ion goian raisa grigor vasile marin florentin …smarandache/algebraicproblemsexercises.pdf · ion...

Click here to load reader

Post on 08-Mar-2018

224 views

Category:

Documents

2 download

Embed Size (px)

TRANSCRIPT

  • The Educational Publisher

    Columbus, 2015

    ALGEBRAIC PROBLEMS AND EXERCISES

    FOR HIGH SCHOOL

    Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

  • Algebraic problems and exercises for high school

    1

    Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    ALGEBRAIC PROBLEMS AND EXERCISES FOR HIGH SCHOOL

    Sets, sets operations Relations, functions Aspects of combinatorics

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    2

    First imprint: Algebra n exerciii i 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 Raisa Grigor Vasile Marin Florentin Smarandache

    ALGEBRAIC PROBLEMS

    AND EXERCISES

    FOR HIGH SCHOOL

    Sets, sets operations

    Relations, functions

    Aspects of combinatorics

    The Educational Publisher

    Columbus, 2015

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    4

    Ion Goian, Raisa Grigor, Vasile Marin, Florentin Smarandache,

    and the Publisher, 2015.

  • Algebraic problems and exercises for high school

    5

    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 Raisa Grigor Vasile Marin Florentin Smarandache

    6

    Contents

    Foreword / 5

    Notations / 7

    1. Sets. Operations with sets / 9

    1.1. Definitions and notations / 9

    1.2. Solved exercises / 16

    1.3. Suggested exercises / 32

    2. Relations, functions / 43

    2.1. Definitions and notations / 43

    2.2. Solved exercises / 59

    2.3. Suggested exercises / 86

    3. Elements of combinatorics / 101

    3.1. Permutations. Arrangements. Combinations.

    Newtons binomial / 101

    3.2. Solved problems / 105

    3.3. Suggested exercises / 125

    Answers / 137

    References / 144

  • Algebraic problems and exercises for high school

    7

    Notations

    = equality;

    inequality;

    belongs to;

    doesnt belong to;

    subset;

    superset;

    union;

    intersection;

    empty set;

    (or) disjunction;

    (and) conjunction;

    = {0, 1, 2, 3, 4, }

    = { ,2,1, 0, 1, 2, }

    p equivalent with ; natural numbers

    integer numbers set;

    = {

    |, , 0} rational numbers set;

    real numbers set;

    = { + |, , 2 = 1} complex numbers set;

    + = { | > 0}, {,, };

    = { | < 0}, {,,};

    = { | 0} = {0}, {, , ,, };

    ||

    []

    {}

    absolute value of ; integer part of x R; fractional part of , 0 {} < 1;

    (, ) pair with the first element

    and the second element

    (also called "ordered pair");

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    8

    (, , ) triplet with corresponding elements

    , , ;

    = {, | , } Cartesian product of set

    and set ;

    = {, , | , , }

    Cartesian product of sets , , ;

    universal set;

    () = {| } set of parts (subsets) of set ;

    = () ( )

    equality of sets and ;

    () ( )

    is included in ;

    = { | } union of sets and ;

    = { | } intersection of sets and ;

    = { | } difference between sets and ;

    = ( ) ( ) symmetrical difference;

    () = = complement of set relative to ;

    relation defined on sets and ;

    : function (application) defined on

    with values in ;

    () domain of definition of function :

    () domain of values of function .

  • Algebraic problems and exercises for high school

    9

    1. Sets. Operations with sets

    1.1. Definitions and notations

    It is difficult to give an account of the axiomatic theory of sets at

    an elementary level, which is why, intuitively, we shall define a set as a

    collection of objects, named elements or points of the set. A set is

    considered defined if its elements are given or if a property shared by

    all of its elements is given, a property that distinguishes them from the

    elements of another set. Henceforth, we shall assign capital letters to

    designate sets: ,, , , , , , and small letters for elements in sets:

    , , , , , , etc.

    If is an element of the set , we will write and we will

    read " belongs to " or " is an element of ". To express that is not

    an element of the set , we will write and we will read " does

    not belong to ".

    Among sets, we allow the existence of a particular set, noted as

    , called the empty set and containing no element.

    The set that contains a sole element will be noted with {}.

    More generally, the set that doesnt contain other elements except the

    elements 1, 2, , will be noted by {1, 2, , }.

    If is a set, and all of its elements have the quality , then we

    will write = {| verifies } or = {|()} and we will read: "

    consists of only those elements that display the property (for which

    the predicate ( ) is true)."

    We shall use the following notations:

    = {0, 1, 2, 3, } the natural numbers set;

    = {0, 1, 2, 3, } the natural numbers set without zero;

    = { ,2,1, 0, 1, 2, } the integer numbers set;

    = {1,2,3 } the integer numbers set without zero;

    = {

    |, , } the rational numbers set;

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    10

    = the rational numbers set without zero;

    = the real numbers set;

    = the real numbers set without zero;

    + = { | 0}; + = { | > 0};

    = { + |, , 2 = 1} = the complex numbers set;

    = the complex numbers set without zero;

    {1, 2, , } = 1, ;

    () = { | } = the set of all integer divisors of number

    ;

    () = || = the number of the elements of finite set A.

    Note. We will consider that the reader is accustomed to the

    symbols of Logic: conjunction (and), disjunction (or),

    implication , existential quantification () and universal quantification

    ().

    Let and be two sets. If all the elements of the set are also

    elements of the set , we then say that is included in , or that is

    a part of , or that is a subset of the set and we write . So

    ()( ).

    The properties of inclusion a. () , (reflexivity);

    b. ( ) (transitivity);

    c. (), .

    If is not part of the set , then we write , that is,

    ()( ).

    We will say that the set is equal to the set , in short = ,

    if they have exactly the same elements, that is

    = ( ).

    The properties of equality Irrespective of what the sets , and may be, we have:

    a. = (reflexivity);

    b. ( = ) ( = ) (symmetry);

  • Algebraic problems and exercises for high school

    11

    c. ( = = ) ( = ) (transitivity).

    With () we will note the set of all parts of set , in short

    () .

    Obviously, , ().

    The universal set, the set that contains all the sets examined

    further, which has the containing elements nature one and the same,

    will be denoted by .

    Operations with sets Let and be two sets, , ().

    1. Intersection.

    = { | },

    i.e.

    ( ), (1)

    ( ), (1)

    2. Union.

    = { | },

    i.e.

    ( ), (2)

    ( ), (2)

    3. Difference.

    = { | },

    i.e.

    ( ), (3)

    ( ), (3)

    4. The complement of a set. Let (). The difference

    is a subset of , denoted () and called the complement of

    relative to , that is

    () = = { | }.

    In other words,

    () , (4)

    () . (4)

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    12

    Properties of operations with sets = , = (idempotent laws)

    = , = (commutative laws)

    ( ) = ( );

    ( ) = ( ) (associativity laws)

    ( ) = ( ) ( );

    ( ) = ( ) ( ) (distributive laws)

    ( ) = ;

    ( ) = (absorption laws)

    ( ) = () ();

    ( ) = () () (Morgans laws)

    Two "privileged" sets of are and .

    For any (), we have:

    ,

    = , = , () = ,

    = , = , () = ,

    () = , () = ,

    (()) = (principle of reciprocity).

    Subsequently, we will use the notation () = .

    5. Symmetric difference.

    = ( ) ( ).

    Properties. Irrespective of what the sets , and are, we

    have:

    a. = ;

    b. = (commutativity);

    c. = = ;

    d. () = ;

    e. () = () (associativity);

    f. () = ( )( );

    g. = ( ) ( ).

  • Algebraic problems and exercises for high school

    13

    6. Cartesian product. Let and be two objects. The set

    {{}, {, }}, whose elements are the sets {} and {, }, is called an

    ordered pair (or an ordered couple) with the first component and

    the second component and is denoted as (, ). Having three objects

    , and , we write (, , ) = ((, ), ) and we name it an ordered

    triplet.

    Generally, having objects 1, 2, , we denote

    (1, 2, , ) = (((1, 2)3))

    and we name it an ordered system of elements (or a cortege of

    length ).

    We have

    (1, 2, , ) = (1, 2, , )

    (1 = 1 2 = 2 = ).

    Let , (). The set

    = {(, )| }

    is called a Cartesian product of sets and . Obviously, we can define

    = {(, , )| }.

    More generally, the Cartesian product of the sets 1, 2, ,

    1 2 = {(1, 2, , )| , = 1, }.

    For = = = 1 = 2 = . . . = , we have

    2, 3, . ori

    For example, 3 = {(, , )|, , }. The subset

    = {(, )| } 2

    is called the diagonal of set 2.

    Examples.

