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An investigation of the heuristics used by selected grade eleven academic algebra students in the solution… Dinsmore, Laurie Annette 1971

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AN INVESTIGATION OF THE HEURISTICS USED BY SELECTED GRADE ELEVEN ACADEMIC ALGEBRA STUDENTS IN THE SOLUTION OF MATHEMATICAL PROBLEMS by LAURIE ANNETTE DINSMORE B.Sc., The U n i v e r s i t y of B r i t i s h Columbia, 1965 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF ARTS i n the F a c u l t y of Education Department of Mathematics Education We accept t h i s t h e s i s as conforming to the re q u i r e d standard THE UNIVERSITY OF BRITISH COLUMBIA December, 1971 In presenting t h i s t h e s i s i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r reference and study. I f u r t h e r agree that permission f o r extensive copying of t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head of my Department or by h i s r e p r e s e n t a t i v e s . I t i s understood that copying or p u b l i c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my w r i t t e n permission. Department of 0^ajcX^u>^t>CLc^ c ^ l c * Ot^P^n The U n i v e r s i t y of B r i t i s h Columbia Vancouver 8, Canada Date ABSTRACT This normative survey investigated the question, "What general h e u r i s t i c s are used by selected grade eleven academic algebra students i n the so l u t i o n of mathematical problems?" The investigator was interested i n determining i f students who had either A or B mathematics eleven grades used any h e u r i s t i c s . Forty-two students, who were enrolled i n nine schools, were interviewed. Each student was given two mathematical problems to solve. These problems could be solved using two of nine general h e u r i s t i c s namely, cases, deduction, inverse deduction, i n v a r i a t l o n , analogy, symmetry, preservation of rul e s , v a r i a t i o n , and extension. The researcher requested the students to think aloud. The student was encouraged to attempt the problems any way he chose. They were asked to be more concerned with revealing as much of t h e i r thought processes as possible, as with the accuracy of t h e i r solution. A l l the interviews were taped. The investigator found evidence that eight of the nine h e u r i s t i c s were used. The h e u r i s t i c s were cases, deduction, inverse deduction, i n v a r i a t l o n , analogy, preservation of rules, v a r i a t i o n , and extension. T h i r t y -eight of the forty-two students interviewed showed evidence of using one or more of the h e u r i s t i c s . Eighteen of the students used cases, seven used deduction, three used i n v a r i a t i o n , two used inverse deduction, seven used analogy, two used preservation of rules, three used v a r i a t i o n , and seven used extension. The investigator also found evidence that a h e u r i s t i c which was not mentioned previously was used by eleven of the students. For the purpose of t h i s i n v e s t i g a t i o n the h e u r i s t i c was ca l l e d "successive v a r i a t i o n . " •tWhen the h e u r i s t i c of successive v a r i a t i o n i s used a possible •rsolution to the given problem i s chosen at random. I f the answer i s not correct, the student determines what changes must be made. Then the possible so l u t i o n Is varied successively u n t i l the correct answer i s found. The students* command of the h e u r i s t i c s was not developed and therefore they could not use these techniques e f f i c i e n t l y and e f f e c t i v e l y to solve the problems they were given. TABLE OP CONTENTS CHAPTER PAGE I. THE PROBLEM AND DEFINITION OF TERMS 1 The Problem 1 Background of the problem 1 Statement of the problem . . . . . 5 Importance of the study 5 D e f i n i t i o n of Terms . . . . . . 6 H e u r i s t i c 6 Cases 8 Deduction . . . . . 10 Inverse Deduction . . . . . . . . . 12 Invariation 13 Analogy , 17 Symmetry . . . . . . . . . . . . . 19 Preservation of Rules 20 V a r i a t i o n 21 Extension . 23 I I . REVIEW OF THE LITERATURE 25 Studies Having General Relevance to the Problem . . 25 Studies Having S p e c i f i c Relevance to the Problem . , 31 Opinions Having General Relevance to the Problem . . 3k Opinions Having S p e c i f i c Relevance to the Problem . 36 Summary , 36 L i m i t a t i o n of Previous Studies 37 i i i i i CHAPTER P A G E I I I . THE DESIGN AND STATISTICAL TREATMENT 38 The Design 3 8 The Problems Used ^0 Cases kl Deduction ^1 Inverse Deduction k2 Invar i a t ion k3 Analogy 4 3 Symmetry . . kk Preservation of Rules kk V a r i a t i o n k5 Extension 4 5 The Group Used ^ 5 S t a t i s t i c a l Treatment . . k6 Limitations of the Study k6 IV. RESULTS 4 8 I d e n t i f i c a t i o n and Frequency of Student Usage of He u r i s t i c s k8 Cases kQ Deduction 52 Inverse Deduction 5 5 I n v a r i a t i o n 5 5 Analogy 56 Symmetry 58 i v CHAPTER PAGE Preservation of Rules 58 V a r i a t i o n . . . . . . . 6 l Extension 62 The Students' Use of Other H e u r i s t i c s 6k Successive V a r i a t i o n . . . . . . . . . . 65 V. SUMMARY AND CONCLUSIONS 69 Summary 69 Need f o r Further Study 70 Conclusions 70 BIBLIOGRAPHY 72 APPENDIX A. Student Information Form 76 APPENDIX B. Interview Format 77 APPENDIX C. Solutions to the Problems Used i n the Investigation 80 APPENDIX D. Sample Interviews and Analyses . . . . . . . . 95 APPENDIX E. Information on the Students Interviewed . . . . 1^9 APPENDIX F. The Students' Use of H e u r i s t i c s 153 APPENDIX G. Information on the/Schools Sampled 158 LIST OF TABLES TABLE PAGE I. The H e u r i s t i c s Used For Each Problem 68 v CHAPTER I THE PROBLEM AND DEFINITION OF TERMS I. THE PROBLEM Background of the problem. Most students encounter d i f f i c u l t y when solving problems in mathematics. Washburne and Osborne stated that, "training children to solve arithmetic problems is one of the hardest and most discouraging tasks of the teacher."-*-Techniques have been devised to help students learn to solve mathematical problems. Some of these procedures are formal analysis, selecting correct procedures, and solving a wide variety of problems. These adult-devised techniques generally have l i t t l e effect in helping students to become better problem solvers. Allowing the student to choose his own method when solving a problem (student-generated sequences) is sometimes better than a specific adult-devised method,^ Kaplan suggested that the classroom teacher should spend time investigating student-generated sequences.3 -kjarleton W. Washburne and Raymond Osborne, "Solving Arithmetic Problems," The Elementary School Journal, XXVII (November, 1926), 219. 2Paul R. Hanna, "Methods of Arithmetic Problem Solving, The Mathematics Teacher, XXIII (November, 1930), kkZ-k^O. -^ Jerome D. Kaplan, "An Example of Student-Generated Sequences in Mathematics Instruction," The Mathematics  Teacher. LVII (May, 196*0, 298-302. 1 2 Grossnickle stated that: Before giving s p e c i f i c helps l n problem solving, i t i s necessary to consider d i f f e r e n t l e v e l s of maturity which characterize problem by employing d i f f e r e n t patterns of mathematical thinking.4 Grossnickle further stated that, " i t should not be i n f e r r e d that a p u p i l should always follow one p a r t i c u l a r pattern i n sol v i n g a problem."-' In h i s doctoral d i s s e r t a t i o n , K i l p a t r i c k s a i d thatj To make our knowledge of problem-solving behavior relevant to education, we must eventually study how students solve problems of the sort they meet i n the classroom. I f at present there are too many uncontrol-l a b l e or even unknown sources of v a r i a t i o n f o r c a r e f u l experimental studies, we should at le a s t attempt analyses of behavior. Such analyses . . . a r e necessary f o r d i r e c t i n g future experimentation.° The investigator hypothesizes that the adult-devised methods r e f e r r e d to e a r l i e r are not general enough to be used to attack a l l types of mathematics problems. According to Luchlns' research, d r i l l i n one s p e c i f i c method of problem solving may have a tendency to b l i n d the student to other approaches to solv i n g problems. The p u p i l may develop a set and be unable to solve the problems that do not f i t the ^Foster E. Grossnickle, "Verbal Problem Solving," The Arithmetic Teacher. XI (January, 1964), 12. 5 I b i d . , p. 14. ^Jeremy K i l p a t r i c k , "Analyzing the So l u t i o n of Word Problems i n Mathematics: An Exploratory Study," (unpublished doctoral d i s s e r t a t i o n , Stanford University, Stanford, 196?), p. 2. 3 type encountered previously. 7 Hence i t would appear useful to teach methods of problem sol v i n g that are so general that they can be used to attempt to solve many d i f f e r e n t types of problems. The use of c e r t a i n general h e u r i s t i c s appears to produce t h i s wide a p p l i c a b i l i t y . ' Several writers have discussed h e u r i s t i c s and a few studies have been done on some aspect of the topic. Polya urged teachers to make use of s p e c i f i c h e u r i s t i c questions such as, "What i s the unknown?" and "What are the data?"^ Ashton^ t r i e d to apply Polya 1 s suggestions while Kilpatrick-*- 0 attempted to discover i f students use h e u r i s t i c questions when they solve mathematical problems. W i l l s , Larson, ^Abraham Luchins, "Mechanization i n Problem Solving," Psychological Monographs. LIV (whole no. 2*4-8, 19*1-2) . p George Polya, How to Solve I t (second e d i t i o n ; New York: Doubleday and Company, Inc., 1 9 5 7 ) , P. 7 . ^ S i s t e r Madeleine Rose Ashton, "Heuristic Methods i n Problem Solving i n Ninth Grade Algebra," (unpublished doctoral d i s s e r t a t i o n , Stanford University, Stanford, 1 9 6 2 ) . 1 0Jeremy K i l p a t r i c k , op_. c i t . ^Herbert W i l l s I I I , "Transfer of Problem Solving A b i l i t y Gained Through Learning by Discovery," D i s s e r t a t i o n  Abstracts. XXVIII ( 1 9 6 7 ) , 1319-A. x c C h a r l e s McLoud Larsen, "The H e u r i s t i c Standpoint i n the teaching of Elementary Calculus," D i s s e r t a t i o n  Abstracts. XXI ( 1 9 6 1 ) , 2 6 3 2 - 2 6 3 3 . '• 4 Wilson, 13 Washburne and Osborne,14 and Schaaf 1$ attempted to teach various heuristics to students by having the teacher use these strategies. Henderson,x^ Jones-*-? and Bruner 1^ urged teachers to use stratagems in the hope that students would learn how to use these methods. None of these studies investigated the question , "Do students use any general heuristics, such as analogy or inverse deduction, when they solve problems using any method they choose?" The purpose of this investigation is to see i f students use any of the nine heuristics proposed by MacPherson,x9 when they attempt to solve x 3 j a m e s William Wilson, "Generality of Heuristics as an Instructional Variable," Dissertation Abstracts, XXVIII (January, 1968), 2575-A. I^Carleton W. Washburne and Raymond Osborne, op_. c i t . , p. 219-226, 296-304. 1 5 0 scar Schaaf, "Student Discovery of Algebraic Principles as a Means of Developing Ability to Generalize," The Mathematics Teacher, XLVIII (May, 1 9 5 5 ) , 324-327. x^Kenneth B. Henderson, "Strategies for Teaching by the Discovery Method," Updating Mathematics (high school teachers edition; New London, Connecticut! Croft Educational Services, 1 9 6 4 ) , p. 57-64. - ^ P h i l l i p S. Jones, "Discovery Teaching From Socrates to Modernity," The Mathematics Teacher. LXIII (October, 1 9 7 0 ) , 501-508. x®Jerome S. Bruner, The Process of Education (New York* Vintage Books, i 9 6 0 ) . x 9 E r i c D. MacPherson, (unpublished paper on heuristics, The University of B r i t i s h Columbia, Vancouver, 1 9 6 9 ) . (mimeographed). 5 mathematical problems that can be solved by using at l e a s t one of these h e u r i s t i c s . The type of i n v e s t i g a t i o n i s a normative survey. Statement of the problem. What general h e u r i s t i c s are used by selected grade eleven academic algebra students i n the s o l u t i o n of mathematical problems? Importance of the study. MacPherson's2° l i s t of general h e u r i s t i c s i s an adult model of h e u r i s t i c s that can be used i n mathematical problem solving. Research has shown, however, that students who are taught adult-devised methods seldom become better problem solvers. 2-*- I f t h i s normative survey finds evidence that students use c e r t a i n of the h e u r i s t i c s i n attempting to solve mathematical problems, then the Investigator proposes that an e f f o r t be made to teach students these general h e u r i s t i c procedures i n the hope that they w i l l become better problem solvers. Since there are always new problems to be solved, i t i s advantageous f o r the students to know several general techniques that can be used to attempt to f i n d solutions to novel problems. E r i c D. MacPherson, l o c . c i t . 21 Carleton W. Washburne and Raymond Osborne, op_. c i t . , p. 219-226; Latney Conrad Welker, "A Study of I n t e r r e l a t i o n -ships l n Arithmetical Problem Solving," (unpublished doctoral d i s s e r t a t i o n , The University of Southern M i s s i s s i p p i , Hattiesburg, 1 9 6 2 ) . 6 The investigator w i l l also observe i f any other general h e u r i s t i c s , which could be useful i n solving a wide v a r i e t y of problems, are used by the students. II . DEFINITION OF TERMS He u r i s t i c . The term h e u r i s t i c has been variously defined i n the l i t e r a t u r e . Bruner defined a h e u r i s t i c as an approach one follows to solve problems.22 Henderson r e f e r r e d to a h e u r i s t i c as a stratagem, which he defined as a plan of action.^3 Polya c a l l e d h e u r i s t i c , "The study of means and methods of problem s o l v i n g . A s h t o n ' s d e f i n i t i o n , based on Polya's, Is: H e u r i s t i c methods are based on the mental operations involved i n the a r t of invention. They do not consist of a series of formal steps, since these might well hinder rather than a i d thought. They are intended rather to lead the problem solver to recognize and then consciously apply himself to c e r t a i n modes of thinking about problems, 2"5 Newell et a l . used the term h e u r i s t i c "to denote any p r i n c i p l e or device that contributes to the reduction i n ?6 the average search to solution." MacPherson defined 22 Jerome S. Bruner, op_. c i t . , p. 27. ^Kenneth B. Henderson, op_. c i t . , p. 57. oh, George Polya, Mathematical Discovery (New Yorki John Wiley and Sons, Inc., 1 9 6 2 ) , I, p. v i . ^ S i s t e r Madeleine Rose Ashton, op_. c i t . , p. 1. ?6 c A l l e n Newell, J.C. Shaw, and Herbert A. Simon, "The Process of Creative Thinking," Contemporary Approaches £o Creative Thinking (New York: Atherton Press, 1 9 6 4 ) , p. 78. 7 h e u r i s t i c as, "an a l o g i c a l algorithm which i s used f o r the purpose of discovering some order of mathematical generalization i n a novel situation," 2'' 7 B a s i c a l l y a l l these d e f i n i t i o n s described a h e u r i s t i c procedure as a general or s p e c i f i c plan of ac t i o n which i s used to attempt to solve a problem. In t h i s paper h e u r i s t i c i s defined as a general plan of attack or process of reasoning that can be employed i n the search f o r a soluti o n to a p a r t i c u l a r problem, PR The nine general h e u r i s t i c s proposed by MacPherson, ° namely, cases, deduction, inverse deduction, i n v a r i a t i o n , analogy, symmetry, preservation of rul e s , v a r i a t i o n , and extension, w i l l subsequently be defined. A problem and i t s s o l u t i o n also w i l l be given to i l l u s t r a t e the use of each h e u r i s t i c . Some of these procedures have been discussed i n the l i t e r a t u r e as techniques that the teacher can use i n discovery teaching. Henderson stated that the stratagems he l i s t s , "can be used to plan the teaching of a mathematical concept or p r i n c i p l e so that the student discovers the knowledge rather than l i s t e n s while the teacher t e l l s i t to him." 2^ None of the h e u r i s t i c s 2 ? E r i c D. MacPherson, op_. c i t . 2 8 I b i d . ^Kenneth B. Henderson, op_. c i t . , p. 6 l . 8 appeared to have been f u l l y defined in the literature from the point of view of the use of the heuristic in the solution of a problem. Heuristic I: Gases. Leask defined simple enumeration as: The presentation of many instances of the generali-zation to be discovered. The students form hypotheses based on the example and test those to determine which is correct. One counter-example i s sufficient to warrant rejection of a hypothesis. 30 This definition applies to the heuristic as i t could be used by the teacher in presenting his lesson. The heuristic, cases, w i l l now be defined as i t applies to the student's use of the technique. The heuristic of cases is a method for finding the solution to a problem by considering a l l or some of the possible solutions, and testing them until the correct solution i s discovered. Either a random or a systematic sampling of some of the solutions can be considered. If a systematic examination of cases is tried, i t can be a c r i t i c a l search which involves looking for the exception to the rule, or a sequential search through the possible solutions. The ^I s a b e l Campbell Leask, "The Effectiveness of Simple Enumeration as a Strategy for Discovery," (unpublished Master's thesis, The University of British Columbia, Vancouver, 1968), p. 8. following problem illustrates how the heuristic of cases could be used in i t s solution! How many tangents common to both circles can two equal circles in the same plane have?-*1 Consider two disjoint circles moving together to become a single c i r c l e . The following three cases result. Case 1 ^Charles T. Salkind, The Contest Problem Book (New York: Random House, Inc., 1961), p. lb. 10 Case 3 2 tangents I H e u r i s t i c I I : Deduction. The h e u r i s t i c of deduction can be sub-classed into the two h e u r i s t i c s , d i r e c t deduction and hypothetical deduction. Direct deduction i s the procedure of discovering a consequence from a set of premises, by l o g i c a l reasoning. In hypothetical deduction i t i s assumed that a sub-problem of the o r i g i n a l problem can be solved. Then a c t i n g on t h i s assumption, the consequences that follow are determined. I f these consequences are valuable i n that they lead to a so l u t i o n of the o r i g i n a l problem, then an attempt i s made to solve the sub-problem, as a f i r s t step i n f i n d i n g the so l u t i o n to the o r i g i n a l problem. 'Deduction could be used i n the s o l u t i o n of the following problem: Triangle ABC i s isosceles with side AC equal to 12 inches and side AB equal to 6 inches. What i s the length of the radius CD of the c i r c l e drawn through the v e r t i c e s of t r i a n g l e ABC? 3 3 ^ I b l d . , p. 93. 3 3 i b l d ., p. 28. \ 11 If the length of CE was known, the right triangle AED could be solved for AD. Since AD = CD, f i r s t consider the sub-problem; find the length of CE. Consider triangle CEB o 1. angle CEB = 90 2. CB = AC = 12 inches 3. AE = EB = 3 inches 4. (CE) 2 = (CB)2 - (EB)2 CE = 7(CB)2 - (EB)2 = 7(12)2 - (3) 2 = jiM - 9 1. by construction, angle AEB = angle AEC + angle CEB o angle AEB = 180 angle AEC = 90° 2. triangle ABC is isosceles AC = 12 inches 3. AB = 6 inches triangle ABC is isosceles Pythagorean theorem 12 Consider t r i a n g l e ADE 5. (AD) 2= ( A E ) 2 + (ED) 2 ' 5. Pythagorean theorem AD =7(3)2+ (JUJ - CD)"2 AD = CD therefore =79 + 135 - 2/135" CD + (CD) 2 (AD) 2 = '(CD)2 =yi44 - 2yi35~AD + (AD) 2 2VT35AD = 144 , 3Z+ AD = 144/2/133" = 72/V135 H e u r i s t i c I I I : Inverse Deduction. Polya r e f e r r e d to t h i s h e u r i s t i c as working backwards. He stated, "We concentrate upon the desired end, we v i s u a l i z e the f i n a l p o s i t i o n i n which we would l i k e to be. Prom what foregoing p o s i t i o n could we get there?"35 Newell e_t. a l . discussed the strategy of working backwards and stated i t s usefulness i n a problem s i t u a t i o n that has "many possible s t a r t i n g points versus a single end." 3^ Inverse deduction i s defined i n t h i s paper as a method of working from a known or an assumed conclusion backwards to the o r i g i n a l statement of the problem. The following problem i l l u s t r a t e s the use of the h e u r i s t i c of inverse deduction: How could you bring up from a r i v e r exactly 6 quarts of water using only two containers, a 3 ^ I b i d . . p. 103. ^George Polya, How to Solve I t (second e d i t i o n ; New York: Doubleday and Company, Inc., 1957), p. 230. 3 % e w e l l , op_. c i t . , p. 80. ! 3 k quart p a i l and a 9 quart p a i l . 