determine the functions: = + , = − , = . , = / .
d) Is function bijective?
Solution. a), b) The graphic of the function is represented in
Fig. 2.13 with a continuous line, and the graphic of −
is represented
with a dotted line.
Figure 2.13.
We have:
−
= {
− , if ,
− , if > ,
= {
+ , if < ,
+ ,if .

76
c) We draw the following table:
We have
=
{
, � ,
+ , � < < ,
− ,� ,
=
{
, � ,
− ,� < < ,
− , � ,
=
{
− , � ,
+ − , � < < ,
+ − , � ,
=
{
+
−
, � ,
+
−
, � < < ,
+
−
, � .

77
d) We have = , ⩝ −∞, ], therefore is not
injective, in other words, is not bijective. This proves that the
difference sum of two bijective functions isn’t necessarily a bijective
function.
16. Let’s consider the function : ℝ {− / } ⟶ ℝ { / },
a) Show that function is irreversible.
b) Determine −
.
c) In what case do we have = −
?
Solution:
⇔ − − = ⇔ = ⇒ is injective.
2) Let ℝ { / }. We examine the equation = . Then
If
Imposibile. So − / − ℝ {− / }, which proves
that the equation = has solutions in ℝ {− / } for any ℝ
{ / }. This proves that is a surjective function.
From 1) - 2), it follows that is a bijective function, hence it is
also inversibile.
b) We determine

78
c) In order to have = −
, it is necessary that the functions
and −
have the same definition domain and so = − . This
condition is also sufficient for the equality of the functions and −
.
It is true that, if = − , the functions and −
are defined on ℝ
{ / }, take values in ℝ { / }, and −
=
�+
�+
= , ⩝
ℝ { }, namely = −
.
17. Represent the function = √ − + as a
compound of two functions.
Solution. We put
Then
Answer:
18. Calculate , , and , where:
a) = �−
and = √ ;
b) = √ + and = − ;
c) = {
+ , � x −
− − , � −
and = {
− , � ,
− + , � > .
Indicate and .
Solution: a) We have

79
So ≠ , which proves that the comutative
law of function composition, generally, does not take place: ≠
.
c) To determine the functions and , in this particular
case we proceed as it follows:
= ( ) == {
+ , � < −
− − , � −
=
= {
+ � < − ,
− − , � − and ,
− − + − , � − and >
=
{
− + − � − < − ,
− − − , � − − and ,
− − + − , � − − and >
=
= {
+ − � < − / ,
− − , � − / ,
+ − , � > .
= ( ) = {
, � ,
− + , � > .
We notice that

80
So for , we obtain
= {
+ − , if [− − √ , −
− − − , if [− ,+∞ .
Similarly,
and that is why for > , we obtain
for −∞, − − √ ).
To sum it up, we have
=
= {
+ + − + , � −∞, − − √ ,
+ − , if [− √ , − ,
− − , if − .
= − . − = − .
19. Solve the following equations:
a) = , if = + , =
+ ;
b) = , if =
�+
�+
, ℝ, ℝ {− };
c) = , if = − , = − .
Solution. We determine the functions and
:
The equation becomes

81
Answer: = .
b) We calculate :
Answer: {− , }
c) We determine and :
meaning the equality is true for any ℝ.
Answer: ℝ.
20. Let , : ℝ ⟶ ℝ given by = + + and =
− + . Show there is no function �: ℝ ⟶ ℝ, so that

82
Solution. The relation can also be written
We suppose there is a function �: ℝ ⟶ ℝ that satisfies the
relation ′ . We put in ′ = and = − . We obtain:
which implies � +
� ≠ � + � , contradiction with the hypothesis. So there is
no function � with the property from the enunciation.
21. Determine all the values of the parameters , ℝ for
which = , ⩝ ℝ , where = − and
= + + .
Solution. We determine and :
Then
Answer: = − , = , = − = .
22. Given the functions , , ℎ ∶ ℝ ⟶ ℝ,
.
Show that:
a) is not injective;

83
b) is injective;
c) ℎ is bijective and determine ℎ−
.
Solution. a) Let = .
Then
For example, = + + , taking = and = − ,
we obtain = − = , ≠ .
because
so ℎ is an injective function.
We prove that ℎ is surjective. Let ℝ. We solve the equation
(the root of an uneven order exists out of any real number). So ℎ is a
surjection, therefore, ℎ is a bijective function.
We determine ℎ−
:
23. Let : [ , +∞ ⟶ [ ,+∞ with
a) Prove that is a bijective function.
b) Determine −
.
Solution. a) We have = − + . We represent this
function as a compound of two functions:
, : [ ,+∞ ⟶ [ , +∞ , where = , = − + .
Then = , which implies that = is bijective,
being the compound of two bijective functions.