    1. Let = {1,2} and = {1, 2, 3}. Then

    = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3)}

    and

    = {(1,1), (1,2), (2,1), (2,2), (3,1), (3,2)}.

    We notice that .

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    14

    2. The Cartesian product 2 = can be geometrically

    represented as the set of all the points from a plane to wich a

    rectangular systems of coordinates has been assigned,

    associating to each element (, ) 2 a point (, ) from the plane

    of the abscissa and the ordinate .

    Let = [2; 3] and = [1; 5], (, ). Then can be

    geometrically represented as the hatched rectangle (see Figure

    1.1), where (2,1), (2,5),(3,5), (3,1).

    Figure 1.1.

    The following properties are easily verifiable:

    a. ( ) ;

    b. ( ) = ( ) ( ),

    ( ) = ( ) ( );

    c. = ( = = ),

    ( ).

  • 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 {| } = {} whose elements are the sets ,

    , (). We say that , , () is a family of sets

    indexed with the set .

    Let there be a family of sets {| }. Its union (or the union of

    the sets , ) is the set

    .

    The intersection of the given family (or the intersection of the

    sets , ) is the set

    In the case of = {1,2, , }, we write

    8. Euler-Wenn Diagrams. We call Euler Diagrams (in USA

    Wenns 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 Morgans law.

    Solution.

    In fig. 1.2.a, the hatched part is ; the uncrossed one

    (except ) represents ( ).

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    16

    Figure 2.2.a Figure 3.2.b

    In fig. 1.2b the side of the square hatched with \ \ \ \ is equal

    to () and the one hatched with / / / / is equal to (). All the

    hatched side forms () () (the uncrossed side is exactly

    ).

    From these two figures it can be seen that ( ) (the

    uncrossed side of the square in fig. 1.2.a coincides with ()

    () (the hatched side of fig. 1.2.b), meaning

    ( ) = () ().

    1.2. Solved exercises 1. For any two sets and , we have

    = \ (\)

    Solution. Using the definitions from the operations with sets, we

    gradually obtain:

    From this succession of equivalences it follows that:

    which proves the required equality.

  • Algebraic problems and exercises for high school

    17

    Remark. The equality can also be proven using Eulers diagrams.

    So = \(\).

    2. Whatever , are, the following equality takes place:

    Solution. The analytical method. Using the definitions from

    the operations with sets, we obtain:

    This succession of equivalences proves that the equality from

    the enunciation is true.

    The graphical method. Using Eulers circles, we have

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    18

    Figure 4.3.a Figure 5.3.b

    In fig. 1.3.a, we have ( ) ( ), which represents the

    hatched side of the square. From fig. 1.3.b it can be seen that

    ( ) ( ) = ( ) ( ).

    3. For every two sets , , the equivalence stands true:

    Solution. Let \ = \. We assume that . Then there

    is with or with .

    In the first case we obtain \ and \ , which

    contradicts the equality \ = \. In the second case, we obtain

    the same contradiction.

    So, if \ = \ = .

    Reciprocally, obviously.

    4. Sets = {1,2,3,4,5,6,7,8,9,10}, = {2,4,6,8,10} and =

    {3,6,9} are given. Verify the following equalities:

    a) \( ) = (\) (\);

    b) \( ) = (\) (\).

    Solution. a) We have = {2,3,4,6,8,9,10} , \( ) =

    {1,5,7}, \ = {1,3,5,7,9}, \ = {1,2,4,5,7,8,10}, (\) (\) =

    {1,5,7} = \( ).

  • Algebraic problems and exercises for high school

    19

    b) For the second equality, we have

    = {6} , \( ) = {1,2,3,4,5,6,7,8,9,10} , (\)

    (\) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = \( ).

    5. Determine the sets A and B that simultaneously satisfy the

    following the conditions:

    1) = {1,2,3,4,5};

    2) = {3,4,5};

    3)1 \;

    4) 2 \.

    Solution. From 1) and 2) it follows that {3,4,5} and

    {3,4,5} . From 3) it follows that 1 or 1 . If 1 ,

    then from = {1,2,3,4,5} it follows that 1 . But, if 1 ,

    then 1 , because, on the contrary it would follow that 1 =

    {3,4,5}. So 1 and 1 remain. Similarly, from 4) it follows that

    2 and so 2 . In other words, {3,4, 5} {2,3,4,5} and

    {3, 4, 5} {1,3,4,5} with 2 , 1 and that is why

    = {2,3,4,5}, and = {1,3,4,5}.

    Answer: = {2,3,4,5}, = {1,3,4,5}.

    6. The sets = {11 + 8 }, = {4 } and =

    {11(4 + 1) 3 }, are given. Prove that = .

    Solution. To obtain the required equality, we will prove the truth

    of the equivalence:

    .

    Let . Then and and that is why two

    integer numbers , , exist, so that = 11 + 8 = 4

    11 = 4( 2). In this equality, the right member is divisible by 4,

    and 11, 4 are prime. So, from 11 4 it follows that 4, giving =

    4 for one . Then

    = 11 + 8 = 11 4 + 8 = 11 4 + 11 3 = 11(4 + 1) 3

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    20

    which implies , in other words, we have proven the implication

    . (1)

    Similarly, let . Then exists with

    = 11(4 + 1) 3 = 11 4 + 11 3 = 11 4 + 8 = 4(11 + 2)

    Taking 4 = and 11 + 2 = , we obtain

    = 11 + 8 = 4 ,

    which proves the truth of the implication

    . (2)

    From (1) and (2) the required equality follows.

    7. The following sets are given

    and = .

    Prove that:

    a) all sets with = {3,4,5,6,7,8,9};

    b) all = { 2 }, so that = {3}.

    Solution. We determine the set :

    Then = (3; 6) = {3,4,5}.

    a) All possible subsets of are

    , {3}{4}{5}{3,4}{3,5}{4,5}{3,4,5} =

    The required sets are such that = {3,4,5,6,7,8,9} and

    will thus be like = {6,7,8,9}, where (), namely, the sets

    required at point a) are:

    1{6,7,8,9} , 2{3, 6, 7,8, 9} , 3{4, 6, 7,8, 9} , 4{5,6, 7,8,9} ,

    5{3,4,6,7,8,9}, 6{3,5,6,7,8,9}, 7{4,5,6, 7,8,9}, 8{3,4,5,6, 7,8,9}..

    b) Because , then and vice versa. Considering

    that 2 = {3,4,5,6,7,8,9}, we obtain 2 {4,9}, namely 2

    {3,2,2,3} = . The parts of the set are:

  • Algebraic problems and exercises for high school

    21

    , {3} , {2} , {2} , {3} , {3,2} , {3,2} , {3,3} , {2,2} ,

    {2,3}, {2,3}, {3,2,2}, {3,2,3}, {3,2,3}, {2,2,3}, .

    From the condition = {3} it follows that is one of the

    sets

    1 = {3}, 2 = {3,3}, 3 = {2,3}, 4 = {2,3}, 5 = {3,2,3},

    6 = {3,2,3}, 7 = {2,2,3}, 8 = = {3,2,2,3}.

    Answer: a) {1, 2, 3, 4, 5, 6, 7, 8};

    b) {1, 2, 3, 4, 5, 6, 7, 8}.

    8. Determine ,, and , if

    = {1,2,3,4,5,6} , = {1,2} , = {5,6} , =

    = {3,4}.

    Solution. From = = {3,4}, it follows that {3,4}

    .

    We know that

    = ( ) ( ) = ( ) ( ),

    = ( ) ( ) = ( ) ( ).

    Then:

    1 (1 1 ) ((1 1 )

    1 ).

    The following cases are possible:

    a) 1 and1 ;

    b) 1 and 1

    (case no. 3, 1 and 1 1 = {3,4}, is impossible).

    In the first case 1 and 1 , from = {5,6} it follows

    that 1 , because otherwise 1 and 1 1

    = {5,6}. So, in this case we have 1 = {3,4}which is

    impossible, so it follows that 1 , 1 . Similarly, we obtain 2

    and 2 and 2 , 5 and 5 ,5 and 6 , 6 , 6

    .

    In other words, we have obtained:

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    22

    = {1,2,3,4}, = {3,4,5,6}, = {3,4} and = {1,2,5,6}.

    Answer: = {1,2,3,4}, = {3,4,5,6}, = {3,4} and

    = {1,2,5,6}.

    9. The sets = {1,2}, = {2,3} are given. Determine the fol-

    lowing sets:

    a) ; b) ; c) 2;

    d) 2; e) ( ) ( ); f) ( ) ;

    g) ( ) ( ).

