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Teaching a compulsory mathematics program to unselected students in this

and many other countries is first of all a matter of current practice and also a

matter of broader perspective. Currently a teacher's abilities are rated by

the grades earned by his students in the official examinations and competitions.

We must in all honesty recognize, however, that only in very few cases

there is any strong relationship between the training for these competitions

and that necessary for future mathematical activity.

Our discussion will be focused on standard teaching procedures that, to

a greater or lesser degree, contain the elements of modern mathematics.

Experiments in the teaching of this new approach have exerted some influence

on the traditional program, but the effects have been minimal. Although

many methods have been tested, it seems that the solving of problems, both

during class periods and especially in home assignments, remains the most

effective basic activity.

It is an incontrovertible fact that most textbooks are written for the

students. However, we are kidding ourselves if we think that, regardless of

whether the mathematics is traditional or modern, the students look at

anything but the problems. Lessons are taught in class, and students do not

feel the necessity to read the book again. Thus the book is useful only for the

problems at the end of the chapter. This seems to us to be normal because

most students have not learned how to read 'mathematical prose'. To teach

students to read a mathematical text is an important but difficult charge.

Even the International Commission for Mathematical Teaching has dealt

with the matter of problem solving. In 1966 Mme. A. Z. Krygowska, as a

representative of that organization, presented a report before the International

Congress of Mathematicians in Moscow entitled, 'The Development

of the Pupil's Mathematical Activity; the Role of Problems in this Development'.

In those situations where the teaching of mathematics is based exclusively

on problem solving, the selection of the problems themselves is of the utmost

importance. Every generation has had its own collection of problems. Many

mathematicians had their first contact with mathematics in Exercises de

Gdometrie (F.G.M.). The supreme aim of almost all teachers in this field

has been to equip their students to solve the currently fashionable selection.

This is often justified by saying that one proceeds in this way to accustom

Educational Studies in Mathematics 2 (1970) 430-437. Copyright © 1970 by D. Reidel

Publishing Company, Dordrecht-Holland. All Rights Reserved.


the pupils to the examination problems. But as a matter of fact, it is not the

normal pattern for textbook problem collections to be based on the expected

examinations. Quite the contrary, examinations usually contain problems

in the spirit of the standard textbooks. It has thus come about that instead

of subordinating the collections of problems to mathematics teaching,

teaching has become subordinated to the collections of problems. This

points up the necessity for collections of problems corresponding to the

actual development of mathematics. Those problems selected must have

the pedagogical objective of conducting the students, in small steps, to an

understanding of the essential ideas. They should proceed from the simple to

the complex, from the concrete to the abstract. We shall enumerate some

essential points for those who might want to construct such problems either

for a traditional mathematics program or a modern one. Problems should

refer to essential aspects, definitions, and fundamental properties in mathematics

and should help the students recognize all of these in a wide variety

of situations. Furthermore, we are very much opposed to problems, unfortunately

very common, that require unimportant secondary information.

We are also convinced that every idea must be illustrated by examples and

counterexamples. A student who can define a convergent sequence but who

does not know what a divergent one is, and is therefore unable to give an

example, does not really understand convergence. Problems stating 'Show

that proposition ... is right' must be accompanied by those stating 'Show

that proposition ... is wrong' or those asking 'Is proposition ... right or

wrong?' Many propositions that are part of the theory may be given as

problems, the expectation being that the results must be remembered. This

applies to those propositions that are easier than problems of intermediate

difficulty, In this way students would learn many important results through

independent activity. As is well known, such effort produces better results

than comparable activity in the classroom.

Even rather difficult propositions, presented as model demonstrations in

class, can be assigned as problems. This enables the teacher to spend more

time checking the knowledge and discussing the shades and subtleties of the

theory acquired at home.

The majority of the problems, however, must be of medium level because,

if they are too easy, they do not stimulate interest, and if too difficult, they

demoralize. Unfortunately such problems between the simple and familiar

and the difficult and unfamiliar are hard to find.

We shall give an example from the teaching of some elements of set theory.

Many teachers regard this subject as defining union and intersection and

then fixing these ideas in the students' minds with banal examples, perhaps

with some schemes. After such preliminary instruction they feel they can


proceed to the demonstration of the distributivity of intersection over union.

