The Educational Publisher
Columbus, 2015
ALGEBRAIC PROBLEMS AND EXERCISES
FOR HIGH SCHOOL
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
Algebraic problems and exercises for high school
1
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
ALGEBRAIC PROBLEMS AND EXERCISES FOR HIGH SCHOOL
Sets, sets operations โ Relations, functions โ Aspects of combinatorics
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
2
First imprint: Algebra รฎn exerciศii ศi probleme pentru liceu (in Romanian),
Cartier Publishing House, Kishinev, Moldova, 2000
Translation to English by Ana Maria Buzoianu
(AdSumus Cultural and Scientific Society)
The Educational Publisher
Zip Publishing
1313 Chesapeake Ave.
Columbus, Ohio 43212, USA
Email: [email protected]
AdSumus Cultural and Scientific Society
Editura Ferestre
13 Cantemir str.
410473 Oradea, Romania
Email: [email protected]
ISBN 978-1-59973-342-5
Algebraic problems and exercises for high school
3
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
ALGEBRAIC PROBLEMS
AND EXERCISES
FOR HIGH SCHOOL
โ Sets, sets operations
โ Relations, functions
โ Aspects of combinatorics
The Educational Publisher
Columbus, 2015
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
4
ยฉ Ion Goian, Raisa Grigor, Vasile Marin, Florentin Smarandache,
and the Publisher, 2015.
Algebraic problems and exercises for high school
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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.
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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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
2.3. Suggested exercises / 86
3. Elements of combinatorics / 101
3.1. Permutations. Arrangements. Combinations.
Newtonโs binomial / 101
3.2. Solved problems / 105
3.3. Suggested exercises / 125
Answers / 137
References / 144
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Notations
= equality;
โ inequality;
โ belongs to;
โ doesnโt belong to;
โ subset;
โ superset;
โช union;
โฉ intersection;
โ empty set;
โจ (or) disjunction;
โง (and) conjunction;
๐๐๐๐โ ๐
โ = {0, 1, 2, 3, 4,โฆ }
โค = {โฆ ,โ2,โ1, 0, 1, 2,โฆ }
p equivalent with ๐;
natural numbers
integer numbers set;
โ = {๐
๐|๐, ๐ โ โค, ๐ โ 0} rational numbers set;
โ real numbers set;
โ = {๐ + ๐๐|๐, ๐ โ ๐ , ๐2 = โ1} complex numbers set;
๐ด+ = {๐ฅ โ ๐ด|๐ฅ > 0}, ๐ด โ {โค,โ, โ};
๐ดโ = {๐ฆ โ ๐ด|๐ฆ < 0}, ๐ด โ {โค,โ,โ};
๐ดโ = {๐ง โ ๐ด|๐ง โ 0} = ๐ด โ {0}, ๐ด โ {โ, โค, โ,โ, โ};
|๐ฅ|
[๐ฅ]
{๐ฅ}
absolute value of ๐ฅ โ โ;
integer part of x โ R;
fractional part of ๐ฅ โ โ,
0 โค {๐ฅ} < 1;
(๐, ๐) pair with the first element ๐
and the second element ๐
(also called "ordered pair");
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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(๐, ๐, ๐) triplet with corresponding elements
๐, ๐, ๐;
๐ด ร ๐ต = {๐, ๐|๐ โ ๐ด, ๐ โ ๐ต} Cartesian product of set
๐ด and set ๐ต;
๐ด ร ๐ต ร ๐ถ = {๐, ๐, ๐|๐ โ ๐ด, ๐ โ ๐ต, ๐ โ ๐ถ}
Cartesian product of sets ๐ด, ๐ต, ๐ถ;
๐ธ universal set;
๐(๐ธ) = {๐|๐ โ ๐ธ} set of parts (subsets) of set ๐ธ;
๐ด = ๐ต๐๐๐โ (โ) ๐ฅ โ ๐ธ(๐ฅ โ ๐ด โ ๐ฅ โ ๐ต)
equality of sets ๐ด and ๐ต;
๐ด โ ๐ต๐๐๐โ (โ) ๐ฅ โ ๐ธ(๐ฅ โ ๐ด โ ๐ฅ โ ๐ต)
๐ด is included in ๐ต;
๐ดโ๐ต = {๐ฅ โ ๐ธ|๐ฅ โ ๐ด โจ ๐ฅ โ ๐ต} union of sets ๐ด and ๐ต;
๐ดโ๐ต = {๐ฅ โ ๐ธ|๐ฅ โ ๐ด โง ๐ฅ โ ๐ต} intersection of sets ๐ด and ๐ต;
๐ด โ ๐ต = {๐ฅ โ ๐ธ|๐ฅ โ ๐ด โง ๐ฅ โ ๐ต} difference between sets ๐ด and ๐ต;
๐ดโ๐ต = (๐ด โ ๐ต) โช (๐ต โ ๐ด) symmetrical difference;
โ๐(๐ด) = ๏ฟฝฬ ๏ฟฝ = ๐ธ โ ๐ด complement of set ๐ด relative to ๐ธ;
๐ผ โ ๐ด ร ๐ต relation ๐ผ defined on sets ๐ด and ๐ต;
๐: ๐ด โ ๐ต function (application) defined on ๐ด
with values in ๐ต;
๐ท(๐) domain of definition of function ๐:
๐ธ(๐) domain of values of function ๐.
Algebraic problems and exercises for high school
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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: ๐ด,๐ต, ๐ถ, โฆ , ๐, ๐, ๐, and small letters for elements in sets:
๐, ๐, ๐, โฆ , ๐ฅ, ๐ฆ, ๐ง etc.
If ๐ is an element of the set ๐ด, we will write ๐ โ ๐ด and we will
read "๐ belongs to ๐ด" or "๐ is an element of ๐ด". To express that ๐ is not
an element of the set ๐ด, we will write ๐ โ ๐ด and we will read "๐ does
not belong to ๐ด".
Among sets, we allow the existence of a particular set, noted as
โ , called the empty set and containing no element.
The set that contains a sole element will be noted with {๐}.
More generally, the set that doesnโt contain other elements except the
elements ๐1, ๐2, โฆ , ๐๐ will be noted by {๐1, ๐2, โฆ , ๐๐}.
If ๐ด is a set, and all of its elements have the quality ๐, then we
will write ๐ด = {๐ฅ|๐ฅ verifies ๐} or ๐ด = {๐ฅ|๐(๐ฅ)} and we will read: "๐ด
consists of only those elements that display the property ๐ (for which
the predicate ๐( ๐ฅ) is true)."
We shall use the following notations:
โ = {0, 1, 2, 3,โฆ } โ the natural numbers set;
โโ = {0, 1, 2, 3,โฆ } โ the natural numbers set without zero;
โค = {โฆ ,โ2,โ1, 0, 1, 2,โฆ } โ the integer numbers set;
โค = {ยฑ1,ยฑ2,ยฑ3โฆ } โ the integer numbers set without zero;
โ = {๐
๐|๐, ๐ โ โค, ๐ โ โโ} โ the rational numbers set;
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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โโ = the rational numbers set without zero;
โ = the real numbers set;
โโ = the real numbers set without zero;
โ+ = {๐ฅ โ โ|๐ฅ โฅ 0}; โ+โ = {๐ฅ โ โ|๐ฅ > 0};
โ = {๐ + ๐๐|๐, ๐ โ ๐ , ๐2 = โ1} = the complex numbers set;
โโ = the complex numbers set without zero;
๐ โ {1, 2,โฆ , ๐} โบ ๐ = 1, ๐ฬ ฬ ฬ ฬ ฬ ;
๐ท(๐) = {๐ โ โคโ|๐ โฎ ๐} = the set of all integer divisors of number
๐ โ โค;
๐(๐ด) = |๐ด| = the number of the elements of finite set A.
Note. We will consider that the reader is accustomed to the
symbols of Logic: conjunction โง (โฆandโฆ), disjunction โจ (โฆorโฆ),
implication โน, existential quantification (โ) and universal quantification
(โ).
Let ๐ด and ๐ต be two sets. If all the elements of the set ๐ด are also
elements of the set ๐ต, we then say that ๐จ is included in ๐ฉ, or that ๐จ is
a part of ๐ฉ, or that ๐จ is a subset of the set ๐ฉ and we write ๐ด โ ๐ต. So
๐ด โ ๐ต โ (โ)๐ฅ(๐ฅ โ ๐ด โน ๐ฅ โ ๐ต).
The properties of inclusion a. (โ) ๐ด, ๐ด โ ๐ด (reflexivity);
b. (๐ด โ ๐ต โง ๐ต โ ๐ถ) โน ๐ด โ ๐ถ (transitivity);
c. (โ)๐ด, โ โ ๐ด.
If ๐ด is not part of the set ๐ต, then we write ๐ด โ ๐ต, that is, ๐ด โ
๐ต โ (โ)๐ฅ(๐ฅ โ ๐ด โง ๐ฅ โ ๐ต).
We will say that the set ๐ด is equal to the set ๐ต, in short ๐ด = ๐ต,
if they have exactly the same elements, that is
๐ด = ๐ต โบ (๐ด โ ๐ต โง ๐ต โ ๐ด).
The properties of equality Irrespective of what the sets ๐ด, ๐ต and ๐ถ may be, we have:
a. ๐ด = ๐ด (reflexivity);
b. (๐ด = ๐ต) โน (๐ต = ๐ด) (symmetry);
Algebraic problems and exercises for high school
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c. (๐ด = ๐ต โง ๐ต = ๐ถ) โน (๐ด = ๐ถ) (transitivity).
With ๐(๐ด) we will note the set of all parts of set ๐, in short
๐ โ ๐(๐ด) โบ ๐ โ ๐ด.
Obviously, โ , ๐ด โ ๐(๐ด).
The universal set, the set that contains all the sets examined
further, which has the containing elementsโ nature one and the same,
will be denoted by ๐ธ.
Operations with sets Let ๐ด and ๐ต be two sets, ๐ด, ๐ต โ ๐(๐ธ).
1. Intersection.
๐ด โฉ ๐ต = {๐ฅ โ ๐ธ|๐ฅ โ ๐ด โง ๐ฅ โ ๐ต},
i.e.
๐ฅ โ ๐ด โฉ ๐ต โบ (๐ฅ โ ๐ด โง ๐ฅ โ ๐ต), (1)
๐ฅ โ ๐ด โฉ ๐ต โบ (๐ฅ โ ๐ด โง ๐ฅ โ ๐ต), (1โ)
2. Union.
๐ด โช ๐ต = {๐ฅ โ ๐ธ|๐ฅ โ ๐ด โจ ๐ฅ โ ๐ต},
i.e.
๐ฅ โ ๐ด โช ๐ต โบ (๐ฅ โ ๐ด โจ ๐ฅ โ ๐ต), (2)
๐ฅ โ ๐ด โช ๐ต โบ (๐ฅ โ ๐ด โจ ๐ฅ โ ๐ต), (2โ)
3. Difference.
๐ด โ ๐ต = {๐ฅ โ ๐ธ|๐ฅ โ ๐ด โง ๐ฅ โ ๐ต},
i.e.
๐ฅ โ ๐ด โ ๐ต โบ (๐ฅ โ ๐ด โง ๐ฅ โ ๐ต), (3)
๐ฅ โ ๐ด โ ๐ต โบ (๐ฅ โ ๐ด โจ ๐ฅ โ ๐ต), (3โ)
4. The complement of a set. Let ๐ด โ ๐(๐ธ). The difference ๐ธ โ
๐ด is a subset of ๐ธ, denoted ๐ถ๐ธ(๐ด) and called โthe complement of ๐ด
relative to ๐ธโ, that is
๐ถ๐ธ(๐ด) = ๐ธ โ ๐ด = {๐ฅ โ ๐ธ|๐ฅ โ ๐ด}.
In other words,
๐ฅ โ ๐ถ๐ธ(๐ด) โบ ๐ฅ โ ๐ด, (4)
๐ฅ โ ๐ถ๐ธ(๐ด) โบ ๐ฅ โ ๐ด. (4โ)
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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Properties of operations with sets ๐ด โฉ ๐ด = ๐ด,๐ด โช ๐ด = ๐ด (idempotent laws)
๐ด โฉ ๐ต = ๐ต โฉ ๐ด, ๐ด โช ๐ต = ๐ต โช ๐ด (commutative laws)
(๐ด โฉ ๐ต) โฉ ๐ถ = ๐ด โฉ (๐ต โฉ ๐ถ);
(๐ด โช ๐ต) โช ๐ถ = ๐ด โช (๐ต โช ๐ถ) (associativity laws)
๐ด โช (๐ต โฉ ๐ถ) = (๐ด โช ๐ต) โฉ (๐ด โช ๐ถ);
๐ด โฉ (๐ต โช ๐ถ) = (๐ด โฉ ๐ต) โช (๐ด โฉ ๐ถ) (distributive laws)
๐ด โช (๐ด โฉ ๐ต) = ๐ด;
๐ด โฉ (๐ด โช ๐ต) = ๐ด (absorption laws)
๐ถ๐ธ(๐ด โช ๐ต) = ๐ถ๐ธ(๐ด) โฉ ๐ถ๐ธ(๐ต);
๐ถ๐ธ(๐ด โฉ ๐ต) = ๐ถ๐ธ(๐ด) โช ๐ถ๐ธ(๐ต) (Morganโs laws)
Two "privileged" sets of ๐ธ are โ and ๐ธ.
For any ๐ด โ ๐(๐ธ), we have:
โ โ ๐ด โ ๐ธ,
๐ด โช โ = ๐ด, ๐ด โฉ โ = โ , ๐ถ๐ธ(โ ) = ๐ธ,
๐ด โช ๐ธ = ๐ธ, ๐ด โฉ ๐ธ = ๐ด, ๐ถ๐ธ(๐ธ) = โ ,
๐ด โช ๐ถ๐ธ(๐ด) = ๐ธ, ๐ด โฉ ๐ถ๐ธ(๐ด) = โ ,
๐ถ๐ธ(๐ถ๐ธ(๐ด)) = ๐ด (principle of reciprocity).
Subsequently, we will use the notation ๐ถ๐ธ(๐ด) = ๏ฟฝฬ ๏ฟฝ.
5. Symmetric difference.
๐ดโ๐ต = (๐ด โ ๐ต) โช (๐ต โ ๐ด).
Properties. Irrespective of what the sets ๐ด, ๐ต and ๐ถ are, we
have:
a. ๐ดโ๐ด = โ ;
b. ๐ดโ๐ต = ๐ตโ๐ด (commutativity);
c. ๐ดโโ = โ โ๐ด = ๐ด;
d. ๐ดโ (๐ดโ๐ต) = ๐ต;
e. (๐ดโ๐ต)โ๐ถ = ๐ดโ(๐ตโ๐ถ) (associativity);
f. ๐ด โฉ (๐ตโ๐ถ) = (๐ด โฉ ๐ต)โ(๐ด โฉ ๐ถ);
g. ๐ดโ๐ต = (๐ด โช ๐ต) โ (๐ด โฉ ๐ต).
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6. Cartesian product. Let ๐ฅ and ๐ฆ be two objects. The set
{{๐ฅ}, {๐ฅ, ๐ฆ}}, whose elements are the sets {๐ฅ} and {๐ฅ, ๐ฆ}, is called an
ordered pair (or an ordered couple) with the first component ๐ฅ and
the second component ๐ฆ and is denoted as (๐ฅ, ๐ฆ). Having three objects
๐ฅ, ๐ฆ and ๐ง, we write (๐ฅ, ๐ฆ, ๐ง) = ((๐ฅ, ๐ฆ), ๐ง) and we name it an ordered
triplet.
Generally, having ๐ objects ๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐ we denote
(๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐) = (โฆ((๐ฅ1, ๐ฅ2)๐ฅ3)โฆ๐ฅ๐)
and we name it an ordered system of ๐ elements (or a cortege of
length ๐).
We have
(๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐) = (๐ฆ1, ๐ฆ2, โฆ , ๐ฆ๐)
โบ (๐ฅ1 = ๐ฆ1 โง ๐ฅ2 = ๐ฆ2 โง โฆโง ๐ฅ๐ = ๐ฆ๐).
Let ๐ด, ๐ต โ ๐(๐ธ). The set
๐ด ร ๐ต = {(๐, ๐)|๐ โ ๐ด โง ๐ โ ๐ต}
is called a Cartesian product of sets ๐ด and ๐ต. Obviously, we can define
๐ด ร ๐ต ร ๐ถ = {(๐ฅ, ๐ฆ, ๐ง)|๐ฅ โ ๐ด โง ๐ฆ โ ๐ต โง ๐ง โ ๐ถ}.
More generally, the Cartesian product of the sets ๐ด1, ๐ด2, โฆ , ๐ด๐
๐ด1 ร ๐ด2 รโฆร ๐ด๐ = {(๐ฅ1, ๐ฅ2, โฆ , ๐ฅ๐)|๐ฅ๐ โ ๐ด๐ , ๐ = 1, ๐ฬ ฬ ฬ ฬ ฬ }.
For ๐ด = ๐ต = ๐ถ = ๐ด1 = ๐ด2 = . . . = ๐ด๐, we have
๐ด ร ๐ด โ ๐ด2, ๐ด ร ๐ด ร ๐ด โ ๐ด3, ๐ด ร ๐ด ร โฆร ๐ดโ โ ๐ด๐.๐ ori
For example, โ3 = {(๐ฅ, ๐ฆ, ๐ง)|๐ฅ, ๐ฆ, ๐ง โ โ}. The subset
ฮ = {(๐, ๐)|๐ โ ๐ด} โ ๐ด2
is called the diagonal of set ๐ด2.
Examples.
1. Let ๐ด = {1,2} and ๐ต = {1, 2, 3}. Then
๐ด ร ๐ต = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}
and
๐ต ร ๐ด = {(1,1), (1,2), (2,1), (2,2), (3,1), (3,2)}.
We notice that ๐ด ร ๐ต โ ๐ต ร ๐ด.
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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2. The Cartesian product โ2 = โ รโ can be geometrically
represented as the set of all the points from a plane to wich a
rectangular systems of coordinates ๐ฅ๐๐ฆ has been assigned,
associating to each element (๐ฅ, ๐ฆ) โ โ2 a point ๐(๐ฅ, ๐ฆ) from the plane
of the abscissa ๐ฅ and the ordinate ๐ฆ.