3 7 The aim i s to end with 6 quarts of water i n the 9 quart p a i l . I f there were 1 quart of water i n the k quart p a i l , i t would be possible to f i l l up the 9 quart p a i l and pour out 3 quarts of water into the k quart p a i l and then be l e f t with 6 quarts of water l n the 9 quart p a l l . Suppose i t i s possible to get 1 quart of water i n the k quart p a i l . How could t h i s be done? P i l l the 9 quart p a i l and pour from i t k quarts into the smaller p a i l , dump i t out, and f i l l i t again. Then there w i l l be 1 quart of water l e f t i n the 9 quart p a i l . The 1 quart can be poured into the k quart p a i l and the sol u t i o n i s accomplished.3 8 H e u r i s t i c IV: Invariatlon . When t h i s h e u r i s t i c i s employed i n attempting to solve a problem, either some va r i a b l e In the problem i s f i x e d or else some of the data i s ignored?. Then an attempt i s made to see i f the problem can be solved under these new conditions. The s o l u t i o n of the s i m p l i f i e d problem may provide insight into the sol u t i o n of the o r i g i n a l problem. The following problem w i l l i l l u s t r a t e tine heuristic« Construct a t r i a n g l e ABC given the length of the -^George Polya, op_. c i t . , p. 2 2 6 . 3 8 I b i d . , p. 2 2 7 - 2 3 0 . Ik base BC, the measure of angle A, and the length of the altitude h.39 B C / \ h In order to solve the problem, consider the two conditions separately. F i r s t ignore the measure of the angle, and consider constructing a triangle given the length of the base BC and the length of the altitude h. The vertex A of the triangle would l i e along a line x which is parallel to side BC and at a distance h from i t . B C Second consider constructing a triangle using the length of the base BC and the measure of angle A. Several triangles can be constructed with base BC and angle A as the angle opposite this base. Since the sum of the measures of the o angles of a triangle i s 180 , extend BC through C to D. 3 9 I b i d . , p. 80. Construct angle XCD = angle A. Divide angle BCX into any two angles BCY and YCX. Construct angle CBZ = angle YCX. Rays BZ and CY w i l l i n t e r s e c t at point A. Now t r i a n g l e ABC has base BC and angle A as the angle opposite BC. .A B C D By d i v i d i n g angle BCX into d i f f e r e n t parts, the following six t r i a n g l e s can be drawn. vA B C The locus of the vertex A of the t r i a n g l e ABC i s a c i r c l e with chord BC. The centre of the c i r c l e can be constructed using the three points B, C, and A and. the f a c t that the perpendicular b i s e c t o r of a chord of a c i r c l e passes through the centre of the c i r c l e . The centre of the c i r c l e i s point 0. Now the i n t e r s e c t i o n of the c i r c l e w ith the l i n e x shows the two s o l u t i o n s to the problem.^ The f o l l o w i n g diagrams i l l u s t r a t e these two answers to the problem! A I b i d . , p. 81-82. 17 H e u r i s t i c Vt Analogy. Henderson described the stratagem of analogy as a teaching technique which i s useful i n discovery learning. He planned the lesson so that the new problem s i t u a t i o n was very s i m i l a r to a problem that the student already knew how to s o l v e , ^ Jones i l l u s t r a t e d analogy as the perception of a new problem as a v a r i a t i o n of another problem that one already could solve. He implied that the problem previously solved acts as a mental model i n suggesting a s o l u t i o n f o r the newer problem^ 2 Polya on the other hand, attempted a d e f i n i t i o n of the h e u r i s t i c of analogy as "a sort of s i m i l a r i t y . " 3^ Analogy i s also defined as a sort of isomorphism between structures. MacPherson stated that t h i s h e u r i s t i c enables one to know what questions to ask i n a new s i t u a t i o n because there i s a s i m i l a r i t y between t h i s new question and a problem that was previously s t u d i e d . ^ Analogy w i l l be defined i n t h i s study as the perception of a s i m i l a r i t y between the problem to be solved and one that has already been solved. The same methods of attack may be t r i e d on the new problem as were used previously. The 4lKenneth B. Henderson, op_. c i t . , p. 57-58. 4 2 p h i l l i p S. Jones, op_. c i t . , p. 507-508. 43George Polya, op_. c i t . . p. 37. LL Statement by E r i c D. MacPherson, personal interview. 18 following problem i l l u s t r a t e s how the h e u r i s t i c of analogy-can be used i n the solu t i o n of the problem: Find the length of the diagonal of the rectangular box (the length of the l i n e j o i n i n g points A and Z ) . ^ A 1 inch Y 2 inches Z Consider the two-dimensional case of a rectangle. To f i n d the length of the diagonal (the length of the l i n e j o i n i n g points A and C) use the Pythagorean theorem. 1 inch 2 inches AC =J(AB) 2 + (CD) 2 = Vl + k = f~5~ AC = inches In the three-dimensional case of a rectangular box, there are two t r i a n g l e s , t r i a n g l e XYZ and t r i a n g l e AZX, • Consider t r i a n g l e XYZ, Since angle Y = 9 0 ° , XZ =7(XY)2 + (YZ) 2 eorge Polya, op. c i t . , p. 7. XZ = V l + 4 = 7 5 7 Consider t r i a n g l e AZX. Since angle X = 90°, AZ =7(XZ)2 + (AX) 2. AZ = V5 + 1 =V67 Therefore AZ = 76~ inches. H e u r i s t i c VI; Symmetry. The use of the h e u r i s t i c of symmetry involves an attempt to find, some inherent symmetry i n the problem i t s e l f . The perception of symmetry may lead to an add i t i o n of data that w i l l now help to solve the new problem. It i s hoped that by omitting the added information, a so l u t i o n to the o r i g i n a l problem w i l l c l e a r l y be seen. Polya said that a solut i o n to the problem may depend on n o t i c i n g that some parts of the data are interchangeable. He said that intexchangeable parts should be treated i n the same Lf, manner. The following problem i l l u s t r a t e s the use of symmetry: Find the shortest path between points X and Y that passes throught point X, touches l i n e AB, and then passes through point Y. <r~ > A 'B * Y « X Suppose there i s a point P located the same distance from l i n e AB as point Y, but In the other h a l f plane. The shortest distance between points X and P would be the Lf George Polya, op. c i t . , p. 199. 20 s t r a i g h t l i n e j o i n i n g them which passes through l i n e AB at point i . Triangle YiP i s isosceles with Y i = iP, therefore the s o l u t i o n to the o r i g i n a l problem i s the l i n e j o i n i n g points X, i , and Y. X H e u r i s t i c VII: Preservation of Rules. The h e u r i s t i c preservation of rules would be used when one system i s extended to create a new system. The use of the h e u r i s t i c would involve asking a question such as, how many of the rules of the previous system apply to the new system? , If the rul e s of the previous system are preserved, then i,t may-be possible to solve the problem i n the new system by an • analogy to the method of so l u t i o n i n the previous systent. In other words the h e u r i s t i c , preservation of rules, provides an i n i t i a l s t a r t i n g point i n attempting to solve the problem. The problem i l l u s t r a t e d below could be used to tes t the use of the h e u r i s t i c of preservation of ru l e s : Consider a pseudo-ternary system with the three numberals -1, 0, and +1. Now replace -1 by -, +1 by + and leave 0 unchanged. 21 Consequently, -, 0, and + are the only permissible digits in pseudo-ternary arithmetic. Given the fact that 8ten = (+0-)p,t,, change 5^ten to i t s equivalent representation in 47 pseudo-ternary arithmetic. ' Suppose that the rules follow for the representation of numbers in pseudo-ternary arithmetic as they would in other base systems. In other words the representation of 8ten should be (+1 x 3 ) + (0 x 3 ) + (-1 x 3°)» which does in fact equal 8 t e n . Therefore following the method used for converting base ten numbers to their equivalents in other systems, 54 18 0 ~6~ 0 -1 + 1 48 And therefore 5 4 T E N = + - 0 0 0 P T T . Heuristic V i l l i Variation. In using variation some fact i n the problem i s changed. The attempt i s made to find out what other facts are now either variant or invariant. This heuristic could extend the knowledge of a particular topic since i t s use involves asking the question, What would happen i f . . . ? Any answers either to the new problems suggested by the use of variation, or to the original problem ^Winifred Asprey, "Pseudo-Ternary Arithmetic," The Mathematics Teacher, LIX (March, 1966), 241. ^ 8 I b i d . . p. 241. 2 2 would "be generated by other h e u r i s t i c s . This h e u r i s t i c provides an i n i t i a l s t a r t i n g point i n the search f o r a solu t i o n to the problem. Polya said that i f the s o l u t i o n to a problem can not be r e a d i l y seen, the data should be varied.^9 The following problem i l l u s t r a t e s how v a r i a t i o n could be used. Investigate the graph of y = a s i n bx. The graph of the equation y = a s i n bx could be investigated as f o l l o w s i Suppose b = 1 , then y = a s i n x. What happens as a takes on d i f f e r e n t values? 0 0 TT 2 TT 3 n 2 - 1 2TT TT 2 TT 3jJ 2 - 2 2 u 0 a = 2 y = 2 s i n x As a changes the amplitude of the wave var i e s . Suppose ^George Polya, op_. c i t . , p. 2 1 0 - 2 1 1 . 23 a = 1, then y = s i n bx. What happens as b takes on d i f f e r e n t values? b = 2 y = s i n 2x 1 As b changes the wavelength v a r i e s . H e u r i s t i c IX: Extension. Polya's h e u r i s t i c of generalization, which he defined as a "passing from the consideration of a given set of objects to that of a larger set, containing the given one," could be classed here.5° This h e u r i s t i c i s used i n the generation of new problems. Some f a c t i s added to or subtracted from the present system and then the question, What can now be discovered? i s asked. Other h e u r i s t i c s would then be used to solve the new problem that had been suggested by the use of the h e u r i s t i c of extension. The following problem i l l u s t r a t e s the use of the h e u r i s t i c of extension! Solve x 2 + 1 = 0. The solut i o n to the equation x 2 + 1 = 0 remained unsolved u n t i l the r e a l number system was extended to include the complex number i , which was defined as 5°ibld. , p. 12. 24 p The s o l u t i o n to the problem i s x = i , s i n c e i = -1. The h e u r i s t i c s , p r e s e r v a t i o n of r u l e s , v a r i a t i o n , and ex t e n s i o n are used when i n i t i a l l y a t t a c k i n g a new problem or f o r m u l a t i n g a new problem. The problem w i l l u s u a l l y be sol v e d u s i n g one of the other h e u r i s t i c s , most probably, cases, deduction, i n v e r s e deduction, or analogy. In other words the three h e u r i s t i c s , p r e s e r v a t i o n of r u l e s , v a r i a t i o n , and extension suggest an i n i t i a l q u e s t i o n t o be asked when faced with the problem, but the a c t u a l s o l u t i o n w i l l u s u a l l y i n v o l v e the use of the other h e u r i s t i c s . CHAPTER II REVIEW OF THE LITERATURE The purpose of t h i s study i s to investigate the use of general h e u r i s t i c s by high school students when they solve mathematical problems. Some authors have urged teachers to use various h e u r i s t i c s when they demonstrate the solution to problems or have urged them to teach these procedures to t h e i r students. Several studies were reviewed i n which c e r t a i n h e u r i s t i c procedures were taught to students i n order to improve t h e i r problem-solv i n g s k i l l s . One study was concerned with the h e u r i s t i c s students use when asked to solve a problem using any method they chose. This chapter reviews some of the more important studies and opinions r e l a t e d to the topic of t h i s i n v e s t i g a t i o n . For organizational purposes the l i t e r a t u r e i s c l a s s i f i e d according to studies having general relevance to the problem, studies having s p e c i f i c relevance to the problem, opinions having general relevance to the problem, and opinions having s p e c i f i c relevance to the problem. Studies having general relevance to the problem. The Investigator reviewed several studies i n which attempts were made to improve the student's problem-solving a b i l i t y by r e q u i r i n g the in s t r u c t o r to use a h e u r i s t i c method when teaching solutions to problems. 2 5 26 Ash-ton* s research tested the effectiveness of teaching problem solving i n ninth grade algebra by a h e u r i s t i c method as opposed to a conventional method where the so l u t i o n to a p a r t i c u l a r type of problem i s demonstrated and then s i m i l a r problems are given f o r the students to solve. The h e u r i s t i c method used was based on Polya's writings and Involved frequent questioning techniques such a s i What i s the unknown?, What are the data?, and What are the conditions?^ A t o t a l of 1 3 4 students were taught using the h e u r i s t i c method and 1 1 0 were taught by the conventional method. Each teacher taught two classes, one with the h e u r i s t i c method and the other using the conventional method. A f t e r the ten-week teaching period, the r e s u l t s showed that a l l classes gained i n t h e i r scores from pre-test to post-test. However, the gains of the h e u r i s t i c group were greater, and were r s i g n i f i c a n t at the 0 . 0 1 l e v e l f o r classes i n four of the f i v e p a r t i c i p a t i n g s c h o o l s . 2 W i l l s t r i e d to determine i f the student•s Jproblem-sol v i n g a b i l i t y was improved i f a p a r t i c u l a r set of discovery stratagems were used by the teacher. The procedure "'•George Polya, How to Solve I t (second e d i t i o n ; New York: Doubleday and Company, Inc., 1 9 5 7 ) . p S i s t e r Madeleine Rose Ashton, "Heuristic Methods i n Problem Solving i n Ninth Grade Algebra," (unpublished doctoral d i s s e r t a t i o n , Stanford University, Stanford, 1 9 6 2 ) . 2? followed was: (a) Students were presented with a d i f f i c u l t problem which could be solved e a s i l y by anyone aware of a c e r t a i n generali zat i on. (b) Simpler problems s i m i l a r to the i n i t i a l task were presented developmentally. (c) The r e s u l t s of these simpler tasks were organized i n the form of a table which served to reveal a pattern to the students. (d) The generalization suggested by the pattern was applied to the i n i t i a l task. (e) The r e s u l t s were checked. 3 The pre-test and post-test contained six t y problems that involved a wider range of mathematical topics than the current i n s t r u c t i o n a l u n i t . The experimental group generally doubled t h e i r pre-test r e s u l t s , while the control group made only a s l i g h t score gain. W i l l s found no difference between the experimental group that was taught by the teacher who merely used the stratagems, as opposed to the group who took an active r o l e i n discussing the stratagems and t h e i r relevance to problem solving. Larsen wished to determine i f general h e u r i s t i c suggestions, which would f a c i l i t a t e the student's s o l u t i o n of other mathematical problems, could be given independently of the d e t a i l of a p a r t i c u l a r calculus problem. The r e s u l t s suggested that students can learn to use h e u r i s t i c methods when taught h e u r i s t i c s , but that t h i s learning may be at the J Herbert W i l l s I I I , "Transfer of Problem Solving A b i l i t y Gained Through Learning by Discovery," D i s s e r t a t i o n  Abstracts. XXVIII ( 1 9 6 7 ) , 1319-A. ( 28 expense of normal course content.** Wilson investigated the problem-solving performance of lkU- high school students on learning and transfer tasks, a f t e r i n s t r u c t i o n i n which the l e v e l of the generality of the problem-solving h e u r i s t i c s was varied. Three l e v e l s of generality were used; means-ends h e u r i s t i c s , task s p e c i f i c h e u r i s t i c s , and planning. Results showed that f o r the learning tasks the performance tended to be independent of the l e v e l of generality of h e u r i s t i c s used. However, f o r the transfer tasks the subjects seemed to benefit from having a wide range of h e u r i s t i c s a v a i l a b l e . ^ Schaaf's i n v e s t i g a t i o n was made to determine i f students could l e a r n to be better generalizers i f the teaching procedure was a discovery method which emphasized such procedures of generalizing as simple enumeration, analogy, continuity of form, s t a t i s t i c a l procedures, deduction, v a r i a t i o n , formal analogy, and inverse deduction. Results showed that most of the students made an improvement i n t h e i r a b i l i t y to generalize and that t h i s learning was ^Charles McLoud Larsen, "The H e u r i s t i c Standpoint i n the Teaching of Elementary Calculus," D i s s e r t a t i o n Abstracts. x x i ( 1 9 6 1 ) , 2632-2633. 5james William Wilson, "Generality of H e u r i s t i c s as an I n s t r u c t i o n a l Variable," D i s s e r t a t i o n Abstracts. XXVIII (January, 1 9 6 8 ) , 2575-A. J 29 6 not at the expense of the mastery of algebraic concepts, Washburne and Osborne attempted to t r a i n students i n grade six and seven to see the analogy between a d i f f i c u l t problem and a simple problem of the same type. The pupils were required to write a simple problem that was analogous to the more d i f f i c u l t problem they were required to solve. The students then observed how they solved the simple problem and applied t h i s process to the o r i g i n a l problem. As contrasted with the other two methods tested (solving many problems and formal analysis) t h i s method resu l t e d i n approximately a 2% gain i n the f i n a l t e s t scores,'''' R u s s e l l discussed a study by Doty who interviewed 151 grade four students and thirty-one grade six students to determine what methods they used to solve verbal arithmetic problems. Among the procedures which l e d to correct answers was r e c a l l i n g analogous problems.8 5Oscar Schaaf, "Student Discovery of Algebraic P r i n c i p l e s as a Means of Developing A b i l i t y to Generalize," The Mathematics Teacher, XLVIII (May, 1 9 5 5 ) , 324-327. ?Carleton W. Washburne and Raymond Osborne, "Solving Arithmetic Problems," The Elementary School Journal, XXVII (November, 1 9 2 6 ) , 219-2267 (December, 1926), 296-304. ^David H. R u s s e l l , Children's Thinking (Toronto: B l a i s d e l l Publishing Company, 1956), p. 275. 30 C r u t c h f i e l d attempted to.teach creative problem solv i n g by using a series of a u t o - i n s t r u c t i o n a l materials. "Each lesson i s a complete problem-solving episode, containing a l l of the p r i n c i p l e steps and processes inherent i n creative problem-solving." 9 The lessons are constructed not only to give the reader repeated experiences i n the s o l u t i o n of i n t e r e s t i n g problems, but also d i r e c t l y to i n s t r u c t him i n h e l p f u l strategies or h e u r i s t i c procedures f o r creative problem solving, by showing him how he can use them i n the concrete problems. The procedures pert a i n to the formulation of the problem, the asking of relevant questions, the l a y i n g out of a plan of attack, the generation of many ideas, the search f o r uncommon ideas, the transformation of the problem i n new ways, the evaluation of hypotheses, the s e n s i t i v i t y to odd and discrepant f a c t s , and the openness to metaphorical and analogical hints leading to solution.10 C r u t c h f i e l d c a r r i e d out two major studies using the materials i n f i f t h and s i x t h grade classes. The designs of the two studies were e s s e n t i a l l y the same, except that the t r a i n i n g and t e s t i n g materials were revised and lengthened f o r the second'study. A t o t a l of 267 c h i l d r e n were given the t r a i n i n g materials and another 214 formed the control group. The f i r s t experiment took three weeks and the second l a s t e d four weeks. ^Richard S. C r u t c h f i e l d , "Creative Thinking i n Children: I t ' s Teaching and Testing," I n t e l l i g e n c e : Perspective 1965 (New York: Harcourt, Brace, and World, Inc., 1966), p. 43. Ibid., p. 44. 31 Each student worked i n d i v i d u a l l y at h i s own pace. The post-te s t r e s u l t s showed that the problem-solving a b i l i t y of the trained c h i l d r e n surpassed that of the control group. According to C r u t c h f i e l d : The trained c h i l d r e n asked a greater number of relevent questions while working on the problems, were more sen s i t i v e to s i g n i f i c a n t clues and f a c t u a l discrepancies, generated more ideas and better ideas, and were better able to u t i l i z e clues and hints as help l n getting to solutions. l x Studies having s p e c i f i c relevance to the.problem. Only one study attempted to discover which h e u r i s t i c s the students used to solve problems. K i l p a t r i c k ' s main aim was to develop a system to analyze the h e u r i s t i c s students used i n solving word problems. F i f t y - s i x grade nine students were i n d i v i d u a l l y interviewed and asked to think aloud as they solved mathematical problems. The h e u r i s t i c processes i d e n t i f i e d were based on Polya*s w r i t i n g s . x 2 K i l p a t r i c k designed four d i f f e r e n t systems f o r coding the student's response. The f i r s t system attempted to record instances of the p u p i l asking himself any of t h i r t y - s i x questions. Some of the questions were; what i s the unknown?, what i s the condition?, and "Bave I seen the problem before?" K i l p a t r i c k concluded that t h i s l i s t was not useful f o r characterizing the student's Ibid., p. 51. •George Polya, l o c . c i t . 