We confess to having had the same pedagogical naivet6 ourselves, but the

students' lack of understanding indicated we had left out an important

stage; intermediate exercises are necessary to adequately familiarize them

with the concepts. The problems should be so contrived as to arouse their

curiosity and give them at least the illusion of creative effort.

This lack of intermediate-level problems is especially felt in modern

mathematics texts where most of the problems are either quite commonplace

(sometimes solvable by drawings) or too difficult and requiring too high a

level of abstraction. The language in which problems are written must be

rich in terms of high frequency in mathematics such as terms of logic and

of set theory. Moreover one must create an abundance of problems with the

purpose of accustoming the students to the terms of logic (exists, anything,

or, and) and of the theory of sets. The opinion that the use of such problems

is a matter of common sense and should be obvious is contradicted in

numerous examples of school practice.

Finite models should be used as much as possible. Ideas are thus more

accessible, more easily checked, and more convincing. More abstract

problems, whose solutions are based on difficult demonstrations, should be

preceded by familiar special cases, preferably finite, where results can be


We have been using arguments which relate particularly to the teaching of

modern mathematics in which structures play an important role, finite sets

being a rich source of suggestive and accessible examples. The real number

that reigns over elementary mathematics does not quite become real. Many

ideas presented in school are learned only superficially because they are

justified and illustrated using the properties of sets of real numbers. We

shall stress some of the aspects that lead to difficulties in understanding real


(a) The features of the real number system are derived from the properties

of three structures: algebraic structure of commutative fields, topological

structure, and structure of ordered sets.

(b) The set of real numbers is infinite and of the power of the continuum

- a concept that needs an advanced stage of abstraction to be understood.

Specialized symbols do not exist for every number; the validity of some

symbols is based directly on an existence theorem.

For example, the symbol for x/2 appears as an affirmation that the

equation x2= 2 has a unique solution, but this number is probably not very

real in the student's mind. We are in favor of using, as much as possible,

models based on real numbers, and wherever we have applicable and suggestive

models, we prefer them.


It is known that if new mathematics is to be learned, it must be associated,

at least tacitly, with familiar examples. In the same view, it is also well

known that an idea cannot be considered to have been mastered if the student

cannot recognize and apply it to a particular situation. In many textbooks

and problem collections, exercises for firmly establishing theories in the

students' minds are almost entirely lacking.

In particular, in almost all textbooks, the chapter on Integral Calculus

begins in the following manner: "Compute the integral of ...". We asked

the students to find the first three terms of a sequence of Riemann sums (of

their choice) corresponding to the given function and interval. We were

surprised to discover that the students could not solve the problem correctly.

The pedagogical deficiency that we stress here is the lack of exercises of this

type through which the students should have learned the theory.

The conclusion is that in textbooks and collections of problems there

should be numerous examples as preparation for subsequent theoretical

problems as well as exercises associated with the fundamental definitions

and with the essence of the important theorems on which the teacher has


To remedy this situation we have been assembling a collection of problems

wherein these principles have been applied to some of the less traditional

themes (problems which are sadly lacking in current teaching materials). The

first volume, tentatively entitled Problems in Set Theory, is designed specifically

for students. The collection is keyed to our teaching program and,

although it refers to set theory, it contains problems to help the students

understand some of the elements of traditional mathematics as well. Exercises

of intermediate level have been emphasized, although neither the simple nor

the difficult have been neglected.

The problems that follow are taken from this book and are presented with

their solutions to indicate the relationships and the principles they illustrate.

The first examples are problems designed to strengthen ideas of membership,

set inclusion, equality, union, intersection, cartesian product, and subset.

They are problems of intermediate difficulty, presupposing that the students

have already worked on simple problems with direct applications of principles.

(1) Let X and Y be two subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9} and the


(1) Xc~ Y= {4, 6, 9}

(2) Xw {3, 4, 5} = {1, 3, 4, 5, 6, 8, 9},

(3) ru {4, 8} = {2, 3, 4, 5, 6, 7, 8, 9}.


(a) Assume (1) and (2). Prove that:

leX and I~Y,

8eX and 86Y,


(b) Determine the sets Xand Ysatisfying the conditions (1), (2), and (3).