Let ๐ด = [2; 3] and ๐ต = [1; 5], (๐ด, ๐ต โ โ). Then ๐ด ร ๐ต can be
geometrically represented as the hatched rectangle ๐พ๐ฟ๐๐ (see Figure
1.1), where ๐พ(2,1), ๐ฟ(2,5),๐(3,5), ๐(3,1).
Figure 1.1.
The following properties are easily verifiable:
a. (๐ด โ ๐ถ โง ๐ต โ ๐ท) โน ๐ด ร ๐ต โ ๐ถ ร ๐ท;
b. ๐ด ร (๐ต โช ๐ถ) = (๐ด ร ๐ต) โช (๐ด ร ๐ถ),
๐ด ร (๐ต โฉ ๐ถ) = (๐ด ร ๐ต) โฉ (๐ด ร ๐ถ);
c. ๐ด ร ๐ต = โ โบ (๐ด = โ โจ ๐ต = โ ),
๐ด ร ๐ต โ โ โบ (๐ด โ โ โง ๐ต โ โ ).
Algebraic problems and exercises for high school
15
7. The intersection and the union of a family of sets. A family of
sets is a set {๐ด๐|๐ โ ๐ผ} = {๐ด๐}๐โ๐ผ whose elements are the sets ๐ด๐ , ๐ โ
๐ผ, ๐ด๐ โ ๐(๐ธ). We say that ๐ด๐ , ๐ โ ๐ผ, ๐ด๐ โ ๐(๐ธ) is a family of sets
indexed with the set ๐ผ.
Let there be a family of sets {๐ด๐|๐ โ ๐ผ}. Its union (or the union of
the sets ๐ด๐ , ๐ โ ๐ผ) is the set
. ๐ โ ๐ผ
The intersection of the given family (or the intersection of the
sets ๐ด๐, ๐ โ ๐ผ) is the set
In the case of ๐ผ = {1,2,โฆ , ๐}, we write
8. Euler-Wenn Diagrams. We call Euler Diagrams (in USA โ
Wennโs 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 Morganโs law.
Solution.
In fig. 1.2.a, the hatched part is ๐ด โฉ ๐ต; the uncrossed one
(except ๐ด โฉ ๐ต) represents ๐ถ๐ธ(๐ด โฉ ๐ต).
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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Figure 2.2.a Figure 3.2.b
In fig. 1.2b the side of the square hatched with โ \ \ \ \ โ is equal
to ๐ถ๐ธ(๐ด) and the one hatched with โ / / / / โ is equal to ๐ถ๐ธ(๐ต). All the
hatched side forms ๐ถ๐ธ(๐ด) โช ๐ถ๐ธ(๐ต) (the uncrossed side is exactly ๐ด โฉ
๐ต).
From these two figures it can be seen that ๐ถ๐ธ(๐ด โฉ ๐ต) (the
uncrossed side of the square in fig. 1.2.a coincides with ๐ถ๐ธ(๐ด) โช
๐ถ๐ธ(๐ต) (the hatched side of fig. 1.2.b), meaning
๐ถ๐ธ(๐ด โฉ ๐ต) = ๐ถ๐ธ(๐ด) โช ๐ถ๐ธ(๐ต).
1.2. Solved exercises 1. For any two sets ๐ด and ๐ต, we have
๐ด โฉ ๐ต = ๐ด \ (๐ด\๐ต)
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 Eulerโs diagrams.
So ๐ด โฉ ๐ต = ๐ด\(๐ด\๐ต).
2. Whatever ๐ด,๐ต โ ๐ธ 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 Eulerโs circles, we have
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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Figure 4.3.a Figure 5.3.b
In fig. 1.3.a, we have (๐ด โฉ ๐ต) โช ( ๐ด โฉ ๐ต), which represents the
hatched side of the square. From fig. 1.3.b it can be seen that
(๐ด โฉ ๐ต) โช ( ๐ด โฉ ๐ต) = (๐ด โช ๐ต) โฉ (๐ด โฉ ๐ต).
3. For every two sets ๐ด, ๐ต โ ๐ธ, the equivalence stands true:
Solution. Let ๐ด\๐ต = ๐ต\๐ด. We assume that ๐ด โ ๐ต. Then there
is ๐ โ ๐ด with ๐ โ ๐ต or ๐ โ ๐ต with ๐ โ ๐ด.
In the first case we obtain ๐ โ ๐ด\๐ต and ๐ โ ๐ต\๐ด , which
contradicts the equality ๐ด\๐ต = ๐ต\๐ด. In the second case, we obtain
the same contradiction.
So, if ๐ด\๐ต = ๐ต\๐ด โน ๐ด = ๐ต.
Reciprocally, obviously.
4. Sets ๐ด = {1,2,3,4,5,6,7,8,9,10}, ๐ต = {2,4,6,8,10} and ๐ถ =
{3,6,9} are given. Verify the following equalities:
a) ๐ด\(๐ต โช ๐ถ) = (๐ด\๐ต) โฉ (๐ด\๐ถ);
b) ๐ด\(๐ต โฉ ๐ถ) = (๐ด\๐ต) โช (๐ด\๐ถ).
Solution. a) We have ๐ต โช ๐ถ = {2,3,4,6,8,9,10} , ๐ด\(๐ต โช ๐ถ) =
{1,5,7}, ๐ด\๐ต = {1,3,5,7,9}, ๐ด\๐ถ = {1,2,4,5,7,8,10}, (๐ด\๐ต) โฉ (๐ด\๐ถ) =
{1,5,7} = ๐ด\(๐ต โช ๐ถ).
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b) For the second equality, we have
๐ต โฉ ๐ถ = {6} , ๐ด\(๐ต โฉ ๐ถ) = {1,2,3,4,5,6,7,8,9,10} , (๐ด\๐ต) โช
(๐ด\๐ถ) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = ๐ด\(๐ต โฉ ๐ถ).
5. Determine the sets A and B that simultaneously satisfy the
following the conditions:
1) ๐ด โช ๐ต = {1,2,3,4,5};
2) ๐ด โฉ ๐ต = {3,4,5};
3)1 โ ๐ด\๐ต;
4) 2 โ ๐ต\๐ด.
Solution. From 1) and 2) it follows that {3,4,5} โ ๐ด โ ๐ด โช ๐ต and
{3,4,5} โ ๐ต โ ๐ด โช ๐ต. From 3) it follows that 1 โ ๐ด or 1 โ ๐ต. If 1 โ ๐ด,
then from ๐ด โช ๐ต = {1,2,3,4,5} it follows that 1 โ ๐ต. But, if 1 โ ๐ต,
then 1 โ ๐ด, because, on the contrary it would follow that 1 โ ๐ด โฉ ๐ต =
{3,4,5}. So 1 โ ๐ต and 1 โ ๐ด remain. Similarly, from 4) it follows that
2 โ ๐ต and so 2 โ ๐ด . In other words, {3,4, 5} โ ๐ด โ {2,3,4,5} and
{3, 4, 5} โ ๐ต โ {1,3,4,5} with 2 โ ๐ด โช ๐ต , 1 โ ๐ด โช ๐ต and that is why
๐ด = {2,3,4,5}, and ๐ต = {1,3,4,5}.
Answer: ๐ด = {2,3,4,5}, ๐ต = {1,3,4,5}.
6. The sets ๐ด = {11๐ + 8โ๐ โ โค}, ๐ต = {4๐โ๐ โ โค} and ๐ถ =
{11(4๐ + 1) โ 3โ๐ โ โค}, are given. Prove that ๐ด โฉ ๐ต = ๐ถ.
Solution. To obtain the required equality, we will prove the truth
of the equivalence:
๐ฅ โ ๐ด โฉ ๐ต โบ ๐ฅ โ ๐ถ.
Let ๐ฅ โ ๐ด โฉ ๐ต . Then ๐ฅ โ ๐ด and ๐ฅ โ ๐ต and that is why two
integer numbers ๐,๐ โ โค , exist, so that ๐ฅ = 11๐ + 8 = 4๐โบ
11๐ = 4(๐ โ 2). In this equality, the right member is divisible by 4,
and 11, 4 are prime. So, from 11๐ โฎ 4 it follows that ๐ โฎ 4, giving ๐ =
4๐ก for one ๐ก โ โค. Then
๐ฅ = 11๐ + 8 = 11 โ 4๐ก + 8 = 11 โ 4๐ก + 11 โ 3 = 11(4๐ก + 1) โ 3
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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which implies ๐ฅ โ ๐ถ, in other words, we have proven the implication
๐ฅ โ ๐ด โฉ ๐ต โ ๐ โ ๐ถ. (1)
Similarly, let ๐ฆ โ ๐ถ. Then ๐ โ โค exists with
๐ฆ = 11(4๐ + 1) โ 3 = 11 โ 4๐ + 11 โ 3 = 11 โ 4๐ + 8 = 4(11๐ + 2)
Taking 4๐ = ๐ข โ โค and 11๐ + 2 = ๐ฃ โ โค, we obtain
๐ฆ = 11๐ข + 8 = 4๐ฃ โ ๐ด โฉ ๐ต,
which proves the truth of the implication
๐ฆ โ ๐ถ โ ๐ฆ โ ๐ด โฉ ๐ต. (2)
From (1) and (2) the required equality follows.
7. The following sets are given
and ๐ต = ๐ด โฉ โ.
Prove that:
a) all ๐ sets with ๐ต โช ๐ = {3,4,5,6,7,8,9};
b) all ๐ = {๐ฆ โ โคโ๐ฆ2 โ ๐ต โช ๐}, so that ๐ต โฉ ๐ = {3}.
Solution. We determine the set ๐ด:
Then ๐ต = (3; 6) โฉ ๐ผ๐ = {3,4,5}.
a) All possible subsets of ๐ต are
โ , {3}{4}{5}{3,4}{3,5}{4,5}{3,4,5} = ๐ต
The required ๐ sets are such that ๐ โช ๐ต = {3,4,5,6,7,8,9} and
will thus be like ๐ = ๐ถ โช {6,7,8,9}, where ๐ถ โ ๐(๐ต), namely, the sets
required at point a) are:
๐1{6,7,8,9} , ๐2{3, 6, 7,8, 9} , ๐3{4, 6, 7,8, 9} , ๐4{5,6, 7,8,9} ,
๐5{3,4,6,7,8,9}, ๐6{3,5,6,7,8,9}, ๐7{4,5,6, 7,8,9}, ๐8{3,4,5,6, 7,8,9}..
b) Because ๐ฆ โ โค, then and vice versa. Considering
that ๐ฆ2 โ ๐ต โช ๐ = {3,4,5,6,7,8,9}, we obtain ๐ฆ2 โ {4,9}, namely ๐ฆ2 โ
{โ3,โ2,2,3} = ๐. The parts of the set ๐ are:
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21
โ , {โ3} , {โ2} , {2} , {3} , {โ3,โ2} , {โ3,2} , {โ3,3} , {โ2,2} ,
{โ2,3}, {2,3}, {โ3,โ2,2}, {โ3,โ2,3}, {โ3,2,3}, {โ2,2,3}, ๐.
From the condition ๐ต โฉ ๐ = {3} it follows that ๐ is one of the
sets
๐1 = {3}, ๐2 = {โ3,3}, ๐3 = {โ2,3}, ๐4 = {2,3}, ๐5 = {โ3,โ2,3},
๐6 = {โ3,2,3}, ๐7 = {โ2,2,3}, ๐8 = ๐ = {โ3,โ2,2,3}.
Answer: a) ๐ โ {๐1, ๐2, ๐3, ๐4, ๐5, ๐6, ๐7, ๐8};
b) ๐ โ {๐1, ๐2, ๐3, ๐4, ๐5, ๐6, ๐7, ๐8}.
8. Determine ๐ด,๐ต, ๐ถ โ ๐ and ๐ด โ ๐ต, if
๐ = {1,2,3,4,5,6} , ๐ด โ ๐ถ = {1,2} , ๐ต โ ๐ถ = {5,6} , ๐ด โฉ ๐ถ = ๐ต โฉ
๐ถ = {3,4}.
Solution. From ๐ด โฉ ๐ถ = ๐ต โฉ ๐ถ = {3,4}, it follows that {3,4} โ
๐ด โฉ ๐ต โฉ ๐ถ.
We know that
๐ด โ ๐ถ = (๐ด โ ๐ถ) โช (๐ถ โ ๐ด) = (๐ด โช ๐ถ) โ (๐ด โฉ ๐ถ),
๐ด โ ๐ถ = (๐ต โ ๐ถ) โช (๐ถ โ ๐ต) = (๐ต โช ๐ถ) โ (๐ต โฉ ๐ถ).
Then:
1 โ ๐ด โ ๐ถ โ (1 โ ๐ด โช ๐ถ โง 1 โ ๐ด โฉ ๐ถ) โบ ((1 โ ๐ด โจ 1 โ ๐ถ) โง
1 โ ๐ด โฉ ๐ถ).
The following cases are possible:
a) 1 โ ๐ด and1 โ ๐ถ;
b) 1 โ ๐ด and 1 โ ๐ถ
(case no. 3, 1 โ ๐ด and 1 โ ๐ถ โ 1 โ ๐ด โฉ ๐ถ = {3,4}, is impossible).
In the first case 1 โ ๐ด and 1 โ ๐ถ, from ๐ต โ ๐ถ = {5,6} it follows
that 1 โ ๐ต , because otherwise 1 โ ๐ต and 1 โ ๐ถ โ 1 โ ๐ถ โ ๐ต โ
๐ตโ ๐ถ = {5,6}. So, in this case we have 1 โ ๐ต โฉ ๐ถ = {3,4}which is
impossible, so it follows that 1 โ ๐ด, 1 โ ๐ถ. Similarly, we obtain 2 โ
๐ด and 2 โ ๐ต and 2 โ ๐ถ , 5 โ ๐ต and 5 โ ๐ด ,5 โ ๐ถ and 6 โ ๐ต , 6 โ ๐ด, 6 โ
๐ด.
In other words, we have obtained:
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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๐ด = {1,2,3,4}, ๐ต = {3,4,5,6}, ๐ถ = {3,4} and ๐ด โ ๐ต = {1,2,5,6}.
Answer: ๐ด = {1,2,3,4}, ๐ต = {3,4,5,6}, ๐ถ = {3,4} and
๐ด โ ๐ต = {1,2,5,6}.
9. The sets ๐ด = {1,2}, ๐ต = {2,3} are given. Determine the fol-
lowing sets:
a) ๐ด ร ๐ต; b) ๐ต ร ๐ด; c) ๐ด2;
d) ๐ต2; e) (๐ด ร ๐ต) โฉ (๐ต ร ๐ด); f) (๐ด โช ๐ต) ร ๐ต;
g) (๐ด ร ๐ต) โช (๐ต ร ๐ต).
Solution. Using the definition for the Cartesian product with
two sets, we obtain:
a) ๐ด ร ๐ต = {(1,2), (1,3), (2,2), (2,3)};
b) ๐ต ร ๐ด = {(2,1), (2,2), (3,1), (3,2)};
c) ๐ด2 = {(1,1), (1,2), (2,1), (2,2)};
d) ๐ต2 = {(2,2), (2,3), (3,2), (3,3)};
e) (๐ด ร ๐ต) โฉ (๐ต ร ๐ด) = {(2,2)};
f)๐ด โช ๐ต = {1,2,3}; (๐ด โช ๐ต) ร ๐ต = {(1,2), (1,3), (2,2)
(2,3), (3,2), (3,3)};
g) (๐ด ร ๐ต) โช (๐ต ร ๐ต) = {(1,2), (1,3), (2,2), (2,3),
(3,2), (3,3)} = (๐ด โช ๐ต) ร ๐ต.
10. The sets ๐ด = {1,2, ๐ฅ}, ๐ต = {3,4, ๐ฆ} are given. Determine ๐ฅ
and ๐ฆ, knowing that {1,3} ร {2,4} โ ๐ด ร ๐ต.
Solution. We form the sets ๐ด ร ๐ต and {1,3} ร {2,4} :
Because {1,3} ร {2,4} โ ๐ด ร ๐ต, we obtain
For ๐ฅ = 3 and ๐ฆ = 2, we have (3,2) โ ๐ด ร ๐ต.
Answer: ๐ฅ = 3, ๐ฆ = 2.
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23
11. If ๐ด โ ๐ต, then
Prove.
Solution. ๐ต โ ๐ด โ (๐ด โ ๐ต) โช ๐ต = ๐ด and that is why
(we have used the equality (๐ด โช ๐ต) ร ๐ถ = (๐ด ร ๐ถ) โช (๐ต ร ๐ถ)).
12. How many elements does the following set have:
Solution. The set ๐ด has as many elements as the fraction
(๐2 + 1) โ (2๐2 + ๐ + 1) has different values, when ๐ takes the values
1,2,3,โฆ , 1000. We choose the values of ๐ for which the fraction takes
equal values.
Let ๐, ๐ โ โโ, ๐ < 1 with
Then
But ๐ โ โโ and consequently
For ๐ = 2, we obtain ๐ = 3, and for ๐ = 3, we have ๐ = 2.
Considering that ๐ < ๐, we obtain ๐ = 2 and ๐ = 3. So, only for ๐ =
2 and ๐ = 3, the same element of the set ๐ด: ๐ฅ = 5 โ 11 is obtained.
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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Answer: the set ๐ด has 999 elements, namely ๐(๐ด) = 999.
13. Determine the integer numbers ๐ฅ, ๐ฆ that make the following
statement true (๐ฅ โ 1) โ (๐ฆ โ 3) = 13.
Solution. We write
As 13 is a prime number, and ๐ฅ, ๐ฆ โ โค, the following cases are
possible:
meaning the proposition ๐(๐ฅ) 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 ๐, ๐ and ๐ are different, we
cannot have the equality:
namely in this case ๐ด = โ ;
b) if ๐ = ๐ = ๐, then ๐ฅ = โ๐ and ๐ด = {โ๐}.
Answer: 1) for ๐ = ๐ = ๐, we have ๐ด = {โ2}; 2) if at least two of
the numbers ๐, ๐ and ๐ are different, then ๐ด = โ .
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15. Determine the set
Solution. Possible situations:
๐ฅ + 2 โค 4 โ ๐ฅ/3 or ๐ฅ + 2 > 4 โ ๐ฅ/3.