84
b) −
= −
= − −
. We determine −
and −
.
Because:
We solve this equation according to [ ,+∞ .
The discriminant is:
The equation has only one root in [ , +∞ :
Then
So,
= + √ √ − / , with −
: [ ,+∞ ⟶ [ , +∞ .
24. Considering the function : ℝ ⟶ ℝ with the properties:
a) Determine the function .
b) Calculate (√ ).
Solution. a) For = from 1), it follows that = +
, which implies that = . For = − from 1), we obtain
Then

85
Let { , }. Then
On the other hand,
−�
=
−�+�
−�
= +
�
−�
implies
From (2) and (3) it follows that
So = , ⩝ ℝ.
Answer: a) = ; b) (√ ) = √ .
25. Using the properties of the characteristic function, prove the
equality:
Solution. Using the properties = ⇔ = , we will
prove the required equality by calculating with the help of , and
the characteristic functions of sets and
:

86
which implies and .
1. Determine the definition domains and the values domains of
the following relations:
2. Let = { , , , } and = { , , , }, × .
a) Determine the graphic of the relation .
b) Construct the schema of the relation :
1) = { , │ < and > };
2) = { , │ > and < };
3) = { , │ > or > };

87
3. Let = { , , , } and = { , , , , }. Write the graphic of
the relation × , if:
4. Let = { , , , }, = { , , , } and be the graphic of
the relation . Write the relation with propositions containing letters
and , with and :
5. Let A = { , . , }. Analyze the properties of the relation
(var. 1-6) and ℝ (var. 7-14):

88
7) = { , ℝ │ > and > };
8) = { , ℝ │ {
> ,
>
or {
< ,
<
};
9) = { , ℝ │ or };
10) = { , ℝ │ or };
6. For each binary relation defined on set ℕ:
a) determine the definition domain and the values domain ;
b) establish the properties (reflexivity, ireflexivity, symmetry,
anti-symmetry, transitivity);
c) determine the inverse relation −
, ℕ :
7. Given the set and the binary relation , prove that is
an equivalence relation and determine the factor set / .

89
8. Given the set = { , , , , , , , , } and the subset system
= { � , = ,
̅̅̅̅̅}, prove that defines a partition on and build
the equivalence relation .
9. We define on ℝ the binary relation :

90
a) Show that is an equivalence relation on ℝ.
b) Determine the equivalence classes.
10. We define on ℝ the binary relation :
.
a) Show that is an equivalence relation.
b) Determine the equivalence classes.
11. Let , where = {− , − ,− , − , − , , , , , , , ,
, } with
a) Is an equivalence relation?
b) If so, determine the equivalence classes.
12. Determine: , , −
, , −
, −
, if:
13. Determine the relations
, , −
, −
, −
, −
;
5) = { , ℤ │ − �s even},
= { , ℤ │ − is uneven};

91
14. Let = { , , , }, = { , , , } . Determine the
definition domain and the values domain of each of the following
relations , . Which ones are applications? Determine the application
type. Determine the relation −
, −
, −
, −
. Are
they applications?
15. We consider the application �: ℝ ⟶ ℝ , � = sin .
Determine:
16. The application �: ⟶ is given by the table below, where
= { , , , , , , , , } and = { , , , , , �}.
Determine:
17. We consider the application �: ℝ ⟶ ℤ, � = [ ] [ ] is
the integer part of . Determine
�− { , , } and �−
− .

92
18. The following application is given: �: ℕ ⟶ ℕ, � = .
Determine � and �−
, if = { , , , , , , , , , }.
19. Let and be two finite sets, | | = . | |. How many
surjective applications �: ⟶ are there, if:
20. Given the graphic of the relation , establish if is a func-
tion. Determine and .
Changing and , make become an application injective,
surjective and bijective.
Draw the graphic for the relation −
. Is relation −
a function?
What type is it?

93
21. Let = { , , , }, = { , , , } and = { , , }.
a) Draw the diagrams of two injective and two non-injective
functions defined on with values in .
b) Draw the diagrams of two surjective functions and two non-
surjective applications defined on with values in .
c) Draw the diagrams of two bijective and two non-bijective
functions defined on with values in .