(a) The elements of the set X must be sought among the elements 1, 2, 3,

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

As 1 e X u {3, 4, 5} (condition (2)), it follows that 1 e X and 1 e X u Y

and we conclude (condition (1)) that 1 ¢ Y. In the same way one obtains

8 eXand86 Y.

As 7 ~ X u {3, 4, 5} it follows that 7

(b) We have:


1 e X and 1 ~ Y(point (a)),

2 ~ X (condition (2)) and 2 e Y (condition (3)),

3 e Y(condition (3)) and 3 ~ X(condition (1)),

4 e X and 4 e Y(condition (1)),

5 e Y(condition (3)) and 5 ¢ X(condition (1)),

6 e X and 6 e Y(condition (1)),

7 e Y(condition (3)) and 7 ¢ X(condition (1)),

8 e X and 8 ~ Y (point (a)),

9 e X and 9 e Y(condition (1)),

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

(2) Determine the set E satisfying the conditions:

(1) E c {1,2, 3,4, 5, 6},

(2) E c {1, 3,4, 6,7, 8},

(3) {1, 3,4,6} c E.


E={1, 2, 3, 4, 5, 6} and Ec{1, 3, 4, 6, 7, 8}=~Ec{1, 2, 3, 4, 5, 6}tn

{1, 3, 4, 6, 7, 8}, thus E~{1, 3, 4, 6}. From Ec{1, 3, 4, 6} and {1, 3, 4, 6}

cE (condition (3)), it follows that E= {1, 3, 4, 6}.

(3) Determine the set E such that:

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

(2) {4, 5} =E,

(3) E = {1,2, 3,4, 5}.


{1, 2, 3}~e and {4, 5}=E=~{1, 2, 3}w {4, 5}cS, thus {1, 2, 3, 4, 5}cS

and, as E={1, 2, 3, 4, 5} (condition (3)), it follows that E={1, 2, 3, 4, 5}.


(4) Determine the sets A= {1, 3, x, y} andB= {2, x, z} suehthat{(1, 3) (2, 4)}

A x B (one agrees to write an element only once in braces).


Since, with the above notations each element is written only once in the

set to which it belongs, we have:

x~{1,2,3}, y~{1,3}, z#2, x¢y.

From (1, 3) e d x B results 3 s B, hence one of the elements x, z is 3, but

as x # 3, we obtain z = 3.

As (2, 4) s A x B we have 2 s A and 4 s B; since in B only x has remained

undetermined, it follows that x=4. Since in A only y is unknown and

2 e A, it follows that y = 2. We have obtained x---4, y = 2, z = 3; and therefore

A = {1, 2, 3, 4}, B= {2, 3, 4}.

The following exercises, also for sets, combine the ideas of sets with terms

of logic. (They should be presented after the basic definitions have been

assimilated.) These exercises also are of medium difficulty, keyed to the

students' ability to understand.

(5) Determine the sets A and B which satisfy the conditions;

(1) AuB={1,2,3,4,5,6},

(2) Ac~B=0,

(3) Whatever x ~ A is, there exists y ~ B

so that x - y = 1. Whatever y ~ B is, there exists x ~ A

so thaty - x = I.


1 ¢ A for, in the contrary case 1 e A, there exists y ~ B (therefore y ~ A~JB)

such that 1-y=l. But this equation has the solution y=0 and 0¢B

(because 0 ~ A w B). From 1 ¢ A and from condition (1) there results 1 ~ B.

2 ¢ B for, in the contrary case 2 ~ B, and there exists x e A with 2-x= 1,

thus 1 E A, which was proved wrong. Hence 2 ~ A.

In the same way one continues to show that 3 ¢ A, and so on.

We obtain A = {2, 4, 6} and B= {1, 3, 5}.

(6) Let the set E={x,y, z, t}. Determine x, y, z, t such that the following

conditions are fulfilled simultaneously.

(1) {x,y, 1} ~ ~(E),

(2) {1, 2, z} ~ ~(E),

(3) {x, y, 3} ~ ~(E),

(4) {y, t, 4} ~ ~(~).


(The elements written within each pair of brackets are supposed distinct.)