We examine each case:
In this case we have:
All integer numbers ๐ฅ for which โ1 โค ๐ฅ โค 3 โ 2 applies, are:
โ1,0,1.
Then
All integer numbers ๐ฅ for which 3 โ 2 < x โค 9 applies, are:
2,3,4,5,6,7,8,9.
We thus obtain:
Answer: ๐ด = {โ1,0,1,2,3,4,5,6,7,8,9}.
16. Determine the values of the real parameter ๐ for which the
set has:
a) one element;
b) two elements;
c) is null.
Solution. The set ๐ด 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 ๐ โ โ for which the equation has one solution, two
different solutions or no solution.
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a) The equation (1) has one (two equal) solution, if and only if
๐ท = 0 for ๐ = ๐ โ 1 โ 0 or if ๐ = ๐ โ 1 = 0.
We examine these cases:
2) For ๐ = 1, the equation (1) becomes โ5๐ฅ + 15 = 0 โ ๐ฅ = 3. So,
the set ๐ด consists of one element for
b) The equation (1) has two different roots, if and only if ๐ท > 0,
namely
c) The equation (1) doesnโt have roots ๐ท < 0 โ 39๐2 โ
60๐โ 28 >
Answer:
17. Let the set ๐ด = {3,4,5} and ๐ต = {4,5,6}. Write the elements
of the sets ๐ด2 โฉ ๐ต2 and (๐ด โ (๐ต โ ๐ด)) ร (๐ต โฉ ๐ด).
Solution. a) For the first set we have:
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๐ด2 = {(3,3), (3,4), (3,5), (4,3), (4,4), (4,5), (5,3), (5,4), (5,5)},
๐ต2 = {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6)},
๐ด2 โฉ ๐ต2 = {(4,4), (4,5), (5,4), (5,5)}.
b) For the second set we have:
Then
Answer:
18. Given the sets ๐ด = {1,2,3,4,5,6,7} and ๐ต = {2,3,4}, solve
the equations ๐ด โ ๐ = ๐ด โ ๐ต and (๐ด โ ๐)โ ๐ต = ๐ถ๐ด(๐ต).
Solution. In order to solve the enunciated equations, we use the
properties of the symmetrical difference: ๐ด โ(๐ด โ ๐ต) = ๐ต , its
associativity and commutativity .
But
Consequently,
We calculate
Thus
Answer: ๐ = ๐ต, ๐ = โ .
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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19. The following sets are given
Determine the sets ๐ด โช ๐ต, ๐ด โฉ ๐ต, ๐ด, ๐ต, ๐ด โ ๐ต, ๐ต โ ๐ด, (๐ด โช ๐ต) โ
(๐ต โ ๐ด) and ฤ ร (๐ต โ ๐ด).
Solution. 1) We determine the sets ๐ด and ๐ต.
The inequation (โ) 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 โ 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?
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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Solution. Let ๐ด be the set of students who have correctly solved
only the problem, ๐ต โ only the inequation, ๐ถ โ that have proven only
the identity, ๐ท โ the set of students that have solved only the problem
and the inequation, ๐ธ โ the set of students that have solved only the
problem and have proven the identity, ๐น โ the set of students who
have solved only the inequation and have proven the identity, and ๐บ โ
the set of students that have successfully solved everything.
From the given conditions, it follows that ๐ (๐ด) = 3, ๐ (๐ต) = 5,
๐ (๐ถ) = 4, ๐ (๐ท) = 8, ๐ (๐ธ) = 7, ๐ (๐น) = 9.
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 ๐ด, ๐ต, ๐ถ, ๐ท, ๐ธ, ๐น, ๐บ have only elements of the null set in common,
the union of the sets ๐ด, ๐ต, ๐ถ, ๐ท, ๐ธ, ๐น, ๐บ 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 Dodjsonโs problem)
In a tense battle, 72 out of 100 pirates have lost one eye, 75 โ
one ear, 80 โ one hand and 85 โ 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 ๐ด the set of one-eyed pirates, with ๐ต โ
the set of one eared pirates, with ๐ถ โ the set of pirates with one hand
and with ๐ท โ the set of the pirates with one leg.
The problems requires to appreciate the set ๐ด โฉ ๐ต โฉ ๐ถ โฉ ๐ท.
Algebraic problems and exercises for high school
31
It is obvious that the universal set ๐ธ is composed from the set
๐ด โฉ ๐ต โฉ ๐ถ โฉ ๐ท 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 ๐ธ is not smaller (doesnโt have fewer
elements) than the sum of the elements of the sets ๏ฟฝฬ ๏ฟฝ, ๏ฟฝฬ ๏ฟฝ, ๏ฟฝฬ ๏ฟฝ, ๏ฟฝฬ ๏ฟฝ and ๐ด โฉ
๐ต โฉ ๐ถ โฉ ๐ท (equality would have been the case only if the sets ๐ด, ๐ต, ๐ถ
and ๐ท, two by two, do not intersect). But,
Substituting, we have ๐(๐ธ) = 100 namely 100 โค ๐(๐ด โฉ ๐ต โฉ
๐ถ โฉ ๐ท) + 30 + 25 + 20 + 15.
Consequently, ๐(๐ด โฉ ๐ต โฉ ๐ถ โฉ ๐ท) โฅ 100โ 30 โ 25 โ 20 = 10.
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 โ English and
German, 10 - English and French, 5 โ French and German, 3 students -
all three languages. How many study only one language? How many
not even one language?
Solution. Let ๐ด be the set of students that study English, ๐ต -
German, ๐ถ - French.
Then the set of students attending both English and German is
๐ด โฉ ๐ต, English and French - ๐ด โฉ ๐ถ, French and German - ๐ต โฉ ๐ถ,
English, German and French - ๐ด โฉ ๐ต โฉ ๐ถ, and the set of students
that study at least one language is ๐ด โช ๐ต โช ๐ถ.
From the conditions above, it follows that the students who only
study English form the set ๐ด\(๐ด โฉ ๐ต) โช (๐ด โฉ ๐ถ) , only German -
๐ต\(๐ด โฉ ๐ต) โช (๐ต โฉ ๐ถ), only French - ๐ถ\(๐ด โฉ ๐ถ) โช (๐ต โฉ ๐ถ).
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But, because ๐ด โฉ ๐ต โ ๐ด, we have ๐(๐ด\((๐ด โฉ ๐ต) โช (๐ด โฉ ๐ถ))) =
๐(๐ด) โ ๐((๐ด โฉ ๐ต) โช (๐ด โฉ ๐ถ)) = ๐(๐ด) โ (๐(๐ด โฉ ๐ต) + ๐(๐ด โฉ ๐ถ) โ
๐(๐ด โฉ ๐ต โฉ ๐ถ)) = ๐(๐ด) โ ๐(๐ด โฉ ๐ต) โ ๐(๐ด โฉ ๐ถ) + ๐(๐ด โฉ ๐ต โฉ ๐ถ).
Similarly, ๐ (๐ต\((๐ด โฉ ๐ต) โช (๐ต โฉ ๐ถ))) = ๐(๐ต) โ ๐(๐ด โฉ ๐ต) โ
๐(๐ต โฉ ๐ถ) + ๐(๐ด โฉ ๐ต โฉ ๐ถ); ๐(๐ถ\((๐ด โฉ ๐ถ) โฉ (๐ต โฉ ๐ถ)) = ๐(๐ถ) โ
๐(๐ด โฉ ๐ถ) โ ๐(๐ต โฉ ๐ถ) + ๐(๐ด โฉ ๐ต โฉ ๐ถ).
Let ๐ท be the set of students who only study one language, then
Consequently,
= 28 + 30 + 42 โ 2.8 โ 2.10 โ 2.5 + 3.3 = 63. ๐(๐ท) = 63.
The number of students who donโt 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 ๐(๐ป) =
๐(๐) โ ๐(๐ด โช ๐ต โช ๐ถ), where ๐ป is the set of students that donโt study
any language and ๐ the set of all 100 students.
From problem 20 we have
So ๐(๐ป) = 100 โ 80 = 20.
Answer: 63 students study only one language, 20 students donโt
study any language.
Algebraic problems and exercises for high school
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 (๐ด,๐ต and ๐ถ are arbitrary sets)?
3. Let ๐ด = {๐ฅ โ โโ๐ฅ2 โ 12๐ฅ + 35) = 0},
๐ต = {๐ฅ โ โโ๐ฅ2 + 2๐ฅ + 35) = 0};๐ถ = {๐ฅ โ โโ๐ฅ2 + 2๐ฅ + 35) = 0}
a) Determine the sets ๐ด, ๐ต and ๐ถ.
b) State if the numbers 1/5, 5, 7, 1/2 belong to these sets or
not.
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4. Determine the sets:
5. Let ๐ด = {๐ฅ โ โโx = 4n + 6m,n,m โ โโ};
a) Write three elements belonging to set ๐ด.
b) Determine if 26, 28, 33 belong to ๐ด.
6. Indicate the characteristic properties of the elements
belonging to the sets: ๐ด = {4, 7, 10}, ๐ต = {3, 6, 12} , ๐ถ =
{1, 4, 9, 16, 25}, ๐ท = {1, 8, 27, 64, 125}.
7. How many elements do the following sets have?
8. Let there be the set ๐ด = {โ3,โ2,โ1,1,2,3}. Determine the
subsets of ๐ด:
9. Determine the sets:
Algebraic problems and exercises for high school
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10. Compare (โ,โ,=,โ,โ ) the sets ๐ด and ๐ต, if:
11. Let ๐ด = {1, 2, 3, 4, 5, 6, 7}, ๐ต = {5, 6, 7, 8, 9, 10}. Using the
symbols โช,โฉ,โ, ๐ถ (complementary), express (with the help of ๐ด, ๐ต and
โโ) the sets:
.
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12. Determine the set ๐ธ, in the case it is not indicated and its
parts ๐ด, ๐ต, ๐ถ simultaneously satisfy the conditions:
๐ด has more elements than ๐ต;
๐ด has more elements than ๐ต;
Algebraic problems and exercises for high school
37
13. Determine the sets ๐ด, ๐ต, ๐ด โช ๐ต, ๐ด โฉ ๐ต, ๐ต โ ๐ด, ๐ด โ ๐ต, ๐ดโ๐ต,
๐ด โช ( ๐ต โ ๐ด), , ๐ด โ (๐ต โ ๐ด), ๐ด ร ๐ต, ๐ต ร ๐ด, (๐ด โช ๐ต) ร ๐ด, ๐ต ร ( ๐ด โ ๐ต), if:
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14. Let
a) Determine the sets:
b) How many elements does the set ๐ท = {๐ฅ โ ๐โ๐ฅ โค
(699 100โ )} have?
15. Determine the sets ๐ด, ๐ต โ ๐ธ, if
16. Determine the set ๐ธ and its parts ๐ด and ๐ต, if
17. Compare the sets ๐ด and ๐ต, if
Algebraic problems and exercises for high school
39
18. Determine the values of the real parameter ๐ for which the
set ๐ด has one element, two elements or it is void, if:
19. Determine the number of elements of the set ๐ด:
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20. The sets ๐ด,๐ต, ๐ถ are given. Determine ๐ด โฉ ๐ต โฉ ๐ถ.
21. Determine ๐ด โฉ ๐ต, if:
22. Prove the properties of the operations with sets.
23. Determine the equalities (๐ด, ๐ต, ๐ถ etc. are arbitrary sets):
Algebraic problems and exercises for high school
41
24. Given the sets ๐ด = {1, 2, 3}, ๐ต = {2, 3, 4}, ๐ถ = {3, 4, 5} and
๐ท = {4, 5, 6}, write the elements of the following sets:
25. Given the sets ๐ด, ๐ต and ๐ถ, solve the equations (๐ด๐๐ต)โ๐ =
๐ถ, where ๐ โ {โช,โฉ,โ โ}:
26. Determine the sets ๐ด โฉ ๐ต, ๐ด โช ๐ต, ฤ, ๐ต,ฬ ๐ด โ ๐ต, ๐ต โ ๐ด, ๐ด โช
(๐ต โ ฤ), ๐ด โฉ (ฤ โ ๐ต), if:
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Algebraic problems and exercises for high school
43
2. Relations, functions
2.1. Definitions and notations
Relations, types. Relations formation
Let ๐ด and ๐ต be two non-empty sets, and ๐ด ร ๐ต their Cartesian
product. Any subset ๐ โ ๐ด ร ๐ต is called a relation between the
elements of ๐ด and the elements of ๐ต. When ๐ด = ๐ต, a relation like ๐ โ
๐ด ร ๐ด is called a binary relation on set ๐ด.
If there exists a relation like ๐ โ ๐ด ร ๐ต, then for an ordered pair
(๐, ๐) โ ๐ด ร ๐ต we can have (๐, ๐) โ ๐ or (๐, ๐) โ ๐ . In the first case,
we write ๐๐ ๐ and we read "๐ is in relation ๐ with ๐", and in the second
case โ ๐๐ ๐, that reads "๐ is not in relation ๐ with ๐". We hold that
๐๐ ๐ โบ (๐, ๐) โ ๐ .
By the definition domain of relation ๐ โ A X B, we understand
the subset ๐ก๐ โ ๐ด consisting of only those elements of set ๐ด that are in
relation ๐ with an element from ๐ต, namely
By values domain of relation ๐ โ ๐ด ร ๐ต we understand the
subset ฯ๐ โ ๐ต consisting of only those elements of set ๐ต that are in
relation ๐ with at least one element of ๐ด, namely
The inverse relation. If we have a relation ๐ โ ๐ด ร ๐ต, then by
the inverse relation of the relation ๐ we understand the relation
๐ โ1 โ ๐ต ร ๐ด as defined by the equivalence
namely
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Example 1. Let us have ๐ด = {1, 2}, ๐ต = {4, 5, 6} 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 ๐ด, ๐ต, ๐ถ be three sets and let us
consider the relations ๐ โ ๐ด ร ๐ต, ๐ โ ๐ต ร ๐ถ. The relation ๐ โ ๐ โ ๐ด ร
๐ถ formed in accordance with the equivalence
namely
is called the formation of relation ๐ and ๐.
Example 2. Let us have ๐ด, ๐ต, ๐ผ, ๐ฝ, ๐พ, from Example 1.
Determine
๐พโ1 โ ๐ฝโ1 and (๐ฝ โ ๐พ)โ1.
Solution. We point out that the compound of the relations ๐ผ โ
๐ถ ร ๐ท with ๐ฝ โ ๐ธ ร ๐ธ exists if and only if ๐ท = ๐ธ.
Because ๐ผ โ ๐ด ร ๐ต, ๐ฝ โ ๐ด ร ๐ต, it follows that ๐ผ โ ๐ผ and ๐ผ โ ๐ฝ
doesnโt exist.
We determine ๐ผ โ ๐พ:
Algebraic problems and exercises for high school
45
(2,4) โ ๐ผ and {(4,1), (4,2)} โ ๐พ โ {(2,1), (2,2)} โ ๐ผ โ ๐พ;
(2,6) โ ๐ผ , but in ๐พ we donโt have a pair with the first
component 6. It follows that
c) ๐ฝ โ ๐ด ร ๐ต , ๐พ โ ๐ด ร ๐ต โ ๐ฝ โ ๐พ exists. By repeating the
reasoning from b),
and ๐ฝ โ ๐ฝโ1 = {(2,2)}.
and ๐พโ1 โ ๐ฝโ1 = {(1,2), (2,2)}.
The equality relation. Let ๐ด be a set. The relation
is called an equality relation on ๐ด. Namely,
Of a real use is:
1stTheorem. Let ๐ด, ๐ต, ๐ถ, ๐ท be sets and ๐ โ ๐ด ร ๐ต , ๐ โ ๐ต ร ๐ถ ,
๐ โ ๐ถ ร ๐ท relations. Then,
1) (๐ โ ๐) โ ๐ = ๐ โ (๐ โ ๐) (associativity of relations formation)
The equivalence Relations. The binary relation ๐ โ ๐ด2 is called:
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a) reflexive, if ๐ฅ๐ ๐ฅ whatever ๐ฅ โ ๐ด is;
b) symmetrical, if (๐ฅ๐ ๐ฆ โ ๐ฆ๐ ๐ฅ), (โฉ)๐ฅ, ๐ฆ โ ๐ด;
c) transitive, if ((๐ฅ๐ ๐ฆ โ ๐ฆ๐ ๐ง) โ ๐ฅ๐ ๐ง), (โฉ) ๐ฅ, ๐ฆ, ๐ง โ ๐ด ;
d) anti-symmetrical, if ((๐ฅ๐ ๐ฆ โ ๐ฆ๐ ๐ฅ) โ ๐ฅ = ๐ฆ), (โฉ) ๐ฅ, ๐ฆ, ๐ง โ ๐ด;
e) equivalence relation on A, if it is reflexive, symmetrical and
transitive;
f) irreflexive, if ๐ฅ๐ ๐ฅ whatever ๐ฅ โ ๐ด is.
Let ๐ be an equivalence relation on set ๐ด. For each element ๐ฅ โ
๐ด, the set
is called the equivalence class of ๐ฅ modulo ๐ (or in relation to R), and
the set
is called a factor set (or quotient set) of ๐ด through ๐ .
The properties of the equivalence classes. Let ๐ be an
equivalence relation on set ๐ด and ๐ฅ, ๐ฆ โ ๐ด . Then, the following
affirmations have effect:
Partitions on a set. Let ๐ด be a non-empty set. A family of
subsets {๐ด๐โ๐ โ ๐ผ} of ๐ด is called a partition on ๐ด (or of ๐ด), if the
following conditions are met:
2nd Theorem. For any equivalence relation ๐ on set ๐ด, the factor
set ๐ด/ ๐ = {๐ ๐ฅโ๐ฅ โ ๐ด} is a partition of ๐ด.
Algebraic problems and exercises for high school
47
3rd Theorem. For any partition ๐ = {๐ด๐โ๐ โ ๐ผ} of ๐ด, there exists
a sole equivalence relation ๐ผ๐ on ๐ด, hence
The relation ๐ผ๐ โ ๐ด is formed according to the rule
It is easily established that ๐ผ๐ is an equivalence relation on ๐ด and
the required equality.