    Solution. Using the definition for the Cartesian product with

    two sets, we obtain:

    a) = {(1,2), (1,3), (2,2), (2,3)};

    b) = {(2,1), (2,2), (3,1), (3,2)};

    c) 2 = {(1,1), (1,2), (2,1), (2,2)};

    d) 2 = {(2,2), (2,3), (3,2), (3,3)};

    e) ( ) ( ) = {(2,2)};

    f) = {1,2,3}; ( ) = {(1,2), (1,3), (2,2)

    (2,3), (3,2), (3,3)};

    g) ( ) ( ) = {(1,2), (1,3), (2,2), (2,3),

    (3,2), (3,3)} = ( ) .

    10. The sets = {1,2, }, = {3,4, } are given. Determine

    and , knowing that {1,3} {2,4} .

    Solution. We form the sets and {1,3} {2,4} :

    Because {1,3} {2,4} , we obtain

    For = 3 and = 2, we have (3,2) .

    Answer: = 3, = 2.

  • Algebraic problems and exercises for high school

    23

    11. If , then

    Prove.

    Solution. ( ) = and that is why

    (we have used the equality ( ) = ( ) ( )).

    12. How many elements does the following set have:

    Solution. The set has as many elements as the fraction

    (2 + 1) (22 + + 1) has different values, when takes the values

    1,2,3, , 1000. We choose the values of for which the fraction takes

    equal values.

    Let , , < 1 with

    Then

    But and consequently

    For = 2, we obtain = 3, and for = 3, we have = 2.

    Considering that < , we obtain = 2 and = 3. So, only for =

    2 and = 3, the same element of the set : = 5 11 is obtained.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    24

    Answer: the set has 999 elements, namely () = 999.

    13. Determine the integer numbers , that make the following

    statement true ( 1) ( 3) = 13.

    Solution. We write

    As 13 is a prime number, and , , the following cases are

    possible:

    meaning the proposition () is true only in these situations:

    14. Determine the set

    Solution. Because:

    it follows that that equality

    is possible, if and only if we simultaneously have:

    Then:

    a) if at least two of the numbers , and are different, we

    cannot have the equality:

    namely in this case = ;

    b) if = = , then = and = {}.

    Answer: 1) for = = , we have = {2}; 2) if at least two of

    the numbers , and are different, then = .

  • Algebraic problems and exercises for high school

    25

    15. Determine the set

    Solution. Possible situations:

    + 2 4 /3 or + 2 > 4 /3.

    We examine each case:

    In this case we have:

    All integer numbers for which 1 3 2 applies, are:

    1,0,1.

    Then

    All integer numbers for which 3 2 < x 9 applies, are:

    2,3,4,5,6,7,8,9.

    We thus obtain:

    Answer: = {1,0,1,2,3,4,5,6,7,8,9}.

    16. Determine the values of the real parameter for which the

    set has:

    a) one element;

    b) two elements;

    c) is null.

    Solution. The set coincides with the set of the solutions of

    the square equation

    and the key to the problem is reduced to determining the values of the

    parameter for which the equation has one solution, two

    different solutions or no solution.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    26

    a) The equation (1) has one (two equal) solution, if and only if

    = 0 for = 1 0 or if = 1 = 0.

    We examine these cases:

    2) For = 1, the equation (1) becomes 5 + 15 = 0 = 3. So,

    the set consists of one element for

    b) The equation (1) has two different roots, if and only if > 0,

    namely

    c) The equation (1) doesnt have roots < 0 392

    60 28 >

    Answer:

    17. Let the set = {3,4,5} and = {4,5,6}. Write the elements

    of the sets 2 2 and ( ( )) ( ).

    Solution. a) For the first set we have:

  • Algebraic problems and exercises for high school

    27

    2 = {(3,3), (3,4), (3,5), (4,3), (4,4), (4,5), (5,3), (5,4), (5,5)},

    2 = {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6), (6,4), (6,5), (6,6)},

    2 2 = {(4,4), (4,5), (5,4), (5,5)}.

    b) For the second set we have:

    Then

    Answer:

    18. Given the sets = {1,2,3,4,5,6,7} and = {2,3,4}, solve

    the equations = and ( ) = ().

    Solution. In order to solve the enunciated equations, we use the

    properties of the symmetrical difference: ( ) = , its

    associativity and commutativity .

    But

    Consequently,

    We calculate

    Thus

    Answer: = , = .

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    28

    19. The following sets are given

    Determine the sets , , , , , , ( )

    ( ) and ( ).

    Solution. 1) We determine the sets and .

    The inequation () is equivalent to the totality of the three

    systems of inequations:

    So

    In other words,

    2) We determine the required sets with the help of the graphical

    representation

  • Algebraic problems and exercises for high school

    29

    20. 40 students have a mathematics test paper to write, that

    contains one problem, one inequation and one identity. Three

    students have successfully solved only the problem, 5 only the

    inequation, 4 have proven only the identity, 7 have solved not only the

    problem, 6 students not only the inequation, and 5 students have

    proven not only the identity. The other students have solved

    everything successfully. How many students of this type are there?

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    30

    Solution. Let be the set of students who have correctly solved

    only the problem, only the inequation, that have proven only

    the identity, the set of students that have solved only the problem

    and the inequation, the set of students that have solved only the

    problem and have proven the identity, the set of students who

    have solved only the inequation and have proven the identity, and

    the set of students that have successfully solved everything.

    From the given conditions, it follows that () = 3, () = 5,

    () = 4, () = 8, () = 7, () = 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 , , , , , , have only elements of the null set in common,

    the union of the sets , , , , , , is the set of students that have

    written the paper.

    Consequently,

    So

    Answer: 4 students out of all that have taken the test have

    solved everything successfully.

    21. (Mathematician Dodjsons problem)

    In a tense battle, 72 out of 100 pirates have lost one eye, 75

    one ear, 80 one hand and 85 one leg. What is the minimum number

    of pirates that have lost their eye, ear, hand and leg at the same time?

    Solution. We note with the set of one-eyed pirates, with

    the set of one eared pirates, with the set of pirates with one hand

    and with the set of the pirates with one leg.

    The problems requires to appreciate the set .

  • Algebraic problems and exercises for high school

    31

    It is obvious that the universal set is composed from the set

    and the number of pirates that have kept either both

    eyes, or both ears, or both hands, or both legs.

    Consequently,

    It follows that the set is not smaller (doesnt have fewer

    elements) than the sum of the elements of the sets , , , and

    (equality would have been the case only if the sets , ,

    and , two by two, do not intersect).

    But,

    Substituting, we have () = 100 namely 100 (

    ) + 30 + 25 + 20 + 15.

    Consequently, ( ) 100 30 25 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 English and

    German, 10 - English and French, 5 French and German, 3 students -

    all three languages. How many study only one language? How many

    not even one language?

    Solution. Let be the set of students that study English, -

    German, - French.

    Then the set of students attending both English and German is

    , English and French - , French and German - ,

    English, German and French - , and the set of students

    that study at least one language is .

    From the conditions above, it follows that the students who only

    study English form the set \( ) ( ) , only German -

    \( ) ( ), only French - \( ) ( ).

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    32

    But, because , we have (\(( ) ( ))) =

    () (( ) ( )) = () (( ) + ( )

    ( )) = () ( ) ( ) + ( ).

    Similarly, (\(( ) ( ))) = () ( )

    ( ) + ( ); (\(( ) ( )) = ()

    ( ) ( ) + ( ).

    Let be the set of students who only study one language, then

    Consequently,

    = 28 + 30 + 42 2.8 2.10 2.5 + 3.3 = 63. () = 63.

    The number of students who dont study any language is equal

    to the difference between the total number of students and the

    number of students that study at least one language, namely () =

    () ( ), where is the set of students that dont study

    any language and the set of all 100 students.

    From problem 20 we have

    So () = 100 80 = 20.

    Answer: 63 students study only one language, 20 students dont

    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 (, and are arbitrary sets)?

    3. Let = { 2 12 + 35) = 0},

    = { 2 + 2 + 35) = 0}; = { 2 + 2 + 35) = 0}

    a) Determine the sets , and .

    b) State if the numbers 1/5, 5, 7, 1/2 belong to these sets or

    not.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    34

    4. Determine the sets:

    5. Let = { x = 4n + 6m,n,m };

    a) Write three elements belonging to set .

    b) Determine if 26, 28, 33 belong to .

    6. Indicate the characteristic properties of the elements

    belonging to the sets: = {4, 7, 10}, = {3, 6, 12} , =

    {1, 4, 9, 16, 25}, = {1, 8, 27, 64, 125}.