94
22. Let = { , , , } and = { , , , , }, , .
Which of the relations stated below represents a function defined on
with values in ? And one defined on with values in ? For functions,
indicate their type:
23. Using the graphic of the function : ⟶ , show that is
injective, surjective or bijective. If bijective, determine −
and draw
the graphic of and −
in the same register of coordinates..
24. Which of the following relations ℝ are functions?
Indicate the definition domain of the functions. Establish their type:

95
25. Given the relation with = [− ; ] and = [− ; ]:
a) Does pair − , belong to the relation ? Why?
b) Indicate all the ordered pairs , with = . Explain.
26. Given the function , calculate its values in the indicated
points.
4) = {
− , � > ,
+ , � ,
− , , ;
5) = {
+ , � >
− , � =
− , � <
, − , / ;
27. Determine , if:
28. Given the functions and , determine the functions
+ , − , · and / , indicating their definition domain.

96
29. Represent the function as a compound of some
functions:
30. Given the functions and , determine the
functions and and calculate and in
options − and − and ( )( -1) in options −
.
31. Given the functions = , = and ℎ = −
, calculate:

97
32. Determine if in the function pairs and one is the inverse
of the other:
33. The value of function is given. Calculate the value of the
function −
, if:
34. Determine −
, if:
35. The function: : ℝ ⟶ ℝ is given:
a) Prove that is bijective.
b) Determine −
c) Calculate −
�� −
.

98
36. Given the functions and , determine the functions
and , : ℝ ⟶ ℝ :
37. Prove the equality of the functions and :
38. Determine the functions = + , = − , = .
and = / :

99
39. Let = { , , , }, = { , , , } and the functions
Can the functions , be defined? If so, determine
these functions. Make their diagrams.
40. a) Show that the function is bijective.
b) Determine −
.
c) Graphically represent the functions and −
in the same
coordinate register.

100
41. Show that the function : ℝ ⟶ ℝ, = − +
admits irreversible restrictions on:
Determine the inverses of these functions and graphically
represent them in the same coordinate register.
42. Using the properties of the characteristic function, prove the
following equalities (affirmations):

101
3. Elements of combinatorics
3.1. Permutations. Arrangements.
Combinations. Newton’s Binomial
In solving many practical problems (and beyond), it is necessary
to:
1) be able to assess the number of different combinations
compound with the elements of a set or a multitude of sets;
2) choose (select) from a set of objects the subsets of elements
that possess certain qualities;
3) arrange the elements of a set or a multitude of sets in a
particular order etc.
The mathematical domain that deals with these kind of
problems and with the corresponding solving methods is called
combinatorics. In other words, combinatorics studies certain
operations with finite sets. These operations lead to notions of
permutations, arrangements and combinations.
Let = { , , … , } be a finite set that has elements. Set
is called ordered, if each of its elements associates with a certain
number from to , named the element rank, so that different
elements of associate with different numbers.
Definition 1. All ordered sets that can be formed with elements
of given set = are called permutations of elements.
The number of all permutations of elements is denoted by the
symbol and it is calculated according to the formula:
By definition, it is considered = ! = = ! = .
Definition 2. All ordered subsets that contain elements of set
with elements are called arrangements of elements taken at
a time.

102
The number of all arrangements of elements taken at a
time is denoted by the symbol and it is calculated with the formula
Definition 3. All subsets that contain elements of set with n
elements are called combinations of elements taken at a time.
The number of all combinations of elements taken at a time
is denoted by the symbol and it is calculated with the formula
where , ℕ; .
Remark. In all the subsets mentioned in definitions 1 - 3, each
element of the initial set appears only once.
Along with the combinations in which each of the different
elements of a set participates only once, we can also consider
combinations with repetitions, that is, combinations in which the same
element can participate more than once.
Let groups of elements be given. Each group contains certain
elements of the same kind.
Definition ′. Permutations of elements each one containing
elements � , elements � , elements � , where + +
+ are called permutations of elements with repetitions.
The number of all permutations with repetitions is denoted by
the symbol ̅ , ,…, �
and it is calculated using the formula:
Definition ′ . The arrangements of elements, each one
containing elements, and each one being capable of repeating itself
in each arrangement for an arbitrary number of times, but not more
than times, are called arrangements of elements taken at a
time with repetitions.