Example 3. We define on set โค the binary relation ๐ผ according to
the equivalence ๐๐ผ๐ โ (๐ โ ๐) โฎ ๐, where ๐ โ โโ, ๐ fixed, (โฉ)๐, ๐ โ
โค.
a) Prove that ๐ผ is an equivalence relation on โค.
b) Determine the structure of the classes of equivalence.
c) Form the factor set โค/๐ผ. Application: ๐ = 5.
Solution. a) Let ๐, ๐, ๐ โ โค. Then:
1) reflexivity ๐ โ ๐ = ๐ โฎโ ๐ ๐๐ผ๐;
2) symmetry
3) transitivity
From 1) - 3) it follows that ๐ผ is an equivalence relation on โค.
b) Let ๐ โ โค. Then
In accordance with the theorem of division with remainder for
integer numbers ๐ and ๐, we obtain
Then
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where ๐ = ๐ก + ๐ โ โค, and therefore
where ๐๐ is the remainder of the division of ๐ through ๐. But,
In other words, the equivalence class of ๐ โ โค coincides with the
class of the remainder of the division of ๐ through ๐.
c) Because by the division to ๐ only the remainders 0, 1, 2, . . .,
๐ โ 1, can be obtained, from point b) it follows that we have the
exact ๐ different classes of equivalence:
The following notation is customarily used
. Then
where ๐ฬ consists of only those integer numbers that divided by ๐ yield
the remainder
.
For n = 5, we obtain
with
Definition. The relation ๐ผ is called a congruency relation
modulo ๐ on โค, and class รข = ๐ผ๐ is called a remainder class modulo
๐ and its elements are called the representatives of the class.
The usual notation:
๐๐ผ๐ โ (๐ โ ๐) โฎ ๐ โ ๐ โก ๐ (mode ๐) (๐ is congruent with ๐
modulo ๐),and
Algebraic problems and exercises for high school
49
is the set of all remainder classes modulo ๐.
Example 4 (geometric). Let ๐ be a plane and ๐ฟ the set of all the
straight lines in the plane. We define ๐ฟ as the binary relation ๐ฝ in
accordance with
a) Show that ๐ฝ is an equivalence relation on ๐ฟ.
b) Describe the equivalence classes modulo ๐ฝ.
c) Indicate the factor set.
Solution. a) It is obvious that ๐ฝ is an equivalence relation (each
straight line is parallel to itself; if ๐1 โฅ ๐2 โ ๐2 โฅ ๐1 and
b) Let ๐ โ ๐ฟ. Then the class
consists only of those straight lines from ๐ฟ that are parallel with the
straight line ๐.
c) ๐ฟ/๐ฝ = {๐ฝ๐โ๐ โ ๐ฟ} is an infinite set, because there are an
infinity of directions in plane ๐.
Example 5. The set๐ด = {1,2,3,4,5,6, 7,8,9}is given and its parts
a) Show that ๐ = {๐ด1, ๐ด2, ๐ด3, ๐ด4 } is a partition of ๐ด.
b) Determine the equivalence relation ๐ผ๐ on ๐ด.
c) Is ๐ = {๐ต1, ๐ต2, ๐ต3 } a partition on ๐ด?
Solution. a) 1) We notice that
meaning the system ๐ of the subsets of the set A defines a partition on
๐ด.
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b) In accordance to the 3rd Theorem, we have
So:
๐ผ๐= {( 1,1), (1,2), (2,1), (2,2), (3,3), (3,4), (3,6), (4,3), (4,4), (4,6), (6,3),
(6,4), (6,6), (5,5), (5,7), (5,8), (8,5), (8,8), (8,7), (7,5), (7,7), (7,8), (9,9)}.
which proves that system ๐ does not define a partition on ๐ด.
Order relations. A binary relation ๐ on the set ๐ด is called an
order relation on ๐ด, if it is reflexive, anti-symmetrical and transitive.
If ๐ is an order relation on ๐ด, then ๐ โ1 is also an order relation
on ๐ด (verify!). Usually, the relation ๐ is denoted by "โค" and the
relation ๐ โ1 by "โฅ", hence
With this notation, the conditions that "โค" is an order relation
on the set ๐ด are written:
reflexivity ๐ฅ โ ๐ด โ ๐ฅ โค ๐ฅ;
asymmetry (๐ฅ โค ๐ฆโ๐ฆ โค ๐ฅ) โ ๐ฅ = ๐ฆ;
transitivity (๐ฅ โค ๐ฆโ๐ฆ โค ๐ง) โ ๐ฅ โค ๐ง.
Example 6. On the set โ we define the binary relation ๐พ in
accordance with
Show that ๐พ is an order relation on โ.
Solution. We verify the condition from the definition of order
relations.
1) Reflexivity
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51
2) Asymmetry. Let ๐, ๐ โ โ with ๐๐พ๐ and ๐๐พ๐. It follows that
there are the natural numbers ๐, ๐ โ โ with ๐ = ๐ . ๐ and ๐ = ๐ . ๐.
Then
which implies that
3) Transitivity. Let ๐, ๐, ๐ โ โ with ๐๐พ๐ and ๐๐พ๐ . Then there
exists ๐ข, ๐ฃ โ โ with ๐ = ๐๐ข and ๐ = ๐๐ฃ hence,
Because ๐ข, ๐ฃ โ โ, the implication is true
which proves that ๐พ is an order relation on the set โ .
Remark. The order relation ๐พ is called a divisibility relation on โ
and it is written as ๐ โฎ ๐, namely
(๐โ๐ reads "b divides a", and ๐ โฎ ๐ "a is divisible through b").
Functional relations Let ๐ด and ๐ต be two sets. A relation ๐ โ ๐ด ร ๐ต is called an
application or a functional relation, if the following conditions are met:
1)โฉ ๐ฅ โ ๐ด(โ)๐ฆ โ ๐ต, so as ๐ฅ๐ ๐ฆ;
An application (or function) is a triplet ๐ = (๐ด, ๐ต, ๐ ), where ๐ด
and ๐ต are two sets and ๐ โ ๐ด ร ๐ต is a functional relation.
If ๐ โ ๐ด ร ๐ต is an application, then for each ๐ฅ โ ๐ด there exists,
according to the conditions 1) and 2) stated above, a single element
๐ฆ โ ๐ต, so that ๐ฅ๐ ๐ฆ; we write this element ๐ฆ with ๐(๐ฅ). Thus,
The element ๐(๐ฅ) โ ๐ต is called the image of element ๐ฅ โ ๐ด
through application ๐, set ๐ด is called the definition domain of ๐ noted
by ๐ท(๐) = ๐ด, and set ๐ต is called the codomain of ๐ and we usually say
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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that ๐ is an application defined on ๐ด with values in ๐ต. The functional
relation ๐ is also called the graphic of the application (function) ๐,
denoted, later on, by ๐บ๐. To show that ๐ is an application defined on ๐ด
with values in ๐ต , we write ๐: ๐ด โถ ๐ต or ๐ด๐โ๐ต , and, instead of
describing what the graphic of ๐ (or ๐โฒ๐ ) is, we indicate for each ๐ฅ โ ๐ด
its image ๐(๐ฅ). Then,
i.e.
The equality of applications. Two applications ๐ = (๐ด, ๐ต, ๐ )
and ๐ = (๐ถ,๐ท, ๐) are called equal if and only if they have the same
domain ๐ด = ๐ถ, the same codomain ๐ต = ๐ท and the same graphic
๐ = ๐. If ๐, ๐: ๐ด โถ ๐ต, the equality ๐ = ๐ is equivalent to ๐(๐ฅ) =
๐(๐ฅ), (โฉ)๐ฅ โ ๐ด, that is to say
The identical application. Let ๐ด be a set. The triplet (๐ด, ๐ด, 1๐ด) is,
obviously, an application, denoted by the same symbol 1๐ด (or ํ) and it
is called the identical application of set ๐ด.
We have
Consequently,
1๐ด: ๐ด โถ ๐ด and 1๐ด(๐ฅ) = ๐ฅ for (โฉ)๐ฅ โ ๐ด.
By ๐น(๐ด, ๐ต) we will note the set of functions defined on ๐ด with
values in ๐ต. For ๐ต = ๐ด, we will use the notation ๐น(๐ด) instead of
๐น(๐ด, ๐ด).
In the case of a finite set ๐ด = {๐1, ๐2, โฆ , ๐๐}, a function
๐: ๐ด โถ ๐ด is sometimes given by means of a table as is:
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53
where we write in the first line the elements ๐1, ๐2, โฆ , ๐๐ of set ๐ด, and
in the second line their corresponding images going through ๐, namely
๐(๐1), ๐(๐2), โฆ , (๐๐).
When ๐ด = {1,2,โฆ , ๐}, in order to determine the application,
we will use ๐: ๐ด โถ ๐ด and the notation
more frequently used to write the bijective applications of the set ๐ด in
itself. For example, if ๐ด = {1,2}, then the elements of ๐น(๐ด) are:
If ๐: ๐ด โถ ๐ต, ๐ โ ๐ด, ๐ โ ๐ต, then we introduce the notations:
- the image of subset ๐ through the application ๐.
Particularly, ๐(๐ด) = ๐ผ๐๐ , the image of application ๐ ;
is the preimage of subset ๐ through the application ๐.
Particularly, for ๐ฆ โ ๐ต, instead of writing ๐โ1({๐ฆ}), we simply
note ๐โ1(๐ฆ), that is
- the set of all preimages of ๐ฆ through the application ๐, and
- the complete preimage of set ๐ต through the application ๐.
The formation of applications. We consider the applications
๐ = (๐ด, ๐ต, ๐ ) and ๐ = (๐ต, ๐ถ, ๐), so as the codomain of ๐ to coincide
with the domain of ๐. We form the triplet ๐ โ ๐ = (๐ด, ๐ถ, ๐ โ ๐).
Then ๐ โ ๐ is also an application, named the compound of the
application ๐ with the application ๐, and the operation "โ" is called the
formation of applications. We have
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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namely
4th Theorem. Given the applications
. It follows that a) (โ โ ๐) โ ๐ = โ โ (๐ โ ๐)
(associativity of the formation of applications); b) ๐ โ 1๐ด = 1๐ต โ ๐ = ๐.
Example 7. We consider the relations ๐ โ โ รโ and ๐ โ
[0,+โ) ร [0,+โ), ๐ โ โ รโ, defined as it follows: ๐ฅ๐ ๐ฆ โ ๐ฅ2 = ๐ฆ,
๐ฅ๐๐ฆ โ ๐ฅ = ๐ฆ2, ๐ฅ๐๐ฆ โ ๐ฆ = ๐ฅ + 1.
A. Determine which of the relations ๐ , ๐ โ1, ๐, ๐โ1, ๐, ๐โ1 are
functional relations.
B. Find the functions determined at point A.
C. Calculate ๐ โ ๐, ๐ โ ๐, ๐ โ โ, โ โ ๐, โ โ โโ1, โโ1 โ โ (f, g, h are
the functions from point B.)
D. Calculate (๐ โ โ)( โ3), (โโ1 โ โ)(1/2), (โ โ ๐)(1/3).
Solution. We simultaneously solve ๐ด and ๐ต.
a) We examine the relation ๐ .
๐ฅ โ โ โ ๐ฅ2 โ โ โ ๐ฆ โ โ, i.e.
1)โฉ ๐ฅ โ โ(โ)๐ฆ = ๐ฅ2 โ โ, so that ๐ฅ๐ ๐ฆ;
which means that ๐ is a functional relation. We write the application
determined by ๐ through ๐, ๐ = ( โ,โ,๐ ).
b) We examine the relation ๐ โ1 .
which means that if ๐ฅ โ โ = {๐ โ โโ๐ < 0}, then there is no ๐ฅ โ โ,
so that ๐ฆ๐ โ1๐ฅ , which proves that ๐ โ1 is not a functional relation.
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55
c) For the relation ๐, we have: ๐ and ๐โ1 are functional relations.
We denote by ๐ = ([0,+โ), [0,+โ), ๐),and ๐โ1 = ([0,+โ), [0,+โ),
๐โ1), the functions (applications) defined by ๐ and ๐โ1, respectively.
d) We examine the relation ๐:
1) ๐ฅ โ โ โ ๐ฅ + 1 โ โ and so
so that ๐ฅ๐๐ฆ;
which proves that ๐ is a functional relation.
e) For the relation ๐โ1, we obtain:
1)(โฉ)๐ฅ โ โ (โ)๐ฅ = ๐ฆ โ 1 โ โ, so that y๐โ1๐ฅ;
meaning that ๐โ1 is also a functional relation. We denote by โ =
(โ,โ, ๐) and โโ1 = (โ,โ, ๐โ1).
C. From A and B, it follows that:
a) Then ๐ โ ๐, ๐ โ ๐ and ๐ โ โ donโt exist, because the domains
and codomains do not coincide:
๐ = (โ,โ, ๐ ), ๐ = ([0,+โ), [0,+โ), ๐), โ = (โ,โ,๐)).
b) We calculate ๐ โ โ, โ โ ๐, โ โ โโ1 and โโ1 โ โ.
So ๐ โ โ โ โ โ ๐ and โ โ โโ1 = โโ1 โ โ = 1โ.
D. We calculate:
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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Injective, surjective and bijective applications. An application
๐: ๐ด โถ ๐ต is called:
1) injective, if
(equivalent: ๐1 โ ๐2 โ ๐(๐1) โ ๐(๐2));
2) surjective, if
(any element from B has at least one preimage in ๐ด);
3) bijective, if ๐ is injective and surjective.
Remark. To prove that a function ๐: ๐ด โถ ๐ต is a surjection, the
equation ๐( ๐ฅ) = ๐ must be solvable in ๐ด for any ๐ โ ๐ต.
Of a real use are:
5th Theorem. Given the applications ๐ด๐โ๐ต
๐โ๐ถ, we have:
a) if ๐ and ๐ are injective, then ๐ โ ๐ is injective;
b) if ๐ and ๐ are surjective, then ๐ โ ๐ is surjective;
c) if ๐ and ๐ are bijective, then ๐ โ ๐ is bijective;
d) if ๐ โ ๐ is injective, then ๐ is injective;
e) if ๐ โ ๐ is surjective, then ๐ is surjective.
6th Theorem. The application ๐: ๐ด โถ ๐ต with the graphic ๐บ๐ =
๐ is a bijective application, if and only if the inverted relation ๐ โ1 is a
functional relation (fโ1is an application).
This theorem follows immediately from
Inverted application. Let ๐: ๐ด โถ ๐ต be a bijective application
with the graphic ๐บ๐ = ๐ . From the 6th Theorem, it follows that the
triplet ๐โ1 = (๐ต, ๐ด, ๐ โ1) is an application (function). This function is
called the inverse of function ๐.
We have
๐โ1: ๐ต โถ ๐ด, and for ๐ฆ โ ๐ ,
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57
i.e.
7th Theorem. Application ๐: ๐ด โถ ๐ต is bijective, if and only if
there exists an application ๐: ๐ต โถ ๐ด with ๐ โ ๐ = 1๐ด and ๐ โ ๐ = 1๐ต.
In this case, we have ๐ = ๐โ1.
Real functions The function ๐: ๐ด โถ ๐ต is called a function of real variable, if
๐ด = ๐ท(๐) โ โ. The function of real variable ๐: ๐ด โถ ๐ต is called a real
function, if ๐ต โ โ. In other words, the function ๐: ๐ด โถ ๐ต is called a
real function, if Aโ โ and Bโ โ . The graphic of the real
function ๐: ๐ด โถ ๐ต is the subset ๐บ๐ of โ2 , formed by all the
pairs (๐ฅ, ๐ฆ) โ โ2 with ๐ฅ โ ๐ด and ๐ฆ = ๐(๐ฅ), i.e.:
If the function ๐ is invertible, then
Traditionally, instead of ๐โ1(๐ฆ) = ๐ฅ we write ๐ฆ = ๐โ1(๐ฅ). Then
the graphic of the inverted function ๐โ1 is symmetrical to the graphic
function ๐ in connection to the bisector ๐ฆ = ๐ฅ.
Example 8. For the function
the inverted function is
The graphic of function ๐ is the branch of the parabola ๐ฆ = ๐ฅ2
comprised in the quadrant I, and the graphic of function ๐โ1 is the
branch of parabola ๐ฅ = ๐ฆ2 comprised in the quadrant I (see below Fig.
2.1).
Algebra operations with real functions Let ๐, ๐: ๐ท โถ โ be two real functions defined on the same set.
We consider the functions:
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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๐ = ๐ + ๐: ๐ท โถ โ ,
defined through ๐ (๐ฅ) = ๐(๐ฅ) + ๐(๐ฅ), (โฉ)๐ฅ โ ๐ท;
๐ = ๐ + g โ sum-function.
๐ = ๐ โ ๐: ๐ท โถ โ ,
defined through ๐(๐ฅ) = ๐(๐ฅ) ยท ๐(๐ฅ), (โฉ)๐ฅ โ ๐ท;
๐ = ๐ . ๐ โ product-function.
๐ = ๐ โ ๐: ๐ท โถ โ,
defined through (๐ฅ) = ๐(๐ฅ) โ ๐(๐ฅ), (โฉ)๐ฅ โ ๐ท;
๐ = ๐ โ ๐ โ difference-function.
๐ =๐
๐: ๐ท1โถโ,
defined through ๐(๐ฅ) =๐(๐ฅ)
๐(๐ฅ), (โฉ)๐ฅ โ ๐ท1
where ๐ท1 = {๐ฅ โ ๐ทโ๐(๐ฅ) โ ๐};
๐ =๐
๐โ quotient-function.
โ๐โ: ๐ท โถ โ,
defined through โ๐โ (๐ฅ) = โ๐ (๐ฅ) = {๐(๐ฅ), ๐๐ ๐(๐ฅ) โฅ 0,
โ๐(๐ฅ), ๐๐ ๐(๐ฅ) < 0,}
for (โฉ)๐ฅ โ ๐ท;
โ๐โ- module-function.
Figure 2.1.