    7. How many elements do the following sets have?

    8. Let there be the set = {3,2,1,1,2,3}. Determine the

    subsets of :

    9. Determine the sets:

  • Algebraic problems and exercises for high school

    35

    10. Compare (,,=,,) the sets and , if:

    11. Let = {1, 2, 3, 4, 5, 6, 7}, = {5, 6, 7, 8, 9, 10}. Using the

    symbols ,,, (complementary), express (with the help of , and

    ) the sets:

    .

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    36

    12. Determine the set , in the case it is not indicated and its

    parts , , simultaneously satisfy the conditions:

    has more elements than ;

    has more elements than ;

  • Algebraic problems and exercises for high school

    37

    13. Determine the sets , , , , , , ,

    ( ), , ( ), , , ( ) , ( ), if:

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    38

    14. Let

    a) Determine the sets:

    b) How many elements does the set = {

    (699 100 )} have?

    15. Determine the sets , , if

    16. Determine the set and its parts and , if

    17. Compare the sets and , if

  • Algebraic problems and exercises for high school

    39

    18. Determine the values of the real parameter for which the

    set has one element, two elements or it is void, if:

    19. Determine the number of elements of the set :

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    40

    20. The sets ,, are given. Determine .

    21. Determine , if:

    22. Prove the properties of the operations with sets.

    23. Determine the equalities (, , etc. are arbitrary sets):

  • Algebraic problems and exercises for high school

    41

    24. Given the sets = {1, 2, 3}, = {2, 3, 4}, = {3, 4, 5} and

    = {4, 5, 6}, write the elements of the following sets:

    25. Given the sets , and , solve the equations () =

    , where {,, }:

    26. Determine the sets , , , , , ,

    ( ), ( ), if:

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    42

  • Algebraic problems and exercises for high school

    43

    2. Relations, functions

    2.1. Definitions and notations

    Relations, types. Relations formation

    Let and be two non-empty sets, and their Cartesian

    product. Any subset is called a relation between the

    elements of and the elements of . When = , a relation like

    is called a binary relation on set .

    If there exists a relation like , then for an ordered pair

    (, ) we can have (, ) or (, ) . In the first case,

    we write and we read " is in relation with ", and in the second

    case , that reads " is not in relation with ". We hold that

    (, ) .

    By the definition domain of relation A X B, we understand

    the subset consisting of only those elements of set that are in

    relation with an element from , namely

    By values domain of relation we understand the

    subset consisting of only those elements of set that are in

    relation with at least one element of , namely

    The inverse relation. If we have a relation , then by

    the inverse relation of the relation we understand the relation

    1 as defined by the equivalence

    namely

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    44

    Example 1. Let us have = {1, 2}, = {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 , , be three sets and let us

    consider the relations , . The relation

    formed in accordance with the equivalence

    namely

    is called the formation of relation and .

    Example 2. Let us have , , , , , from Example 1.

    Determine

    1 1 and ( )1.

    Solution. We point out that the compound of the relations

    with exists if and only if = .

    Because , , it follows that and

    doesnt exist.

    We determine :

  • Algebraic problems and exercises for high school

    45

    (2,4) and {(4,1), (4,2)} {(2,1), (2,2)} ;

    (2,6) , but in we dont have a pair with the first

    component 6. It follows that

    c) , exists. By repeating the

    reasoning from b),

    and 1 = {(2,2)}.

    and 1 1 = {(1,2), (2,2)}.

    The equality relation. Let be a set. The relation

    is called an equality relation on . Namely,

    Of a real use is:

    1stTheorem. Let , , , be sets and , ,

    relations. Then,

    1) ( ) = ( ) (associativity of relations formation)

    The equivalence Relations. The binary relation 2 is called:

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    46

    a) reflexive, if whatever is;

    b) symmetrical, if ( ), (), ;

    c) transitive, if (( ) ), () , , ;

    d) anti-symmetrical, if (( ) = ), () , , ;

    e) equivalence relation on A, if it is reflexive, symmetrical and

    transitive;

    f) irreflexive, if whatever is.

    Let be an equivalence relation on set . For each element

    , the set

    is called the equivalence class of modulo (or in relation to R), and

    the set

    is called a factor set (or quotient set) of through .

    The properties of the equivalence classes. Let be an

    equivalence relation on set and , . Then, the following

    affirmations have effect:

    Partitions on a set. Let be a non-empty set. A family of

    subsets { } of is called a partition on (or of ), if the

    following conditions are met:

    2nd Theorem. For any equivalence relation on set , the factor

    set / = { } is a partition of .

  • Algebraic problems and exercises for high school

    47

    3rd Theorem. For any partition = { } of , there exists

    a sole equivalence relation on , hence

    The relation is formed according to the rule

    It is easily established that is an equivalence relation on and

    the required equality.

    Example 3. We define on set the binary relation according to

    the equivalence ( ) , where , fixed, (),

    .

    a) Prove that is an equivalence relation on .

    b) Determine the structure of the classes of equivalence.

    c) Form the factor set /. Application: = 5.

    Solution. a) Let , , . Then:

    1) reflexivity = ;

    2) symmetry

    3) transitivity

    From 1) - 3) it follows that is an equivalence relation on .

    b) Let . Then

    In accordance with the theorem of division with remainder for

    integer numbers and , we obtain

    Then

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    48

    where = + , and therefore

    where is the remainder of the division of through . But,

    In other words, the equivalence class of coincides with the

    class of the remainder of the division of through .

    c) Because by the division to only the remainders 0, 1, 2, . . .,

    1, can be obtained, from point b) it follows that we have the

    exact different classes of equivalence:

    The following notation is customarily used

    . Then

    where consists of only those integer numbers that divided by yield

    the remainder

    .

    For n = 5, we obtain

    with

    Definition. The relation is called a congruency relation

    modulo on , and class = is called a remainder class modulo

    and its elements are called the representatives of the class.

    The usual notation:

    ( ) (mode ) ( is congruent with

    modulo ),and

  • Algebraic problems and exercises for high school

    49

    is the set of all remainder classes modulo .

    Example 4 (geometric). Let be a plane and the set of all the

    straight lines in the plane. We define as the binary relation in

    accordance with

    a) Show that is an equivalence relation on .

    b) Describe the equivalence classes modulo .

    c) Indicate the factor set.

    Solution. a) It is obvious that is an equivalence relation (each

    straight line is parallel to itself; if 1 2 2 1 and

    b) Let . Then the class

    consists only of those straight lines from that are parallel with the

    straight line .

    c) / = { } is an infinite set, because there are an

    infinity of directions in plane .

    Example 5. The set = {1,2,3,4,5,6, 7,8,9}is given and its parts

    a) Show that = {1, 2, 3, 4 } is a partition of .

    b) Determine the equivalence relation on .

    c) Is = {1, 2, 3 } a partition on ?

    Solution. a) 1) We notice that

    meaning the system of the subsets of the set A defines a partition on

    .

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    50

    b) In accordance to the 3rd Theorem, we have

    So:

    = {( 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 does not define a partition on .

    Order relations. A binary relation on the set is called an

    order relation on , if it is reflexive, anti-symmetrical and transitive.

    If is an order relation on , then 1 is also an order relation

    on (verify!). Usually, the relation is denoted by "" and the

    relation 1 by "", hence

    With this notation, the conditions that "" is an order relation

    on the set are written:

    reflexivity ;

    asymmetry ( ) = ;

    transitivity ( ) .

    Example 6. On the set we define the binary relation in

    accordance with

    Show that is an order relation on .

    Solution. We verify the condition from the definition of order

    relations.

    1) Reflexivity

  • Algebraic problems and exercises for high school

    51

    2) Asymmetry. Let , with and . It follows that

    there are the natural numbers , with = . and = . .

    Then

    which implies that

    3) Transitivity. Let , , with and . Then there

    exists , with = and = hence,

    Because , , the implication is true

    which proves that is an order relation on the set .

    Remark. The order relation is called a divisibility relation on

    and it is written as , namely

    ( reads "b divides a", and "a is divisible through b").

    Functional relations Let and be two sets. A relation is called an

    application or a functional relation, if the following conditions are met:

    1) () , so as ;

    An application (or function) is a triplet = (, , ), where

    and are two sets and is a functional relation.

    If is an application, then for each there exists,

    according to the conditions 1) and 2) stated above, a single element

    , so that ; we write this element with (). Thus,

    The element () is called the image of element

    through application , set is called the definition domain of noted

    by () = , and set is called the codomain of and we usually say

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    52

    that is an application defined on with values in . The functional

    relation is also called the graphic of the application (function) ,

    denoted, later on, by . To show that is an application defined on

    with values in , we write : or , and, instead of

    describing what the graphic of (or ) is, we indicate for each

    its image (). Then,

    i.e.