32 problem-solving behaviour, f o r two reasons; (1) there was not enough time to complete the form and s t i l l maintain rapport with the student and (2) the catagories were not p r e c i s e l y defined. The second system was a l i s t of eleven h e u r i s t i c s that were put as actions the student might take, rather than as questions that he might ask himself. Four of these h e u r i s t i c s were; drawing a fi g u r e , rephrasing the problem, using successive approximations, and checking that the r e s u l t i s reasonable. Although t h i s l i s t was more manageable, the r e s u l t s were also unsatisfactory. According to the study, two of the main reasons why the second l i s t was not successful were; (1) many actions of the student were not being recorded and (2) no i n d i c a t i o n of the sequence of the h e u r i s t i c s used by the student was indicated. The t h i r d system was an attempt to i d e n t i f y s i x t y - f o u r processes and the sequence used during the student's s o l u t i o n to a problem. This system proved to be unmanageable. K i l p a t r i c k then developed h i s fourth and f i n a l system. He t r i e d to combine the ideas of system two and system three Into a manageable l i s t . The processes used were reading and t r y i n g to understand the problem, deduction from the condition, s e t t i n g up an equation, t r i a l and error, and checking the solution. The subject's a b i l i t y was rated as (1) incomplete (2) impasse (3) intermediate r e s u l t (4) i n correct r e s u l t (5) correct r e s u l t . The r e s t of the 33 coding system dealt with t h i r t y items of supplementary information such as; draws a fig u r e , uses a r e l a t e d problem i n the solution, and admits confusion. The r e s u l t s indicated that the students used f i f t e e n of the items but only eight were frequently used, K i l p a t r i c k concludes, "The check l i s t y i e l d s few variables that can be measured 13 r e l i a b l y . " y Prom the process-sequence l i s t K i l p a t r i c k observed that deduction was l i k e l y to r e s u l t i n an Incorrect s o l u t i o n whereas t r i a l and error was l i k e l y to r e s u l t i n a correct s o l u t i o n . The main conclusion K i l p a t r i c k a r r i v e d at seemed to be that mathematics teachers should devote more time to t r i a l and error as a legitimate problem-solving method. K i l p a t r i c k was mainly interested i n analyzing the problem-solving process. The present study i s not concerned with the process or sequence taken i n the s o l u t i o n of the problem, but only interested i n determining i f c e r t a i n general h e u r i s t i c s are used by students when they are asked to attempt to solve a problem f o r which they do not know an algorithm. eremy K i l p a t r i c k , "Analyzing the Solution of Word Problems i n Mathematics: An Exploratory Study," (unpublished doctoral d i s s e r t a t i o n , Stanford University, Stanford, I 9 6 7 ) , P. 63. 34 Opinions having general relevance to the problem. Several writers have urged that teachers employ a h e u r i s t i c method when teaching. Polya stressed a h e u r i s t i c teaching procedure. He described such h e u r i s t i c s as synthesis, analysis, analogy, cases, and working backwards and showed how they could be used when teaching mathematical problem s o l v i n g . ^ Henderson discussed seven stratagems which he f e l t can be employed by the teacher so that students w i l l be helped to discover mathematical truths. The stratagems are; (1) analogy ( 2 ) simple enumeration (3) agreement (4) difference (5) difference and agreement ( 6 ) concomitant v a r i a t i o n and (7) independent action. Although he urged the use of the stratagems (1) to (6) by the teacher, i t would seem as i f Henderson believed that the stratagems would be picked up and used by the students from observing the teacher. His l a s t stratagem, independent action, allows the student to attempt to f i n d an answer to a problem on h i s own. Henderson stated, "Quite possibly t h i s stratagem would not be used except with very bright students, or George Polya, l o c . c i t . 35 students who have had considerable experience with other strategies."15 Jones i l l u s t r a t e s the use of h e u r i s t i c techniques that could be used by teachers i n planning discovery-type lessons. The techniques are induction, permanence of form, structure and deduction, models, l o g i c a l form of statements, generalization and s p e c i a l i z a t i o n , and analogy and mental models.1^ Bruner stated: I t might be wise to assess what atti t u d e s or h e u r i s t i c devices are most pervasive and useful, and then an e f f o r t should be made to teach c h i l d r e n a rudimentary version of them that might be further r e f i n e d as they progress through school.17 Bruner f e l t that i n t u i t i v e thinking which he defines as I p "the t r a i n i n g of hunches" should be taught. He said that h e u r i s t i c rules such as the use of analogy, the appeal to symmetry, the examination of l i m i t i n g conditions, and the 15Kenneth B. Henderson, "Strategies f o r Teaching by the Discovery Method," Updating Mathematics (high school teachers ed i t i o n ; New London, Connecticut: C r o f t Educational Services, 1 9 6 4 ) , p. 64. l ^ p h i l i p S, Jones, "Discovery Teaching Prom Socrates to Modernity," The Mathematics Teacher, LXIII (October, 1 9 7 0 ) , 501-508. ~ ~ l?Jerome S. Bruner, The Process of Education (New York: Vintage Books, I960), p. 27. l 8 I b i d . . p. 13. 36 v i s u a l i z a t i o n of the solution, when used frequently by the teacher should promote i n t u i t i v e thinking i n students. 1^ _Opinions having s p e c i f i c relevance to the problem. K i l p a t r i c k saw the necessity f o r studying how students solve problems and f o r determining what h e u r i s t i c s were used. We have so l i t t l e knowledge of how problem sol v i n g i s and could be taught that we must do more, however, than simply exhort teachers to pay greater a t t e n t i o n to i t . We should encourage them to study the problem-solving behavior of t h e i r students.20 Summary. Ashton, W i l l s , Larsen, Wilson, and Schaaf ^1 a l l urge teachers to use h e u r i s t i c methods when teaching mathematics. In a l l cases the researchers found that i f h e u r i s t i c teaching methods were used, the problem-solving a b i l i t y of the students was improved. Washburne and Osborne 2 2 attempted to t r a i n students to use analogy. They found that the students could solve problems better a f t e r they had been taught to use the h e u r i s t i c of analogy. Several writers have urged that teachers employ a h e u r i s t i c method when teaching. 23 K i l p a t r i c k , however, f e l t that the 1 9 l b l d . , p. 6k. 2 0 K i l p a t r i c k , op_, c i t . , p. 105. 2 1Ashton, l o c . c i t . ; W i l l s , l o c . cit.;< Larsen, l o c . c i t . ; Wilson, l o c . c i t . ; Schaaf, l o c . c i t . 2 2Washburne and Osborne, l o c . c i t . 2 3 p o l y a , l o c . c i t . ; Henderson, l o c . c i t . ; Jones, l o c . c i t . ; Bruner. l o c . c i t . 37 investigator should attempt to determine what h e u r i s t i c s 'the students use and then an attempt should be made by the oh. teacher to improve the pupil's problem-solving a b i l i t y . L i m i t a t i o n of previous studies. The main l i m i t a t i o n of previous research i s that no attempt was made to investigate whether or not the students use any general type h e u r i s t i c s such as, analogy.or inverse deduction, to solve an unfamiliar problem. Since h e u r i s t i c s are useful and worthwhile tools to use when attempting to solve a var i e t y of problems, students should be taught these methods. However i f the students use some h e u r i s t i c s and not others, the students should f i r s t be taught to use the h e u r i s t i c s that they understand. 'Kilpatrick, l o c . c i t . CHAPTER III THE DESIGN AND STATISTICAL TREATMENT I. THE DESIGN Each of the forty-two grade eleven students interviewed was given two mathematical problems to solve. These problems can be solved using two of the nine general heuristics discussed earlier. Although the students were allowed to use pencil and paper, the researcher requested the students to think aloud. Various methods are used to obtain information on the method of solution of problems. They range from a study of the written work of a student, retrospection, introspection, and thinking aloud. In retrospection studies, the subject is asked to give an account of his thinking after he has completed the solution to the problem. Two disadvantages to the use of this technique appear to be, (1) the subject may edit his report and omit errors and false leads, (2) the subject may not remember a l l the steps and the order in which they occurred. Introspection attempts to analyze thought processes by obtaining reports from the student as he i s in the process of solving a problem. Since this technique requires the student to observe and analyze his thinking, the student would have to be aware of his use of general heuristics. 38 39 The investigator f e e l s that few students, i f any, are aware of t h e i r use of any general h e u r i s t i c s . Duncker stated that the i n s t r u c t i o n "think aloud," . . . Is not i d e n t i c a l with the i n s t r u c t i o n to introspect. . . . While the introspector makes himself as thinking the object of h i s attention, the subject who i s thinking aloud remains immediately d i r e c t e d to the problem, so to speak allowing h i s a c t i v i t y to become verbal.1 When the student thinks aloud he does not have to be aware of the methods he i s using to solve the problem. I t i s up to the researcher to analyze the response to determine i f the student shows that he used c e r t a i n h e u r i s t i c s . Another objection to the thinking aloud technique i s that the student may change hi s method of attack on the problem because he i s asked to vocalize h i s thinking. However, K i l p a t r i c k stated, "we can ignore questions about the changes i n thinking produced by v o c a l i z a t i o n . . . , i f we r e s t r i c t our use of the method to producing hypotheses rather than to v a l i d a t i n g them." 2 The researcher was only attempting to discover i f the p u p i l knew any general h e u r i s t i c s and not whether he would always attack the given problems i n t h i s way. Taylor stated that, "thinking aloud" has repeatedly . K a r l Duncker, "On Problem Solving," trans. Lynne S. Lees, Psychological Monographs. LVIII (whole no. 270, 1945), p. 2. 2 K i l p a t r i c k , op_. c i t . , p. 8. proved f r u i t f u l i n the analysis of process."3 A sample of the form which the student was required to f i l l out Is shown i n Appendix A. The two problems were typed on separate sheets of paper. A f t e r the student had attempted the f i r s t problem, he was given the second one. The standard interview format that was used with every student i s shown i n Appendix B. The student was encouraged to attempt the problems any way he chose. They were asked to be more concerned with revealing as much of t h e i r thought processes as possible, as with the accuracy of t h e i r s o l u t i o n . A l l interviews were taped. When the student requested assistance, he was aided only i f the investigator f e l t that the required information would not help the student to plan hi s method of solution. The student was given no i n d i c a t i o n of the correctness of any of the steps i n h i s sol u t i o n . A l l subjects were i n d i v i d u a l l y interviewed by the writer, and worked i n an area where d i s t r a c t i o n s were minimized. - The Problems Used. The following i s a l i s t of the nine problems used i n t h i s study. Each one i s c l a s s i f i e d as to the method by which i t could be solved. Each i n d i v i d u a l student was presented with two of these nine problems. The t o t a l number of t h i r t y - s i x d i f f e r e n t -^Donald W. Taylor, "Discussion of Papers by Adrian de Groot and by J e f f e r y M. Paige and Herbert A. Simon," Problem Solving; Research, Method, and Theory (New York» John Wiley and Sons, Inc., 1966), p. 123. 41 combinations of the nine problems were d i s t r i b u t e d among the forty-two subjects. A complete so l u t i o n which shows how a given h e u r i s t i c technique could be used to solve the problem i s shown i n Appendix C. 1. Cases. Find a l l r e a l numbers x which s a t i s f y the following equation^ x + 1 + 3 x - 1 - 2 x - 2 = x + 2 2. Deduction. Triangle ABC i s isosceles (the two sides AB and BC have the same length) with base AC. Point P i s i n CB and point Q i s i n AB such that AC = AP = PQ = QB. What i s the number of degrees i n angle B? 5 Steven J . Bryant, George E. Graham, and Kenneth G. Wiley, Nonroutlne Problems i n Algebra Geometry and Trlgonom-etry (New York: McGraw-Hill Book Company, 196377 p. 5. ^ Charles T. Salkind, The MAA Problem Book II (New York: Random House, Inc., 1966), p. 13. 42 3. Inverse Deduction. B i l l , Ron, and Ted each have some money of t h e i r own but they do not a l l have the same amount of money. Now, B i l l decides to give some of h i s money to Ron and some of h i s money to Ted. He gives Ron as much money as Ron already has and also he gives Ted as much money as Ted already has. A f t e r t h i s Ron decides to give some of h i s money to B i l l and to Ted. He gives each of them as much money as each of them now has. Then Ted decides that he w i l l give B i l l and Ron as much money as each of them has. Now B i l l has 16^ and Ron has l6fi and Ted has 16^. How much money did B i l l o r i g i n a l l y have before he gave any of h i s money away? The problem was o r i g i n a l l y presented ass A gives B as many cents as B has and C as many cents as C has. S i m i l a r l y , B then gives A and C as many cents as each of them has. C, s i m i l a r l y , then gives A and B as many cents as each of them has. I f each f i n a l l y has 16 cents, with how many cents does A start? Student 1, student 2, and student 3 answered the o r i g i n a l problem. I t was observed that these three students took too much time to comprehend the problem. Since most of the time a l l o t t e d each student should be spent on the solution, i t was necessary to minimize the comprehension time. The revised form reduced the time the students spent Ibid., p. 26. 43 in comprehending the problem. 4. Invariation. Using a compass and a straightedge construct a square inside the given triangle ABC such that two of the vertices of the square are on the base CB of the triangle and the other two vertices of the square are on the other two sides AC and AB of the triangle, one vertex on each side.'' A 5. Analogy. Find the values of the five unknowns x, y, z, u, and t that satisfy the following set of five simultaneous linear equations.^ 4y - 3x + 2 U a 5 9x - 2z + u = 41 7y - 5z - t = 12 3y - 4u + 3t = 7 7z - 5u = 11 ^George Polya, How to Solve It (second edition; New York; Doubleday and Company, Inc., 1957), p. 23. °Reverend Alfred Wrigley, A Collection of Examples  and Problems in Pure and Mixed Mathematics with Answers and  Occasional Hints [London; Longmans, Green, and Company, 1875), P. W. 44 6 . Symmetry. Using a compass and a straightedge construct an i s o s c e l e s t r i a n g l e (two of the sides have the same length) given the a l t i t u d e (the perpendicular distance from the vertex to the side opposite) equal to l 1 - inches and the perimeter equal to 5 inches.9 7 . Preservation of Rules. Using the following mathematical system containing s i x dabas d l , d 2 , d 3 , di)., d 5 , and dg and the d e f i n i t i o n "an aba i s a set containing three dabas" ( i . e . ^d-^, dg, d^ j and dgi dj^ ,, dg][ are abas) , (i) L i s t a l l the abas ( i i ) Decide i f the following statements are true or f a l s e Statement i I f p and q are two dabas, then there e x i s t s one and only one aba containing both p and q. Statement 2 I f L i s an aba, there exists a daba not In L. Statement 3 I f L i s an aba, and p i s a daba not i n L, then there exists one and only one aba containing p and not containing any daba that i s i n L . 1 0 Edmund C. Plant, Geometrical Drawing (London: MacMillan and Company, Limited, 1 8 9 9 ) , p. 1 0 3 . 1 0 B e t t y Plunkett, "Aba Daba Daba," The Mathematics  Teacher, LIX (March, 1 9 6 6 ) , 2 3 6 - 2 3 7 , c i t i n g Howard Eves, and C.V. Newsom. An Introduction to the Foundations and  Fundamental Concepts of Mathematics (New York: Holt, Rinehart and Winston, 1 9 6 4 ) , p. 8 0 . 8. Variation. Construct a trapezoid (a closed plane figure with four sides, two of which are parallel) being given the lengths a, b, c, and d of i t s four s i d e s , 1 1 a b e d 9. Extension. 1 + 8 + 27 + 64 = 100 1^ + 8 + 27 + 64 = 100 What can you discover? 1 2 The problem was originally presented asi 1 + 8 + 27 + 64 = 100 What can you discover? Student 13, student 14, and student 15 answered the original problem. It was observed that the students did not notice that the left-hand side was 1^ + 2r> + 3^ + 4^  or that the right-hand p side was 10 . Since the problem was not to discover the fact that 1^ + 2^ + 3^ + 4^  = 102, but to see i f the student would extend the problem, to see i f , for example, 1^ + 2} + 3 3 3 2 3 + 4^  + 5 = (x) where x is a positive integer, the problem was revised. The Group Used. This investigation was a normative survey of forty-two subjects enrolled i n nine academic grade x x Polya, op_. c i t . , p. 211. 1 2 Ibid., p. 108. 46 eleven algebra classes i n nine schools. Based on t h e i r Christmas and Easter mathematics eleven grades, the students', mathematical a b i l i t y was A or B. The investigator was interested i n determining i f the student who had a high mathematical a b i l i t y used any h e u r i s t i c s . I t i s f e l t that t h i s type of study was j u s t i f i e d since as K i l p a t r i c k statedt The researcher . . . who chooses to investigate problem solving i n mathematics i s probably best advised to undertake c l i n i c a l studies of i n d i v i d u a l subjects . . .. because our ignorance i n t h i s area demands c l i n i c a l studies as precursors to larger e f f o r t s . x 3 I I . STATISTICAL TREATMENT This normative survey was concerned with the i d e n t i f i c a t i o n of any h e u r i s t i c which the students used i n attempting to solve the mathematical problems used 'in*. t h i s i n v e s t i g a t i o n . The survey was also concerned with s the frequency of the use of the h e u r i s t i c s . A tabulation; of the h e u r i s t i c s used by the students and a count of the©: number of h e u r i s t i c s each student used i s presented. Each h e u r i s t i c and i t s use by the students i s discussed. Limitations of the Study. This study attempted to f i n d out what h e u r i s t i c s were used by grade eleven academic mathematics students. The pupil's use of h e u r i s t i c s may d i f f e r at various grade l e v e l s . Consequently a l l . h e u r i s t i c s -^Jeremy K i l p a t r i c k , "Problem Solving i n Mathematics," Review of Educational Research. XXXIX (October, 1969), 523-534. 