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Example 9. Let ๐, ๐: [0,+โ) โถ โ with ๐(๐ฅ) = ๐ฅ2 and ๐(๐ฅ) =
โ๐ฅ. Determine ๐ ยฑ ๐, ๐ โ ๐ and ๐/๐ยท
Solution. As ๐, ๐ share the same definition domain, the
functions ๐ ยฑ ๐, ๐ โ ๐ and ๐/๐ make sense and
Example 10. Let ๐:โ โถ [0,+โ), ๐(๐ฅ) = ๐ฅ2and ๐: [0,+โ)โถ
โ,๐(๐ฅ) = โ๐ฅ. Determine a) ๐ โ ๐ and b) ๐ โ ๐.
Solution. a) ๐ โ ๐ = ๐ makes sense and we obtain the function
๐:โโถ โ with
b) ๐ = ๐ โ ๐: [0,+โ)โถ [0,+โ), also makes sense, with
2.2. Solved exercises 1. Let ๐ด = {2,4,6,8} and ๐ต = {1,3,5,7}, ๐ โ ๐ด ร ๐ต, ๐ผ =
{(๐ฅ, ๐ฆ)|๐ฅ โฅ 6 ๐๐ ๐ฆ โค 1} :
a) What is the graphic of the relation ๐ผ?
b) Make the schema of the relation ๐ผ.
Solution.
a) Because ๐ฅ โ ๐ด, and ๐ฆ โ ๐ต, it follows that
So:
b) See Fig. 2.2 below.
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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Figure 2.2.
2. Let ๐ด = {1,2,3,4} and ๐ต = {1,3,5,7,8}. Write the graphic of
the relation ๐ผ = {(๐ฅ, ๐ฆ)โ3๐ฅ + 4๐ฆ = 10} โ ๐ด ร ๐ต.
Solution. We observe that 3๐ฅ + 4๐ฆ = 10 โ 3๐ฅ = 2(5 โ
2๐ฆ) โ ๐ฅ is even and (5 โ 2๐ฆ) โฎ 3. Considering that ๐ฅ โ ๐ด, and ๐ฆ โ ๐ต,
we obtain: ๐ฅ โ (2,4), and ๐ฆ โ (1,7). The equality 3๐ฅ + 4๐ฆ = 10 is
met only by ๐ฅ = 2 and ๐ฆ = 1. So ๐บ๐ผ = {(2,1)}.
3. Let ๐ด = {1,3,4,5} and ๐ต = {1,2,5,6}. Write the relation ๐ผ
using letters ๐ฅ โ ๐ด, and ๐ฆ โ ๐ต, if the graphic of the relation ๐ผ is known.
Solution. (๐ฅ, ๐ฆ) โ ๐บ๐ผ โ ๐ฆ โฎ ๐ฅ.
4. On the natural numbers set โ we consider the
relations ๐ผ, ๐ฝ, ๐พ, ๐ โ โ2 , defined as it follows: (๐ฅ, ๐ฆ โ โ): ๐ผ =
{(3,5), (5,3), (3,3), (5,5)} ; ๐ฅ๐ฝ๐ฆ โ ๐ฅ โค ๐ฆ ; ๐ฅ๐พ๐ฆ โ ๐ฆ โ ๐ฅ = 12 ;
๐ฅ๐๐ฆ โ ๐ฅ = 3๐ฆ.
a) Determine ๐ฟ๐ผ , ๐๐ผ , ๐ฟ๐ฝ , ๐๐ฝ , ๐ฟ๐พ , ๐๐พ , ๐ฟ๐, ๐๐.
b) What properties do each of the relations ๐ผ, ๐ฝ, ๐พ, ๐ have?
c) Determine the relations ๐ผโ1, ๐ฝโ1, ๐พโ1 and ๐โ1 .
d) Determine the relations ๐ฝ โ ๐พ, ๐พ โ ๐ฝ, ๐พโ1 โ ๐ฝโ1, (๐ฝ โ ๐พ)โ1,
๐พ โ ๐ and ๐โ1 โ ๐.
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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 2 = ๐ฅ + 12 doesnโt have solutions in โ).
(for any ๐ โ โ, we have ๐ = 3๐/๐).
b) 1) ๐ผ is symmetrical, transitive, but is not reflexive. For
example (2,2) โ ๐ผ ; because (3,3) โ ๐ผ :, it follows that ๐ผ isnโt
antireflexive either.
2) ๐ฝ is - reflexive(๐ฅ โค ๐ฅ, (โฉ) ๐ฅ โ โ),
- anti-symmetrical ((๐ฅ๐ฝ๐ฆ โ ๐ฆ๐ฝ๐ฅ) โ
(๐ฅ โค ๐ฆ โ ๐ฆ โค ๐ฅ) โ ๐ฅ = ๐ฆ),
- transitive ((๐ฅ โค ๐ฆ โ ๐ฆ โค ๐ง) โ ๐ฅ โค ๐ง).
So ๐ฝ is an order relation on set โ.
3) ๐พ is antireflexive (๐ฅ โค ๐ฅ, (โฉ) ๐ฅ โ โ), anti-symmetrical
((๐ฅ๐พ๐ฆ โง ๐ฆ๐พ๐ฅ) โ (๐ฆ โ ๐ฅ = 12 and ๐ฅ โ ๐ฆ = 12) โ ๐ฅ โ ๐ฆ = ๐ฆ โ ๐ฅ โ
๐ฅ = ๐ฆ), but it is not symmetrical (๐ฅ๐พ๐ฆ โ ๐ฆ โ ๐ฅ = 12 โ ๐ฅ โ ๐ฆ =
โ12 โ ๐ฅ๐พ๐ฆ), and it is not transitive ((๐ฅ๐พ๐ฆ โ ๐ฆ๐พ๐ง) โ (๐ฆ โ ๐ฅ =
12 โ ๐ง โ ๐ฆ = 12) โ ๐ง โ ๐ฅ = 24 โ ๐ฅ๐พ๐ง ).
4) ๐ is not antireflexive (๐ฅ โ 3๐ฅ, (โฉ) ๐ฅ โ โโ, 0 = 3.0), it is anti-
symmetrical ((๐ฅ๐๐ฆ โ ๐ฆ๐๐ฅ) โ (๐ฅ = 3๐ฆ โ ๐ฆ = 3๐ฅ) โ ๐ฅ = 9๐ฅ โ ๐ฅ =
0 = ๐ฆ ), and for ๐ฅ โ 0 โ ๐ฆ, the equalities ๐ฅ = 3๐ฆ and ๐ฆ = 3๐ฅ cannot
take place), it is not reflexive (for example, 1 โ 3.1 โ 1๐1), it is not
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transitive ((๐ฅ๐๐ฆ โ ๐ฆ๐๐ง) โ (๐ฅ = 3๐ฆ โ ๐ฆ = 3๐ง) โ ๐ฅ = 9๐ง โ 3๐ง ,
generally โ ๐ฅ๐๐ง).
So
i.e.
i.e.
i.e.
which implies
i.e.
In other words,
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โ ๐ฅ + 12 = 3๐ฆ, which implies
which implies
5. We consider the binary relation defined on โ as it follows:
a) Prove that ๐ผ is an equivalence relation.
b) Determine the factor set โ/๐ผ.
Solution.
1) Because ๐ฅ = ๐ฅ, (โฉ) ๐ฅ โ โ, we have ๐ฅ๐ผ๐ฅ, (โฉ) ๐ฅ โ โ.
3) Let ๐๐ผ๐ and ๐๐ผ๐, ๐, ๐, ๐ โ โ. Then
In other words, the binary relation ๐ผ is reflexive, symmetrical
and transitive, which proves that ๐ผ is an equivalence relation on โ.
b) Let ๐ผ โ โ. We determine the class ๐ผ๐of equivalence of ๐ in
connection to ๐ผ.
Then the set factor is
We observe that
In other words, for ๐ โ 1, we have ๐ โ 2 โ ๐ and therefore
๐ผ๐ = ๐ผ2โ๐, and the factor can also be written
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6. Let ๐ด = {1,2,3,4,5} and ๐ต = {๐, ๐, ๐, ๐, ๐}. Which of the
diagrams in Fig. 2.3 represents a function defined on ๐ด with values in ๐ต
and which doesnโt?
Solution. The diagrams a), c), e) reveal the functions defined on
๐ด with values in ๐ต. Diagram b) does not represent a function defined
on ๐ด with values in ๐ต, because the image of 2 is not indicated.
Diagram d) also does not represent a function defined on ๐ด with values
in ๐ต, because 1 has two corresponding elements in ๐ต, namely ๐, ๐.
7. Let A = {1, 2, 3} and ๐ต = {๐, ๐}. Write all the functions
defined on ๐ด with values in ๐ต, indicating the respective diagram.
Solution. There are eight functions defined on ๐ด with values in ๐ต.
Their diagrams are represented in Fig. 2.4.
Figure 2.3.
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65
Figure 2.4.
8. Let ๐ด = {1, 2, 3, 4}, ๐ต = {๐, ๐, ๐}, ๐ถ = {๐ฅ, ๐ฆ, ๐ง, ๐ก} and ๐ท =
{1, 2}.
a) Draw the corresponding diagrams for two surjective functions
defined on ๐ด with values in๐ต.
b) Draw the diagram of the injective functions defined on ๐ท with
values in ๐ต.
c) Draw the diagram of a function denied on ๐ต with values in ๐ถ,
that is not injective.
d) Draw the corresponding diagrams for two bijective functions
defined on ๐ด with values in ๐ถ.
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Solution. a) The diagrams of two surjective and two non-
surjective functions defined on ๐ด with values in ๐ต are represented in
Fig. 2.5 a) and b).
Figure 2.5.
b) The diagram of the injective functions defined on ๐ท with
values in ๐ต are represented in Fig. 2.6.
c) The diagram of a non-injective function defined on ๐ต with
values in ๐ถ is represented in Fig. 2.7.
d) The diagrams of the two bijective functions defined on ๐ด with
values in ๐ถ are represented in Fig. 2.8.
Algebraic problems and exercises for high school
67
Figure 2.5.
Figure 2.6.
Figure 2.7.
Figure 2.8.
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
68
9. Let ๐ด = {0, 1,5, 6}, ๐ต = {0, 3, 8, 15} and the function
๐: ๐ด โถ ๐ต as defined by the equality ๐( ๐ฅ) = ๐ฅ 2 โ 4๐ฅ + 3, (โฉ) ๐ฅ โ
๐ด.
a) Is the function ๐ surjective?
b) Is ๐ injective?
c) Is ๐ bijective?
Solution. a) We calculate ๐(๐ด) = {๐(0), ๐(1), ๐(5), ๐(6)} =
{3,0, 8, 15} = ๐ต. So, ๐ is surjective.
b) The function ๐ is also injective, because ๐(0) = 3, ๐(1) = 0,
๐(5) = 8 and ๐(6) = 15, i.e. ๐ฅ1 โ ๐ฅ2 โ ๐(๐ฅ1) โ ๐(๐ฅ2)
c) The function ๐ is bijective.
10. Using the graphic of the function ๐: ๐ด โถ ๐ต, ๐: โ โถ โ,
show that the function ๐ is injective, surjective or bijective.
a) ๐(๐ฅ) = {๐ฅ โ 2, ๐๐ ๐ฅ โ (โโ, 2),
3๐ฅ โ 6, ๐๐ ๐ฅ โ [2,+โ).
c) ๐: [โ1,+โ] โถ โ,๐(๐ฅ) = {
โ3๐ฅ, ๐๐ โ 1 < ๐ฅ โค 0,โ๐ฅ, ๐๐ 0 < ๐ฅ โค 1,
โ0.5(๐ฅ + 1), ๐๐ ๐ฅ > 1.
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 abscisesโ axis that intersects the graphic of the function in two or
more points, the function in not injective. If ๐ธ(๐) is the values set of
the function ๐ and any parallel to the abscisesโ axis, drawn through the
ordinates axis that are contained in ๐ธ(๐), cuts the graphic of the
function ๐ in at least one point, the function is surjective. It follows that
a function ๐ is bijective if any parallel to the abscisesโ axis, drawn
through the points of ๐ธ(๐), intersects the graphic of ๐ in one point.
a) The graphic of the function ๐ is represented in Fig. 2.9.
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69
We have ๐ธ(๐) = (โโ,+โ) and any parallel ๐ฆ = ๐, ๐ โ โ to
the abscisesโ axis intersects the graphic of the function ๐ in one point.
Consequently, ๐ is a bijectve function.
๐(๐ฅ) = {๐ฅ2 โ 6๐ฅ + 8, ๐๐ ๐ฅ2 โ 6๐ฅ + 8 โฅ 0,
โ๐ฅ2 + 6๐ฅ โ 8, ๐๐ ๐ฅ2 โ 6๐ฅ + 8 < 0=
= {๐ฅ2 โ 6๐ฅ + 8, if ๐ฅ โ (โโ,2] โช [4,+โ),
โ๐ฅ2 + 6๐ฅ โ 8, if ๐ฅ โ (2,4).
Figure 2.9.
The graphic of the function ๐ is represented in Fig. 2.10.
We have ๐ธ(๐) = [0; 3] โ [โ1; 3]. Any parallel ๐ฆ = ๐, ๐ โ
(0,1) intersects the graphic of the function ๐ in four points; the
parallel ๐ฆ = 1 intersects the graphic in three points; any parallel ๐ฆ =
๐, ๐ โ (1; 3] intersects the graphic of the function ๐ in two points.
So the function ๐ is neither surjective, nor injective.
c) The graphic of the function is represented in Fig. 2.11. We
have
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Any parallel ๐ฆ = ๐ to the abscisesโ axis cuts the graphic of the
function ๐(๐ฅ) in one point, at most (๐ฆ = โ3, ๐ฆ = 2, ๐ฆ = 4) and that
is why ๐(๐ฅ) is injective. The equation ๐(๐ฅ) = ๐ โ โ has a solution
only for ๐ โค 3, hence ๐(๐ฅ) is not a surjective function.
d) We explain the function ๐:
๐(๐ฅ) = max(๐ฅ + 1, 1 โ ๐ฅ) = {๐ฅ + 1, ๐๐ 1 โ ๐ฅ โค ๐ฅ + 1,1 โ ๐ฅ, ๐๐ ๐ฅ + 1 < 1 โ ๐ฅ
=
= {๐ฅ + 1, ๐๐ ๐ฅ โฅ 0,1 โ ๐ฅ, ๐๐ ๐ฅ < 0.
Figure 2.10.
Figure 2.11.
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71
The graphic of the function ๐ is represented in Fig. 2.12.
We have ๐ท(๐) = โ, ๐ธ(๐) = [1,+โ). Any parallel๐ฆ = ๐ ,
๐ โ (1,+โ) intersects the graphic in two points, therefore ๐ is not
injective. Straight line ๐ฆ = 1/2 โ [0,+โ) doesnโt intersect the
graphic of the function ๐, so ๐ 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 2 = ๐ฆ + 3 doesnโt have solutions in โ, it follows that ๐
is not a functional relation.
Then, because ๐ท(๐) = ๐ฟ๐ = [โ1; 0], and
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72
with โ3/2 โ โโ3/2, and ๐ is not a functional relation.
c) We have: (๐ฅ, ๐ฆ) โ ๐ โ ๐ฅ๐๐ฆ โ ๐ฆ = ๐(๐ฅ) = ๐ฅ2. Then:
because ๐ฅ1, ๐ฅ2 โ (0,+โ);
4) The equation ๐(๐ฅ) = ๐ โ โ has solutions in [0,+โ), if and
only if ๐ โฅ 0, which proves that ๐ is not a surjection (number -2, for
example, doesnโt have a preimage in [0,+โ).
d) We have:
Repeating the reasoning from c), we obtain that ๐ is an
injection. Furthermore, the equation ๐(๐ฅ) = ๐ โ [0,+โ) has the
solution ๐ฅ = โ๐ โ [0,+โ) , and that is why ๐ is a surjective
application. Then ๐ is a bijective application and, in accordance with
the 6th Theorem, the relation ๐โ1 is also an application (function).
In accordance with the same 6th Theorem, we have
12. Let {๐ด = 1,2,3,4,5,6,7,8,9} and ๐ โ ๐น(๐ด) given; using the
following table:
determine:
a)
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73
b)
c) Calculate ๐โ1(1); ๐โ1(4); ๐โ1(7).
Solution.
13. Let ๐ด = {1,2,3} and ๐ต = {๐, ๐, ๐}. We examine the
relations
a) Determine ๐ฟ๐ผ , ๐ฟ๐ฝ , ๐ฟ๐พ, and ๐๐ผ, ๐๐ฝ, ๐๐พ.
b) Which of the relations ๐ผ, ๐ฝ ศ๐ ๐พ are applications? Indicate the
application type.
c) Determine the relations ๐ผโ1, ๐ฝโ1 and ๐พโ1. Which one is a
function?
d) Determine the relations ๐ผ โ ๐ฝโ1, ๐ฝ โ ๐ผโ1, ๐ผ โ ๐พโ1, ๐พ โ ๐ผโ1, ๐ฝ โ
๐พโ1 and ๐พ โ ๐ฝโ1. Which ones are applications?
Solution.
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74
b) ๐ผ is not an application, because element 3 has two images ๐
and ๐. ๐ฝ is a bijective application; ๐พ is an application, nor injective (the
elements 3 โ 1 have the same image c), nor surjective (b doesnโt have
a preimage).
We have:
๐ผโ1 is not an application, because the element ๐ has two
images 1 and 3; ๐ฝโ1 is a bijective application; ๐พโ1is not an application,
because ๐ฟ๐พโ1 โ ๐ต (the element ๐ doesnโt have an image);
Only the relation ๐พ โ ๐ฝโ1is an application, neither injective, nor
surjective.
14. Given the functions: ๐, ๐:โ โถ โ,
๐(๐ฅ) = {3 โ ๐ฅ, ๐๐ ๐ฅ โค 2,๐ฅ โ 1, ๐๐ ๐ฅ > 2,
๐(๐ฅ) = max(๐ฅ โ 2, 3 โ ๐ฅ),
show that ๐ = ๐.
Solution. The functions ๐ and ๐ are defined on โ and have
values in โ. Letโs show that ๐(๐ฅ) = ๐(๐ฅ), (โฉ)๐ฅ โ โ.