    The equality of applications. Two applications = (, , )

    and = (,, ) are called equal if and only if they have the same

    domain = , the same codomain = and the same graphic

    = . If , : , the equality = is equivalent to () =

    (), () , that is to say

    The identical application. Let be a set. The triplet (, , 1) is,

    obviously, an application, denoted by the same symbol 1 (or ) and it

    is called the identical application of set .

    We have

    Consequently,

    1: and 1() = for () .

    By (, ) we will note the set of functions defined on with

    values in . For = , we will use the notation () instead of

    (, ).

    In the case of a finite set = {1, 2, , }, a function

    : is sometimes given by means of a table as is:

  • Algebraic problems and exercises for high school

    53

    where we write in the first line the elements 1, 2, , of set , and

    in the second line their corresponding images going through , namely

    (1), (2), , ().

    When = {1,2, , }, in order to determine the application,

    we will use : and the notation

    more frequently used to write the bijective applications of the set in

    itself. For example, if = {1,2}, then the elements of () are:

    If : , , , then we introduce the notations:

    - the image of subset through the application .

    Particularly, () = , the image of application ;

    is the preimage of subset through the application .

    Particularly, for , instead of writing 1({}), we simply

    note 1(), that is

    - the set of all preimages of through the application , and

    - the complete preimage of set through the application .

    The formation of applications. We consider the applications

    = (, , ) and = (, , ), so as the codomain of to coincide

    with the domain of . We form the triplet = (, , ).

    Then is also an application, named the compound of the

    application with the application , and the operation "" is called the

    formation of applications. We have

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    54

    namely

    4th Theorem. Given the applications

    . It follows that a) ( ) = ( )

    (associativity of the formation of applications); b) 1 = 1 = .

    Example 7. We consider the relations and

    [0,+) [0,+), , defined as it follows: 2 = ,

    = 2, = + 1.

    A. Determine which of the relations , 1, , 1, , 1 are

    functional relations.

    B. Find the functions determined at point A.

    C. Calculate , , , , 1, 1 (f, g, h are

    the functions from point B.)

    D. Calculate ( )( 3), (1 )(1/2), ( )(1/3).

    Solution. We simultaneously solve and .

    a) We examine the relation .

    2 , i.e.

    1) () = 2 , so that ;

    which means that is a functional relation. We write the application

    determined by through , = ( ,,).

    b) We examine the relation 1 .

    which means that if = { < 0}, then there is no ,

    so that 1 , which proves that 1 is not a functional relation.

  • Algebraic problems and exercises for high school

    55

    c) For the relation , we have: and 1 are functional relations.

    We denote by = ([0,+), [0,+), ),and 1 = ([0,+), [0,+),

    1), the functions (applications) defined by and 1, respectively.

    d) We examine the relation :

    1) + 1 and so

    so that ;

    which proves that is a functional relation.

    e) For the relation 1, we obtain:

    1)() () = 1 , so that y1;

    meaning that 1 is also a functional relation. We denote by =

    (,, ) and 1 = (,, 1).

    C. From A and B, it follows that:

    a) Then , and dont exist, because the domains

    and codomains do not coincide:

    = (,, ), = ([0,+), [0,+), ), = (,,)).

    b) We calculate , , 1 and 1 .

    So and 1 = 1 = 1.

    D. We calculate:

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    56

    Injective, surjective and bijective applications. An application

    : is called:

    1) injective, if

    (equivalent: 1 2 (1) (2));

    2) surjective, if

    (any element from B has at least one preimage in );

    3) bijective, if is injective and surjective.

    Remark. To prove that a function : is a surjection, the

    equation ( ) = must be solvable in for any .

    Of a real use are:

    5th Theorem. Given the applications

    , we have:

    a) if and are injective, then is injective;

    b) if and are surjective, then is surjective;

    c) if and are bijective, then is bijective;

    d) if is injective, then is injective;

    e) if is surjective, then is surjective.

    6th Theorem. The application : with the graphic =

    is a bijective application, if and only if the inverted relation 1 is a

    functional relation (f1is an application).

    This theorem follows immediately from

    Inverted application. Let : be a bijective application

    with the graphic = . From the 6th Theorem, it follows that the

    triplet 1 = (, , 1) is an application (function). This function is

    called the inverse of function .

    We have

    1: , and for ,

  • Algebraic problems and exercises for high school

    57

    i.e.

    7th Theorem. Application : is bijective, if and only if

    there exists an application : with = 1 and = 1.

    In this case, we have = 1.

    Real functions The function : is called a function of real variable, if

    = () . The function of real variable : is called a real

    function, if . In other words, the function : is called a

    real function, if A and B . The graphic of the real

    function : is the subset of 2 , formed by all the

    pairs (, ) 2 with and = (), i.e.:

    If the function is invertible, then

    Traditionally, instead of 1() = we write = 1(). Then

    the graphic of the inverted function 1 is symmetrical to the graphic

    function in connection to the bisector = .

    Example 8. For the function

    the inverted function is

    The graphic of function is the branch of the parabola = 2

    comprised in the quadrant I, and the graphic of function 1 is the

    branch of parabola = 2 comprised in the quadrant I (see below Fig.

    2.1).

    Algebra operations with real functions Let , : be two real functions defined on the same set.

    We consider the functions:

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    58

    = + : ,

    defined through () = () + (), () ;

    = + g sum-function.

    = : ,

    defined through () = () (), () ;

    = . product-function.

    = : ,

    defined through () = () (), () ;

    = difference-function.

    =

    : 1,

    defined through () =()

    (), () 1

    where 1 = { () };

    =

    quotient-function.

    : ,

    defined through () = () = {(), () 0,

    (), () < 0,}

    for () ;

    - module-function.

    Figure 2.1.

  • Algebraic problems and exercises for high school

    59

    Example 9. Let , : [0,+) with () = 2 and () =

    . Determine , and /

    Solution. As , share the same definition domain, the

    functions , and / make sense and

    Example 10. Let : [0,+), () = 2and : [0,+)

    ,() = . Determine a) and b) .

    Solution. a) = makes sense and we obtain the function

    : with

    b) = : [0,+) [0,+), also makes sense, with

    2.2. Solved exercises 1. Let = {2,4,6,8} and = {1,3,5,7}, , =

    {(, )| 6 1} :

    a) What is the graphic of the relation ?

    b) Make the schema of the relation .

    Solution.

    a) Because , and , it follows that

    So:

    b) See Fig. 2.2 below.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    60

    Figure 2.2.

    2. Let = {1,2,3,4} and = {1,3,5,7,8}. Write the graphic of

    the relation = {(, )3 + 4 = 10} .

    Solution. We observe that 3 + 4 = 10 3 = 2(5

    2) is even and (5 2) 3. Considering that , and ,

    we obtain: (2,4), and (1,7). The equality 3 + 4 = 10 is

    met only by = 2 and = 1. So = {(2,1)}.

    3. Let = {1,3,4,5} and = {1,2,5,6}. Write the relation

    using letters , and , if the graphic of the relation is known.

    Solution. (, ) .

    4. On the natural numbers set we consider the

    relations , , , 2 , defined as it follows: (, ): =

    {(3,5), (5,3), (3,3), (5,5)} ; ; = 12 ;

    = 3.

    a) Determine , , , , , , , .

    b) What properties do each of the relations , , , have?

    c) Determine the relations 1, 1, 1 and 1 .

    d) Determine the relations , , 1 1, ( )1,

    and 1 .

  • Algebraic problems and exercises for high school

    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 = + 12 doesnt have solutions in ).

    (for any , we have = 3/).

    b) 1) is symmetrical, transitive, but is not reflexive. For

    example (2,2) ; because (3,3) :, it follows that isnt

    antireflexive either.

    2) is - reflexive( , () ),

    - anti-symmetrical (( )

    ( ) = ),

    - transitive (( ) ).

    So is an order relation on set .

    3) is antireflexive ( , () ), anti-symmetrical

    (( ) ( = 12 and = 12) =

    = ), but it is not symmetrical ( = 12 =

    12 ), and it is not transitive (( ) ( =

    12 = 12) = 24 ).

    4) is not antireflexive ( 3, () , 0 = 3.0), it is anti-

    symmetrical (( ) ( = 3 = 3) = 9 =

    0 = ), and for 0 , the equalities = 3 and = 3 cannot

    take place), it is not reflexive (for example, 1 3.1 11), it is not

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    62

    transitive (( ) ( = 3 = 3) = 9 3 ,

    generally ).

    So

    i.e.

    i.e.

    i.e.

    which implies

    i.e.