From (1) it follows {x, y, 1)cE and hence: 1 e E and x~l, y~ 1, thus

z=l or t=l.

From (2) it follows {1, 2, z}cE and hence 1 ~ E and 2 ~ E and z~ 1 and


Since x# 1, y~ 1, z~ 1, it results that t= 1. But z e E and z~2 and t= l,

implies that x = 2 or y = 2.

From (3) it follows that {x, y, 3}cE and therefore 3 ~ E. As x~3, y~3,

and t ~ 3 (t = 1) it follows that z = 3.

From (4) it follows that {y, t, 4}cE, hence 4~E. As z~4(z=2),y¢4,

t ~ 4, it follows that x = 4. As 2 ~ E (condition (2)) and x ~ 2 (x = 4), z ~ 2 (z = 3),

t~ 2(t = 1), we have y = 2.

So that x=4, y= 2, z=3, t=l.

(7) Prove that the following propositions are false, showing that the negation of

each is a true proposition.

P: any composite natural number n admits a divisor between

n/4 and n/3.

P2: for any composite natural number n, there exists a prime

number between n and ~n.

P3: any composite natural number n has at least one divisor

different from 1 and less than ~/4n.


Non P1 : there exists a composite natural number n, so that none of its

divisors is between n/4 and n/3. Non P1 is true because, for instance, n = 36

does not admit any divisor comprised between -~ = 9 and-~ = 12.

Non P2: there exists a composite natural number n such that no prime

number is between n and 4n/3. Non Pz is true because, if n = 8, between 8

and 4.8/3 = 10~3 there does not exist any prime number.

Non P3: there exists a composite number n which admits no divisor

different from 1 and inferior to ff4n.

Non P3 is true because n = 49 does not admit any divisor different from 1

and less than ~/4.49 = ~/196 (a number less than 6).

(8) Find the sets A, B, C, fulfilling the conditions:

(1) 1 ~A,

(2) {2, 4}rnB = 0,

(3) 3~Ac~BnC,


(4) 4sAcaC,

(5) AcaB¢C,

(6) BuCCA,

(7) AuBuC={1,2,3,4}.


By writing (6) as a negation of the proposition Bu C=A, we obtain

Bw CCA =~ Not (Bw CcA) =~ There exists x ~ Bu C and x ~ A =~ There

exists x so that: (x e B or x ~ C) and x ¢ A~ There exists (x e B and x ¢ A)

or (x ~ C and x ¢ A). We shall show that only the element 2 satisfies condition

(6). Indeed, this element cannot be 1, 3, or 4 because 1 e A (condition

(1)), 3 ~ A (condition (3)), 4 e A (condition (4)). As 2 • B (condition (2)),

2~ Canal2 ¢ A.

From condition (5) it follows that there exists x such that x s A and x e B

and x ¢ C.

This element cannot be anything else but 1. Indeed, 2, 3, or 4 cannot be

x, since from (6) it follows 2 s A, 3 e C (condition (3)) and 4 ¢ B (condition

(2)). Therefore 1 e A and 1 ~ B and 1 ¢ C.

Thus: lea and leB and I¢C,

2¢A and 2¢B and 2eC,

3cA and 3~B and 3eC,

4cA and 4¢B and 4eC.

Hence A = {1, 3, 4} B={1,3}, C={2,3,4}.

The sets A, B, C evidently verify the conditions (1), (2), (3), (4), (5), (6), (7).

(9) Determine the sets A and B fulfilling the conditions:

(1) AuB= {1, 2, 3, 4, 5, 6}.

(2) Neither A nor B is empty.

(3) The proposition 'There exists x ~ A and y ~ B such that

{x} w {y} ¢ (.4 ca B)' is wrong.


If the proposition of assumption (2) is wrong, its negation is true.

'Whatever x ~ A and y ~ B might be, we have {x} w {y} = (.4 ca B).'

Since Bis not empty, thereis anyo ~ B. Letx ~ A. By(3) {x} u {Yo} e A riB,

hence A = A ca B. Likewise B c A ca B. It follows that A = B= (1, 2, 3, 4, 5, 6}.


[ شنبه 28 آذر1388 ] [ 7:53 بعد از ظهر ] [ محمد امین ناصری ] [ ]
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