We explain the function ๐:
๐(๐ฅ) = {3 โ ๐ฅ, ๐๐ ๐ฅ โ 1 โค 3 โ ๐ฅ,๐ฅ โ 1, ๐๐ 3 โ ๐ฅ < ๐ฅ โ 1,
= {3 โ ๐ฅ, ๐๐ ๐ฅ โค 2,๐ฅ โ 1, ๐๐ ๐ฅ > 2,
= ๐(๐ฅ)
no matter what ๐ฅ โ โ is. So ๐ = ๐.
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75
15. The functions ๐, ๐:โ โถ โ,
๐(๐ฅ) = {๐ฅ + 2, if ๐ฅ โค 2,๐ฅ+10
3, if ๐ฅ > 2,
๐(๐ฅ) = {๐ฅ โ 2, if ๐ฅ < 3,2๐ฅ โ 5, if ๐ฅ โฅ 3
are given.
a) Show that ๐ and ๐ are bijective functions.
b) Represent graphically the functions ๐ and ๐โ1 in the same
coordinate register.
c) Determine the functions: ๐ = ๐ + ๐, ๐ = ๐ โ ๐, ๐ = ๐. ๐, ๐ = ๐/๐.
d) Is function ๐ bijective?
Solution. a), b) The graphic of the function ๐ is represented in
Fig. 2.13 with a continuous line, and the graphic of ๐โ1 is represented
with a dotted line.
Figure 2.13.
We have:
๐โ1(๐ฅ) = {๐ฅ โ 2, if ๐ฅ โค 4,3๐ฅ โ 10, if ๐ฅ > 4,
๐(๐ฅ) = {๐ฅ + 2, if ๐ฅ < 1,1
2(๐ฅ + 5), if ๐ฅ โฅ 1.
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76
c) We draw the following table:
We have
๐ (๐ฅ) =
{
2๐ฅ, ๐๐ ๐ฅ โค 2,4
3(๐ฅ + 1), ๐๐ 2 < ๐ฅ < 3,
1
3(7๐ฅ โ 5), ๐๐ ๐ฅ โฅ 3,
๐(๐ฅ) =
{
4
2, ๐๐ ๐ฅ โค 2,
2
3(8 โ ๐ฅ), ๐๐ 2 < ๐ฅ < 3,
5
3(5 โ ๐ฅ), ๐๐ ๐ฅ โฅ 3,
๐(๐ฅ) =
{
๐ฅ2 โ 4, ๐๐ ๐ฅ โค 2,1
3(๐ฅ2 + 8๐ฅ โ 20), ๐๐ 2 < ๐ฅ < 3,
1
3(2๐ฅ2 + 15๐ฅ โ 50), ๐๐ ๐ฅ โฅ 3,
๐(๐ฅ) =
{
๐ฅ + 2
๐ฅ โ 2, ๐๐ ๐ฅ โค 2,
๐ฅ + 10
3(๐ฅ โ 2), ๐๐ 2 < ๐ฅ < 3,
๐ฅ + 10
3(2๐ฅ โ 5), ๐๐ ๐ฅ โฅ 3.
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77
d) We have ๐(๐ฅ) = 4, (โฉ)๐ฅ โ (โโ,2], therefore ๐ is not
injective, in other words, ๐ is not bijective. This proves that the
difference (sum) of two bijective functions isnโt necessarily a bijective
function.
16. Letโs consider the function ๐:โ โ {โ๐/๐} โถ โ โ {๐/๐},
a) Show that function ๐ is irreversible.
b) Determine ๐โ1.
c) In what case do we have ๐ = ๐โ1?
Solution:
โ (๐๐ โ ๐๐)(๐ฅ1 โ ๐ฅ2) = 0 โ ๐ฅ1 = ๐ฅ2 โ ๐ is injective.
2) Let ๐ฅ โ โ โ {๐/๐}. We examine the equation ๐(๐ฅ) = ๐. Then
If
Imposibile. So (๐๐ โ ๐)/(๐ โ ๐๐) โ โ โ {โ๐/๐}, which proves
that the equation ๐(๐ฅ) = ๐ has solutions in โ โ {โ๐/๐} for any โ โ
{๐/๐}. This proves that ๐ is a surjective function.
From 1) - 2), it follows that ๐ is a bijective function, hence it is
also inversibile.
b) We determine
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78
c) In order to have ๐ = ๐โ1, it is necessary that the functions ๐
and ๐โ1 have the same definition domain and so ๐ = โ๐ . This
condition is also sufficient for the equality of the functions ๐ and ๐โ1.
It is true that, if ๐ = โ๐, the functions ๐ and ๐โ1are defined on โ โ
{๐/๐}, take values in โ โ {๐/๐}, and ๐โ1(๐ฅ) =๐๐ฅ+๐
๐๐ฅ+๐= ๐(๐ฅ), (โฉ)๐ฅ โ
โ โ {๐
๐}, namely ๐ = ๐โ1.
17. Represent the function ๐(๐ฅ) = โ5๐ฅ2 โ 2๐ฅ + 8 as a
compound of two functions.
Solution. We put
Then
Answer:
18. Calculate ๐ โ ๐, ๐ โ ๐, (๐ โ ๐)(4) and (๐ โ ๐)(4), where:
a) ๐(๐ฅ) =3
๐ฅโ1 and ๐(๐ฅ) = โ๐ฅ;
b) ๐(๐ฅ) = โ๐ฅ + 3 and ๐(๐ฅ) = ๐ฅ2 โ 4;
c) ๐(๐ฅ) = {๐ฅ2 + 6๐ฅ, ๐๐ x โค โ3โ2๐ฅ โ 5, ๐๐ ๐ฅ โฅ โ3
and ๐(๐ฅ) = {5๐ฅ โ 2, ๐๐ ๐ฅ โค 1,
๐ฅ2 โ 2๐ฅ + 4, ๐๐ ๐ฅ > 1.
Indicate ๐ท(๐ โ ๐) and ๐ท(๐ โ ๐).
Solution: a) We have
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79
So (๐ โ ๐)(4) โ (๐ โ ๐)(4), which proves that the comutative
law of function composition, generally, does not take place: ๐ โ ๐ โ
๐ โ ๐.
c) To determine the functions ๐ โ ๐ and ๐ โ ๐, in this particular
case we proceed as it follows:
(๐ โ ๐)(๐ฅ) = (๐ โ ๐)(๐(๐ฅ)) == {๐2(๐ฅ) + 6๐(๐ฅ), ๐๐ ๐(๐ฅ) < โ3
โ2๐(๐ฅ) โ 5, ๐๐ ๐(๐ฅ) โฅ โ3=
= {
๐2(๐ฅ) + 6๐(๐ฅ) ๐๐ ๐(๐ฅ) < โ3,
โ2๐(๐ฅ) โ 5, ๐๐ ๐(๐ฅ) โฅ โ3 and ๐ฅ โค 1,
โ2(๐ฅ2 โ 2๐ฅ + 4) โ 5, ๐๐ (๐ฅ) โฅ โ3 and ๐ฅ > 1
=
{
(5๐ฅ โ 2)2 + 6(5๐ฅ โ 2) ๐๐ 5๐ฅ โ 2 < โ3,
โ2(5๐ฅ โ 2) โ 5, ๐๐ 5๐ฅ โ 2 โฅ โ3 and ๐ฅ โค 1,
โ2(๐ฅ2 โ 2๐ฅ + 4) โ 5, ๐๐ 5๐ฅ โ 2 โฅ โ3 and ๐ฅ > 1
=
= {
25๐ฅ2 + 10๐ฅ โ 8 ๐๐ ๐ฅ < โ1/5,โ10๐ฅ โ 1, ๐๐ โ 1/5 โค ๐ฅ โค 1,
2๐ฅ2 + 4๐ฅ โ 13, ๐๐ ๐ฅ > 1.
(๐ โ ๐)(๐ฅ) = ๐(๐(๐ฅ)) = {5๐(๐ฅ), ๐๐ ๐(๐ฅ) โค 1,
๐2(๐ฅ) โ ๐ ๐(๐ฅ) + 4, ๐๐ ๐(๐ฅ) > 1.
We notice that
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So for ๐(๐ฅ) โค 1, we obtain
(๐ โ ๐)(๐ฅ) = {5(๐ฅ2 + 6๐ฅ) โ 2, if ๐ฅ โ [โ3 โ โ10,โ3)
5(โ2๐ฅ โ 5) โ 2, if ๐ฅ โ [โ3,+โ).
Similarly,
and that is why for ๐(๐ฅ) > 1, we obtain
for ๐ฅ โ (โโ,โ3 โ โ10).
To sum it up, we have
(๐ โ ๐)(๐ฅ) =
= {
๐ฅ4 + 12๐ฅ3 + 34๐ฅ2 โ 12๐ฅ + 4, ๐๐ ๐ฅ โ (โโ,โ3 โ โ10),
5๐ฅ2 + 30๐ฅ โ 2, if ๐ฅ โ [โ3โ10,โ3),
โ10๐ฅ โ 7, if ๐ฅ โฅ โ3.
(๐ โ ๐)(4) = โ10.4 โ 27 = โ67.
19. Solve the following equations:
a) (๐ โ ๐ โ ๐)(๐ฅ) = (๐ โ ๐ โ ๐)(๐ฅ) , if ๐(๐ฅ) = 3๐ฅ + 1, ๐(๐ฅ) =
๐ฅ + 3;
b) (๐ โ ๐ โ ๐)(๐ฅ) = ๐ฅ, if ๐(๐ฅ) =๐๐ฅ+1
๐ฅ+๐, ๐ โ โ, ๐ฅ โ โ โ {โ๐};
c) (๐ โ ๐)(๐ฅ) = (๐ โ ๐)(๐ฅ), if ๐(๐ฅ) = 2๐ฅ2 โ 1, ๐(๐ฅ) = 4๐ฅ3 โ 3๐ฅ.
Solution. We determine the functions (๐ โ ๐ โ ๐)(๐ฅ) and (๐ โ
๐ โ ๐)(๐ฅ):
The equation becomes
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81
Answer: ๐ฅ = 2.
b) We calculate (๐ โ ๐ โ ๐)(๐ฅ):
The equation becomes
Answer: ๐ฅ โ {โ1,1}
c) We determine (๐ โ ๐) and (๐ โ ๐)(๐ฅ):
The equation becomes
meaning the equality is true for any ๐ฅ โ โ.
Answer: ๐ฅ โ โ.
20. Let ๐, ๐:โ โถ โ given by ๐(๐ฅ) = ๐ฅ2 + ๐ฅ + 12 and ๐(๐ฅ) =
๐ฅ2 โ ๐ฅ + 2. Show there is no function ๐:โโถ โ, so that
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Solution. The relation (๐ด) can also be written
We suppose there is a function ๐:โโถ โ that satisfies the
relation (๐ดโฒ). We put in (๐ดโฒ) ๐ฅ = 1 and ๐ฅ = โ1. We obtain:
which implies ๐(4) +
๐(2) โ ๐(2) + ๐(4), contradiction with the hypothesis. So there is
no function ๐ with the property from the enunciation.
21. Determine all the values of the parameters ๐, ๐ โ โ for
which (๐ โ ๐) = (๐ โ ๐)(๐ฅ) , (โฉ) โ โ , where ๐(๐ฅ) = ๐ฅ2 โ ๐ฅ and
๐(๐ฅ) = ๐ฅ2 + ๐๐ฅ + ๐.
Solution. We determine ๐ โ ๐ and ๐ โ ๐:
Then
Answer: ๐ = โ1, ๐ = 0, ๐(๐ฅ) = ๐ฅ2 โ ๐ฅ = ๐(๐ฅ).
22. Given the functions ๐, ๐, โ โถ โ โถ โ,
.
Show that:
a) ๐ is not injective;
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83
b) ๐ is injective;
c) โ is bijective and determine โโ1.
Solution. a) Let ๐(๐ฅ1) = ๐(๐ฅ2).
Then
For example, ๐(๐ฅ) = ๐ฅ3(๐ฅ + 4) + 3, taking ๐ฅ1 = 0 and ๐ฅ2 = โ4,
we obtain ๐(0) = ๐(โ4) = 3, ๐ฅ1 โ ๐ฅ2.
because
so โ is an injective function.
We prove that โ is surjective. Let ๐ โ โ. We solve the equation
(the root of an uneven order exists out of any real number). So โ is a
surjection, therefore, โ is a bijective function.
We determine โโ1:
23. Let ๐: [1,+โ) โถ [1,+โ) with
a) Prove that ๐ is a bijective function.
b) Determine ๐โ1.
Solution. a) We have ๐(๐ฅ) = (๐ฅ2 โ ๐ฅ + 1) 3. We represent this
function as a compound of two functions:
๐ข, ๐ฃ: [1,+โ) โถ [1,+โ), where ๐ข(๐ฅ) = ๐ฅ3, ๐ฃ(๐ฅ) = ๐ฅ2 โ ๐ฅ + 1.
Then ๐(๐ฅ) = (๐ข โ ๐ฃ)(๐ฅ) , which implies that ๐ = ๐ข โ ๐ฃ is bijective,
being the compound of two bijective functions.
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b) ๐โ1 = (๐ข โ ๐ฃ)โ1 = ๐ฃโ1 โ ๐ขโ1. We determine ๐ฃโ1 and ๐ขโ1.
Because:
We solve this equation according to ๐ฅ โ [1,+โ).
The discriminant is:
The equation has only one root in [1,+โ):
Then
So,
= (1 + โ4โ๐ฅ โ 33
)/2, with ๐โ1: [1,+โ) โถ [1,+โ).
24. Considering the function ๐:โ โถ โ with the properties:
a) Determine the function ๐.
b) Calculate ๐(โ1998).
Solution. a) For ๐ฅ2 = 0 from 1), it follows that ๐(๐ฅ1) = ๐(๐ฅ1) +
๐(0), which implies that ๐(0) = 0. For ๐ฅ2 = โ๐ฅ1 from 1), we obtain
Then
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Let ๐ฅ โ {0, 1}. Then
On the other hand, 1
1โ๐ฅ=1โ๐ฅ+๐ฅ
1โ๐ฅ= 1 +
๐ฅ
1โ๐ฅ implies
From (2) and (3) it follows that
So ๐(๐ฅ) = ๐ฅ, (โฉ) โ โ.
Answer: a) ๐(๐ฅ) = ๐ฅ; b) ๐(โ1998) = โ1998.
25. Using the properties of the characteristic function, prove the
equality:
Solution. Using the properties ๐ด = ๐ต โ ๐๐ด = ๐๐ต , we will
prove the required equality by calculating with the help of ๐๐ด, ๐๐ต and
๐๐ถ the characteristic functions of sets ๐ด โช (๐ต โฉ ๐ถ) and (๐ด โช ๐ต) โฉ
(๐ด โช ๐ถ):
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which implies ๐ด โช (๐ต โฉ ๐ถ) and (๐ด โช ๐ต) โฉ (๐ด โช ๐ถ).
2.3. Suggested exercises 1. Determine the definition domains and the values domains of
the following relations:
2. Let ๐ด = {2,4,6,8} and ๐ต = {1,3,5,7}, ๐ผ โ ๐ด ร ๐ต.
a) Determine the graphic of the relation ๐ผ.
b) Construct the schema of the relation ๐ผ:
1) ๐ผ = {(๐ฅ, ๐ฆ)โ๐ฅ < 3 and ๐ฆ > 3};
2) ๐ผ = {(๐ฅ, ๐ฆ)โ๐ฅ > 2 and ๐ฆ < 5};
3) ๐ผ = {(๐ฅ, ๐ฆ)โ๐ฅ > 6 or ๐ฆ > 7};
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3. Let ๐ด = {1,2,3,4} and ๐ต = {1,3,5,7,8}. Write the graphic of
the relation ๐ผ โ ๐ด ร ๐ต, if:
4. Let ๐ด = {๐, 3, 4, 5}, ๐ต = {๐, 2, 5, 6} and ๐บ be the graphic of
the relation ๐ผ. Write the relation ๐ผ with propositions containing letters
๐ฅ and ๐ฆ, with ๐ฅ ๐ธ ๐ด and ๐ฆ โ ๐ต:
5. Let A = {๐, 2.3,4}. Analyze the properties of the relation ๐ผ โ
๐ด2(var. 1-6) and ๐ผ โ โ2 (var. 7-14):
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7) ๐ผ = {(๐ฅ, ๐ฆ) โ โ2โ๐ฅ > 1 and ๐ฆ > 1 };
8) ๐ผ = {(๐ฅ, ๐ฆ) โ โ2โ{๐ฅ > 0,๐ฆ > 0
or {๐ฅ < 0,๐ฆ < 0
};
9) ๐ผ = {(๐ฅ, ๐ฆ) โ โ2โ๐ฅ โฅ 0 or ๐ฆ โฅ 0 };
10) ๐ผ = {(๐ฅ, ๐ฆ) โ โ2โ๐ฅ โฅ 1 or ๐ฆ โฅ 1 };
6. For each binary relation ๐ผ defined on set โ:
a) determine the definition domain ๐ฟ๐ผ and the values domain ๐๐ผ;
b) establish the properties (reflexivity, ireflexivity, symmetry,
anti-symmetry, transitivity);
c) determine the inverse relation ๐ผโ1(๐ฅ, ๐ฆ โ โ):
7. Given the set ๐ด and the binary relation โ ๐ด2 , prove that ๐ผ is
an equivalence relation and determine the factor set ๐ด/๐ผ.
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89
8. Given the set ๐ด = {1, 2,3,4,5,6,7,8, 9} and the subset system
๐ = {๐ด๐ โ ๐ด, ๐ผ = 1, ๐ฬ ฬ ฬ ฬ ฬ }, prove that ๐ defines a partition on ๐ด and build
the equivalence relation ๐ผ๐.
9. We define on โ the binary relation ๐ผ:
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a) Show that ๐ผ is an equivalence relation on โ.
b) Determine the equivalence classes.
10. We define on โ the binary relation ๐ฝ:
.
a) Show that ๐ฝ is an equivalence relation.
b) Determine the equivalence classes.
11. Let โ ๐ด2 , where ๐ด = {โ5,โ4,โ3,โ2,โ1, 0, 1, 2, 3, 4, 5, 6,
7, 8} with
a) Is ๐ผ an equivalence relation?
b) If so, determine the equivalence classes.