103
The number of all arrangements with repetitions of elements
taken at a time is denoted by the symbol ̅̅̅̅ and it is calculated with
the formula
Definition 3'. The combinations of elements each one
containing elements, and each one being capable of repeating itself
more than once, but not more than times, are called combinations
of elements taken at a time with repetitions.
The number of all combinations with repetitions is denoted by
the symbol ̅̅̅̅ and it is calculated with the formula
In the process of solving combinatorics problems, it is
important to firstly establish the type (the form) of combination. One
of the rules used to establish the type of combination could be the
following table
It is often useful to use the following two rules:
Caution is advised to the order the elements are arranged in:
Arrangements if not all of
the elements from the
initial set participate;
Permutations if all
elements of the initial
set participate;
If the order isn’t important,
we have combinations;
If the order is important,
we have arrangements
or permutations:

104
The sum rule. If the object can be chosen in ways and the
object in n ways, then the choice "either A or B" can be made in +
ways.
The multiplication rule. If the object can be chosen in
ways and after each such choice the object can be chosen in ways,
then the choice "A and B" in this order can be made in . ways.
We will also mention some properties of combinations, namely:
The formula
is called Newton’s Binomial formula ( ℕ ).
The coefficients , , , … , from Newton’s Binomial
formula are called binomial coefficients; they possess the following
qualities:
V. The binomial coefficients from development (7), equally set
apart from the extreme terms of the development, are equal among
themselves.
VI a. The sum of all binomial coefficients equals .
VI b. The sum of the binomial coefficients that are on even
places equals the sum of the binomial coefficients that are on uneven
places.
VII. If is an even number (i.e. = ), then the binomial
coefficient of the middle term in the development (namely ) is the
biggest. If is uneven (i.e. = + ), then the binomial

105
coefficients of the two middle terms are equals (namely = +
)
and are the biggest.
VIII. Term −
, namely the + term in equality ,
is called term of rank + (general term of development) and it is
denoted by + . So,
IX. The coefficient of the term of rank + in the
development of Newton’s binomial is equal to the product of the
coefficient term of rank multiplied with the exponent of in this
term and divided by , namely
3.2. Solved problems
1. In how many ways can four books be arranged on a shelf?
Solution. As order is important and because all the elements of
the given set participate, we are dealing with permutations.
So
Answer: 24.
2. A passenger train has ten wagons. In how many ways can the
wagons be arranged to form the train?
Solution. As in the first problem, we have permutations of 10
elements of a set that has 10 elements. Then the number of ways the
train can be arranged is
Answer: 3 628 800.

106
3. In how many ways can 7 students be placed in 7 desks so that
all the desks are occupied?
Solution.
Answer: 5 040.
4. How many phone numbers of six digits can be dialed:
1) the digit participates in the phone number only once;
2) the digit participates more than once?
(The telephone number can also start with .)
Solution. We have 10 digits on the whole: , , , , , , , , , .
As the phone number can also start with , we have:
1) arrangements of digits taken at a time, i.e.
2) because the digit can repeat in the number, we have
arrangements with repetitions, i.e.
Answer: 1) ; 2) .
5. The volleyball team is made out of 6 athletes. How many
volleyball teams can a coach make having 10 athletes at his disposal?
Solution. As the coach is only interested in the team’s
composition it is sufficient to determine the number of combinations
of 10 elements taken 6 at a time, i.e.
Answer: .
6. How many 5 digit numbers can be formed with digits and ?
Solution. As the digits are repeating and their order is important,
we consequently have arrangements with repetitions. Here, = ,
= .

107
From digits and we can form ̅̅̅ numbers of five digits.
But we have to take into consideration that the number cannot
begin with the digit zero. So, from the number ̅̅̅ we have to begin
with zero. Numbers of this type are ̅̅̅ . Therefore, the number we
seek is
Answer: 16.
7. How many three digit numbers can be formed with the digits
, , , , , if the digits can be repeated?
Solution. As the digits repeat, we obviously have arrangements
with 5 digits taken three times. Therefore, there can be formed
=
̅̅̅̅̅̅̅̅̅̅ numbers of three digits.
Answer: 125.
8. Using roses and Bedding Dahlias, bunches are made that
contain roses and Bedding Dahlias. How many bunches of this kind
can be formed?
Solution. Two roses (out of the we have) can be chosen in
ways, and three Bedding Dahlias (out of can be taken in ways.
Applying the rule of multiplication, we have: the total number of
bunches that can be formed is . = .
Answer: .
9. 12 young ladies and 15 young gentlemen participate at a
dancing soirée. In how many ways can four dancing pairs be chosen?
Solution. The young ladies can be distributed in four person
groups in ways, and the young gentlemen - in ways.
Because in each group of young ladies (or young gentlemen)
order is an issue, each of these groups can be ordered in ways.