47 may not be able to be taught e f f e c t i v e l y at a l l l e v e l s , but may have to be s i m p l i f i e d or even ignored. I t may be that i t i s only p r a c t i c a l to teach c e r t a i n h e u r i s t i c s at one l e v e l and others i n higher grades. Only a l i m i t e d sample of students was used and therefore i t may be that other h e u r i s t i c s would be uncovered i f the sample was larger. This i n v e s t i g a t i o n says nothing about which h e u r i s t i c s the student does not know or i f he knows a h e u r i s t i c but can not use i t , but only which h e u r i s t i c s he happened to use when solving the problems used i n t h i s study. Nine problems were used i n t h i s study. I f d i f f e r e n t problems were presented to the students, i t may be that other h e u r i s t i c s would have been discovered or some of the h e u r i s t i c s may have been used more frequently. Suppose a student attempted to solve a problem using a h e u r i s t i c which was inappropriate. He then could have been given another problem where the use of t h i s h e u r i s t i c was appropriate. It could then be determined i f the student usually used t h i s h e u r i s t i c when attempting problems and whether he could use i t c o r r e c t l y . CHAPTER IV RESULTS I d e n t i f i c a t i o n and frequency of student usage of  h e u r i s t i c s . The i n v e s t i g a t i o n found evidence that eight of the nine h e u r i s t i c s were used by the forty-two students interviewed. The students used cases, deduction, inverse deduction, i n v a r i a t i o n , analogy, preservation of ru l e s , v a r i a t i o n , and extension. No student used the h e u r i s t i c of symmetry. Three students noticed symmetry i n the problem they solved, but they did not use symmetry to help them i n t h e i r s o l u t i o n of the problem. Eighteen of the students t r i e d to use cases to solve the problems they were presented. Table 1, page 68, shows the number of students who used h e u r i s t i c s when attempting to solve a problem and the number of students who were presented with each of the problems. Appendix D shows sample interviews and analyses. Appendix E gives information on the students* use of the general h e u r i s t i c s , Appendix P gives information on the students included i n t h i s survey, and Appendix G gives information on the schools sampled i n t h i s survey. The students' use of each of the nine h e u r i s t i c s w i l l now be di scussed. Cases. Eighteen students used the h e u r i s t i c of cases. Most students considered a random sample of the possible 48 4 9 s o l u t i o n s t o t h e p r o b l e m . U s u a l l y t h e s t u d e n t s f e l t t h a t t h i s m e t h o d was t o o t i m e c o n s u m i n g a n d d i d n o t f e e l i t was a n e f f e c t i v e a p p r o a c h t o t h e p r o b l e m . P r o b l e m 1 w h i c h c o u l d be s o l v e d e f f i c i e n t l y by t h e u s e o f c a s e s was n o t s o l v e d c o r r e c t l y b y a n y o f t h e s i x p u p i l s who t r i e d t o s o l v e t h e p r o b l e m b y u s i n g t h i s h e u r i s t i c . T h e y a p p e a r e d t o h a v e no m e t h o d f o r e f f i c i e n t l y s e l e c t i n g t h e c a s e s t o b e t r i e d , a n d f o r n a r r o w i n g down t h e i r s e a r c h t h r o u g h t h e p o s s i b l e s o l u t i o n s . T h r e e o f t h e s t u d e n t s c a l l e d t h i s m e t h o d t r i a l a n d e r r o r . T h e way e a c h o f t h e s t u d e n t s u s e d c a s e s i n t h e p r o b l e m s t h e y a t t e m p t e d w i l l now be d i s c u s s e d . S t u d e n t 2 P r o b l e m 1 A r a n d o m s a m p l e o f p o s i t i v e a n d n e g a t i v e i n t e g e r s was s u b s t i t u t e d f o r x . S e e A p p e n d i x D. S t u d e n t j l P r o b l e m 1 Two c a s e s w e re c o n s i d e r e d — x a p o s i t i v e v a l u e a n d x a n e g a t i v e v a l u e . S t u d e n t j j P r o b l e m 4 A f i g u r e was d r a w n i n s i d e t h e t r i a n g l e b u t s i n c e i t a p p e a r e d t o be a r e c t a n g l e , t h e u s e o f t h e h e u r i s t i c o f c a s e s was a b a n d o n e d . S t u d e n t 7 P r o b l e m j> The s t u d e n t s u g g e s t e d t h a t n u m bers be s u b s t i t u t e d f o r t h e d i f f e r e n t v a r i a b l e s u n t i l t h e e q u a t i o n s w e r e s a t i s f i e d . 50 Student 9 Problem 8' The student randomly varied the placement of the sides and the measure of the angles i n order to draw a trapezoid. The student r e f e r r e d to t h i s method as t r i a l and error. Student 13 Problem j> Two d i f f e r e n t sets of values f o r y, x, and u that s a t i s f i e d the f i r s t equation were chosen at random. The t h i r d equation was then used to f i n d values f o r z and t. These values were checked i n equation 2. Since the problem was not solved c o r r e c t l y , another approach was chosen. Later the student returned to using cases. The student c a l l e d t h i s method t r i a l and error. Student 15 Problem 6 Various figures were drawn at random i n an attempt to solve the problem. Student 16 Problem J5 The student suggested that the problem could be solved by s u b s t i t u t i n g numbers f o r the var i a b l e s , but that t h i s procedure would be time consuming. Student 17_ Problem 1 Two values f o r x were considered—x negative and x p o s i t i v e . See Appendix D. 51 Student 21 Problem 8 Several figures were drawn. Each time the placement of the sides was randomly varied. The student c a l l e d the procedure guessing and stated that there was no method to the procedure chosen. Student 23 Problem 8 The sides of the trapezoid were placed i n d i f f e r e n t positions at random i n an attempt to construct the correct f i g u r e . Student 26 Problem 1 The student f i r s t changed some of the plus signs to minus and the minus signs to plus, and then a l l of the signs were changed to the opposite sign. Student 29 Problem 8 Various positions f o r the sides and the angles were t r i e d i n the attempt to draw the trapezoid. The student f e l t that the various arrangements f o r the sides and angles would have to be t r i e d u n t i l a s o l u t i o n was found. Student 30 Problem 1 Various values were substituted into part of the equation, x + 1 x - 1 , to see i f there was any r e l a t i o n s h i p between the values substituted and the answers obtained. Values were then substituted into the left-hand side of an equation that the student had i n c o r r e c t l y derived, x - x - 2 = ( x + 2 ) / 6 . 52 Student 31 Problem 8 Several figures were drawn at random to attempt to solve the problem. The student r e f e r r e d to t h i s method as t r i a l and error. Student 32 Problem 4 The student randomly t r i e d various lengths to see i f a square could be constructed. Student 35 Problem 1 Two cases were considered—x a p o s i t i v e value and x a negative value. Student 37 Problem 8 Various figures were drawn. The positions of the sides and the measures of the angles were randomly vari e d i n order to get the required trapezoid. Deduction. Seven students used d i r e c t deduction to attempt to solve problem 2. Three of the seven students, who thought they found the solution, c o r r e c t l y derived f a l s e conclusions from f a l s e premises. A l l the students seemed reasonably confident of t h e i r method of attack. Two students attempted to use hypothetical deduction i n the sol u t i o n of problem 2 and problem 3(b). Problem 2 which could be solved using the technique of deduction was not solved c o r r e c t l y by any of the students. The analysis of the solutions i n which deduction was used w i l l now be discussed. 53 Student 1 Problem 2 Prom the following three facts, (i) AC = AP = PQ = QB ( i i ) triangle ABC, triangle APC and triangle PQB are isosceles and ( i i i ) the base angles of an isosceles triangle are equal, the student incorrectly concluded that angle A = angle C = angle APC = angle QPB = angle B. Since o o angle A + angle C + angle B = 180 , angle B = 60 . See Appendix D. Student 4 Problem 2 Prom the following three incorrect statements (i) o angle QPB + angle APQ = 90 , ( i i ) triangle QPA i s o equilateral, therefore angle QPA = 60 and ( i i i ) angle QPB = t> angle B, the student concluded that angle B = 30 . Student 11 Problem 3(b) The student f e l t that the amount of money B i l l had after he gave some of his money to Ron and Ted, could be found. Then the student said i t would be possible to work back to the original amount B i l l had. The investigator would classify the student's technique as hypothetical deduction. Student 18 Problem 2 Using the fact that two angles were supplementary or that the sum of the angles in a triangle was equal to o 180 , an attempt was made to deduce the measure of angle B. 5 4 Student 24 Problem 2 The student f e l t that the answer could be obtained by adding or subtracting angles. The student noticed that angle QAP + angle PAC = angle QAC, angle QPB + angle APQ = angle APB, and angle PAG = angle BAG - angle QAP. Student 31 Problem 2 The student stated that, he would work from angle APQ o i f i t was 90 . The student said that i f angle A or angle C could be found, then angle B could be obtained since the o sum of the angles in a triangle is 180 . The student called this procedure working backwards. The investigator c l a s s i f i e d the method as hypothetical deduction. Student 33 Problem 2 An attempt to find angle B was made by combining angles that were supplementary or using the fact that the o sum of the angles in a triangle equals 180 . The student f e l t that angle B could be expressed in terms of other angles in various ways and that these statements could be combined by addition or subtraction, to get the value for angle B. See Appendix D. Student 42 Problem 2 The student expressed angle B incorrectly in terms of other angles. The student thought that the sum of the o angles in a triangle was 360 . The student also said that one of the measures of the angles would have to be given 55 before the number of degrees i n angle B could be determined. Inverse Deduction. Two students suggested using the h e u r i s t i c of inverse deduction to solve problem 3« Problem 3 could be solved by t h i s h e u r i s t i c , but neither student was able to use the technique. The students both f e l t that the problem could be solved by working backwards, but they d i d not know how to proceed. Student 2 Problem 3(a) The student suggested that the problem could probably be solved by working backwards. Student 11 Problem 3(b) } The student attempted to f i n d a number clo s e r to the o r i g i n a l amount of money that B i l l had. The student i n c o r r e c t l y found the amount of money B i l l had a f t e r he had given money to Ron and Ted and then wanted to work back to the amount B i l l had o r i g i n a l l y . Appendix D contains the interviews f o r the two students who used inverse deduction. Invariation. Three students used the h e u r i s t i c of i n v a r i a t i o n . In a l l cases problem 1 was attempted by ignoring some of the data. Problem 4 which could have been solved by using i n v a r i a t i o n was not given to these three students. The investigator f e l t that t h i s h e u r i s t i c was used by students 1 and 30 because they were very uncertain how to solve absolute value questions. 56 Student 1 Problem 1 The student decided to ignore the absolute value signs, rewrite the problem, and then try to solve i t . Student 2 Problem 1 The student decided to rewrite the equation leaving out the absolute value signs. Student 30 Problem 1 The student attempted to ignore the absolute values and work the r e s u l t i n g equation. Analogy. Seven students used the h e u r i s t i c of analogy. Six of the students used the h e u r i s t i c i n attempting to solve problem 5. Problem 5 could be solved by the method used to solve three simultaneous l i n e a r equations i n three v a r i a b l e s . Four of the students suggested elimination of variables and two students suggested solving f o r one of the var i a b l e s and su b s t i t u t i n g i t into the next equation where the same var i a b l e appeared. Student 10 Problem 7 The student could not remember how many sets of three dabas could be constructed from a t o t a l of s ix dabas, so the student attempted to decide what the formula was by considering three numbers grouped i n sets of two and then four numbers grouped i n sets of two. 57 Student 10 Problem 8 The student noticed that the trapezoid formed two t r i a n g l e s . The student sai d that to construct a t r i a n g l e one l i n e and two angles would be necessary. The student then t r i e d to r e l a t e the method f o r constructing the t r i a n g l e to the given problem. Student 13 Problem J? The student thought that i f two of the equations had the same unknowns they could be added or subtracted so that one of the variables could be cancelled. Student 14 Problem j> Equations one and two were solved f o r x and then these two equations were combined i n order to solve f o r y. The student then wished to solve equations four and f i v e f o r u and continue u n t i l one of the variables was known. Student 16 Problem J? The student described how to solve two simultaneous l i n e a r equations i n two variables by solving f o r one va r i a b l e and s u b s t i t u t i n g i t i n the second equation. The student said that she had never seen the three v a r i a b l e case before, but she would try the problem the same way as the two va r i a b l e case. See Appendix D. Student 25 Problem j> Af t e r f i r s t looking f o r two equations i n two unknowns, the student thought that i t didn't matter how many variables 58 there were as long as there were l i k e terms i n two equations. Then the equations could be subtracted or added to eliminate one of the unknowns. Student _3_8 Problem j5 The student thought that there were two equations with two variables the same i n each of the two equations, and therefore the equations could be subtracted to cancel out one of the unknowns. The student f e l t that the same procedure would solve the given problem because i t was the method used to solve two simultaneous l i n e a r equations i n two unknowns. See Appendix D. Student 39 Problem j5 The student thought that i f two equations were combined, one va r i a b l e could be eliminated. Symmetry. None of the forty-two students interviewed used the h e u r i s t i c of symmetry. Only three students appeared to notice symmetry i n a problem, but they d i d not use t h i s symmetry to help f i n d a solu t i o n . Problem 6 which could be solved using symmetry, was usually t r i e d by cases or by the h e u r i s t i c of v a r i a t i o n . Student 25, student 27 and student 34, who were given problem 6, a l l noted that the a l t i t u d e of an isos c e l e s t r i a n g l e b i s e c t s the base of the t r i a n g l e . Preservation of Rules. At the high school l e v e l , the investigator f e l t that the h e u r i s t i c could not be tested i n the sense of giving the students a problem i n another system 59 and determining i f they would see i f various rules i n the f a m i l i a r systems they had studied held, before they proceeded to try the problem i n the new system by a method analogous to the method used f o r s i m i l a r problems i n the f a m i l i a r system. However i t was f e l t that the h e u r i s t i c could be tested to see i f students can remain working i n a new system, A small system was defined and then the student was supposed to determine i f statements about the system were true or f a l s e . They had to preserve the d e f i n i t i o n given and not switch out of the system to more f a m i l i a r systems. Two of the eight students who were given problem 7 showed that they r e a l l y could preserve the rules of the system, i n order to c o r r e c t l y answer the questions. A discussion of the solutions to problem 7 by a l l the students who were given the problem, w i l l follow. Student 8 Problem 2 The student showed no preservation of r u l e s . The student f e l t that statement 1 was f a l s e because there were only two dabas l i s t e d and there must be three i n an aba. The student apparently thought that the statement meant "If p and q are two dabas, then there exists one and only one aba containing only p and q." The student f e l t that statement 2 was f a l s e because there are three dabas i n an aba so there would be none l e f t over. 60 Student 10 Problem 2 The student showed no preservation of r u l e s . The student thought there were more than s i x dabas. Student 19 Problem 2 The student showed no preservation of r u l e s . The student was not c e r t a i n of the difference between a daba and an aba and thought that there could be only one set containing p and q, since there were only two numbers p and q f o r the set. Student 20 Problem 2 The student showed no preservation of r u l e s . The student thought there could be only one set containing both p and q, since there were only two numbers f o r the set. Por statement 2, the student f e l t that there were only two sets, L and the empty set. Student 22 Problem 2 The student showed no preservation of r u l e s . The student said statement 1 was f a l s e because there couldn't be an aba since there were only two dabas, and an aba required three dabas. The student apparently thought the statement was, "If p and q are two dabas, then there exists one and only one aba containing only p and q." Student 26 Problem 7 The student showed preservation of r u l e s . See Appendix D. 61 Student 28 Problem 2 The student showed preservation of rules. See Appendix D. Student _39 Problem 7 The student showed preservation of rules. The student thought there was an i n f i n i t e number of dabas in the system. Variation. Three students used the technique of variation. Two of the students varied the problem that was to be solved, while the other student varied the data that was given. Two of the students used variation as they attempted problem 8. Problem 8 could be solved by variation, but i t was usually attempted by cases. Student 25 and student 31 apparently f e l t that variation was not a legitimate technique of problem solving, but a way of cheating in order to arrive at an answer. Actually the way in x^hich these students employed variation, they did not carry the procedure far enough. They wanted to vary the data in order to get a solution instead of varying the data in order to help in the search for the solution. Student 23 and Student 29 Problem 8 The students both suggested that the problem could be solved by making one of the given lengths longer. The students were thinking of adding to the length of one of the given sides i n order to draw a closed figure, and not 6 2 to determine that i f the side was longer a parallelogram could be drawn that would help i n the so l u t i o n of the problem, using the o r i g i n a l lengths of the sides. See Appendix D. Student 29 Problem 4 The student suggested drawing a c i r c l e i n s i d e the t r i a n g l e . The student then f e l t that i t would be possible to contruct a square using the c i r c l e , because there were l i n e s i n d i c a t i n g equal distances from the middle of the t r i a n g l e . Student 3 2 Problem 6 The student thought that i f a c i r c l e with a 3 / 4 inch radius was drawn i t would help i n the construction of the required t r i a n g l e . Extension. The investigator f e l t that at the high school l e v e l i t was not possible to determine,if the student could add or delete c e r t a i n f a c t s from a given system i n order to see what discoveries could now be made. However, i t was f e l t that the h e u r i s t i c of extension i s employed when a student attempts to generalize a given statement. Seven students used the h e u r i s t i c of extension. Three of the students solved problem 9(b) which was given to see i f any of the students would use extension. A l l three of these students had been exposed to computer programming. The three students who used t h i s h e u r i s t i c f o r problem 4 and 6 3 the one student who attempted problem 2 , were not given problem 9(b). Student 1 Problem 2 Line AP was extended past l i n e BC. The student looked at the problem to see i f the solut i o n was now evident. The student concluded that the l i n e would just add more confusion. Student j> Problem 4 The student decided to draw the three a l t i t u d e s of the t r i a n g l e . The student then attempted to see i f a square could be drawn. Student 25 Problem 4 The student thought that i f the base l i n e CB was extended i t might help to solve the problem. I t became evident l a t e r i n the interview that the student could not v i s u a l i z e a square inside the t r i a n g l e . Student 32 Problem 4 The student drew the bisectors of the angles i n order to f i n d the centre of the t r i a n g l e . The student wanted to construct a square with centre the point i n the t r i a n g l e which i s equidistant from the three v e r t i c e s , since there would be equal space on a l l sides. Student 33 Problem 9(b) The student noticed that l 3 + + 3 3 + 2p -( 1 + 2 + 3 +4) and then extended the series to determine 64 what happened with 1 3+ 2 3+ 3 3 + 4 3 + 5 3. The student - 3 3 3 3 added on 6 , 7 , and then 8 and f i n a l l y concluded Pn = 1 J + 2 3 + 3 ^ + . . .+ n 3 = ( l + 2 + 3 + ' . . .+ n ) 2 . See Appendix D. Student 32 Problem 9(h) The student incorrectly concluded that 1 3+ 2 3+ 3 3 + 3 4 = A/100. The student then hypothesized that the sum of any cubes equals the square root of some integer. For example, the student suggested 5^+ 6 3+ 7 3+ 8 3 = x where 'x i s a positive integer. Student 41 Problem 9(h) The student noticed that 1 3+ 2 3+ 3 3 + 4 3 = (1 + 2 + 3+4) . The student then extended the sum by adding 5 3 and 6 3 and concluded that l 3 + 2 3 + 3 3 + , , . + n 3 (1 + 2 + 3 + . . . + n) 2. He then investigated 2 3 + 3 3 , 2 3+ 3 3 + 43, 3 3 + 43, 4 3 + 5 3 and concluded that n 3 + (n + 1) + . . . + (n + x) 3 = (n + (n + 1) + . . . + (n + x)) ( n + ( n + 1)+ . . .+ ( n + x ) + ( n 2 - n ) ) . The student next looked at V l ~ * + •3/2~+ $3~ to determine i f i t was equal to V T . See Appendix D. The students' use of other heuristics. The investigator found evidence that another heuristic was used by eleven of the forty-two students interviewed. The investigator calls the method successive variation. A definition of the heuristic successive variation follows. 