12. Determine: ๐ฟ๐ผ , ๐๐ผ, ๐ผโ1, ๐ผ โ ๐ผ, ๐ผ โ ๐ผโ1, ๐ผโ1 โ ๐ผ, if:
13. Determine the relations
๐ผ โ ๐ฝ, ๐ฝ โ ๐ผ, ๐ผโ1, ๐ฝโ1, ๐ผโ1 โ ๐ฝ, (๐ฝ โ ๐ผ)โ1 ;
5) ๐ผ = {(๐ฅ, ๐ฆ) โ โค2โ๐ฅ โ ๐ฆ ๐s even},
๐ฝ = {(๐ฅ, ๐ฆ) โ โค2โ๐ฅ โ ๐ฆ is uneven};
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14. Let ๐ด = {1, 2, 3, 4}, ๐ต = {๐, ๐, ๐, ๐} . Determine the
definition domain and the values domain of each of the following
relations ๐ผ, ๐ฝ. Which ones are applications? Determine the application
type. Determine the relation ๐ผโ1 โ ๐ผ, ๐ผ โ ๐ฝโ1, ๐ฝ โ ๐ผโ1, ๐ฝ โ ๐ฝโ1. Are
they applications?
15. We consider the application ๐:โ โถ โ , ๐(๐ฅ) = sin ๐ฅ .
Determine:
16. The application ๐:๐ด โถ ๐ต is given by the table below, where
๐ด = {1, 2, 3, 4, 5, 6, 7, 8, 9} and ๐ต = {๐, ๐, ๐, ๐, ๐, ๐}.
Determine:
17. We consider the application ๐:โโถ โค, ๐( ๐ฅ) = [๐ฅ] ([๐ฅ] is
the integer part of ๐ฅ). Determine
๐โ1({2,4,5}) and ๐โ1(โ1).
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18. The following application is given: ๐:โ โถ โ, ๐(๐ฅ) = ๐ฅ2.
Determine ๐(๐ด) and ๐โ1(๐ด), if ๐ด = {1, 2, 3, 4, 5, 6 , 7, 8, 9, 10}.
19. Let ๐ด and ๐ธ be two finite sets, |๐ด| = ๐. |๐ต|. How many
surjective applications ๐: ๐ด โถ ๐ต are there, if:
20. Given the graphic of the relation ๐ผ, establish if ๐ผ is a func-
tion. Determine ๐ฟ๐ผ and ๐๐ผ.
Changing ๐ฟ๐ผ and ๐๐ผ, make ๐ผ become an application injective,
surjective and bijective.
Draw the graphic for the relation ๐ผโ1. Is relation ๐ผโ1 a function?
What type is it?
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21. Let ๐ด = {1, 2, 3, 4}, ๐ต = {0, 1, 5, 6} and ๐ถ = {7, 8, 9}.
a) Draw the diagrams of two injective and two non-injective
functions defined on ๐ถ with values in ๐ต.
b) Draw the diagrams of two surjective functions and two non-
surjective applications defined on ๐ด with values in ๐ถ.
c) Draw the diagrams of two bijective and two non-bijective
functions defined on ๐ด with values in ๐ต.
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22. Let ๐ด = {1, 3, 5, 6} and ๐ต = {0, 1, 2, 3, 5}, ๐ฅ โ ๐ด , ๐ฆ โ ๐ต .
Which of the relations stated below represents a function defined on ๐ด
with values in ๐ต? And one defined on ๐ต with values in ๐ด? For functions,
indicate their type:
23. Using the graphic of the function ๐: ๐ด โถ ๐ต, show that ๐ is
injective, surjective or bijective. If bijective, determine ๐โ1and draw
the graphic of ๐ and ๐โ1in the same register of coordinates..
24. Which of the following relations ๐ โ โ2 are functions?
Indicate the definition domain of the functions. Establish their type:
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25. Given the relation ๐ผ with ๐ฝ๐ผ = [โ3; 5] and ๐๐ผ = [โ4; 7]:
a) Does pair (โ4, 5) belong to the relation ๐ผ? Why?
b) Indicate all the ordered pairs (๐ฅ, ๐ฆ) โ ๐ผ with ๐ฅ = ๐. Explain.
26. Given the function ๐(๐ฅ), calculate its values in the indicated
points.
4) ๐(๐ฅ) = {๐ฅ2 โ 5, ๐๐ ๐ฅ > 0,2๐ฅ + 3, ๐๐ ๐ฅ โค 0,
๐(โ2), ๐(0), ๐(5);
5) ๐(๐ฅ) = {๐ฅ2 + 1, ๐๐ ๐ฅ > 0 โ4, ๐๐ ๐ฅ = 01 โ 2๐ฅ, ๐๐ ๐ฅ < 0
๐(5), ๐(โ1), ๐(1/2);
27. Determine ๐ท(๐), if:
28. Given the functions ๐(๐ฅ) and ๐(๐ฅ), determine the functions
๐ + ๐, ๐ โ ๐, ๐ ยท ๐ and ๐ / ๐, indicating their definition domain.
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29. Represent the function ๐(๐ฅ) as a compound of some
functions:
30. Given the functions ๐ (๐ฅ) and ๐( ๐ฅ ), determine the
functions ๐ โ ๐ and ๐ โ ๐ and calculate (๐ โ ๐ )(3) and (๐ โ ๐ )(3) in
options 1) โ 6) and (๐ โ ๐ )( โ1) and (๐ โ ๐)( -1) in options 7) โ
12).
31. Given the functions ๐(๐ฅ) = ๐ฅ2, ๐(๐ฅ) = 3๐ฅ and โ(๐ฅ) = ๐ฅ โ
1, calculate:
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32. Determine if in the function pairs ๐ and ๐ one is the inverse
of the other:
33. The value of function ๐ is given. Calculate the value of the
function ๐โ1, if:
34. Determine ๐โ1, if:
35. The function: ๐: โ โโถ โ is given:
a) Prove that ๐ is bijective.
b) Determine ๐โ1
c) Calculate ๐ โ ๐โ1 ศ๐ ๐โ1 โ ๐.
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36. Given the functions ๐( ๐ฅ) and ๐( ๐ฅ), determine the functions
(๐ โ ๐)(๐ฅ) and (๐ โ ๐)(๐ฅ)(๐, ๐: โ โถ โ):
37. Prove the equality of the functions ๐ and ๐:
38. Determine the functions ๐ = ๐ + ๐, ๐ = ๐ โ ๐, ๐ = ๐. ๐
and ๐ = ๐/๐:
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39. Let ๐ด = {1, 2, 3, 4}, ๐ต = {0, 1, 3, 4} and the functions
Can the functions ๐ โ ๐, ๐ โ ๐ be defined? If so, determine
these functions. Make their diagrams.
40. a) Show that the function ๐ is bijective.
b) Determine ๐โ1.
c) Graphically represent the functions ๐ and ๐โ1 in the same
coordinate register.
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41. Show that the function ๐: โ โ โถ โ, ๐(๐ฅ) = ๐ฅ2 โ 6๐ฅ + 2
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):
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3. Elements of combinatorics
3.1. Permutations. Arrangements. Combinations. Newtonโs 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 ๐ = {๐1, ๐2, โฆ , ๐๐} be a finite set that has ๐ elements. Set
๐ is called ordered, if each of its elements associates with a certain
number from 1 to ๐ , named the element rank, so that different
elements of ๐ associate with different numbers.
Definition 1. All ordered sets that can be formed with ๐ elements
of given set ๐ (๐( ๐) = ๐) are called permutations of ๐ elements.
The number of all permutations of ๐ elements is denoted by the
symbol ๐๐ and it is calculated according to the formula:
By definition, it is considered ๐0 = 0! = 1 = 1! = ๐1.
Definition 2. All ordered subsets that contain ๐ elements of set
๐ with ๐ elements are called arrangements of ๐ elements taken ๐ at
a time.
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The number of all arrangements of ๐ elements taken ๐ at a
time is denoted by the symbol ๐ด๐๐ and it is calculated with the formula
Definition 3. All subsets that contain elements of set ๐ with n
elements are called combinations of ๐ elements taken ๐ at a time.
The number of all combinations of ๐ elements taken ๐ at a time
is denoted by the symbol ๐ถ๐๐ and it is calculated with the formula
where ๐,๐ โ โ; 0 โค ๐ โค ๐.
Remark. In all the subsets mentioned in definitions 1 - 3, each
element of the initial set ๐ appears only once.
Along with the combinations in which each of the different ๐
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 ๐ groups of elements be given. Each group contains certain
elements of the same kind.
Definition 1โฒ. Permutations of ๐ elements each one containing
๐ผ1 elements ๐๐1, ๐ผ2 elements ๐๐2 , ๐ผ๐ elements ๐๐๐ , where ๐ผ1 + ๐ผ2 +
โฏ+ ๐ผ๐ are called permutations of ๐ elements with repetitions.
The number of all permutations with repetitions is denoted by
the symbol ๏ฟฝฬ ๏ฟฝ๐ผ1,๐ผ2,โฆ,๐ผ๐ and it is calculated using the formula:
Definition 2โฒ . The arrangements of ๐ elements, each one
containing ๐ elements, and each one being capable of repeating itself
in each arrangement for an arbitrary number of times, but not more
than ๐ times, are called arrangements of ๐ elements taken ๐ at a
time with repetitions.
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The number of all arrangements with repetitions of ๐ elements
taken ๐ at a time is denoted by the symbol ๐ด๐๐ฬ ฬ ฬ ฬ and it is calculated with
the formula
Definition 3'. The combinations of ๐ elements each one
containing ๐ elements, and each one being capable of repeating itself
more than once, but not more than ๐ times, are called combinations
of ๐ elements taken ๐ at a time with repetitions.
The number of all combinations with repetitions is denoted by
the symbol ๐ถ๐๐ฬ ฬ ฬ ฬ 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 isnโt important,
we have combinations;
If the order is important,
we have arrangements
or permutations:
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The sum rule. If the object ๐ด can be chosen in ๐ ways and the
object ๐ต in n ways, then the choice "either A or B" can be made in ๐ +
๐ ways.
The multiplication rule. If the object ๐ด can be chosen in ๐
ways and after each such choice the object ๐ต can be chosen in ๐ ways,
then the choice "A and B" in this order can be made in ๐ . ๐ ways.
We will also mention some properties of combinations, namely:
The formula
is called Newtonโs Binomial formula (๐ โ โโ).
The coefficients ๐ถ๐0, ๐ถ๐
1, ๐ถ๐2, โฆ , ๐ถ๐
๐ from Newtonโs 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 2๐.
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 ๐ is an even number (i.e. ๐ = 2๐), then the binomial
coefficient of the middle term in the development (namely ๐ถ๐๐) is the
biggest. If ๐ is uneven (i.e. ๐ = 2๐ + 1 ), then the binomial
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105
coefficients of the two middle terms are equals (namely ๐ถ๐๐ = ๐ถ๐
๐+1)
and are the biggest.
VIII. Term ๐ถ๐๐๐ฅ๐โ๐๐๐, namely the (๐ + 1) term in equality (7),
is called term of rank ๐ + 1 (general term of development) and it is
denoted by ๐๐+1. So,
IX. The coefficient of the term of rank ๐ + 1 in the
development of Newtonโs binomial is equal to the product of the
coefficient term of rank ๐ multiplied with the exponent of ๐ฅ in this
term and divided by ๐, 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.
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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 0.)
Solution. We have 10 digits on the whole: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
As the phone number can also start with 0, we have:
1) arrangements of 10 digits taken 6 at a time, i.e.
2) because the digit can repeat in the number, we have
arrangements with repetitions, i.e.
Answer: 1) 151 200; 2) 1 000 000.
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 teamโs
composition it is sufficient to determine the number of combinations
of 10 elements taken 6 at a time, i.e.
Answer: 210.
6. How many 5 digit numbers can be formed with digits 0 and 1?
Solution. As the digits are repeating and their order is important,
we consequently have arrangements with repetitions. Here, ๐ = 5,
๐ = 2.
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107
From digits 0 and 1 we can form ๐ด25ฬ ฬ ฬ numbers of five digits.
But we have to take into consideration that the number cannot
begin with the digit zero. So, from the number ๐ด25ฬ ฬ ฬ we have to begin
with zero. Numbers of this type are ๐ด24ฬ ฬ ฬ . Therefore, the number we
seek is
Answer: 16.
7. How many three digit numbers can be formed with the digits
1, 2, 3, 4, 5, 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
๐ด53 = 53ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ numbers of three digits.
Answer: 125.
8. Using 10 roses and 8 Bedding Dahlias, bunches are made that
contain 2 roses and 3 Bedding Dahlias. How many bunches of this kind
can be formed?
Solution. Two roses (out of the 10 we have) can be chosen in
๐ถ102 ways, and three Bedding Dahlias (out of 8) can be taken in ๐ถ8
3 ways.
Applying the rule of multiplication, we have: the total number of
bunches that can be formed is ๐ถ102 . ๐ถ8
3 = 1 890.
Answer: 1890.
9. 12 young ladies and 15 young gentlemen participate at a
dancing soirรฉe. In how many ways can four dancing pairs be chosen?
Solution. The 12 young ladies can be distributed in four person
groups in ๐ถ124 ways, and the 15 young gentlemen - in ๐ถ15
4 ways.
Because in each group of young ladies (or young gentlemen)
order is an issue, each of these groups can be ordered in ๐4ways.
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As a result (we apply the multiplication rule), we will have
๐ถ124 . ๐4. ๐ถ15
4 = ๐ด124 . ๐ถ15
4 = ๐ถ124 . ๐ด15
4 = 16216200.
Answer: 16216200.
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 โ 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 tripletโs
members is equally important. So, the commanding triplet can be
formed in ๐ด253 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 ๐ถ202 ways. Regarding the doctor โ the situation is the
same, namely the doctor can be chosen in ๐ถ81 ways.
Using the multiplication rule, we have: a pair of engineers can be
assigned in ๐ถ202 ways to each commanding triplet. We will have, in
total, ๐ด253 . ๐ถ20
2 quintets. To each quintet a doctor can be associated in
๐ถ81 ways.
As a result, the spacecraftโs crew can be assembled in
๐ด253 . ๐ถ20
2 . ๐ถ81 ways, or
Answer: 20 976 000.
Algebraic problems and exercises for high school
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 11 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 11 types is equal to the number of the combinations with
repetitions of 11 elements taken five at a time, i.e.
Answer: 3003.
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 ๐ถ21 ways, the swimmer in
๐ถ31 ways, and the athlete in ๐ถ5
4 ways. Using the multiplication rule, we
have ๐ถ21. ๐ถ3
1. ๐ถ54 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 ๐ถ21. ๐ถ3
1. ๐ถ54 .
c) In the team there can be one rower, three swimmers and two
athletes. The number of teams in this case will be ๐ถ21. ๐ถ3
3. ๐ถ52.
d) In the team there can be two rowers, one swimmer and three
athletes. We will have ๐ถ22. ๐ถ3
1. ๐ถ53 of these teams.
e) In the team there can be two rowers, two swimmers and two
athletes. The number of teams will be ๐ถ22. ๐ถ3
2. ๐ถ52.
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f) In the team there can be two rowers, three swimmers and one
athlete. The number of teams will be ๐ถ22. ๐ถ3
3. ๐ถ51.
Using the addition rule, the total number of teams that can be
assembled is:
Answer: 175.
13. Given ๐ = 15 capital letters, ๐ = 10 vowels and ๐ = 11
consonants (in total ๐ + ๐ + ๐ = 36 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 ๐ = 4 different vowels (out of the ๐ = 10 given)
and ๐ฃ = 6 different consonants (out of the ๐ = 11 given).
Solution. We choose a capital letter. This choice can be made in
๐ ways. Then from ๐ vowels we choose ๐ letters. This can be done in
๐ถ๐๐
ways. Finally, we choose ๐ฃ consonants, which can be made in ๐ถ๐๐ฃ
ways. Using the multiplication rule, we can choose the required letters
to form the word in ๐. ๐ถ๐๐. ๐ถ๐๐ฃ ways.
After placing the capital letter at the beginning, with the other
๐ + ๐ฃ letters we can form (๐ + ๐ฃ)! permutations. Each such
permutation yields a new word. So, in total, a number of ๐. ๐ถ๐๐. ๐ถ๐๐ฃ(๐ +
๐ฃ)! different words can be formed, namely 15. ๐ถ104 ๐ถ11
6 . 10!
Answer: 15. ๐ถ104 ๐ถ11
6 . 10!
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 11).
Answer: 21.
Algebraic problems and exercises for high school
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 6 categories of officers. Repeating the
reasoning from problem 10, we have to calculate combinations with
repetitions of 6 elements taken 10 times each, as follows
Answer: 3 003.
16. On a shelf there are ๐ + ๐ different books. Among them ๐
have a blue cover, and ๐ 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 ๐ places;
b) the books in the yellow covers sit beside them?
Solution. a) The books with blue covers can be placed in the first
๐ places in ๐๐ = ๐! ways. With each such allocation, the books with
yellow covers can be placed in ๐๐ = ๐! ways. Using the multiplication
rule, we have in total ๐! ยท ๐! positions in which the blue covers books
occupy the first ๐ places.
b) Let the books in blue covers sit beside. It follows that right
beside them on the shelf there can be either ๐ books with yellow
covers or ๐ โ 1, or ๐ โ 2, ... , or no book with yellow covers. We can
thus place the books with blue covers so that they follow each other in
๐ + 1 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 ๐! . ๐! . (๐ + 1)
different positions for the books.
Answer: a) ๐! ยท ๐!; b) ๐! ยท ๐! ยท (๐ + 1).
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17. Determine the fourth term of the development of Newtonโs
binomial:
Solution. According to the formula (9), the rank 4 term has the
form
Answer: โ1792 ยท ๐ฅ17/2.
18. Determine the biggest coefficient in the development of the
binomial
Solution.
If ๐ is an even number, i.e. ๐ = 2๐ , ๐ โ โโ , then the
development of the binomial contains 2๐ + 1 terms, and according
to the ๐๐ผ๐ผ property, the coefficient ๐ถ2๐ ๐ is the biggest.
If ๐ is an uneven number, i.e. ๐ = 2๐ + 1, ๐ โ โ, according
to the same ๐๐ผ๐ผ property, the development of the binomial contains
two terms have the biggest coefficients ๐ถ2๐ +1๐ , ๐ถ2๐ +1
๐ +1 .