    In other words,

  • Algebraic problems and exercises for high school

    63

    + 12 = 3, which implies

    which implies

    5. We consider the binary relation defined on as it follows:

    a) Prove that is an equivalence relation.

    b) Determine the factor set /.

    Solution.

    1) Because = , () , we have , () .

    3) Let and , , , . Then

    In other words, the binary relation is reflexive, symmetrical

    and transitive, which proves that is an equivalence relation on .

    b) Let . We determine the class of equivalence of in

    connection to .

    Then the set factor is

    We observe that

    In other words, for 1, we have 2 and therefore

    = 2, and the factor can also be written

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    64

    6. Let = {1,2,3,4,5} and = {, , , , }. Which of the

    diagrams in Fig. 2.3 represents a function defined on with values in

    and which doesnt?

    Solution. The diagrams a), c), e) reveal the functions defined on

    with values in . Diagram b) does not represent a function defined

    on with values in , because the image of 2 is not indicated.

    Diagram d) also does not represent a function defined on with values

    in , because 1 has two corresponding elements in , namely , .

    7. Let A = {1, 2, 3} and = {, }. Write all the functions

    defined on with values in , indicating the respective diagram.

    Solution. There are eight functions defined on with values in .

    Their diagrams are represented in Fig. 2.4.

    Figure 2.3.

  • Algebraic problems and exercises for high school

    65

    Figure 2.4.

    8. Let = {1, 2, 3, 4}, = {, , }, = {, , , } and =

    {1, 2}.

    a) Draw the corresponding diagrams for two surjective functions

    defined on with values in.

    b) Draw the diagram of the injective functions defined on with

    values in .

    c) Draw the diagram of a function denied on with values in ,

    that is not injective.

    d) Draw the corresponding diagrams for two bijective functions

    defined on with values in .

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    66

    Solution. a) The diagrams of two surjective and two non-

    surjective functions defined on with values in are represented in

    Fig. 2.5 a) and b).

    Figure 2.5.

    b) The diagram of the injective functions defined on with

    values in are represented in Fig. 2.6.

    c) The diagram of a non-injective function defined on with

    values in is represented in Fig. 2.7.

    d) The diagrams of the two bijective functions defined on with

    values in are represented in Fig. 2.8.

  • Algebraic problems and exercises for high school

    67

    Figure 2.5.

    Figure 2.6.

    Figure 2.7.

    Figure 2.8.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    68

    9. Let = {0, 1,5, 6}, = {0, 3, 8, 15} and the function

    : as defined by the equality ( ) = 2 4 + 3, ()

    .

    a) Is the function surjective?

    b) Is injective?

    c) Is bijective?

    Solution. a) We calculate () = {(0), (1), (5), (6)} =

    {3,0, 8, 15} = . So, is surjective.

    b) The function is also injective, because (0) = 3, (1) = 0,

    (5) = 8 and (6) = 15, i.e. 1 2 (1) (2)

    c) The function is bijective.

    10. Using the graphic of the function : , : ,

    show that the function is injective, surjective or bijective.

    a) () = { 2, (, 2),

    3 6, [2,+).

    c) : [1,+] ,() = {

    3, 1 < 0,, 0 < 1,

    0.5( + 1), > 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 abscises axis that intersects the graphic of the function in two or

    more points, the function in not injective. If () is the values set of

    the function and any parallel to the abscises axis, drawn through the

    ordinates axis that are contained in (), cuts the graphic of the

    function in at least one point, the function is surjective. It follows that

    a function is bijective if any parallel to the abscises axis, drawn

    through the points of (), intersects the graphic of in one point.

    a) The graphic of the function is represented in Fig. 2.9.

  • Algebraic problems and exercises for high school

    69

    We have () = (,+) and any parallel = , to

    the abscises axis intersects the graphic of the function in one point.

    Consequently, is a bijectve function.

    () = {2 6 + 8, 2 6 + 8 0,

    2 + 6 8, 2 6 + 8 < 0=

    = {2 6 + 8, if (,2] [4,+),

    2 + 6 8, if (2,4).

    Figure 2.9.

    The graphic of the function is represented in Fig. 2.10.

    We have () = [0; 3] [1; 3]. Any parallel = ,

    (0,1) intersects the graphic of the function in four points; the

    parallel = 1 intersects the graphic in three points; any parallel =

    , (1; 3] intersects the graphic of the function in two points.

    So the function is neither surjective, nor injective.

    c) The graphic of the function is represented in Fig. 2.11. We

    have

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    70

    Any parallel = to the abscises axis cuts the graphic of the

    function () in one point, at most ( = 3, = 2, = 4) and that

    is why () is injective. The equation () = has a solution

    only for 3, hence () is not a surjective function.

    d) We explain the function :

    () = max( + 1, 1 ) = { + 1, 1 + 1,1 , + 1 < 1

    =

    = { + 1, 0,1 , < 0.

    Figure 2.10.

    Figure 2.11.

  • Algebraic problems and exercises for high school

    71

    The graphic of the function is represented in Fig. 2.12.

    We have () = , () = [1,+). Any parallel = ,

    (1,+) intersects the graphic in two points, therefore is not

    injective. Straight line = 1/2 [0,+) doesnt intersect the

    graphic of the function , so is not surjective either.

    Figure 2.12.

    11. Determine which of the following relations are applications,

    and which ones are injective, surjective, bijective?

    For the bijective function, indicate its inverse.

    Solution.

    Because

    the equation 2 = + 3 doesnt have solutions in , it follows that

    is not a functional relation.

    Then, because () = = [1; 0], and

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    72

    with 3/2 3/2, and is not a functional relation.

    c) We have: (, ) = () = 2. Then:

    because 1, 2 (0,+);

    4) The equation () = has solutions in [0,+), if and

    only if 0, which proves that is not a surjection (number -2, for

    example, doesnt have a preimage in [0,+).

    d) We have:

    Repeating the reasoning from c), we obtain that is an

    injection. Furthermore, the equation () = [0,+) has the

    solution = [0,+) , and that is why is a surjective

    application. Then is a bijective application and, in accordance with

    the 6th Theorem, the relation 1 is also an application (function).

    In accordance with the same 6th Theorem, we have

    12. Let { = 1,2,3,4,5,6,7,8,9} and () given; using the

    following table:

    determine:

    a)

  • Algebraic problems and exercises for high school

    73

    b)

    c) Calculate 1(1); 1(4); 1(7).

    Solution.

    13. Let = {1,2,3} and = {, , }. We examine the

    relations

    a) Determine , , , and , , .

    b) Which of the relations , are applications? Indicate the

    application type.

    c) Determine the relations 1, 1 and 1. Which one is a

    function?

    d) Determine the relations 1, 1, 1, 1,

    1 and 1. Which ones are applications?

    Solution.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    74

    b) is not an application, because element 3 has two images

    and . is a bijective application; is an application, nor injective (the

    elements 3 1 have the same image c), nor surjective (b doesnt have

    a preimage).

    We have:

    1 is not an application, because the element has two

    images 1 and 3; 1 is a bijective application; 1is not an application,

    because 1 (the element doesnt have an image);

    Only the relation 1is an application, neither injective, nor

    surjective.

    14. Given the functions: , : ,

    () = {3 , 2, 1, > 2,

    () = max( 2, 3 ),

    show that = .

    Solution. The functions and are defined on and have

    values in . Lets show that () = (), () .

    We explain the function :

    () = {3 , 1 3 , 1, 3 < 1,

    = {3 , 2, 1, > 2,

    = ()

    no matter what is. So = .

  • Algebraic problems and exercises for high school

    75

    15. The functions , : ,

    () = { + 2, if 2,+10

    3, if > 2,

    () = { 2, if < 3,2 5, if 3

    are given.

    a) Show that and are bijective functions.

    b) Represent graphically the functions and 1 in the same

    coordinate register.

    c) Determine the functions: = + , = , = . , = /.

    d) Is function bijective?

    Solution. a), b) The graphic of the function is represented in

    Fig. 2.13 with a continuous line, and the graphic of 1 is represented

    with a dotted line.

    Figure 2.13.

    We have:

    1() = { 2, if 4,3 10, if > 4,

    () = { + 2, if < 1,1

    2( + 5), if 1.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    76

    c) We draw the following table:

    We have

    () =

    {

    2, 2,4

    3( + 1), 2 < < 3,

    1

    3(7 5), 3,

    () =

    {

    4

    2, 2,

    2

    3(8 ), 2 < < 3,

    5

    3(5 ), 3,

    () =

    {

    2 4, 2,1

    3(2 + 8 20), 2 < < 3,

    1

    3(22 + 15 50), 3,

    () =

    {

    + 2

    2, 2,

    + 10

    3( 2), 2 < < 3,

    + 10

    3(2 5), 3.