65 A problem and i t s s o l u t i o n also i s given to i l l u s t r a t e the use of the h e u r i s t i c . Successive V a r i a t i o n . A probable s o l u t i o n to the given problem i s chosen at random. If the answer i s not correct, the student determines what changes must be made. Then the probable solu t i o n that i s chosen i s va r i e d successively u n t i l the student reaches the correct answer. Higgins r e f e r s to a problem-solving technique that involves "guessing an answer, working out i t s consequences, and— by comparing these with the o r i g i n a l conditions of the problem—improving the o r i g i n a l guess."! The inve s t i g a t o r f e e l s that t h i s h e u r i s t i c has most value i f the problem requires only an approximate answer. The following problem i l l u s t r a t e s how t h i s h e u r i s t i c could be e f f e c t i v e l y used: Find an approximation to three decimal places f o r V27~. The answer can be successively bracketed as follows: To one decimal place 5 < JZ7 < 6. Since 5 2 = 25 and 6 =36, the answer i s closer to 5 than to 6. To two decimal places try ( 5 . 1 ) . The answer to two decimal places i s 5 . 1 < $7 < 5 . 2 . Since ( 5 . 1 ) 2= 26 .01 and ( 5 . 2 ) 2 = 27.04, the answer i s closer to 5 « 2 than to 5 . 1 . Hence to three decimal places try (5.19) 2 The answer to three decimals i s 5.9 < </27 < 5 . 2 0 . The way each of the students used Jon L. Higgins, "A New Look at H e u r i s t i c Teaching," The Mathematics Teacher, LXIV (October, 1971), 489. 6 6 successive variation in the problems they attempted w i l l now be discussed. Student 3 Problem k An estimated guess determined the position of the i n i t i a l figure drawn inside the triangle. Then the placement of the figure was varied unti l a square was obtained. The student referred to this method as t r i a l and error. Student 6 Problem 4 A figure was drawn and then the placement of the lines was varied to attempt to form a square. Student 7 Problem 8 One figure was drawn and then the sides and angles were varied u n t i l the figure was a trapezoid. Student 10 Problem 8 A figure was drawn and the positions of the sides and the measures of the angles were varied u n t i l a solution was obtained. Student 11 Problem 3(b) An estimated guess of the amount of money B i l l , Ron, and Ted originally had was made. The student then suggested that these values be adjusted un t i l the correct answer was obtained. Student 12 Problem k A figure was drawn and then the positions of the sides were moved un t i l a square was constructed. 6 ? Student 1 7 Problem 6 The compass was set at d i f f e r e n t positions and two equal sides were drawn from the a l t i t u d e to the base l i n e . The student drew the t r i a n g l e and then measured the perimeter. If the perimeter was les s than 5» the student lengthened the sides of the t r i a n g l e f o r the next t r i a l . The student r e f e r r e d to the method as t r i a l and error and said i t would take forever. Student 22 Problem 4 An attempt was made to draw a square i n the tr i a n g l e , by varying the placement of the sides of the square. Student 24 Problem 6 The student suggested that some t r i a n g l e s be drawn and then measured to see i f they were correct. The student r e f e r r e d to t h i s technique as t r i a l and error. Student ji4 Problem 6 D i f f e r e n t lengths f o r the sides of the t r i a n g l e were drawn and measurements were taken to determine i f the perimeter was 5 inches. Student J36 Problem 4 An approximate f i g u r e was drawn. Then the positions of the sides were varie d u n t i l a perfect square was obtained. The student c a l l e d t h i s method t r i a l and error. TABLE I THE HEURISTICS USED FOR EACH PROBLEM Problem Number of Students Presented the Problem Number of Students Using the HeurJ LStlCS Cases Deduc-t i o n Inverse Deduction Invariation Analogy Preserva-t i o n of Rules Varia-t i o n Extension Succes-sive V a r i a -t i o n 1 9 5 3 2 9 7 i ! 1 | 1 3(a) 3 1 i 3(b) 9 1 1 i i 1 10 3 1 3 5 9 3 6 -6 8 1 . 1 3 7 8 ! 1 1 2 8 8 6 i i | i 1 ' 1 2 2 9(a) 3 | i i 9(b) 8 ... j _ •• i ! 1 3 CO CHAPTER V SUMMARY AND CONCLUSIONS I. SUMMARY The purpose of t h i s normative survey was to determine i f students used any of the nine h e u r i s t i c s namely, cases, v a r i a t i o n , preservation of ru l e s , deduction, inverse.deduction, i n v a r i a t i o n , extension, analogy, and symmetry, when they attempted to solve mathematical problems that could be solved by using at lea s t one of these h e u r i s t i c s . Each of the forty-two grade eleven students that was interviewed was given two mathematical problems to solve. Although the students were allowed to use p e n c i l and paper, the researcher requested the students to think aloud. The i n v e s t i g a t i o n found evidence that eight of the nine h e u r i s t i c s namely, cases, deduction, inverse deduction, i n v a r i a t i o n , analogy, preservation of rules, extension, and va r i a t i o n , were used. Eighteen students used cases, seven used deduction, three used i n v a r i a t i o n , two used inverse deduction, seven used analogy, two used preservation of rules, three used v a r i a t i o n , and seven used extension. The investigator also found evidence that a h e u r i s t i c which was not mentioned previously i n the study was used by eleven of the students. For the purposes of t h i s i n v e s t i g a t i o n the h e u r i s t i c was c a l l e d "successive v a r i a t i o n . " When the h e u r i s t i c of successive v a r i a t i o n i s used a probable so l u t i o n to the 68 given problem i s chosen at random. If the answer i s not correct, the student determines what changes must be made. Then the probable solu t i o n i s varied successively u n t i l the correct answer i s found. The students' command of the h e u r i s t i c s was not developed and therefore they could not use these techniques e f f i c i e n t l y and e f f e c t i v e l y to solve the problems they were given. The investigator f e l t that i n most cases the students used the h e u r i s t i c s i n a very elementary manner. Need f o r Further Study. The following problems could be investigated, 1. Do students at d i f f e r e n t grade l e v e l s show evidence of using h e u r i s t i c s when solving mathematical problems? 2. Do students who achieve C or lower l e t t e r grades i n mathematics use h e u r i s t i c s ? 3. Can students be taught to use h e u r i s t i c s ? 4. Are any other h e u r i s t i c s used by students? 5. Does computer programming a i d the student to e f f i c i e n t l y use h e u r i s t i c s ? 6. Does flow charting help the student to use h e u r i s t i c methods more e f f e c t i v e l y ? II. CONCLUSIONS This i n v e s t i g a t i o n found evidence that most of the grade eleven algebra students interviewed, used h e u r i s t i c s when solving mathematical problems. Nine h e u r i s t i c s were i d e n t i f i e d namely, cases, deduction, i n v e r s e deduction, analogy, p r e s e r v a t i o n of r u l e s , v a r i a t i o n , extension, i n v a r i a t i o n , and successive v a r i a t i o n . The students' use of these h e u r i s t i c s was g e n e r a l l y very p o o r l y developed. The knowledge of a p a r t i c u l a r h e u r i s t i c u s u a l l y was of l i t t l e help t o the student. However, the knowledge of a h e u r i s t i c enabled the student t o make an e f f e c t i v e attempt to s o l v e the problem even though the student u s u a l l y c ould not e a s i l y and c o r r e c t l y f i n d the s o l u t i o n to the problem. The student who d i d not show the use of any h e u r i s t i c s u s u a l l y was unable to make a c o n s t r u c t i v e attempt to f i n d the s o l u t i o n t o the problem. BIBLIOGRAPHY Ashton, Sister Madeleine Rose. "Heuristic Methods in Problem Solving in Ninth Grade Algebra." Unpublished doctoral dissertation, Stanford University, Stanford, 1962. Asprey, Winifred. "Pseudo-Ternary Arithmetic," The  Mathematics Teacher. LIX (March, 1 9 6 6 ) , 240-253. Bruner, Jerome S. The Process of Education. New York: Vintage Books, I960. Bryant, Steven J., George E. Graham, and Kenneth G. Wiley, Nonroutine Problems in Algebra Geometry and Trigonometry. New York: McGraw-Hill Book Company, 1965. Crutchfield, Richard S. "Creative Thinking in Children: Its Teaching and Testing," Intelligence: Perspective  1965. New York: Harcourt, Brace and World, Inc., 1966. Duncker, Karl. "On Problem Solving," trans. Lynne S. Lees. Psychological Monographs, LVIII (whole no. 270, 1 9 4 5 ) . Grossnickle, Poster E. "Verbal Problem Solving," The  Arithmetic Teacher. XI (January, 1 9 6 4 ) , 12-17. Hanna, Paul R. "Methods of Arithmetic Problem Solving," The Mathematics Teacher. XXIII (November, 1 9 3 0 ) , 442-•430. Henderson, Kenneth B. "Strategies for Teaching by the Discovery Method," Updating Mathematics, high school teachers edition. New London, Connecticut: Croft Educational Services, 1964. 57-64. Higgins, Jon L. "A New Look at Heuristic Teaching," The  Mathematics Teacher. LXIV (October, 1 9 7 1 ) , 487-495. Jones, P h i l l i p S. "Discovery Teaching From Socrates to Modernity," The Mathematics Teacher. LXIII (October, 1 9 7 0 ) , 501-50BT" Kahl, Joseph D. The American Class Structure. New York: Holt, Rinehart and Winston, 1957. Kaplan, Jerome D. "An Example of Student-Generated Sequences in Mathematics Instruction," The Mathematics Teacher. LVII (May, 1 9 6 4 ) , 298-302. * 72 73 Kilpatrick, Jeremy. "Analyzing the Solution of Word Problems in Mathematics: An Exploratory Study." Unpublished doctoral dissertation, Stanford University, Stanford, 1967. . "Problem Solving in Mathematics," Review of Educational Research, XXXIX (October, 1969), 52"3^534. Larsen, Charles McLoud. "The Heuristic Standpoint in the Teaching of Elementary Calculus," Dissertation Abstracts, xxi ( 1 9 6 1 ) , 2632-2633. Leask, Isabel Campbell. "The Effectiveness of Simple Enumeration as a Strategy for Discovery." Unpublished Master's thesis, The University of B r i t i s h Columbia, Vancouver, 1968. Luchins, Abraham. "Mechanization in Problem Solving," Psychological Monographs, LIV (whole no. 248, 1 9 4 2 ) . MacPherson, Eric D. unpublished paper on heuristics, The University of B r i t i s h Columbia, Vancouver,,1969. (mimeographed.) Newell, Allen, J.C. Shaw, and Herbert A. Simon. "The Processes of Creative Thinking," Contemporary Approaches to Creative Thinking. New York: Atherton Press, 1964. 03-119. Nichols, Eugene D., Wagner G. Collins, and Eric E. MacPherson. Modern Elementary Algebra. Toronto: Holt, Rinehart and Winston of Canada, Limited, 1961. Plant, Edmund C. Geometrical Drawing. London: MacMillan and Company, Limited, 1899. Plunkett, Betty. "Aba Daba Daba," The Mathematics Teacher, LIX (March, 1 9 6 6 ) , 236-239.. Polya, George. Mathematical Discovery. 2 vols. New York: John Wiley and Sons, Inc., 1962. . How to Solve It. second edition. New York: Doubleday and Company, Inc., 1957. Russell, David H. Children's Thinking. Toronto: Blaisdel Publishing Company, 1956. 74 S a l k i n d , C h a r l e s T. The MAA Problem Book I I . New York: Random House, Inc., 1 9 6 6 . . The Contest Problem Book. New York: Random House, I n c ~ 1 9 6 1 . Schaaf, Oscar. "Student D i s c o v e r y of A l g e b r a i c P r i n c i p l e s as a Means of D e v e l o p i n g A b i l i t y t o G e n e r a l i z e , " The  Mathematics Teacher, XLVIII (May, 1955), 324-327. T a y l o r , Donald W. " D i s c u s s i o n of Papers by A d r i a n de Groot and by J e f f e r y M. Paige and H e r b e r t A. Simon," Problem  S o l v i n g : Research, Method, and Theory. New York: John Wiley and Sons, Inc., 1966. Washburne, C a r l e t o n W., and Raymond Osborne. " S o l v i n g A r i t h m e t i c Problems," The Elementary S c h o o l J o u r n a l , XXVII (November, 1 9 2 6 ) , 219-226; [December, 1 9 2 6 ) , 2 9 6 - 3 0 4 . Welker, Latney Conrad. "A Study of I n t e r r e l a t i o n s h i p s i n A r i t h m e t i c a l Problem S o l v i n g . " Unpublished d o c t o r a l d i s s e r t a t i o n , The U n i v e r s i t y of Southern M i s s i s s i p p i , H a t t i e s b u r g , 1 9 6 2 . W i l l , H e r b e r t I I I . " T r a n s f e r of Problem S o l v i n g A b i l i t y Gained Through L e a r n i n g by D i s c o v e r y , " D i s s e r t a t i o n  A b s t r a c t s , XXVIII ( 1 9 6 ? ) , 1 3 1 9-A to 13203A";; Wilson, James W i l l i a m . " G e n e r a l i t y of H e u r i s t i c s as an I n s t r u c t i o n a l V a r i a b l e , " D i s s e r t a t i o n A b s t r a c t s , XXVIII (January, 1 9 6 8 ) , 2 5 7 5-A. W r i g l e y , Reverend A l f r e d . A C o l l e c t i o n of Examples and Problems i n Pure and Mixed Mathematics with Answers and O c c a s i o n a l H i n t s . N i n t h E d i t i o n . London: Longmans, Green, and Company, I 8 7 5 . APPENDICES 75 APPENDIX A 76 STUDENT INFORMATION FORM Date Name School Grade Birthdate _ — Month Day Year Mathematics Le t t e r Grade Christmas Easter Mathematics Teacher Do you plan to take Math 12? Yes No Do you plan to take mathematics either i n grade 13 or i n f i r s t year at a junior college or university? Yes No Try your best to solve the following two problems, using any method you wish. You may use pe n c i l and paper, but please think aloud as you work. Be concerned with t e l l i n g a l l your thoughts. Don't worry about the accuracy of your answer. If the method that you f i r s t choose does not solve the problem, you are encouraged to use as many other methods as you can to attempt to solve the problem. APPENDIX B 77 INTERVIEW FORMAT My name i s Laurie Dinsmore. I'm attending U.B.C. where I'm completing my M.A. i n Mathematics Education. I already have a B.Sc. i n Mathematics and Chemistry. I've been Interviewing some grade 11 math students, those who make good marks l i k e yourself, to f i n d out how students solve p a r t i c u l a r math problems that I give them. I'm interested i n how you solve mathematical problems. I'm interested i n your thoughts as you attempt to solve a problem and not i n the f i n a l answer. I want you to think out loud as you work. You may use pe n c i l and paper but always remember to t e l l me your thoughts. Just as a pr a c t i c e i n thinking out loud t e l l me how you would attempt to solve t h i s problem. Try to t e l l me a l l your thoughts. I may ask you some questions while you work. I'm interested i n how you try to solve the problem and not i n the accuracy of your solu t i o n . The student was given one of the following three problems to solve. (a) 215 t i c k e t s were sold f o r a concert. Adult t i c k e t s were 75^ each and children's t i c k e t s were 35^ each. The sale of both kinds of t i c k e t s amounted to $135.25. How 78 many adult t i c k e t s were s o l d ? 1 (b) How long i s a rectangular l o t i f i t s length i s 10 feet longer than i t s width and i t s perimeter i s 52 feet?2 (c) The sum of three consecutive odd Integers i s 273. What are the integers? 3 The student was given encouragement and t r a i n i n g i n thinking aloud, according to the i n d i v i d u a l students, i n order to develop his a b i l i t y f o r t e l l i n g a l l h i s thoughts. The student was then given the form shown i n Appendix A. The student was given the two problems, one at a time. No assistance was given unless the i n v e s t i g a t o r f e l t that the requested information would not help the student to plan h i s solution. Questions such as; What are you thinking now?, Any reason f o r that choice?, Finding anything?, and What did you do there? were frequently asked. An attempt was made to get a l l the student's thoughts, including his reasons f o r c e r t a i n steps or methods of solution. I t was considered important to make c e r t a i n the student understood the problem without giv i n g him any clues as to how he might solve i t . The students were not t o l d i f a p a r t i c u l a r step was correct, ^Eugene D. Nichols, Wagner G. C o l l i n s , and E r i c D. MacPherson, Modern Elementary Algebra (Toronto; Holt, Rinehart and Winston of Canada, Limited, 1 9 6 l ) , p. 194. 2 I b l d . , p. 197. 3 l b i d . , p. 201. 79 but were encouraged to keep going u n t i l they e i t h e r were s a t i s f i e d that they had a s o l u t i o n or u n t i l they gave up. 80 APPENDIX C SOLUTIONS TO THE PROBLEMS USED IN THE INVESTIGATION PAGE Cases 81 Deduction . . . . . . . . . . . . . 83 Inverse Deduction 84 Invariation 85 Analogy 87 Symmetry 90 Preservation of Rules 91 Variation 93 Extension . . . . . 94 APPENDIX C 81 SOLUTIONS TO THE PROBLEMS USED IN THE INVESTIGATION 1. Gases. Find a l l real numbers x which satisfy the following 1 equation x + 1 + 3 x - 1 - 2 x - 2 = x + 2 Suppose that x i s a real number satisfying the equation. Consider the five casesi Case 1 Let x ^  2 then x + 1 - 2 x - 2 + 3 x - 1 x + 1 - x + 3(x - 1) - 2(x - 2) = x + 2 therefore x + 2 = x + 2 and the equation holds for a l l x ^  2 Cases 2 Let 1< x < 2 then x + 1 + 3 x - 1 - 2 x - 2 x + 1 - x + 3(x - 1) - 2(2 - x) = 5x - 6 Steven J. Bryant, George E. Graham, and Kenneth G. Wiley, Nonroutlne Problems in Algebra Geometry and Trigonom-etry (New York: McGraw-Hill Book Company, 1 9 6 3 7 7 P. 5 . therefore 5x - 6 = x + 2 4x = 8 x = 2 so there i s no s o l u t i o n Case 3 Let 0 ^  x < 1 then x + 1 + 3 x - 1 - 2 x - 2 x + 1 - x + 3 ( 1 - x) - 2 ( 2 - x) = -x therefore -x = x + 2 x = -1 so there i s no s o l u t i o n Case k Let -1 ^  • x < 0 then x + 1 - I x + 3 - 2 x - 2 x - 1 x + 1 + x + 3 ( 1 - x) - 2 ( 2 - x) = x therefore x = x + 2 so there i s no s o l u t i o n Case 5 Let x < -1 then x + 1 + 3 x - 1 - 2 x - 2 -x - 1 + x + 3 ( 1 - x) - 2 ( 2 - x) = -x - 2 therefore -x - 2 = x + 2 -2x = 4 x = -2 i s the only s o l u t i o n 2 Thus the solutions are x >^  2 or x = -2 2 I b i d . , p. 33. 2. Deduction. 83 Triangle ABC is isosceles (the two sides AB and BC have the same length) with base AC. Point P i s in CB and point Q is in AB such that AC = AP = PQ = QB. What is the number of degrees in angle B? 3 Statement Let m be the magnitude of angle B Reason 1. angle QPB = m 2. angle BQP = 180 - 2m 3. angle AQP = 2m 1. triangle QBP i s isosceles with side QP equal to side QB 2. sum of the angles in a triangle i s 180° 180° = angle B + angle QPB + angle BQP = m + m + angle BQP 3. angle AQB = angle AQP + angle PQB = 180° angle AQB = angle AQP + 180° - 2m = 180° -^Charles T. Salkind, The MAA Problem Book II (New Yorks Random House, Inc., 1966), p. 13. 84 Statement Reason 4. angle QAP = 2m 4. triangle QAP is isosceles with side AP equal to side QP o 5. angle QPA = 180 - 4m 5. sum of the angles in triangle QAP = 180 = angle AQP + angle QAP + angle QPA = 2m + 2m + angle QPA 6. angle APC = 3m 6. angle BPG = 180° = angle QPB + angle QPA + angle APC = m + (180° - 4m) + angle APC 7. angle C = 3m 7. triangle APC is isosceles with side AP = side AC 8. angle A = 3m 8. triangle ABC is isosceles with side AB = side BC o 9. angle B = 25 5/7 9. sum of the angles of triangle ABC = 180° = angle A + angle B + angle C = 3m + 3m + m The heuristic of hypothetical deduction is used. If the subproblem of finding the base angles of the isosceles triangle ABC can be solved, then angle B can be found from o the relation, the sum of the angles of the triangle is 180 . The solution of the problem involves finding the magnitude of successive angles, in terms of the unknown angle B, u n t i l the magnitude of the base angle i s found.^ 3 . Inverse Deduction. B i l l , Ron, and Ted each have some money of their own but they do not a l l have the same amount of money. Now, Ibid., p. 53. 85 B i l l decides to give some of h i s money to Ron and some of hi s money to Ted. He gives Ron as much money as Ron already has and also he gives Ted as much money as Ted already has. A f t e r t h i s Ron decides to give some of h i s money to B i l l and to Ted. He gives each of them as much money as each of them now has. Then Ted decides that he w i l l give B i l l and Ron as much money a's each of them has. Now B i l l has l6fi and Ron has 16^ and Ted has l6fi. How much money d i d B i l l o r i g i n a l l y have before he gave any of his money away?5 Working backwards from the l a s t condition to the f i r s t , the following table can be derived.^ amounts i n cents step 4 step 3 step 2 step 1 B i l l 16 8 4 26 Ron 16 8 28 14 Ted 16 32 16 8 4, In v a r i a t i o n . Using a compass and a straightedge construct a square inside the given t r i a n g l e ABC such that two of the v e r t i c e s of the square are on the base CB of the t r i a n g l e and the other two v e r t i c e s of the square are on the other two sides ^Ibid., p. 26. 6 I b l d . t p. 75. 86 AC and AB of the t r i a n g l e , one vertex on each side.? A Suppose only a part of the condition i s retained. Draw a square DEFG with two v e r t i c e s on the base of the given t r i a n g l e and one of the other two v e r t i c e s on one of the two sides of the t r i a n g l e . Then draw square IJKL of d i f f e r e n t size with the same requirements. The locus of the fourth vertex of the square i s a str a i g h t l i n e . Draw the str a i g h t l i n e y. Using the point of i n t e r s e c t i o n x of t h i s l i n e with the side of the t r i a n g l e , construct the square abed with side the perpendicular distance from t h i s point to the base of the t r i a n g l e . 8 ^George Polya, How to Solve I t (second e d i t i o n ; New York: Doubleday and Company, Inc., 1 9 5 7 ) , p. 23. 