108
As a result (we apply the multiplication rule), we will have
. . = . = . = .
Answer: .
10. To make a cosmic flight to Mars, it is necessary to constitute
the crew of the spacecraft in the following distribution: the spacecraft
captain, the first deputy of the captain, the second deputy of the
captain, two mechanical engineers, and a doctor. The command triplet
can be selected out of the 25 pilots that are flight-ready, two
mechanical engineers out of the 20 specialists who master the
construction of the spacecraft, and the doctor – out of the 8 available
doctors. In how many ways can the crew be assembled?
Solution. Choosing the captain and his deputies, it is important
to establish which of the pilots would best deal with random steering
commands. So, the distribution of the tasks between the triplet’s
members is equally important. So, the commanding triplet can be
formed in ways.
The tasks of the two engineers are more or less the same. They
can accomplish these tasks consecutively. So the pair of engineers can
be formed in ways. Regarding the doctor – the situation is the
same, namely the doctor can be chosen in ways.
Using the multiplication rule, we have: a pair of engineers can be
assigned in ways to each commanding triplet. We will have, in
total, . quintets. To each quintet a doctor can be associated in
ways.
As a result, the spacecraft’s crew can be assembled in
. . ways, or
Answer: .

109
11. In how many different ways can 5 cakes, of the same type or
different, be chosen in a cake shop where there are types of
different cakes?
Solution. The five cakes can all be of the same type or four of the
same type and one of a different type, or three of the same type and
two of a different type, etc., or all can be of different types.
The number of the possible sets comprised of five cakes out of
the existing types is equal to the number of the combinations with
repetitions of 11 elements taken five at a time, i.e.
Answer: .
12. In a group of 10 sportsmen there are two rowers, three
swimmers and the others are athletes. A 6 person team must be
formed for the upcoming competitions in such a manner that the team
will comprise at least one representative from the three nominated
sport types. In how many ways can such a team be assembled?
Solution. a) In the team there can be one rower, one swimmer
and four athletes. The rower can be chosen in ways, the swimmer in
ways, and the athlete in ways. Using the multiplication rule, we
have . . ways.
b) In the team there can be one rower, two swimmers and three
athletes. According to the aforementioned reasoning, the numbers of
the teams of this type will be . . .
c) In the team there can be one rower, three swimmers and two
athletes. The number of teams in this case will be . . .
d) In the team there can be two rowers, one swimmer and three
athletes. We will have . . of these teams.
e) In the team there can be two rowers, two swimmers and two
athletes. The number of teams will be . . .

110
f) In the team there can be two rowers, three swimmers and one
athlete. The number of teams will be . . .
Using the addition rule, the total number of teams that can be
assembled is:
Answer: .
13. Given = capital letters, = vowels and =
consonants (in total + + = letters) determine: How many
different words can be formed using these letters, if in each word the
first letter has to be a capital letter, among the other different letters
there have to be � = different vowels (out of the = given)
and = different consonants (out of the = given).
Solution. We choose a capital letter. This choice can be made in
ways. Then from vowels we choose � letters. This can be done in
�
ways. Finally, we choose consonants, which can be made in �
ways. Using the multiplication rule, we can choose the required letters
to form the word in .
�
. �
ways.
After placing the capital letter at the beginning, with the other
� + letters we can form � + ! permutations. Each such
permutation yields a new word. So, in total, a number of .
�
. �
� +
! different words can be formed, namely . . !
Answer: . . !
14. In a grocery store there are three types of candy. The candy
is wrapped in three different boxes, each brand in its box. In how many
ways can a set of five boxes be ordered?
Solution (see Problem ).
Answer: .

111
15. In order to form the 10 person guard of honor, officers of the
following troops are invited: infantry, aviation, frontier guards, artillery;
navy officers and rocket officers.
In how many ways can the guard of honor be assembled?
Solution. We have categories of officers. Repeating the
reasoning from problem , we have to calculate combinations with
repetitions of 6 elements taken 10 times each, as follows
Answer: .
16. On a shelf there are + different books. Among them
have a blue cover, and a yellow cover. The books are permutated in
every way possible. How many positions do the books have, if:
a) the books with the blue cover occupy the first places;
b) the books in the yellow covers sit beside them?
Solution. a) The books with blue covers can be placed in the first
places in = ! ways. With each such allocation, the books with
yellow covers can be placed in = ! ways. Using the multiplication
rule, we have in total ! · ! positions in which the blue covers books
occupy the first places.
b) Let the books in blue covers sit beside. It follows that right
beside them on the shelf there can be either books with yellow
covers or − , or − , ... , or no book with yellow covers. We can
thus place the books with blue covers so that they follow each other in
+ ways. In each of these positions, the books with yellow covers
can be permutated in any way, also the books with the blue covers can
be permutated in any way. As a result, we will have ! . ! . +
different positions for the books.
Answer: a) ! · !; b) ! · ! · + .