  • Algebraic problems and exercises for high school

    77

    d) We have () = 4, () (,2], therefore is not

    injective, in other words, is not bijective. This proves that the

    difference (sum) of two bijective functions isnt necessarily a bijective

    function.

    16. Lets consider the function : {/} {/},

    a) Show that function is irreversible.

    b) Determine 1.

    c) In what case do we have = 1?

    Solution:

    ( )(1 2) = 0 1 = 2 is injective.

    2) Let {/}. We examine the equation () = . Then

    If

    Imposibile. So ( )/( ) {/}, which proves

    that the equation () = has solutions in {/} for any

    {/}. This proves that is a surjective function.

    From 1) - 2), it follows that is a bijective function, hence it is

    also inversibile.

    b) We determine

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    78

    c) In order to have = 1, it is necessary that the functions

    and 1 have the same definition domain and so = . This

    condition is also sufficient for the equality of the functions and 1.

    It is true that, if = , the functions and 1are defined on

    {/}, take values in {/}, and 1() =+

    += (), ()

    {

    }, namely = 1.

    17. Represent the function () = 52 2 + 8 as a

    compound of two functions.

    Solution. We put

    Then

    Answer:

    18. Calculate , , ( )(4) and ( )(4), where:

    a) () =3

    1 and () = ;

    b) () = + 3 and () = 2 4;

    c) () = {2 + 6, x 32 5, 3

    and () = {5 2, 1,

    2 2 + 4, > 1.

    Indicate ( ) and ( ).

    Solution: a) We have

  • Algebraic problems and exercises for high school

    79

    So ( )(4) ( )(4), which proves that the comutative

    law of function composition, generally, does not take place:

    .

    c) To determine the functions and , in this particular

    case we proceed as it follows:

    ( )() = ( )(()) == {2() + 6(), () < 3

    2() 5, () 3=

    = {

    2() + 6() () < 3,

    2() 5, () 3 and 1,

    2(2 2 + 4) 5, () 3 and > 1

    =

    {

    (5 2)2 + 6(5 2) 5 2 < 3,

    2(5 2) 5, 5 2 3 and 1,

    2(2 2 + 4) 5, 5 2 3 and > 1

    =

    = {

    252 + 10 8 < 1/5,10 1, 1/5 1,

    22 + 4 13, > 1.

    ( )() = (()) = {5(), () 1,

    2() () + 4, () > 1.

    We notice that

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    80

    So for () 1, we obtain

    ( )() = {5(2 + 6) 2, if [3 10,3)

    5(2 5) 2, if [3,+).

    Similarly,

    and that is why for () > 1, we obtain

    for (,3 10).

    To sum it up, we have

    ( )() =

    = {

    4 + 123 + 342 12 + 4, (,3 10),

    52 + 30 2, if [310,3),

    10 7, if 3.

    ( )(4) = 10.4 27 = 67.

    19. Solve the following equations:

    a) ( )() = ( )() , if () = 3 + 1, () =

    + 3;

    b) ( )() = , if () =+1

    +, , {};

    c) ( )() = ( )(), if () = 22 1, () = 43 3.

    Solution. We determine the functions ( )() and (

    )():

    The equation becomes

  • Algebraic problems and exercises for high school

    81

    Answer: = 2.

    b) We calculate ( )():

    The equation becomes

    Answer: {1,1}

    c) We determine ( ) and ( )():

    The equation becomes

    meaning the equality is true for any .

    Answer: .

    20. Let , : given by () = 2 + + 12 and () =

    2 + 2. Show there is no function : , so that

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    82

    Solution. The relation () can also be written

    We suppose there is a function : that satisfies the

    relation (). We put in () = 1 and = 1. We obtain:

    which implies (4) +

    (2) (2) + (4), contradiction with the hypothesis. So there is

    no function with the property from the enunciation.

    21. Determine all the values of the parameters , for

    which ( ) = ( )() , () , where () = 2 and

    () = 2 + + .

    Solution. We determine and :

    Then

    Answer: = 1, = 0, () = 2 = ().

    22. Given the functions , , ,

    .

    Show that:

    a) is not injective;

  • Algebraic problems and exercises for high school

    83

    b) is injective;

    c) is bijective and determine 1.

    Solution. a) Let (1) = (2).

    Then

    For example, () = 3( + 4) + 3, taking 1 = 0 and 2 = 4,

    we obtain (0) = (4) = 3, 1 2.

    because

    so is an injective function.

    We prove that is surjective. Let . We solve the equation

    (the root of an uneven order exists out of any real number). So is a

    surjection, therefore, is a bijective function.

    We determine 1:

    23. Let : [1,+) [1,+) with

    a) Prove that is a bijective function.

    b) Determine 1.

    Solution. a) We have () = (2 + 1) 3. We represent this

    function as a compound of two functions:

    , : [1,+) [1,+), where () = 3, () = 2 + 1.

    Then () = ( )() , which implies that = is bijective,

    being the compound of two bijective functions.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    84

    b) 1 = ( )1 = 1 1. We determine 1 and 1.

    Because:

    We solve this equation according to [1,+).

    The discriminant is:

    The equation has only one root in [1,+):

    Then

    So,

    = (1 + 4 33

    )/2, with 1: [1,+) [1,+).

    24. Considering the function : with the properties:

    a) Determine the function .

    b) Calculate (1998).

    Solution. a) For 2 = 0 from 1), it follows that (1) = (1) +

    (0), which implies that (0) = 0. For 2 = 1 from 1), we obtain

    Then

  • Algebraic problems and exercises for high school

    85

    Let {0, 1}. Then

    On the other hand, 1

    1=1+

    1= 1 +

    1 implies

    From (2) and (3) it follows that

    So () = , () .

    Answer: a) () = ; b) (1998) = 1998.

    25. Using the properties of the characteristic function, prove the

    equality:

    Solution. Using the properties = = , we will

    prove the required equality by calculating with the help of , and

    the characteristic functions of sets ( ) and ( )

    ( ):

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    86

    which implies ( ) and ( ) ( ).

    2.3. Suggested exercises 1. Determine the definition domains and the values domains of

    the following relations:

    2. Let = {2,4,6,8} and = {1,3,5,7}, .

    a) Determine the graphic of the relation .

    b) Construct the schema of the relation :

    1) = {(, ) < 3 and > 3};

    2) = {(, ) > 2 and < 5};

    3) = {(, ) > 6 or > 7};

  • Algebraic problems and exercises for high school

    87

    3. Let = {1,2,3,4} and = {1,3,5,7,8}. Write the graphic of

    the relation , if:

    4. Let = {, 3, 4, 5}, = {, 2, 5, 6} and be the graphic of

    the relation . Write the relation with propositions containing letters

    and , with and :

    5. Let A = {, 2.3,4}. Analyze the properties of the relation

    2(var. 1-6) and 2 (var. 7-14):

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    88

    7) = {(, ) 2 > 1 and > 1 };

    8) = {(, ) 2{ > 0, > 0

    or { < 0, < 0

    };

    9) = {(, ) 2 0 or 0 };

    10) = {(, ) 2 1 or 1 };

    6. For each binary relation defined on set :

    a) determine the definition domain and the values domain ;

    b) establish the properties (reflexivity, ireflexivity, symmetry,

    anti-symmetry, transitivity);

    c) determine the inverse relation 1(, ):

    7. Given the set and the binary relation 2 , prove that is

    an equivalence relation and determine the factor set /.

  • Algebraic problems and exercises for high school

    89

    8. Given the set = {1, 2,3,4,5,6,7,8, 9} and the subset system

    = { , = 1, }, prove that defines a partition on and build

    the equivalence relation .

    9. We define on the binary relation :

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    90

    a) Show that is an equivalence relation on .

    b) Determine the equivalence classes.

    10. We define on the binary relation :

    .

    a) Show that is an equivalence relation.

    b) Determine the equivalence classes.

    11. Let 2 , where = {5,4,3,2,1, 0, 1, 2, 3, 4, 5, 6,

    7, 8} with

    a) Is an equivalence relation?

    b) If so, determine the equivalence classes.

    12. Determine: , , 1, , 1, 1 , if:

    13. Determine the relations

    , , 1, 1, 1 , ( )1 ;

    5) = {(, ) 2 s even},

    = {(, ) 2 is uneven};

  • Algebraic problems and exercises for high school

    91

    14. Let = {1, 2, 3, 4}, = {, , , } . Determine the

    definition domain and the values domain of each of the following

    relations , . Which ones are applications? Determine the application

    type. Determine the relation 1 , 1, 1, 1. Are

    they applications?

