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THE IMPORTANCE OF APPROPRIATE PROBLEMS IN THE TEACHING OF MATHEMATICS 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) 430437. Copyright © 1970 by D. Reidel Publishing Company, DordrechtHolland. All Rights Reserved. THE IMPORTANCE OF APPROPRIATE PROBLEMS 431 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 432 E. GEORGESCUBUZ/kU ET AL. 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 intermediatelevel 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 verified. 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 numbers: (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. THE IMPORTANCE OF APPROPRIATE PROBLEMS 433 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 lectured. 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 conditions: (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}. 434 E. GEORGESCUBUZA, U ET AL. (a) Assume (1) and (2). Prove that: leX and I~Y, 8eX and 86Y, 7¢X. (b) Determine the sets Xand Ysatisfying the conditions (1), (2), and (3). Answer: (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: ~x. 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. Answer: 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}. Answer: {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}. THE IMPORTANCE OF APPROPRIATE PROBLEMS 435 (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). Answer: 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 x4, 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. Answer: 1 ¢ A for, in the contrary case 1 e A, there exists y ~ B (therefore y ~ A~JB) such that 1y=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 2x= 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} ~ ~(~). 436 n. GEORGESCUBUZAU ET AL. (The elements written within each pair of brackets are supposed distinct.) Answer: 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 z~2. 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. Answer: 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, THE IMPORTANCE OF APPROPRIATE PROBLEMS 437 (4) 4sAcaC, (5) AcaB¢C, (6) BuCCA, (7) AuBuC={1,2,3,4}. Answer: 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. Answer: 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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