112
17. Determine the fourth term of the development of Newton’s
binomial:
Solution. According to the formula , the rank 4 term has the
form
Answer: − · /
.
18. Determine the biggest coefficient in the development of the
binomial
Solution.
If is an even number, i.e. = , ℕ , then the
development of the binomial contains + terms, and according
to the � property, the coefficient �
�
is the biggest.
If is an uneven number, i.e. = + , ℕ, according
to the same � property, the development of the binomial contains
two terms have the biggest coefficients �+
�
, �+
�+
.
Answer: �
�
, if is an even number; �+
�
, �+
�+
if is an
uneven number.
19. Determine which term do not contain in the development
of the binomial:

113
Solution.
The term of rank + in the development of this binomial has
the form
This term does not contain only if − = ⇔ = . So
the term that doesn’t contain is + .
Answer: + .
20. In the development of the binomial
determine the terms that contains to the
power of three, if the sum of the binomial coefficients that occupy
uneven places in the development of the binomial is equal to .
Solution. We will first determine the exponent . Based on the
� th
property, the sum of the binomial coefficients is . Because the
sum of the binomial coefficients that occupy uneven places in the
development of the binomial is equal to , and based on the �
property, it is equal to the sum of the coefficients that occupy even
places in the respective development, we have
Therefore, the degree of the binomial is . The term of rank
+ takes the form
Considering the requirements of the problem, we have
and

114
Answer: − .
21. For what value of the binomial coefficients of the second,
third and fourth term from the development of the binomial
+ forms an arithmetic progression?
Solution. Based on the formula , we have
and from the conditions of the problem the following relations issues
The conditions of the problem are verified by the value = .
Answer: = .
22. Prove that the difference of the coefficients +
and in
the development of the binomial + +
is equal to the difference
of the coefficients of +
and −
I in the development of the
binomial + .
Solution. The coefficients of +
and in the development of
the binomial + +
are +
+
and + respectively. We assess the
difference
In the development of the binomial + , the coefficients of
+
and −
are +
and −
, respectively. We assess the
difference

115
As the members from the right in ( ) and ( are equal, the
equality of the members from the left results, i.e.
which had to be proven.
23. Comparing the coefficients of in both members of the
equality
prove that
Solution.
In the right member of this equality, the coefficient of is
and in the development of the binomial + +
, the term of rank
+ has the form
As the polynomials + . + and + +
are
equal and have the same degree, the equality of the coefficients beside
these powers of result and this also concludes the demonstration of
the equality .

116
24. Prove the equality
Solution. Let:
We multiply both members of the last equality with + .
We obtain
Returning to the initial expression, we have
that had to be proven.
25. Prove that
Solution. We write the development of the binomial − :

117
We derive both members of the equality according to . We
obtain
We place in = . Then
26. Prove that the equality
is true.
Solution. We notice that the expression
represents the development of the binomial − = =
, that had to be proven.
27. Bring the expression + + + to a simpler form.
Solution. We will make the necessary transformations by
applying the method of mathematical induction. Let
For:
= , we have = ⇔ = ;
= , we have + = ⇔ + . ! = ⇔ = ⇔
! − = ⇔ − =
= , we have + + = ⇔ + = ⇔

118
We assume that for = the equality ( ) takes the form
We calculate the value of the expression ( ) for = + . We
have
Based on the principle of mathematical induction, we reach the
conclusion that
Answer:
28. Prove that indifferent of what , ℕ, are ! . ! divides
+ !
Solution. According to the definition 3,
+ !
! . !
= + is the
number of the subsets that have elements of a set with ( +
elements, namely + is a natural number. Consequently,
+ !
! . !
is
an integer number, or this proves that ! . ! divides + !
29. Deduce the equality
Solution. We use property . We have:
As a result,

119
30. Calculate the sum
Solution. We notice that term of this sum can be transformed
as it follows:
Then the sum takes the form
Answer:
31. Solve the equation
Solution.