    15. We consider the application : , () = sin .

    Determine:

    16. The application : is given by the table below, where

    = {1, 2, 3, 4, 5, 6, 7, 8, 9} and = {, , , , , }.

    Determine:

    17. We consider the application : , ( ) = [] ([] is

    the integer part of ). Determine

    1({2,4,5}) and 1(1).

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    92

    18. The following application is given: : , () = 2.

    Determine () and 1(), if = {1, 2, 3, 4, 5, 6 , 7, 8, 9, 10}.

    19. Let and be two finite sets, || = . ||. How many

    surjective applications : are there, if:

    20. Given the graphic of the relation , establish if is a func-

    tion. Determine and .

    Changing and , make become an application injective,

    surjective and bijective.

    Draw the graphic for the relation 1. Is relation 1 a function?

    What type is it?

  • Algebraic problems and exercises for high school

    93

    21. Let = {1, 2, 3, 4}, = {0, 1, 5, 6} and = {7, 8, 9}.

    a) Draw the diagrams of two injective and two non-injective

    functions defined on with values in .

    b) Draw the diagrams of two surjective functions and two non-

    surjective applications defined on with values in .

    c) Draw the diagrams of two bijective and two non-bijective

    functions defined on with values in .

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    94

    22. Let = {1, 3, 5, 6} and = {0, 1, 2, 3, 5}, , .

    Which of the relations stated below represents a function defined on

    with values in ? And one defined on with values in ? For functions,

    indicate their type:

    23. Using the graphic of the function : , show that is

    injective, surjective or bijective. If bijective, determine 1and draw

    the graphic of and 1in the same register of coordinates..

    24. Which of the following relations 2 are functions?

    Indicate the definition domain of the functions. Establish their type:

  • Algebraic problems and exercises for high school

    95

    25. Given the relation with = [3; 5] and = [4; 7]:

    a) Does pair (4, 5) belong to the relation ? Why?

    b) Indicate all the ordered pairs (, ) with = . Explain.

    26. Given the function (), calculate its values in the indicated

    points.

    4) () = {2 5, > 0,2 + 3, 0,

    (2), (0), (5);

    5) () = {2 + 1, > 0 4, = 01 2, < 0

    (5), (1), (1/2);

    27. Determine (), if:

    28. Given the functions () and (), determine the functions

    + , , and / , indicating their definition domain.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    96

    29. Represent the function () as a compound of some

    functions:

    30. Given the functions () and ( ), determine the

    functions and and calculate ( )(3) and ( )(3) in

    options 1) 6) and ( )( 1) and ( )( -1) in options 7)

    12).

    31. Given the functions () = 2, () = 3 and () =

    1, calculate:

  • Algebraic problems and exercises for high school

    97

    32. Determine if in the function pairs and one is the inverse

    of the other:

    33. The value of function is given. Calculate the value of the

    function 1, if:

    34. Determine 1, if:

    35. The function: : is given:

    a) Prove that is bijective.

    b) Determine 1

    c) Calculate 1 1 .

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    98

    36. Given the functions ( ) and ( ), determine the functions

    ( )() and ( )()(, : ):

    37. Prove the equality of the functions and :

    38. Determine the functions = + , = , = .

    and = /:

  • Algebraic problems and exercises for high school

    99

    39. Let = {1, 2, 3, 4}, = {0, 1, 3, 4} and the functions

    Can the functions , be defined? If so, determine

    these functions. Make their diagrams.

    40. a) Show that the function is bijective.

    b) Determine 1.

    c) Graphically represent the functions and 1 in the same

    coordinate register.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    100

    41. Show that the function : , () = 2 6 + 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):

  • Algebraic problems and exercises for high school

    101

    3. Elements of combinatorics

    3.1. Permutations. Arrangements. Combinations. Newtons 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 = {1, 2, , } be a finite set that has elements. Set

    is called ordered, if each of its elements associates with a certain

    number from 1 to , named the element rank, so that different

    elements of associate with different numbers.

    Definition 1. All ordered sets that can be formed with elements

    of given set (( ) = ) are called permutations of elements.

    The number of all permutations of elements is denoted by the

    symbol and it is calculated according to the formula:

    By definition, it is considered 0 = 0! = 1 = 1! = 1.

    Definition 2. All ordered subsets that contain elements of set

    with elements are called arrangements of elements taken at

    a time.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    102

    The number of all arrangements of elements taken at a

    time is denoted by the symbol and it is calculated with the formula

    Definition 3. All subsets that contain elements of set with n

    elements are called combinations of elements taken at a time.

    The number of all combinations of elements taken at a time

    is denoted by the symbol and it is calculated with the formula

    where , ; 0 .

    Remark. In all the subsets mentioned in definitions 1 - 3, each

    element of the initial set appears only once.

    Along with the combinations in which each of the different

    elements of a set participates only once, we can also consider

    combinations with repetitions, that is, combinations in which the same

    element can participate more than once.

    Let groups of elements be given. Each group contains certain

    elements of the same kind.

    Definition 1. Permutations of elements each one containing

    1 elements 1, 2 elements 2 , elements , where 1 + 2 +

    + are called permutations of elements with repetitions.

    The number of all permutations with repetitions is denoted by

    the symbol 1,2,, and it is calculated using the formula:

    Definition 2 . The arrangements of elements, each one

    containing elements, and each one being capable of repeating itself

    in each arrangement for an arbitrary number of times, but not more

    than times, are called arrangements of elements taken at a

    time with repetitions.

  • Algebraic problems and exercises for high school

    103

    The number of all arrangements with repetitions of elements

    taken at a time is denoted by the symbol and it is calculated with

    the formula

    Definition 3'. The combinations of elements each one

    containing elements, and each one being capable of repeating itself

    more than once, but not more than times, are called combinations

    of elements taken at a time with repetitions.

    The number of all combinations with repetitions is denoted by

    the symbol and it is calculated with the formula

    In the process of solving combinatorics problems, it is

    important to firstly establish the type (the form) of combination. One

    of the rules used to establish the type of combination could be the

    following table

    It is often useful to use the following two rules:

    Caution is advised to the order the elements are arranged in:

    Arrangements if not all of

    the elements from the

    initial set participate;

    Permutations if all

    elements of the initial

    set participate;

    If the order isnt important,

    we have combinations;

    If the order is important,

    we have arrangements

    or permutations:

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    104

    The sum rule. If the object can be chosen in ways and the

    object in n ways, then the choice "either A or B" can be made in +

    ways.

    The multiplication rule. If the object can be chosen in

    ways and after each such choice the object can be chosen in ways,

    then the choice "A and B" in this order can be made in . ways.

    We will also mention some properties of combinations, namely:

    The formula

    is called Newtons Binomial formula ( ).

    The coefficients 0,

    1, 2, ,

    from Newtons 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 2.

    VI b. The sum of the binomial coefficients that are on even

    places equals the sum of the binomial coefficients that are on uneven

    places.

    VII. If is an even number (i.e. = 2), then the binomial

    coefficient of the middle term in the development (namely ) is the

    biggest. If is uneven (i.e. = 2 + 1 ), then the binomial

  • Algebraic problems and exercises for high school

    105

    coefficients of the two middle terms are equals (namely =

    +1)

    and are the biggest.

    VIII. Term , namely the ( + 1) term in equality (7),

    is called term of rank + 1 (general term of development) and it is

    denoted by +1. So,

    IX. The coefficient of the term of rank + 1 in the

    development of Newtons binomial is equal to the product of the

    coefficient term of rank multiplied with the exponent of in this

    term and divided by , namely

    3.2. Solved problems 1. In how many ways can four books be arranged on a shelf?

    Solution. As order is important and because all the elements of

    the given set participate, we are dealing with permutations.

    So

    Answer: 24.

    2. A passenger train has ten wagons. In how many ways can the

    wagons be arranged to form the train?

    Solution. As in the first problem, we have permutations of 10

    elements of a set that has 10 elements. Then the number of ways the

    train can be arranged is

    Answer: 3 628 800.

  • Ion Goian Raisa Grigor Vasile Marin Florentin Smarandache

    106

    3. In how many ways can 7 students be placed in 7 desks so that

    all the desks are occupied?

    Solution.

    Answer: 5 040.

    4. How many phone numbers of six digits can be dialed:

    1) the digit participates in the phone number only once;

    2) the digit participates more than once?

    (The telephone number can also start with 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 teams

    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, = 5,

    = 2.

  • Algebraic problems and exercises for high school

    107

    From digits 0 and 1 we can form 25 numbers of five digits.

    But we have to take into consideration that the number cannot

    begin with the digit zero. So, from the number 25 we have to begin

    with zero. Numbers of this type are 24 . 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