8 I b i d . . p. 23-25. 87 5. Analogy. Find the values of the five unknowns x, y, z, u, and t that satisfy the following set of five simultaneous linear equations. 9 4y - 3x + 2u = 5 9x - 2z + u = 41 7y - 5z - t = 12 3y - 4u + 3t = 7 7z - 5u = 11 Consider the set of three simultaneous linear equations 2x + 4y + 5z = 49 (1) 3x + 5y + 6z = 64 (2) 4x + 3y + 4z = 55 (3) According to the algorithm, (a) eliminate the variable x using (i) equations (1) and (2) to obtain equation (4) 3 x [2x + 4y + 5z = 49] 6x + 12y + I5z = 147 -2 x [3x + 5y + 6z = 64] -6x - 10y - 12z = 128 2y + 3z = 19 (4) ( i i ) equations (1) and (3) to obtain equation (5) everend Alfred Wrigley, A Collection of Examples  and Problems in Pure and Mixed Mathematics with Answers and  Occasional Hints [London: Longmans, Green, and Company, 1875), P. 3%~. 8 8 4 x [ 2 x + 4 y + 5 z = ^ 9 ] 8 x + l 6 y + 2 0 z = 1 9 6 - 2 x [ 4 x + 3 y + 4 z = 5 5 ] - 8 x - 6 y - 8 z = - 1 1 0 lOy + 1 2 z = 8 6 ( 5 ) (b) eliminate the v a r i a b l e y using equations ( 4 ) and ( 5 ) - 5 x [ 2 y + 3 z = 1 9 ] - 1 0 y - 1 5 z = - 9 5 - 3 z = - 9 z = 3 lOy + 1 2 z = 8 6 lOy + 1 2 z = 8 6 (o) solve f o r y using equation ( 4 ) 2 y + 3 ( 3 ) = 1 9 2 y + 9 = 1 9 2 y = 1 0 y = 5 (d) solve f o r x using equation ( 1 ) 2 x + 4 ( 5 ) + 5 ( 3 ) = 4 9 2 x + 2 0 + 1 5 = 4 9 2 x = 14 x = 7 Now to consider the problem of f i v e simultaneous l i n e a r equations 9 x - 2 z + u = 41 ( 1 ) 7 y _ 5 z - t = 1 2 ( 2 ) 4 y - 3 * + 2 u = 5 ( 3 ) 3 y - 4 u + 3 t = 7 ( 4 ) 7 z - 5 u = 1 1 ( 5 ) Using the algorithm f o r three simultaneous l i n e a r equations, the variables x, y, z, and t could be eliminated i n the f i v e simultaneous equation case. Applying the algorithm analogously (a) eliminate the variable x using equations ( 1 ) and ( 3 ) to obtain equation ( 6 ) 89 9x - 2z + u = 41 9x - 2z + u = 41 -3 x [4y - 3x + 2u = 5] -9* + 12y + 6u = 15 -2z + 12 y + 7u = 56 (6) (b) eliminate the variable y using (i) equations (6) and (2) to obtain equation (7) 7 x [-2z + 12y + 7u = 56] 84y - l4z + 49u = 392 -12 x [7y - 5z - t = 12] -84y + 60z + 12t = -144 46z + 49u + 12t = 248 (7) ( i l ) equations (6) and (4) to obtain equation (8) -2z + 12y + 7u = 56 -2z + 12y + 7u = 56 -4 x [3y - 4u + 3t = 7] -12t - 12y + l6u = -28 -2z - 12t + 23u =28 (8) (c) eliminate the variable z using (i) equations (7) and (8) to obtain equation (9) 46z + 49u + 12t = 248 46z + 49u + 12t = 248 23 x [-2z - 12t + 23u = 28] -46z + 529u - 276t = 644 578u - 264t = 892 (9) ( i i ) equations (7) and (5) to obtain equation (10) 7 x [46z + 49u + 12t = 248] 322z + 343u + 84t = 1736 -46 x [7z - 5u = 11] -322z + 230u = -506 573u + 84t = 1230 (10) (d) eliminate the variable t using equations (9) and (10) 7 x [578u - 264t = 892] 4o46u - 1848t = 6244 22 x [573u + 84t = 1230] 12606u + 1848t = 1230 l6652u = 33304 u = 2 90 (e) solve for t using equation (9) 578(2) - 264t = 892 1156 - 264t =892 -264t = -264 t = 1 (f) solve for z using equation (7) 46z + 49(2) + 12(1) = 248 46z + 98 + 12 = 248 46z = 1138 z = 3 (g) solve for y using equation (6) -2(3) + 12y + 7(2) = 56 -6 + 12y + 14 = 56 12y = 48 y = 4 (h) solve for x using equation (3) 4(4) - 3x + 2(2) =5 16 - 3x + 4 = 5 -3x = -15 x = 5 6. Symmetry. Using a compass and a straightedge construct an isosceles triangle (two of the sides have the same length) given the altitude (the perpendicular distance from the vertex to the side opposite) equal to 1% inches and the perimeter equal to 5 inches, 1 0 A S G D B T luEdmund G. Plant, Geometrical Drawing (London: MacMIllan and Company, Limited, 1899), p. 103. 91 Draw any i s o s c e l e s t r i a n g l e ABC w i t h a l t i t u d e AD equal t o l£- i n c h e s . S i n c e AC = AB and a n g le C = angle B, the t r i a n g l e i s symmetric w i t h r e s p e c t t o the a l t i t u d e AD. T h e r e f o r e t r i a n g l e ACD i s congruent t o t r i a n g l e ADB and DC = DB. Suppose t h a t t h i s i s the c o r r e c t t r i a n g l e . I n order to p r e s e r v e the symmetry, the p e r i m e t e r ( l e n g t h ST) c o u l d be i n t r o d u c e d i n t o the p i c t u r e a l o n g CB so t h a t SC = AC and BT = AB. N o t i c i n g t h a t t r i a n g l e SAC and t r i a n g l e ABT a r e i s o s c e l e s , the a l t i t u d e s from C and B would b i s e c t SA and AT r e s p e c t i v e l y . T h e r e f o r e the c o n s t r u c t i o n of t r i a n g l e ABC c o u l d now proceed as f o l l o w s . Draw ST equal t o 5 i n c h e s . B i s e c t ST by AD. Draw AD p e r p e n d i c u l a r t o ST and equal t o 1% i n c h e s . J o i n AS# and AT. B i s e c t SA and AT by l i n e s p e r p e n d i c u l a r t o SA amd AT which meet ST i n C and B r e s p e c t i v e l y . J o i n A and^C- and A and B. ABC i s the t r i a n g l e r e q u i r e d . i X 7. P r e s e r v a t i o n of R u l e s . ( i ) L i s t a l l the abas ( i i ) Decide i f the f o l l o w i n g statements a r e t r u e or f a l s e U s i n g the f o l l o w i n g mathematical system c o n t a i n i n g ? 11 Edmund C. P l a n t , l o c . c i t . 92 Statement 1 If p and q are two dabas, then there exists one and only one aba containing both p and q. Statement 2 If L is an aba, there exists a daba not in L. Statement 3 If L i s an aba, and p is a daba not in L, then there exists one and only one aba containing p and not containing any daba that is in L. Preserving the rules using the definition and the elements (dabas) in this system results in the following solution. abas a^ = \dj, d2. d 3) a2 = w, d 3, d ^ a 3 = l d l , dit, a 4 = idl» d5' d6^  a 5 = I d l . d 2, <lk] a6 = i d l . d2, an = {d l f d 2, d6^  a8 = ldl> d 3, d5] ao, = * d l , d3-a10 = {dl» d 4 » d6] a l l = fa>. d 3, d ^ a12 = ld2. d 3, a13 = ld2» d3- d6} a l 4 = ld2. d^, d5  a15 = \d2. d 4 » &i6 - ld2» d 5 , d63 a17 = ld3, d 4 . d5$ a18 = ld3» d 4 » a19 = fa-3' d 5. d6$ a20 = U4, d5' d63 Statement 1 False If p and q are two dabas, then there exists more than one aba containing both p and q. -^Betty Plunkett, "Aba Daba Daba," The Mathematics  Teacher, LIX (March, 1966), 236-237, citing Howard Eves, and CV. Newsom. An Introduction to the Foundations and  Fundamental Concepts of Mathematics (New York: Holt, Rinehart and VJinston, 1964) , p. 80. 93 Let p = di and' q = d2 then a x = \dlt d 2, a 5 = t d l i d 2 » d i J ^ a 6 = t d L d 2 . d 5 ^ a 7 = { d l t d 2 , dg] Statement 2 True Statement 3 T r u e 1 3 8. V a r i a t i o n . Construct a trapezoid (a closed plane f i g u r e with four sides, two of which are p a r a l l e l ) being given the lengths a, b, c, and d of i t s four s i d e s . ± l * a Let a be the base and c the top with a and c p a r a l l e l but unequal. The other two sides b and d are not p a r a l l e l . Suppose we vary c. What happens when c increases u n t i l i t becomes equal to a? The trapezoid becomes a parallelogram. The t r i a n g l e that has now been added has sides b, d, and 1 3 T b l d . , p. 2 3 7 . iDp o l y a , op_. c i t . , p. 2 1 1 . 94 a - c. Since we have the lengths a, b, c, and d we can construct the t r i a n g l e and then by completing the parallelogram the o r i g i n a l problem can be solved. x5 9. Extension. 1 + 8 + 27 + 64 = 1 0 0 l 3 + 8 + 27 + 64 = 1 0 0 What can you discover?1 6 The f i r s t observation i s , l 3 + 2 3 + 3 3 + 4 3 = 1 0 2 = ( 1 + 2 + 3 + 4 ) 2 Now the question can be asked, "Does i t happen that the sum of other consecutive cubes i s a square?" The sum of the consecutive numbers can be extended to, l 3 + 2 3 + 3 3 + 4 3 + 5 3 = 1 + 8 + 27 + 64 + 1 2 5 = 2 2 5 = 1 5 2 + ( 1 + 2 + 3 + 4 + 5 ) 2 Then t h i s can be continued to suggest that, 1"' 7 l 3 + 2 3 + 3 3 + . . . + n 3 + ( 1 + 2 + 3 + . . . + n ) 2 -^Ibid., p. 2 1 1 - 2 1 2 . l 6 I b i d . , p. 1 0 8 . 1 7 I b i d . , p. 1 0 8 . APPENDIX D 95 SAMPLE INTERVIEWS AND ANALYSES Page He u r i s t i c Interview Problem Student Interview Analysis Cases 1 1 2 96 99 2 1 17 100 102 Deduction 3 2 1 103 105 4 2 33 106 108 Inverse Deduction 5 3 2 109 113 6 3 11 114 118 Analogy- 7 5 16 119 122 8 5 38 123 124 Preservation of Rules 9 7 26 125 128 10 7 28 129 130 V a r i a t i o n 11 8 23 131 134 12 8 29 135 138 Extension 13 9 33 139 141 14 9 41 142 144 Successive V a r i a t i o n 15 4 3 145 146 16 4 22 147 148 9 6 Interview 1 Problem 1 Student 2 H e u r i s t i c : Cases Student Well f i r s t I'm t r y i n g to just get the o v e r a l l p i c ture. I think I ' l l take them l i k e out of the absolute value things and just try and work i t out l i k e that. Interviewer Any reason f o r that choice? Student Well you can't r e a l l y work with i t when i t ' s i n the absolute value things, so I ' l l write i t down. Well now I think I ' l l just try and solve f o r x and see what happens. Boy I've got a problem; have 0 = -2 so. Interviewer So what are you thinking about now? What did you think when you saw that? Student Well that can't be. You can't have two numbers equal. So It could mean that there's no so l u t i o n or else I figured i t out wrong. Interviewer Well you have a choice now. Do you think there's no so l u t i o n or do you think you, do you think you fig u r e d i t out wrong? Student I don't know I probably figured i t out wrong. You know there's got to be probably a so l u t i o n . Interviewer Why would there have to be a solution? Student I don't know. Interviewer What are you thinking about now? What are you looking at now? Student Oh what I'm just t r y i n g to figur e i t out i f there's any way, how the absolute value would a f f e c t i t . You know to see i f there's any way we could get around i t . Interviewer What about t h i s absolute value, what, what's i t doing or what are you thinking when you said you said you were t r y i n g to get around the absolute value? Interview 1 (continued) 97 Student Well i f the absolute value wasn't there, well then there*d be no, x would not equal, anything you can have i t unless i t ' s imaginary number or something. But with the absolute value there i t could be either two d i f f e r e n t numbers, so don't know how to do that. Interviewer Is there any other, do you think you could try i t any other way? Do you have any other any other ideas? Student Oh I don't know. I suppose you could t r y sub s t i t u t i n g a number f o r x, you know i f you want to just guess or something. Interviewer Well you can try anything you want. That's what I'd l i k e to, you know, see what you want to tr y . Student I think I ' l l just try su b s t i t u t i n g i n just to see i f anything w i l l work. I think i t ' s probably the n u l l set or something. Interviewer You can, you know, whenever you f i g u r e you're fi n i s h e d , but i f you want to keep on at that that's f i n e . So you t r i e d putting zero i n and that didn't work fo r you? Student No, i t didn't. Interviewer You just, what are you doing now? Putting other numbers i n , or or what? Student No, I'm just kind of thinking i f I could somehow get t h i s equal, the equation, to see i f i t w i l l correspond to x + 2. Interviewer What do you mean by making the equation equal? Student Oh I don't know exactly. Just l i k e say t h i s was x l i k e here, just saying l e t ' s say -x was and just t r y i n g that out. Interviewer So you put a -x i n there f o r the f i r s t one. What are you going to do when you come to that that next absolute value one? You're putting, you're leaving that unchanged now? 98 Interview 1 (continued) Student Cause i f i t was negative then the opposite of i t would be s t i l l p o s i t i v e . I t works out that way you get x + 2 = x + 2. Interviewer And what does that say to you? Student That i f x was p o s i t i v e i t wouldn't work out. I f i t was negative maybe i t would. So I think I ' l l t ry a negative number and see i f i t works out. Interviewer Okay. Student Yes i t works out that way. So I I think I could assume that a l l negative numbers would probably work. Analysis of Interview- 1 Problem 1 Student 2 99 Cases "you could try s u b s t i t u t i n g a number f o r x . . . I ' l l just try s u b s t i t u t i n g i n just to see i f anything w i l l work . . . I ' l l try a negative number and see i f i t works out." The student t r i e d 0, -x, and -2. 100 Interview 2 Problem 1 Student 17 Heuristi c : Gases Student F i r s t I'd, can I write t h i s down? Interviewer Yes. Student Okay f i r s t I'd multiply out the brackets, what's i n front of them. Interviewer Those are absolute values, okay? Student Oh oh st a r t over. Well then f i r s t of a l l I'd add add the x's together, x - x no that doesn't look r i g h t . Interviewer Any reason why you thought that wouldn't work? Student Let's see. I have to make i t equal. There's two values f o r the absolute value. Interviewer Okay. What do you mean by two values f o r absolute value? Student Well one x could equal negative and t h i s absolute value could, makes i t p o s i t i v e and then x could be po s i t i v e and s t i l l might get the same answer. And l e t ' s see, l e t ' s see remove the absolute value and f i n d x. Let's see now, I've, add the x's up 1 + 3 i s 4 - 3 i s 1 + 5 = x + 2 and now I'd br i n g the x»s over to t h i s side and then I get zero equals - 3 . Oh that didn't work. Interviewer Well. What do you think now? Is there any reason why you just l e f t out the absolute values there and solved f o r x? Student No. I thought I was going to make, f i n d one way, f i n d x f i r s t whatever i t came out to be. Like maybe i f x was, equalled -3, then I could substitute i t i n there and see i f i t was true, and then just try the po s i t i v e value of l t . 101 Interview 2 (continued) Interviewer Oh I see. Student But i t didn't work. Interviewer Any other ideas? Student Oh I don't know. Interviewer Think you could do anything else with i t ? Student Oh I can't even remember what I did with, used to do, cause I haven't done these f o r so long. Interviewer You used to do things l i k e that? Student Yes, about three grades ago. Interviewer Is that what you're t r y i n g to do now, remember how you did them? Student Yes. I'm not sure. Would you multiply? I f I'd multiply a negative, the things i n front of these absolute values, i f you'd multiply three times x or just leave i t . Interviewer Well you can try whatever you'd, you know, whatever you'd l i k e to try you can try i t and see what happens. I'm just Interested i n how many ideas you have. You can, you know, when you're ready, when you run out of ideas of course you can stop, but. Student Let's see. Interviewer Thinking of anything now? Student No I don't think so. I can't think of what I'd do. Interviewer Do you want to try another problem or do you think you might have another idea? Student Yes. I don't think I have another one. Analysis of Interview 2 Problem 1 Student 17 102 Gases "there's two values f o r the absolute value . . . x could equal negative . . . and then x could be p o s i t i v e . . . I thought I was going to make, f i n d one way, f i n d x f i r s t whatever i t came out to be. Like maybe i f x was, equalled -3, then I could substitute i t i n there and see i f i t was true, and then just try the p o s i t i v e value of i t . " 103 Interview 3 Problem 2 Student 1 Heuristic'.Deduction Student Well these, t h i s i s an Isosceles according to t h i s and so i s t h i s . So a l l isosceles t r i a n g l e s . This Is i n geometry i n grade 10. So AC = AP. Do you mind i f I mark this? Interviewer No I don't mind. Student So I can f i n d out what t h i s i s l i k e . So now i f I can f i g u r e out which angles are equal. Oh these l f i t ' s i sosceles the angles should be equal, i s that right? Interviewer What angles? Student Well the ones at the bottom. So that should equal to that and t h i s should be equal to that and t h i s should be equal to that and t h i s should be equal to that and t h i s should be equal to that. So that should be 180 divided by three. Sixty degrees? Interviewer That's your, that's good. That's your solution? Student I think so, yes. Let me check that over again f o r mistakes. Angle B. So these two w i l l have to be equal. Now that has to be equal to that. No these two won't necessarily have to be equal. I don't, just t r y i n g to f i g u r e out which one that can be equal to. Interviewer What are you thinking now? Student I'm t r y i n g to fig u r e out which angle w i l l r e l a t e to t h i s somehow. I t won't be that, i t ' s not p a r a l l e l or anything. And that's d e f i n i t e l y out. I f I put a l i n e there. Interviewer So what are you saying now? Student I t ' s not 60 degrees l i k e I thought. These two are equal and that's equal. I ' l l just confuse i t some more i f I bring the l i n e out across P. I t ' l l just confuse i t some more I think so. 104 Interview 3 (continued) Interviewer T e l l me a l l the things that you're thinking about doing. Like you just told me now you, you know, you were thinking maybe you'd bring that line out but that you didn't didn't feel that was going to work. T e l l me a l l these l i t t l e things that you're trying. Student Well I'm just generally looking i t over and trying to relate everything together. Interviewer Anything you've been looking for? You mentioned something awhile ago. Student Oh, here's another, here's another isosceles triangle QPA, that I didn't see before. So this angle here should be equal to angle Q, i f that's of any help. It's probably just more confusion in the end. It's probably just confusion. Let's see, s t i l l ' trying to relate everything together. This w i l l be, what's that 180? Yes. This angle BQP w i l l be P, w i l l be 180 minus the other angle, i f that's of any help. It's crazy. Sometimes at this stage of the game I take a wild guess, so. But I don't think I want.to now. Interviewer Well you can try as many different approaches as you, as you feel you can. And then i f you want to quit, well. Student Let's see. If that equal to that. A i s equal to C and C w i l l be equal to APC and that's fine and dandy to there. And QAP w i l l be equal to AQP and QPB w i l l be equal to QBP. Everything is equal, but i t doesn't come out. I think I'm just wasting time. Interviewer You've got no more ideas? Student No, I'm afraid not. 105 Analysis of Interview 3 Problem 2 Student 1 Deduction " i f i t ' s isosceles the angles should be equal . . . the ones at the bottom. So that should equal to that and this should be equal to that and this should be equal to that and this should be equal to that and this should be equal to that. So that should be 180 divided by three. Sixty degrees? . . . here's another isosceles triangle QPA . . . so this angle here should be equal to angle Q . . . angle BQP w i l l be P, w i l l be 180 minus the other angle." 106 Interview 4 Problem 2 Student 33 Heuristic: Deduction Student Okay. Well f i r s t of a l l , by a l l those sides equal, these are a l l isosceles triangles. And so I haven't been given any numbers, so I ' l l just do i t almost algebraically. Angle C. Let's see now. Yes angle C. Sorry, angle B = angle P, angle C = angle BAC and I can, I see that I can use some possibly supplementary angles. I know the angle sum of a triangle i s 180 degrees and I could use supplementary angles to say that i f B i s equal to B. Two times angle B equals angle PQA because two angle B i s going to be supplementary to that and that's supplementary to that, you know. Interviewer Supplementary to the Q there, yes I see. Student Okay angle PQA i s equal to PAQ so. And angle, so those two angles. It w i l l be angle C, 180 degrees minus angle C and angle CAQ. Or, sorry i t ' l l be 180 minus two times angle CAQ. Interviewer Well, could I ask you why for that? Student Why? Cause I'm going to try and work, dividing CAQ into that angle plus that angle which w i l l I hope isolate angle PAQ. And I've already got a relationship with angle B and PAQ and a I'm just starting to get one now, i f I can just find i t . Angle B = 180°- 2PAC - 2PAQ. That, so two times angle B = angle PAQ and you could say that angle B, 180°- 2PAC - Wangle B. Add Wangle B to both sides. 5B that equals 180°. That's 2PAC. Now I get that angle. I don't really think I, that's going to get me anywhere. Interviewer Could I ask you what your reason was for the 2PAC there? Student 2PAC was there from the time before. PAC plus PAQ is equal to the QAC and. 107 Interview 4 (continued) Interviewer Okay. What, where did that 2PAC come from i n the f i r s t step? _ o Student Oh, angle B was 180 - angle 2 - 2angle CAQ. Now i f you break down angle 2CAQ you get two of these plus two of these, okay? Interviewer I get i t . Okay. Student Now i f I could get, try to work from PAC. Now 2PAC i s going to equal PAG = 180°- 2G and 2PAC = 360°- 4C. Let's see i f I've got angle C up here. Anyways, yes angle PAC. So 5B = 4C. I can substitute -4C f o r that. Now I get -180° - 4C or 5B + 180*= 4C, sorry, negative kC which means that and also 180 degrees, 180 degrees minus 2C = B. So B = 180°- 2C. So I should be able to get i t from t h i s and I, I'm just going to add both equations to both sides. 