120
Because � has meaning only for , it follows that the
solution of the initial equation is = .
Answer: = .
Solution.
Answer: = .
Solution. As:
We have:

121
As ℕ and , we have { , , , , , , , , }.
Answer: = ; { , , , , , , , , }.
34. Determine the values of that verify the equality
Solution.
because the solutions of the equation + − = are irational
numbers.
Answer: = .
Solution.

122
Then { , , , , , , , , }.
Answer: = , { , , , , , , , , }.
36. Solve the system of equations
Solution. From the conditions of the problem, we have , ℕ
with and . Based on the formulas − , we have
The divisors of the free term in are the numbers
± ; ± ; ± ; ± ; ± ; ± ; ± ; ± ; ± ; ± ; . . . .
We use (omer’s scheme and the Bezout theorem to select the
numbers that are solutions of the equation . As ℕ, we will
verify only the natural numbers. It verifies that numbers { , , , , , }
are not solutions of the equation .
We verify = .
So
because for > , the expression − + + + > .

123
Hence, the solution of the system is the pair , .
Answer: { , }.
37. Find x and y, if
Solution. We will first bring to a simpler form the expression
As a result, the system takes the form
Answer: { , }.
38. Determine the values of , so that
Solution. From the enunciation it follows that ℕ. Then

124
and ℕ. So { , , , , , , }.
Answer: { , , , , , , }.
39. Determine the values of that verify the inequation
Solution. has meaning for ℕ and . Because
it follows that
Answer: { , , , }.
40. Solve the system of inequations:
Solution.

125
Answer: { , }.
1. A commission is formed of one president, his assistant and
five other persons. In how many ways can the members of the
commission distribute the functions among themselves?
2. In how many ways can three persons be chosen from a group
of 20 persons to accomplish an assignment?
3. In a vase there are 10 red daffodils and 4 pink ones. In how
many ways can three flowers be chosen from the vase?
4. The padlock can be unlocked only if a three digit number is
correctly introduced out of five possible digits. The number is guessed,
randomly picking 3 digits. The last guess proved to be the only
successful one. How many tries have proceeded success?

126
5. On a shelf there are 30 volumes. In how many ways can the
books be arranged, so that volumes 1 and 2 do not sit beside each
other on the shelf?
6. Four sharpshooters have to hit eight targets (two targets
each). In how many ways can the targets be distributed among the
sharpshooters?
7. How many four digit numbers, made out of digits 0, 1, 2, 3, 4,
5, contain digit 3, if: a) the digits do not repeat in the number; b) the
digits can repeat themselves?
8. In the piano sections there are 10 participating persons, in the
reciter section, 15 persons, in the canto section, 12 persons, and in the
photography section - 20 persons. In how many ways can there a team
be assembled that contains 4 reciters, 3 pianists, 5 singers and a
photographer?
9. Seven apples and three oranges have to be placed in two
bags such that every bag contains at least one orange and the number
of fruits in the bags is the same. In how many ways can this distribution
be made?
10. The matriculation number of a trailer is composed of two
letters and four digits. How many matriculation numbers can there be
formed using 30 letters and 10 digits?
11. On a triangle side there are taken points, on the second
side there are taken points, and on the third side, points.
Moreover, none of the considered points is the triangle top. How many
triangles with tops in these points are there?

127
12. Five gentlemen and five ladies have to be sited around a
table so that no two ladies and no two gentlemen sit beside each other.
In how many ways can this be done?
13. Two different mathematic test papers have to be distributed
to 12 students. In how many ways can the students be arranged in two
rows so that the students that sit side by side have different tests and
those that sit behind one another have the same test?
14. Seven different objects have to be distributed to three
persons. In how many ways can this distribution be made, if one or two
persons can receive no object?
15. How many six digit numbers can be formed with the digits 1,
2, 3, 4, 5, 6, 7, in such a way that the digits do not repeat, and that the
digits at the beginning and at the end of the number are even?
16. How many different four digit numbers can be formed with
the digits 1, 2, 3, 4, 5, 6, 7, 8, if in each one the digit one appears only
once, and the other digits can appear several times?
17. To award the winners of the Mathematics Olympiad, three
copies of a book, two copies of another book and one copy of a third
book have been provided. How many prizes can be awarded if 20
persons have attended the Olympiad, and no one will receive two
books simultaneously? Same problem, if no one will receive two copies
of the same book but can receive two or three different books?
18. The letters of the Morse alphabet is comprised of symbols
(dots and dashes). How many letters can be drawn, if it is required that
each letter contains no more than five symbols?