Answer: ๐ถ2๐ ๐ , if ๐ is an even number; ๐ถ2๐ +1
๐ , ๐ถ2๐ +1๐ +1 if ๐ is an
uneven number.
19. Determine which term do not contain ๐ฅ in the development
of the binomial:
Algebraic problems and exercises for high school
113
Solution.
The term of rank ๐ + 1 in the development of this binomial has
the form
This term does not contain ๐ฅ only if ๐ โ ๐ = 0 โ ๐ = ๐. So
the term that doesnโt contain ๐ฅ is ๐๐+1.
Answer: ๐๐+1.
20. In the development of the binomial
determine the terms that contains ๐ 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 2 048.
Solution. We will first determine the exponent ๐. Based on the
๐๐ผth property, the sum of the binomial coefficients is 2๐ . Because the
sum of the binomial coefficients that occupy uneven places in the
development of the binomial is equal to 2048, and based on the ๐๐ผ ๐
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 12. The term of rank
๐ + 1 takes the form
Considering the requirements of the problem, we have
and
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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Answer: โ264๐3๐7.
21. For what value of ๐ the binomial coefficients of the second,
third and fourth term from the development of the binomial
(๐ฅ + ๐ฆ)๐ forms an arithmetic progression?
Solution. Based on the formula (8), we have
and from the conditions of the problem the following relations issues
The conditions of the problem are verified by the value ๐ = 7.
Answer: ๐ = 7.
22. Prove that the difference of the coefficients ๐ฅ๐+1 and ๐ฅ๐ in
the development of the binomial (1 + ๐ฅ)๐+1 is equal to the difference
of the coefficients of ๐ฅ๐+1 and ๐ฅ๐โ1 I in the development of the
binomial (1 + ๐ฅ)๐.
Solution. The coefficients of ๐ฅ๐+1 and ๐ฅ๐ in the development of
the binomial (1 + ๐ฅ)๐+1 are ๐ถ๐+1๐+1and ๐ถ๐+1
๐ respectively. We assess the
difference
In the development of the binomial (1 + ๐ฅ)๐, the coefficients of
๐ฅ๐+1 and ๐ฅ๐โ1 are ๐ถ๐๐+1 and ๐ถ๐
๐โ1 , respectively. We assess the
difference
Algebraic problems and exercises for high school
115
As the members from the right in (โ) and (โโ) are equal, the
equality of the members from the left results, i.e.
which had to be proven.
23. Comparing the coefficients of ๐ฅ in both members of the
equality
prove that
Solution.
In the right member of this equality, the coefficient of ๐ฅ๐is
and in the development of the binomial (1 + ๐ฅ)๐+๐ , the term of rank
๐ + 1 has the form
As the polynomials (1 + ๐ฅ)๐. (1 + ๐ฅ)๐ and (1 + ๐ฅ)๐+๐ are
equal and have the same degree, the equality of the coefficients beside
these powers of ๐ฅ result and this also concludes the demonstration of
the equality (๐ด).
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24. Prove the equality
Solution. Let:
We multiply both members of the last equality with (๐ + 1).
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 (๐ฅ โ 1)๐:
Algebraic problems and exercises for high school
117
We derive both members of the equality (๐ด) according to ๐ฅ. We
obtain
We place in (โโ) ๐ฅ = 1. Then
that had to be proven.
26. Prove that the equality
is true.
Solution. We notice that the expression
represents the development of the binomial (1 โ 10)2๐ = 92๐ =
(81)๐, that had to be proven.
27. Bring the expression ๐1 + 2๐2 +โฏ+ ๐๐๐ to a simpler form.
Solution. We will make the necessary transformations by
applying the method of mathematical induction. Let
For:
๐ = 1, we have ๐1 = ๐ด1 โ ๐ด1 = 1;
๐ = 2, we have ๐1 + 2๐2 = ๐ด2โ 1+ 2.2! = ๐ด2โ 5 = ๐ด2โ
3! โ 1 = ๐ด2 โ ๐3 โ 1 = ๐ด2
๐ = 3 , we have ๐1 + 2๐2 + 3๐2 = ๐ด3 โ ๐ด2 + 3๐3 = ๐ด3 โ
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We assume that for ๐ = ๐ the equality (โ) takes the form
We calculate the value of the expression (โ) for ๐ = ๐ + 1. We
have
Based on the principle of mathematical induction, we reach the
conclusion that
Answer:
28. Prove that indifferent of what ๐,๐ โ โ, are ๐! . ๐! divides
(๐ + ๐)!
Solution. According to the definition 3, (๐+๐)!
๐! . ๐!= ๐ถ๐+๐
๐ is the
number of the subsets that have ๐ elements of a set with (๐ + ๐)
elements, namely ๐ถ๐+๐๐ is a natural number. Consequently,
(๐+๐)!
๐! . ๐! is
an integer number, or this proves that ๐! . ๐! divides (๐ + ๐)!
29. Deduce the equality
Solution. We use property๐. We have:
As a result,
Algebraic problems and exercises for high school
119
that had to be proven.
30. Calculate the sum
Solution. We notice that term ๐๐ of this sum can be transformed
as it follows:
Then the sum ๐๐ takes the form
Answer:
31. Solve the equation
Solution.
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Because ๐ถ๐ฅ
3 has meaning only for ๐ฅ โฅ 3, it follows that the
solution of the initial equation is ๐ฅ = 7.
Answer: ๐ฅ = 7.
32. Solve the equation
Solution.
Answer: ๐ฅ = 5.
33. Solve the equation
Solution. As:
We have:
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121
As ๐ฆ โ โ and ๐ฆ โค ๐ฅ, we have ๐ฆ โ {0, 1 ,2, 3, 4, 5, 6, 7, 8}.
Answer: ๐ฅ = 8; ๐ฆ โ {0, 1, 2, 3, 4, 5, 6, 7, 8}.
34. Determine the values of ๐ฅ that verify the equality
Solution.
because the solutions of the equation ๐ฅ2 + ๐ฅ โ 5 = 0 are irational
numbers.
Answer: ๐ฅ = 3.
35. Solve the equation
Solution.
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Then ๐ โ {0, 1,2,3,4,5,6, 7,8}.
Answer: ๐ฅ = 9, ๐ โ {0, 1,2,3,4,5,6, 7,8}.
36. Solve the system of equations
Solution. From the conditions of the problem, we have ๐ฅ, ๐ฆ โ โ
with ๐ฅ โฅ 1 and โค. Based on the formulas (1) โ (3), we have
The divisors of the free term in (โ) are the numbers
ยฑ1; ยฑ2; ยฑ3; ยฑ4; ยฑ5; ยฑ6; ยฑ7; ยฑ8; ยฑ9; ยฑ10; . . . .
We use Homerโs scheme and the Bezout theorem to select the
numbers that are solutions of the equation (โ). As ๐ฆ โ โ, we will
verify only the natural numbers. It verifies that numbers {1, 2, 3, 4, 5, 6}
are not solutions of the equation (โ).
We verify ๐ฆ = 7.
So
because for ๐ฆ > 7, the expression ๐ฆ4 โ 3๐ฆ3 + 14๐ฆ2 + 48๐ฆ + 360 > 0.
Algebraic problems and exercises for high school
123
Hence, the solution of the system (๐ต) is the pair (5,7).
Answer: {(5,7)}.
37. Find x and y, if
Solution. We will first bring to a simpler form the expression
As a result, the system (๐ถ) takes the form
Answer: {(7, 3)}.
38. Determine the values of ๐ฅ, so that
Solution. From the enunciation it follows that ๐ฅ โ โ. Then
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and ๐ฅ โ โ. So ๐ฅ โ {0, 1, 2, 3, 4, 5, 6}.
Answer: ๐ฅ โ {0, 1, 2, 3, 4, 5, 6}.
39. Determine the values of ๐ฅ that verify the inequation
Solution. (โ) has meaning for ๐ฅ โ โ and ๐ฅ โฅ 3. Because
it follows that
Answer: ๐ฅ โ {3, 4, 5, 6}.
40. Solve the system of inequations:
Solution.
Algebraic problems and exercises for high school
125
Answer: ๐ฅ โ {2,3}.
3.3. Suggested exercises 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?
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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 ๐ points, on the second
side there are taken ๐ points, and on the third side, ๐ points.
Moreover, none of the considered points is the triangle top. How many
triangles with tops in these points are there?
Algebraic problems and exercises for high school
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?
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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 ๐, ๐ โ โ, ๐ > ๐ + 3);
Algebraic problems and exercises for high school
129
Solve the equations and the systems of equations:
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Solve the inequations and the systems of inequations:
Algebraic problems and exercises for high school
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 ๐ from the development
of the binomial (1 + ๐)๐, if the coefficient of the 5th term is equal to
the coefficient of the 9th term.
110. Determine ๐ด๐2 , if the 5th term from the development of the
binomial doesnโt depend on ๐ฅ.
111. In the development of the binomial
the coefficient of the third term is equal to 28. Determine the middle
term from the development of the binomial.
112. Determine the smallest value of ๐โฒ๐ exponent from the
development of the binomial (1 + ๐)๐ , if the relation of the
coefficients of two neighboring arbitrary is equal to 7: 15.
113. Determine the term from the development of the binomial
that contains ๐ฅ3.
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132
114. Determine the term from the development of the binomial
that doesnโt depend on ๐.
115. Determine the term from the development of the binomial
that doesnโt contain ๐, if the sum of
the coefficients of the first three terms from the development is equal
to 79.
116. Determine the term from the development of the binomial
that doesnโt contain ๐.
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
2048.
119. Determine the term from the development of the binomial
that contains ๐6, if the relation of the
binomial coefficients of the fourth and second term is equal to 187.
120. Determine the term from the development of the binomial
that doesnโt contain ๐ฅ, 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 78.
Algebraic problems and exercises for high school
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 2/7. . Determine the term from the
development of the binomial that contains ๐ฅโ5/2.
122. Determine ๐ฅ, ๐ฆ and ๐ง, if it is known that the 2nd, 3rd and 4th
terms from the development of the binomial are equal to
240, 720, 1080, respectively.
123. For what value of the exponent ๐, 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 (๐ + ๐)23 the coefficients of which form
an arithmetic progression.
127. Determine the term from the development of the binomial
(โ๐ฅ + โ๐ฅโ34
)๐ that contains ๐ฅ6.5, if the 9th term has the biggest
coefficient.
128. The 3rd term from the development of the binomial
(๐ ๐ฅ +1
๐ฅ2)๐ doesnโt contain ๐ฅ. For which value of ๐ฅ is this term equal to
the 2nd term from the development of the binomial (1 + ๐ฅ3)30?
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134
129. For what positive values of ๐ฅ the biggest term from the
development of the binomial (5 + 3๐ฅ)10 is the 4th term?
130. In the development of the binomial (โ๐ฅ +1
2 โ๐ฅ4 )
๐ the first
three terms form an arithmetic progression. Determine all rational
terms from the development of this binomial.
131. Determine the values of ๐ฅ for which the difference between
the 4th and the 6th term from the development of the binomial
(โ2๐ฅ
โ816 +
โ3216
โ2๐ฅ)๐ is equal to 56, if it is known that the exponent ๐ of the
binomial is smaller by 20 than the binomial coefficient of the 3rd term
from the development of this binomial.
132. Given that ๐ is the biggest natural number only if
๐๐๐๐/3 ๐ + ๐๐๐๐๐/3 ๐ > 0, determine the term that contains ๐2 from
the development of the binomial (โ๐ โ โ๐3)๐.
133. Determine ๐ฅ for which the sum of the 3rd and al 5th term
from the development of the binomial (โ2๐ฅ + โ21โ๐ฅ)๐ is equal to
135, knowing that the sum of the last three binomial coefficients is 22.
134. Determine ๐ฅ , knowing that the 6th term from the
development of the binomial (๐ + ๐)๐, where ๐ = โ2lg (10โ3๐ฅ), ๐ =
โ2(๐ฅโ2)๐๐35
is 21, and the coefficients of the binomial of the terms
ranked 2, 3 and 4 are respectively the 1st, the 3rd and the 5th term of an
arithmetic progression.
135. Are there any terms independent of ๐ฅ in the development
of the binomial Write these terms.
Algebraic problems and exercises for high school
135
136. How many terms from the development of the binomial
(โ35+ โ7
3)36 are integer terms?
137. In the development of the binomial
the first three coefficients form an arithmetic
progression. Determine all terms from the development of the
binomial that contain the powers of ๐ฆ with a natural exponent.
138. Determine ๐ฅ, if the 3rd term from the development of the
binomial is equal to 106.
139. In the development of (1 + ๐ฅ โ ๐ฅ2)25 find the term
corresponding to the exponent of ๐ฅ that is 3 times as big as the sum of
all the developmentโs coefficients sum.
140. Determine the rank of the biggest term from the
development of the binomial (๐ + ๐)๐ according to the descending
powers of ๐, assuming that ๐ > 0; ๐ > 0; ๐ + ๐ = 1. 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?
Ion Goian โ Raisa Grigor โ Vasile Marin โ Florentin Smarandache
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Answers
Note. For chapters 1 and 2 only the less โbulkyโ answers have
been provided and those that are relatively more complicated (in our
own opinion).
1. Sets. Operations with sets
(because ๐ด consists of even natural numbers).
a)if ๐๐ โ ๐๐ = 0 โ ๐ถ = {๐
๐} ; b) if ๐๐ โ ๐๐ โ 0 โ ๐ถ โ๐๐ ๐ ๐๐๐๐๐๐๐ก๐
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2. Relations, functions
4. 1) ๐ฅ + ๐ฆ = 6; 2) ๐ฆ = ๐ฅ + 1; 3) ๐ฅ < ๐ฆ; 4) max (๐ฅ, ๐ฆ) > 4; 5) min
(๐ฅ, ๐ฆ) = 2; 6) cmmdc (๐ฅ, ๐ฆ) = 2; 7) ๐ฅ is even or ๐ฆ = 6; 8) ๐ฅ = ๐ฆ2.
5. 1) transitive; 2) symmetrical; 3) symmetrical and transitive; 4)
reflexive, transitive; 5) reflexive; 6) reflexive, symmetrical; 7), 8)
reflexive, symmetrical, transitive; 9) reflexive, transitive; 10) transitive
etc.
6. 1) ๐ฟ๐ผ = ๐๐ผ = โ, symmetrical; 2) ๐ฟ๐ผ = ๐๐ผ = โ, reflexive; 3) ๐ฟ๐ผ =
๐๐ผ = โ, symmetrical, antireflexive; 4) ๐ฟ๐ผ = {1, 4, 9,โฆ , ๐2, โฆ },๐๐ผ = โ
anti-symmetrical; 5) ) ๐ฟ๐ผ = ๐๐ผ = โ,, reflexive, symmetrical, transitive
etc. 6) ๐ฟ๐ผ = ๐๐ผ = {1, 2, 3, 5, 6, 10, 15, 30}, antireflexive, symmetrical;
7) ๐ฟ๐ผ = โ, ๐๐ผ = {3, 4, 5, . . . } , antireflexive, symmetrical; 8) ๐ฟ๐ผ =
๐๐ผ = โ , reflexive, anti-symmetrical, transitive; 9) ๐ฟ๐ผ = โ, ๐๐ผ =
{3,4,5, . . . } , antireflexive, anti-symmetrical, transitive; 10) ๐ฟ๐ผ =
โ, ๐๐ผ = {0, 2, 4, . . . }, anti-symmetrical; 11) ๐ฟ๐ผ = ๐๐ผ = โ , reflexive,
symmetrical, transitive; 12) ๐ฟ๐ผ = ๐๐ผ = โ, symmetrical.
9. 1) รข = {๐, ๐/๐}, ๐ โ โ, ๐ โ ๐
๐; 2) ๐ = โ๐, we have รข = {โ๐}.
10. รข = {๐ฅ โ โโ๐ฅ = ๐ + 2๐๐ ๐๐ ๐ฅ = ๐ โ ๐ + 2๐๐, ๐,๐ โ โค}.
11. a) yes;
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23. 1) bijective; 2) not bijective; 3) bijective; 4) not bijective; 5) bijective:
6) not bijective; 7) bijective; 8) not bijective; 9) injective; 10) not
injective; 11) injective; 12) surjective; 13) not surjective; 14) surjective;
15) surjective.
32. 1) yes; 2) no; 3) yes; 4) yes; 5) yes; 6) no; 7) no; 8) no.
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3. Elements of combinatorics
12. 110. 240. 111. 70(1 โ ๐ฅ2)2. 112. ๐ = 22๐+15
7, the smallest value
๐ = 6, then ๐ = 21. ๐ถ166 . ๐ฅ3. 114. ๐6 = ๐ถ15
6 . 115. ๐5 = ๐ถ125 . 6โ7. 116.
137. ๐ = 8, we have ๐1 = ๐ฆ4, ๐5 =
35
8๐ฆ; ๐ = 4, we have ๐1 = ๐ฆ
2 .
138. 10.
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References
1. Goian I., Grigor R., Marin V.: Baccalaureate "Mathematics" 1993 -
1995. Chisinau: Amato, 1996.
2. Goian I., Grigor R., Marin V.: Baccalaureate "Mathematics" 1996.
Chisinau: Evrica, 1997.
3. Ionescu-ศฤu C., Muลat ลt.: Exercises and mathematics problems for
grades IX - X. Bucharest: Didacticฤ and pedagogicฤ, 1978.
4. Militaru C.: Algebra. Exercises and problems for high school and
university admission. Bucharest: Alux, 1992.
5. Nฤstฤsescu C., Niลฃฤ C., Popa S.: Mathematics. Algebra. Textbook for
grade X-a. Bucharest: Didacticฤ and pedagogicฤ, 1994.
6. Petricฤ I., Lazฤr I.: Math tests for rank I-a and a II-a high school.
Bucharest: Albatros, 1981.
7. Sacter O.: Math problems Workbook suggested at maturity exams and
at admissions in institutions and universities. Bucharest: Tehnicฤ,
1963.
8. Stamate I., Stoian I.: Algebra problems workbook for high schools.
Bucharest: Editura didacticฤ and pedagogicฤ, 1971.
9. ลahno C.I.: Elementary mathematics problems workbook. Translation
from Russian by I. Cibotaru. Chisinau: Lumina, 1968.
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In this book, you will find algebra 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 workout. 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.