6B + 180°= 180° negative 4C plus that. So i t ' s negative 4c plus. So i t ' s 180°- 6c, 6B, negative 6c which means that I've done something wrong with. Interviewer What's your o v e r a l l plan f o r solving i t ? Could you just explain you know, your? Student My o v e r a l l plan's r i g h t . You can get values f o r B by, i n terms of other d i f f e r e n t angles, and you can by working i t , by using equations, you can get a value f o r B i n terms of other angles, you combine that and make i t so that i t could, and could be, i t could be 30 degrees or i t could be some other value. You get i t i n t h i s . A l l these isosceles t r i a n g l e s make i t quite easy to work i t around so you can get i t i n terms of say you get angle B i n terms of C i n two d i f f e r e n t ways, l i k e I d i d there, except I've done something wrong. So I think I was on the r i g h t track there. 108 Analysis of Interview 4 Problem 2 Student 33 Deduction " I ' l l just do i t almost a l g e b r a i c a l l y . . . angle B = angle P, angle C = angle BAG . . . I can use some possibly supplementary angles. I know the angle sum of a t r i a n g l e i s 180 degrees and I could use supplementary angles to say that i f B i s equal to B. Two times angle B = angle PQA because two angle B i s going to be supplementary to that and that's supplementary to that . . . angle PQA i s equal to angle PAQ so and angle. So those two angles. I t w i l l be angle C, 180 degrees minus angle C and angle CAQ . . . you can get values f o r B by, i n terms of other d i f f e r e n t angles, and you can by working i t , by using equations, you can get a value f o r B i n terms of other angles, and you could combine that and make i t so that i t could, and 'could be, i t could be 30 degrees or i t could be some other value. You get i n t h i s . A l l these isosceles t r i a n g l e s make i t quite easy to work i t around so you can get i t i n terms of, say you get angle B i n terms of C i n two d i f f e r e n t ways." 109 Interview 5 Problem 3 Student 2 H e u r i s t i c : Inverse Deduction Student Well f i r s t of a l l I just read through the problem, just to get a general idea. Now I am going to try, you know, fi g u r e i t out. Okay I ' l l say B has, l e t ' s say x cents. So so B has a c e r t a i n amount of cents, c a l l i t x and A w i l l give him x more cents so he w i l l have 2x, Does t h i s mean l i k e ? Okay. Let's say B doesn't have anything yet, but A gives him how much and that's how much he has, or does i t mean he already has something? Interviewer I t means A has something and he gives something to B and to C. Then when that i s done B gives a l i t t l e b i t to A and to C. So t h i s i s done one a f t e r the other. F i r s t of a l l A has something and he gives to B and C. Student Do B and C have anything to begin with? Interviewer Well I don't. Student I guess I have to fi g u r e out. Okay. Interviewer Finding anything? Student No. I'm s t i l l t r y i n g to, you know, just think i t out. Interviewer Seeing anything i n the problem? When you are t r y i n g to figur e out, are you saying anything, to yourself? Student I, I'm just kind of reading the question over and just kind of concentrating on i t , you know. I'm not doing anything. Well I think that seeing that A i s giving each of them something I think he i s givi n g B a c e r t a i n amount and C a c e r t a i n amount so he's just giving back what he got, sort of. So now I'm just going to t ry and fi g u r e out, you know, working with the 1 6 , sort of, and just t r y i n g . 110 Interview 5 (continued) Interviewer Working with the 16? Student Yes. So to begin with A's going to have a c e r t a i n amount that B has, plus a c e r t a i n amount that C has. So we'll say that's C + B. That's the c e r t a i n amount he has. Interviewer What's going through that head now Theresa? Student I don't know, I'm just t r y i n g to f i g u r e i t out, get things together. I t ' s kind of confused, you know. Interviewer You're thinking about A now are you? Student Yes. Interviewer That B + G i s , you're saying that, what's happening, he's getting some from B plus C? Student Well i f A's got B + C a c e r t a i n amount of cents and he gives some to B and he gives some to C well I'm t r y i n g to f i g u r e out i f he has anything l e f t or does he s t i l l have some. Interviewer What do you think? Does he have anything l e f t at a l l ? Student Does t h i s mean that each, both A, B, and C, a l l of them have the 16^? Interviewer Yes. See, you see i t says i f each f i n a l l y has l6fi. So what are you thinking now? Student I'm just t r y i n g to fi g u r e out, somehow make an equation and get. Interviewer Got any new ideas, or are you just s t i l l t r y i n g to make an equation? Student I'm s t i l l just t r y i n g to figur e i t out. Interviewer Make an equation? I l l Interview 5 (continued) Student Yes, Interviewer See, you know, i s there anything l i k e , are you thinking of anything when you're t r y i n g to fi g u r e i t out? Like, you know, i s there anything going through your head? Like, you know, are you t r y i n g d i f f e r e n t things, or are you? Student Yes. I'm just thinking of d i f f e r e n t things to do. Interviewer Well t e l l me some of these d i f f e r e n t things you're thinking of. Student Oh, okay. Interviewer Cause I'd l i k e to know. Student Well f i r s t of a l l I'm thinking that maybe I, I can't just assume that A has nothing to s t a r t with, or that B and C don't. So I was thinking that A gives B, B w i l l have B + x, h e ' l l give him B and he w i l l always, already have x and C w i l l have C + y or something and. Interviewer Well how, how i s that coming? Are you getting, how are you coming along with that? Student I'm thinking that i f I did make an equation anyway, I'm tr y i n g , see I know, cause there's three d i f f e r e n t variables l i k e A, B, and C, So can't just have one equation and have three v a r i a b l e s . I've got something. Okay. So f i r s t of a l l A has a c e r t a i n amount, A has x, and he gives B. A just has a c e r t a i n amount then he gives B, B plus what he already has, say i t ' s B + x, and he gives C, C + x. Then B, which has B + x, gives A, B + x, so A now w i l l have the o r i g i n a l amount plus what B gave him back. And then C i s going to give A, which w i l l be C + y, and so A w i l l have the o r i g i n a l amount plus B + x + C + y. Interviewer And now what happens? 112 Interview 5 (continued) Student Well since I can't work with three v a r i a b l e s , I'm t r y i n g to figur e out i f I can get, maybe put one i n terms of the others, something l i k e that. So I think I, I ' l l just write down, you know, rather than thinking of something, write i t down. Okay. Well say A has a ce r t a i n amount plus what B gave him which would be B + x plus what C gave him, which i s G + y. Just t r y i n g to work backwards. And so you end up with 16. Okay. And some i s going to give to G, and some of that to B, and then h e ' l l end up with the o r i g i n a l amount. Analysis of Interview 5 Problem 3 Student 2 113 Inverse Deduction "just t r y i n g to work backwards. And so you end up with 16. Okay. And some i s going to give to C, and some of that to B, and then h e ' l l end up with the o r i g i n a l amount." 114 Interview 6 Problem 3 Student 11 H e u r i s t i c 1 Inverse Deduction Student Lookee, i t ' s kind of long. A l r i g h t . Well I think I»11 s t a r t at the beginning and work down. They do not have the same amount of money. Well, there's three amounts of money, I know that. I think I ' l l probably give them variables, just f o r the heck of i t . Interviewer So you have x, y, and z. Student Yes. He gives Ron as much money as Ron already has. So that would mean Ron i s going to have twice as much as he has. So that's be 2 y . And t h i s would-be x - y. And be the same f o r Ted. 2 x , 2 z , and y - z. There s t i l l seems funny. Interviewer Now what are you thinking at the moment? Student Well he has t h i s much. Ron has 2 y and he gives him twice t h i s and him twice t h i s and then i t works out the same thing over again. He gives twice t h i s and twice t h i s again f o r , and he would have subtracted from here 2 z and. This i s going to make a l o t of nothing once I get there, but x - y - z, a l o t of va r i a b l e s . And t h i s would be - 2 y - 2 x - 5 z . There I'm done. Interviewer Now, what are you going to do next? Student Now f i n a l l y B i l l has 16, Ron has 16, and Ted has 16. Okay. I'd l i k e to change i t into one var i a b l e , i f I could. Oh, they a l l have 16, Okay. Okay, 16. Before he gave away any of h i s money, d i d B i l l o r i g i n a l l y have. Interviewer What are you thinking now? Student Well I don't know. I just, I can't work three variables, I know that. I can work two vari a b l e s , but a, but I'm just t r y i n g to fi g u r e out how to get r i d of some of them. Interviewer Have you got any ideas there? Interview 6 (continued) 115 Student There must be some way to connect the, the amounts they have by a an a d d i t i o n or something. I don't know what. So he has 16 now. Interviewer Now, what's that idea coming? Student Well, B i l l has 16 now, r i g h t at the end, and a of t h i s 16, part of i t , a well you can f i n d out how much he had a f t e r he gave to Ron and Ted because he has i t doubled over twice. Like a Ron and, or a, yes Ron gave him twice as much as he had, or the same amount over, and Ted gave him the same amount over. Interviewer Now, where i s t h i s you're getting that from? Student From the second thing here. Interviewer The second sentence? Student Or t h i s , or Ron decides to give some of his money to B i l l and to Ted. He gives them as much money as each of them now has. So B i l l has x and he i s given twice that amount. And then Ted gives him twice that which would be 4x and that should be equal to 16. What he has now and that should work out. I hope, I don't think i t w i l l though. I think I d i d something wrong. No. I t should be 2x times 2 which would be 4x = 16 which would be 4. Interviewer Could you explain that, a 2 times the 2x there? Student Well he had, he has x amount of money a f t e r he gives the money to Ron and Ted. And he has t h i s x doubled by Ron, which i s 2x = 16, the number we are looking, working with. And then he has i t doubled again. So i t would be 2 times the 2x = 16 he now has. Which works out to 4. So he had 4 a f t e r he gave the money to Ron and Ted. And that should work out f o r the other two, I think. One kind of equation l i k e that. Interviewer Well, what do you mean by working out f o r the other two? Student Taking t h e i r , the number they now have, and working with that. 116 Interview 6 (continued) Interviewer Oh yes. And then you would know then what each of them had at the beginning? Is that what you're doing or? I'm not sure. Student I f , eventually I should be able to f i n d out. Interviewer Oh. Well what i s that 4 then? Student The 4 i s the number that B i l l had a f t e r he gave his money away and so. Interviewer Well where i n the problem would that be? Student Money to Ted. Interviewer So i t ' s r i g h t there i n that sentence, now B i l l decides to give some of h i s money. Student He gives t h i s to Ron and he gives t h i s to Ted, Interviewer So a f t e r he gives i t to Ron and to Ted he has? Student 4 cents. Interviewer 4 cents. Now where would you go from there Marie? Student Possibly i f I could f i n d out how much Ron and Ted had a f t e r a, oh, or a. Like, well 16 does me no good, that Ted and Ron have r i g h t now, because i t ' s been doubled over and double over. And, and what I want to do i s f i n d a number closer to the o r i g i n a l number they had. So I could f i n d out how much B i l l gave Ron and Ted, and add i t back to what B i l l had. Interviewer Well how might you f i n d a closer number to there? Student Well that's where I'm just looking at. So Ron ends up with 16. They a l l end up with the same amount. Everybody has t h e i r a money four times over I think, l i k e . Interviewer So what could you do with that? What would be, you be thinking to do with that? 117 Interview 6 (continued) Student Well B i l l would have to have more money than anyone to s t a r t with. And there would have to be some kind of a, I don't know how to say t h i s , a d i v i s i b l e | s i c ] of 4, more than l i k e l y , or they a l l are, as i t ends out. And a I was just wondering i f i t , a i f I could work i t out without an equation. I mean not r e a l l y without an equation, but a that i t might work out just by a guess, you know. I mean l i k e an estimated guess of how much more B i l l would have i n the beginning, to have to give that much to Ron and Ted. interviewer And then what would you do with your estimated guess? Student I'd see i f i t works. Like take i t through the equation, the whole problem, and see i f i t worked. And I would be tempted to give B i l l 32 and Ron 8 and Ted 8 f o r the f i r s t , you know. Which would mean, wouldn't work, I know now. Interviewer Okay. Suppose i t didn't work, what would you do? Student I know that B i l l would have to have less than 32 r i g h t now, because the way i t works now, i f he gave 8 to Ted and 8 to Ron he would have 16 l e f t and he's supposed only, w i l l have 4 l e f t . I know that. So I can bring i t down to a possibly a d i v i s i b l e [ s i c ] of 12, which i s a good round number, maybe 24. 118 Analysis of Interview 6 Problem 3 Student 11 Inverse Deduction "what I want to do is find a number closer to the original number they had. So I could find out how much B i l l gave Ron and Ted, and add i t back to what B i l l had." 119 Interview 7 Problem 5 Student 16 H e u r i s t i c : Analogy-Student Well, t h i s i s l i k e f i v e separate questions, i s i t ? Interviewer No. Five equations a l l together. That's why we c a l l them simultaneous. Student Yes. So f i n d these f i v e values that w i l l s a t i s f y a l l these equations? Interviewer Right. Student Oh very well. I ' l l solve each one separately and. then search f o r common values. I guess we used to do t h i s also. We'll take 4y - 3 x , the f i r s t one. We'll try to solve with, l e t me see now. Get both variables equal to 1 , such as 4y a 4 y = .5 + 3 x - 2 u . Take them over to t h i s side and I got a value f o r k-y. And also 2 u would equal 5 + 3 x - 4 y . Oh I'm s t i l l stuck with a y. This i s n ' t going to work. That's not good enough. So maybe i t w i l l , equals 5 - 3 x - 2 u . I think I'm going to get myself mixed up here. The 2 u i s equal to t h i s , which w i l l a l l cancel now. There's some way, there must be some way to get r i d of a l l these variables and be l e f t with the one and solve f o r that. And then work i t back to f i n d a value f o r each of these. Interviewer I see. Well just l e t me make you go on f o r just a second. Suppose you did that, with that one, then what would you do? Student Okay. I wanted to solve here f o r 4y and f i n d the value f o r 4 y , substitute i t l n so that 3x has some value as 4 y . There's some way that i t equals 4 y , some f r a c t i o n perhaps. And 2 u i s also equal i n some way to 4 y which i s equal to 5 . I t ' s a l l , a l l added, subtracted together to give you 5« Interviewer I see. So you, you'd have one var i a b l e up there then? 120 Interview 7 (continued) Student Yes. And then I already know what 4y is somehow related to 3x and relate that back to find the value for x. And do the same thing with u. And I've got there, I've got my one, two, and three values for this equation. Go on and do the same for each equation. And then given, then I ' l l , I ' l l just go through and pick the common values from each one. There w i l l be several values from each one, there must be, I can see how there w i l l be. Interviewer Okay. So that's your, you're sure that would work and that's what you're going to do? Student Yes I think I'd take a lot of time but i t must work that way. I can solve for this one i t ' s shorter. I'm not too sure how we'd go about three variables. Interviewer Try that last one for me. Student Okay, well there's got to be a better way here. Trust me to multiply here. Interviewer You took the last one and solved for 7z, okay, for 7z and then what was that next step there? Student Oh subtract 5« No. That would get r i d of the whole variable. Oh, oh dear heavens. Well I'm solving for 7z. 7z w i l l equal 11 plus 5u. That, that's just by, I forget what we c a l l i t now, taking, subtracting more or less 5u, or adding 5u on each side. That w i l l cancel that off. 7z = 11 + 5u, . substitute this back into this equation. But then again I need another equation. You need two equations. That, that's what I'm, that's where I'm thinking of. You need two equations to solve this and I've only got the one. Interviewer Well how did you come up with the idea that you need two equations to solve it? 121 Interview 7 (continued) Student Well given say x + y equals oh, z, some number. You're given two of these values. No, you're given the one value, t h i s z perhaps. Given that x - z equals another value. Again say, well you'd f i n d a value f o r x and you can substitute i t into here, into t h i s second equation. And solve i t and you get a say that's what you're looking f o r . And you'd get a, by f i n d i n g a value f o r t h i s , f o r a l l these variables i n t h i s f i r s t equation. This one Is going to have an algebraic, l i k e value, i s going to have some l e t t e r i n i t , some va r i a b l e . Substitute i t into here. So l i k e i n here i t ' l l equal z - y. Substitute z - y into t h i s here equation. So you get z - y equals a. I think that's how I go about i t . Interviewer Those are two x's and those are two z's are they, or i s that y and z there? Student No, I take t h i s equation here and I get t h i s one from i t by subtracting y. Interviewer Oh yes, your z i s over there. Pine, yes. Student Yes, and then I take t h i s here, t h i s value f o r x, and take t h i s and put i n . Interviewer Put i t i n there? Yes. Student Solve. This i s z - z, they cancel. And then -y = a. And you've got y, that's a constant. Interviewer You've done these type before? Student Yes. Interviewer When you had two of them? Student Yes. Well I've never seen t h i s three v a r i a b l e s before, but I think that's how I'd go about i t . Yes, I think so. But that's how I would go about i t . I think I'd try to f i n d a value f o r t h i s v a r i a b l e and substitute i t i n , i n each, and then f i n d that value f o r the. The interviewer f a i l e d to notice that the tape had run out as the student made her l a s t comment. The student suggested using cases but said i t would "take a l l day." Analysis of Interview 7 Problem 5 Student 16 122 Analogy "you need two equations to solve t h i s . . . well given say x + y equals oh, z, some number. You're given two of these values. No, you're given the one value, t h i s z perhaps. Given that x - z equals another value. Again say, well you'd f i n d a value f o r x and you can substitute i t into here, into t h i s second equation. And solve i t and you get a say that's what you're looking f o r . And you'd get a, by f i n d i n g a value f o r t h i s , f o r a l l these variables i n t h i s f i r s t equation. This one i s going to have an algebraic, l i k e value, i s going to have some l e t t e r i n i t , some va r i a b l e . Substitute i t into here. So l i k e In here i t ' l l equal z - y substitute z - y into t h i s here equation. So you get z .- y = a. I think that's how I go about i t . . . Then I take t h i s here, t h i s value f o r x, and take t h i s and put i n . . . and solve. This i s z - z, they cancel. And then -y = a. And you've got y, that's a constant . . . well I've never seen t h i s three variables before, but I think that's how I'd go about i t . . . I ' d try to f i n d a value f o r t h i s v a r i a b l e and substitute i t i n , i n each, and then f i n d that value." 