128
19. To find their lost friend, some tourists have formed two
equal groups. Only four persons among them know the surroundings.
In how many ways can the tourists divide so that each group has two
persons that know the surroundings and considering they are 16
persons in total?
20. Each of the 10 radio operators located at point A is trying to
contact each of the 20 radio operators located at point B. How many
different options for making contact are there?
Prove the equalities:
(where , ℕ, > + ;

129
Solve the equations and the systems of equations:

130
Solve the inequations and the systems of inequations:

131
107. Determine the fifth term from the development of the
binomial
108. Determine the middle term from the development of the
binomial
109. Determine the value of exponent from the development
of the binomial + , if the coefficient of the 5th
term is equal to
the coefficient of the 9th
term.
110. Determine , if the 5th
term from the development of the
binomial doesn’t depend on .
the coefficient of the third term is equal to . Determine the middle
term from the development of the binomial.
112. Determine the smallest value of ′ exponent from the
development of the binomial + , if the relation of the
coefficients of two neighboring arbitrary is equal to : .
113. Determine the term from the development of the binomial
that contains .

132
that doesn’t depend on .
that doesn’t contain , if the sum of
the coefficients of the first three terms from the development is equal
to .
that doesn’t contain .
117. Determine the free term from the development of the
binomial
118. Determine the third term from the development of the
binomial if the sum of the binomial coefficients is
.
that contains , if the relation of the
binomial coefficients of the fourth and second term is equal to .
that doesn’t contain , if the sum of the
coefficients of the second term from the beginning of the
development and the third term from the end of the development is
equal to .

133
121. The relation of the coefficient of the third term to the
coefficient of the 5th term in the development of the binomial
is equal to / . . Determine the term from the
development of the binomial that contains − /
.
122. Determine , and , if it is known that the 2nd, 3rd and 4th
terms from the development of the binomial are equal to
, , , respectively.
123. For what value of the exponent , the coefficients of the
2nd, the 3rd and the 4th term from the development of the binomial
form an arithmetic progression?
124. Determine the terms from the development of the
binomial that do not contain irrationalities.
125. How many rational terms does the following development
of the binomial contain
126. Determine the ranks of three consecutive terms from the
development of the binomial + the coefficients of which form
an arithmetic progression.
√ + √ − that contains .
, if the 9th
term has the biggest
coefficient.
128. The 3rd term from the development of the binomial
+
�
doesn’t contain . For which value of is this term equal to
the 2nd
term from the development of the binomial + ?

134
129. For what positive values of the biggest term from the
development of the binomial + is the 4th
term?
130. In the development of the binomial √ +
√�
the first
three terms form an arithmetic progression. Determine all rational
terms from the development of this binomial.
131. Determine the values of for which the difference between
the 4th and the 6th term from the development of the binomial
√ �
√
+
√
√ �
is equal to , if it is known that the exponent of the
binomial is smaller by than the binomial coefficient of the 3rd term
from the development of this binomial.
132. Given that is the biggest natural number only if
/ + / > , determine the term that contains from
the development of the binomial √ − √ .
133. Determine for which the sum of the 3rd and al 5th term
from the development of the binomial √ + √ −� is equal to
, knowing that the sum of the last three binomial coefficients is .
134. Determine , knowing that the 6th term from the
development of the binomial + , where = √ lg − �
, =
√ �− is , and the coefficients of the binomial of the terms
ranked , and are respectively the 1st
, the 3rd
and the 5th
term of an
arithmetic progression.
135. Are there any terms independent of in the development
of the binomial Write these terms.

135
136. How many terms from the development of the binomial
√ + √ are integer terms?
the first three coefficients form an arithmetic
progression. Determine all terms from the development of the
binomial that contain the powers of with a natural exponent.
138. Determine , if the 3rd term from the development of the
binomial is equal to .
139. In the development of + − find the term
corresponding to the exponent of that is 3 times as big as the sum of
all the development’s coefficients sum.
140. Determine the rank of the biggest term from the
development of the binomial + according to the descending
powers of , assuming that > ; > ; + = . In which
conditions:
a) the biggest term will be the first?
b) the biggest term will be the last?
c) the development will contain two consecutive equal terms
that are bigger than all the other terms of the development?

136

137
Answers
Note. For chapters 1 and 2 only the less „bulky answers have
been provided and those that are relatively more complicated (in our
own opinion).
1. Sets. Operations with sets
(because consists of even natural numbers).
a)if − = ⇒ = { } ; b) if − ≠ ⇒ ℎ

138

139

Algebraic Problems And Exercises For High School

Algebraic Problems And Exercises For High School

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