123 Interview 8 Problem 5 Student 38 Heuristic^ Analogy Student I'm thinking, l i k e maybe i f you make them, so that they're equal, and subtract them, so that you can cancel one of the, one of the unknowns out. Interviewer Okay. Could you explain to me what you mean when you say getting them equal? Student Well multiplying the equation so that they, l i k e maybe multiply t h i s one by 2, so that that would be 2y. And then you subtract them so that that would cancel out. Interviewer Any p a r t i c u l a r reason why you decided that that might work? Student I t ' s the only thing I can think of that looks l i k e t h i s , that we've done. Interviewer What kind of things did you do before? Student What? Interviewer Well you said i t reminded you of something you'd done before. What kind of things had you done before? Student Oh yes. Well working with two v a r i a b l e s . I think I'd probably work with the bottom one cause i t ' s just got the two. So take one of the other ones and cancel, so that you can cancel them out. Interviewer So that's going to be your method of attack i s i t ? Student I'd try that f i r s t , yes. 124 Analysis of Interview 8 Problem 5 Student 3 8 Analogy "I'm thinking, l i k e maybe i f you make them, so that they're equal, and subtract them so that you can cancel one of the, one of the unknowns out . . . i t ' s the only thing I can think of that looks l i k e t h i s , that we've done . . . working with two variab l e s . I think I'd probably work with the bottom one cause i t ' s just got the two. So take one of the. other ones and cancel, so that you can cancel them out." 125 Interview 9 Problem 7 Student 26 H e u r i s t i c : Preservation of Rules Student L i s t a l l the abas. Oh, f o r example t h i s , do I have to l i s t t h i s and this? Interviewer No. Student dgd-^ and &h., d^d/^d^. Interviewer Okay, I ' l l l e t you do i t a short way. You can just write down the 2, 3, 4*s, okay? So we can save time. Well I mean just so we can save time. The next one you write down, l i k e just write down the numbers, okay? So we don't have to bother . with the brackets. I t just saves a l i t t l e time. Student Okay. 345, 456, 561, and 612, and 134, 154, 156, 234. I have that already 245, 246, 356, and. Oh can I go back again? I think so. 261, 2o3, 264, oh. Interviewer Now what are you thinking there? Student I was wondering i f I can go back, you know. I was just a, at f i r s t I was just l i s t i n g when I read i t . But I'm, but then I was thinking i f I can l i s t i t l i k e 56 and 1. That means, that's a, order counts. Does i t mean i f I put i t i n another order, i t ' s another aba or whether i t ? Interviewer What do you think about sets? That's a set. Student I think the order matters. Interviewer Okay, you l i s t them. Student 3 l e t me see 356. Did I have 346? 346, 361, 362. Can I use the, l i k e can I use 363, you know, the same number? No? Interviewer Oh, not the same number twice. No. 126 I n t e r v i e w 9 (continued) Student Not the same number. 36"4. I n t e r v i e w e r So what you're d o i n g now i s , you're e x p l a i n i t to me, what you're d o i n g now. Student I'm j u s t going over the number. L i k e , I have a system. L i k e , t h a t I go 123 and the 124 and 125 and then l i k e t h a t and so I . I n t e r v i e w e r Sure, okay. Now what are you d o i n g down here? What's your system f o r g e t t i n g these, these ones here, you're going back you s a i d now? Student Yes. I'm, i f I s t a r t with 3, l i k e I wrote, I wrote 3^5 a l r e a d y , so I s k i p t h i s number and 346 and 341, 342. And then a f t e r I f i n i s h e d with the 4, I use 356 and 351, 352, and l i k e t h a t . And then a f t e r w a r d s , I use the 636 a 1, 362, 364. I n t e r v i e w e r Have you done a l l the, y e t , or? Student I t h i n k so. I n t e r v i e w e r Okay take a l o o k a t statement, the second p a r t of the q u e s t i o n t h e r e . Student Dabas are these number. Only, one and only one aba, P and q are two dabas. No, I t h i n k i t ' s f a l s e . Because, l i k e f o r 1, 2, these are two dabas and i f I want to put i t i n t o an aba I only have t o add one more. And I can use 123, 124, 125. I n t e r v i e w e r Good. What about statement two? Student I f L i s an aba t h e r e e x i s t s a daba not i n L. I f L i s an aba th e r e e x i s t s a daba not i n L. I don't q u i t e get the q u e s t i o n . There e x i s t s a daba not i n L. Yes. I n t e r v i e w e r Okay. I'd l i k e you t o . Student Because i f , l f L i s an aba t h a t means t h e r e ' s only t h r e e i n t h i s case and t h e r e ' s s i x dabas here. And i f t h e r e ' s only t h r e e , w e l l , w e l l an aba, you know, any of the ot h e r s , not, c o u l d not be i n the L. Interview 9 (continued) 127 Interviewer Good reason. Fine. Take a look at three. Student If L i s an aba and p i s a daba not i n L, then there exists one and only one aba, aba containing p and not containing any daba that i s i n L. I f L i s an aba and p i s a daba not i n L, then there exists one and only one aba. Yes. I think i t ' s possible. I f , i f L i s only one an aba, because, and there's six numbers here. And f o r instance, i f 123 i s i n L and 456 i s n ' t , and well, f o r example, 6 i s i s p, then i t can use 4 and 5 just to have another aba which i s not consisted [contained] i n L. Analysis of Interview 9 Problem 7 Student 26 128 Preservation of Rules Statement 1 "I think i t ' s false because . . . i f I want to put i t into an aba I only have to add one more. And I can use 123, 124, 125." Statement 2 "yes because i f , i f L i s an aba that means there's only three in this case and there's six dabas here. And i f there's only three, well, well an aba, you know, any any of the others, not, could not be i n the L." Statement 3 "I think i t ' s possible. If, i f L i s only one an aba, because, and there's six numbers here. And for instance, i f 123 is in L and. 456 isn't, and well, for example, 6 is in p, then i t can use 4 and 5 just to have another aba which is not consisted [contained] in L." 129 Interview 10 Problem 7 Student 28 Heuristic:Preservation of Rules Student This may be a l i t t l e primitive method, but I think I'd just start by using the f i r s t daba and then combining i t with a the two others. And just keep going on like that. So i t ' d be d]_d2d3 and d]_d2dij., d]_d2d*5 . . . 18. If p and q are two dabas I'd say the, oh, two. These things are dabas. Well the f i r s t one's false. Interviewer Okay. You give me a reason why. Student Well, just a, well just sort of the way I did It, Like a you have the di and the d 2 in the f i r s t and a di and d2 in the second so that obviously comes out there being only one set that a contains the two of them. Interviewer Good reason. Take a look at statement two for me. Student A daba. Yes, because there can only be three sets of or three dabas in an aba and there are six dabas. So obviously one of them may not be in a set with the three of them. Analysis of Interview 10 Problem 7 Student 28 130 Preservation of Rules Statement 1 "well the f i r s t one's false . . . like you have the di and the d 2 in the f i r s t and a d]^  and d 2 in the second so that obviously comes out there being only one set that contains the two of them." Statement 2 "there can only be . . . three dabas in an aba and there are six dabas. So obviously one of them may not be in a set with the three of them." 131 Interview 11 Problem 8 Student 23 Heur i s t i c : V a r i a t i o n Student Construct a trapezoid? Just, just draw one? Is that the idea? Interviewer Yes, and those are the lengths of the sides. Student Those are the lengths? Interviewer Do you know what a trapezoid looks l i k e ? Student Yes. Well, so these are the sides you want me to use to? Interviewer Yes Student To use, oh I see. Interviewer I want you to use those sides and make me a trapezoid. Student F i r s t of a l l I»d f i n d out which two were the same length, i f any of them. What happens i f none of them are the same length? Interviewer Okay, then what happens? Student You wouldn't have a trapezoid. Oh, just a minute, p a r a l l e l to i t , p a r a l l e l . Oh, I'd take the two longest ones to a, to a make the two p a r a l l e l sides. And make sure they were p a r a l l e l by using my eyes. And then a, i t would be hard to make t h i s p a r a l l e l . Interviewer The two sides are p a r a l l e l ? Student The two sides are p a r a l l e l . Interviewer Then you have the two sides that are p a r a l l e l now? 132 Interview 11 (continued) Student Yes. But i f I have to use these exact lengths, i t ' s not going to connect. So maybe have to use di f f e r e n t , d i f f e r e n t a two to do i t with. Because i f I use those two, then I use these two f o r my ends. This one would be too short and i t wouldn't, either one go and not the other. You want me to connect? Interviewer Yes. The trapezoid has the two sides p a r a l l e l and then the other two sides. I t ' s closed, that's what I mean by closed. Student Yes, closed. Well, i t wouldn't be closed i f I'd used the sides that I've used. I don't think i t would be. No, i t wouldn't be closed i f I'd used the sides that I've used. I don't think i t would be. No, i t wouldn't. Interviewer So, any ideas? Student Well I could do i t the easy way and add on. Interviewer What's the easy way? Student Well you could always cheat and make t h i s one longer. But no, you've got to do i t that way. Interviewer I see, you could make that one longer. Student Make that one longer or shorten one or put these f a r t h e r apart. But that wouldn't solve i t . Oh, I know what I'm going to do. Interviewer You mean you just wanted to change what I gave you? Student No, No, I know what I'm going to do, I think, I don't know i f t h i s w i l l work. Interviewer No, but the idea. You had an idea there of changing what I gave you? Student Yes. But that wouldn't, you couldn't do that. Easy way to answer f o r mathematics. Okay. Interviewer Now you're doing what at the moment? 133 Interview 11 (continued) Student No. I don't think, that wouldn't do. The trapezoids nave to b e l i k e t h i s or can they be l i k e this? Interviewer Well they have two sides p a r a l l e l and the other two sides can b e anywhere you want. Student Anywhere you want? Interviewer As long as i t ' s closed. Student Okay, well then I ' l l put these two sides p a r a l l e l and see what happens. Interviewer Okay, so you're going to put .b p a r a l l e l to? Student Yes. That w i l l work. Interviewer b p a r a l l e l to what? Student b p a r a l l e l to, oh just a minute, b p a r a l l e l to a. No, yes b p a r a l l e l to a. And I think just have to f i d d l e around with the d i f f e r e n t sides u n t i l you get the a, you know. Let's see, what I'd do i s I'd just f i d d l e around with the d i f f e r e n t sides u n t i l I, I got them i n such a p o s i t i o n where I could maybe one end was closer f o r d i s shorter than a, so that i f , i f , I don't know i f t h i s would work or not. Interviewer Okay. So that's what your plan of attack's going to be then? Student Yes. I'd just sort of f i d d l e around with i t u n t i l i t works. I don't think an equation would get you anywhere. I don't think so. What I'd do, I'd just have to experiment, you know f i d d l e around with each of the d i f f e r e n t sides u n t i l I was able to a f i t them i n so that they worked. By putting, you know, maybe t h i s one farther t h i s way and that one f a r t h e r that way, yet s t i l l being p a r a l l e l . Analysis of Interview 11 Problem 8 Student 23 134 V a r i a t i o n "well I could do i t the easy way and add on . . . Well you could always cheat and make t h i s one longer . . . make that one longer or shorten one or put these farther apart." 135 Interview 12 Problem 8 Student 29 Heuristic'.Variation Student You want to know how I'd go about? Okay. How I'd go about constructing the trapezoid, i f I was given the lengths of a, b, c, and d. Pour sides. Interviewer Yes, These are the lengths of the sides. So I gave you the lengths of the sides. Student Oh. You mean you want me to take, take these there and , and just kind of put them together? Interviewer And make a trapezoid f o r me. Student Well. Interviewer Do you know what a trapezoid looks l i k e ? Student Yes. I t looks, c and b l i k e that. So, I don't know. I think I'd probably measure them f i r s t , maybe. Okay, just, just to see. I just want to see how well. I don't know. I've, look I, I can see the lengths, you know, not exact lengths, but a. I'd probably take b and, b because c i s the smallest. And I'd probably put the c, c on the top. And a i s the longest, so I'd probably put that one on the bottom. And then b and d, b and d on the sides. You want to know why? I don't have any reason, except. Interviewer No. I'd l i k e you to try i t . Student Try to put them there? Interviewer Yes. Student You mean kind of measure i t out and put them? Interviewer Yes. I'd l i k e you to try i t and t e l l me what you'd do now. I want you to try i t . Student Oh, okay. Okay b and d, I think. 136 I n t e r v i e w 12 (continued) I n t e r v i e w e r So you have drawn a f o r . Now you're going to take b and d? Student Well i t might be b e t t e r . Oh, change, I'm s o r r y . I'm t h i n k i n g maybe I should put b and d f i r s t . Okay. So I ' l l do t h a t . I n t e r v i e w e r You've got b? Student Well I don't know which one's going to work r i g h t , so I j u s t do them. In t e r v i e w e r That's b? Student Oh, should be. I t should be more s l a n t e d , okay. I n t e r v i e w e r Why, why d i d you want to s l a n t b more? Student Because I don't t h i n k the angles would come out r i g h t . I t ' d be more l i k e a r i g h t angle t h e r e , okay. And then I'd put a down and then I'd put. I j u s t had b and t h i s i s a, so then I'd put d. I t h i n k . Okay. Not going to come out, I don't t h i n k . I t ' s not going t o , won't come out r i g h t . Oh, w e l l . And then I'd take c and i t doesn't f i t . Go l i k e t h a t , and i t ' s wrong. I n t e r v i e w e r Okay, what would you do now tha t i t doesn't f i t ? Student I don't know, I don't know. I t h i n k I'd take, l e t ' s see. I used. T h i s i s a and t h i s i s c, t h a t ' s d. I t h i n k , w e l l you a c t u a l l y , I don't know. I'd t h i n k that you'd need a longer, a longer base except you can't do t h a t . So I don't know, I t h i n k maybe. I guess I don't know. In t e r v i e w e r What do you mean by making i t much more wider? Student Well l i k e , l i k e you can't j u s t make, i t ' d be a r e c t a n g l e then. I mean the angles here, l i k e you know, you can have i t l i k e t h a t or except I couldn't r e a l l y take i t much, much wider cause i t ' d , i t ' d t u r n i n t o a r i g h t angle, so a I'd a. Interview 12 (continued) 137 Interviewer What was your idea? You said something just "before that idea. You said something about i f you had a longer, but you couldn't do that. What were you thinking? Student A longer? Interviewer Yes, what were you thinking of? What i f a was longer, would that make any difference? Student Well these two would be farther away and therefore that one would f i t . Interviewer Oh, I see. Your b and d would be f a r t h e r away, Student But, but maybe i f you make c longer. I f you had c. I f instead of using c here you used a and d and then put b and c i n the middle. You want me to try that, or? I was saying l i k e i f you had a and d or, and then l e t the b and d or, and then l e t the b and c over here, the shorter ones. But I don't r e a l l y know. Interviewer But that's the way you'd go about solving i t ? Student I think so. A unless, unless a. Yes I think so. But then you'd have to go through a l l the d i f f e r e n t combinations. But I don't see another way. I guess you could measure them exactly and then, I don't know. Maybe you're supposed to know some kind of a, something about trapezoids, that I don't know. 138 Analysis of Interview 12 Problem 8 Student 29 V a r i a t i o n "I'd think that you'd need a longer, a longer base except you can't do that . . . but maybe i f you make c longer." Interview 13 Problem 9 Student 33 Heu r i s t i c . Extension Student Oh well these are the cube roots. I t ' s a 1 cubed plus 2 cubed plus 3 cubed plus 4 cubed equals 100. And so I'd say that a f a u l t y conclusion you might come to i s , i f you added the terms together and made 10 cubed equals 100. Which i s wrong cause you can't combine those. I don't think, l e t ' s see. Interviewer I'd l i k e to know a l l the ideas that occur to you. You know, anything you think of doing or try i n g , or you know. Don't l e t anything s l i p by. Give me an idea what you're looking f o r or t r y i n g . Student Oh, I see. Okay. I ' l l break t h i s down into prime fa c t o r s , I suppose. Equals, l e t ' s see 100 i s 2, 2, 5, 5. Interviewer Now did you do that f o r any reason, or when you got i t did you look f o r anything? Student No, not r e a l l y . A oh well I was sort of thinking of combining to begin with, but a well, l e t ' s see. I'm just t r y i n g to develop a pattern. Just continue i t , 25, 6 cubed, 7 cubed i s . Interviewer So you've written down there a 53, yes 73. And what are you looking f o r now? Student To see i f a any pattern exists i n i t because well f i r s t four. Interviewer The f i r s t four what? Student Well the f i r s t four seem to add up to 100. And see i f I can a, well I ' l l just take a, well you can a divide these l i k e . A 1, well square everything there. No that'd be sort of useless. Let's see a I, I can't seem to a. 140 Interview 13 (continued) Interviewer Now what are you thinking r i g h t now? Student I'm wondering, unless a you just x + y + z a l l those cubed equals a number. But i f you add them a l l together then square i t , i t equals the same number. Interviewer Oh yes, I see. Student So a. But there's no way i t would work i f you continued cause a those numbers, i f you added i t might work. I don't know, l e t ' s see. Interviewer Show me what you mean there. Student Yes. Well I ' l l just use 5, 6, 5, 6, 7, and 8 and see i f i t works f o r the next four numbers. So 8 cubed going to be about 512, I think. So I ' l l add up these. That equals. Interviewer Oh, you're adding up the cubes, yes. Student Yes, 196. 5, 26 squared, 26 times 26. No. So i t doesn't seem to be, i t doesn't work that much. Now I guess I ' l l try i t with any four numbers that add up to 10 and see i f that works. Let's see. Interviewer Why did you pick that p a r t i c u l a r combination, l i k e any four that add up to 10? Student Well these were four numbers that added up tc 10, and I'm just t r y i n g . Well I ' l l try f o r three numbers that add up to 10. 5, 3, and 2. I know i t ' s not going to work because 5 cubed i s over 100. A you see that i t works here i t 1 + 8 = 9 , 1 + 2 squared equals 9 and a i t works f o r the next 3» 1 + 2+ 3 = 6 squared i s 36, 27, 35 and 36. So I think any pattern. Well just add 5 to that and you get 225 and 5 + 10 i s 15 squared i s 225. So one i f you just progress by integers 1 + 2 + 3 . A you could get a general term f o r t h i s . Meaning that Fn equals f o r any, l e t ' s see, equals 1 + 2 cubed + 3 cubed etc. plus n cubed and a equals 1 + 2 + 3 etc. + n a l l squared. Analysis of Interview 13 Problem 9 Student 33 Extension "I'm just t r y i n g to develop the pattern. Just continue 25, 6 cubed, 7 cubed i s . . . I ' l l just use 5, 6, 5» 6, and 8 and see i f i t works f o r the next four numbers." Interview 14 1^2 Problem 9 Student 41 Heuristic*. Extension Student One cubed. Oh looks l i k e that's 2 cubed, 3, 3, that's 3 cubed and 4 cubed equals 100. A 1, 2 equals 1 and and 4 a l l squared. And then just discover from t h i s , or can I jot down some sort of formula? Interviewer I want to know what you can do with i t ? I f there's anything else you can discover. Student I just throw i n 5, 25, 125. Interviewer So now you threw i n a what f o r me? Student I'm throwing i n a 5 cubed, 3 cubed plus, plus 5 cubed equals. What does that equal? 25, 225 and that's 15 a 75, 150. Yes that's, that's 15 squared. And so i t looks l i k e sort of a general. Plus 2 cubed, I ' l l try 6 too. 2 l 6 and that'd give. Interviewer And that i s what? Student 6, 6 cubed and I put that i n with t h i s now. 441. And then a, I ' l l try adding on a, a 6 to the end of the sequence. And a i t ' d give 15 and 20, 21 and I ' l l try squaring that. Yes, that gives the same. So you could probably, that's about a l l you could do I suppose. Let's see i f i t would work f o r , i f you had to s t a r t with one. So I'd just do these. 35, But that's not a square. A divide by 2 + 3, 5 and you'd get something l i k e 7. A 7. Oh I might as well try another. 99. Take another 10, divide by. Interviewer Where did the 99 comes from? I'd l i k e to ask you. Student A 8 + 27 + 64. I'm getting r i d of the 1 now. And I get a 2 + 3 + 4 i s 9, I get 11. Can I just try one more? Interview 14 (continued) 143 Interviewer Sure. You can do anything you want. I want to know what you can discover. Student A 99 + 125. You get 99 and 125, 214, yes, 214 and a 9 plus the l a s t was 5. You'd come up with a boob. I t doesn't work, no. 5, 5, and 14. Sorry, that's once. So 7, 4, 624. No, that doesn't work. Wait, a oh, 99, one twenty. I t ' s 84 divided by 14. A 6 so might be some r e l a t i o n s h i p between 5 and 7, 9, 11, and 14 and 16. Well what's the very f i r s t one. Just take 8 alone, divide by your f i r s t one, two. You'd come 4. Seems to be 2 d i f f e r e n c e . A the f i r s t time i t came out as exact square but a a I guess I ' l l just s t a r t with t h i s one now. A 27 d i v i d i n g by 3 f o r the f i r s t one, come up with 9. 6 difference now these two. A 91 come up with 3 + 4 i s 7. No that's prime, i s n ' t i t . 1, 13, oh, there's 6. Oh, f o r s t a r t i n g with one you'd come up with zero. You come up with a squares. S t a r t i n g with 2 you'd come up with 2 difference between them, the two things. And i f you s t a r t i n g with 3, you'd get 6 difference. And l e t ' s do 4 quickly. 64. The f i r s t would be 4, 16, yes, 12 difference. I f i t were a 189 divided by 4 + 5» 9, you'd come up with 2, 21. That's 12 difference again. Yes. So you'd have a. what can I make out of that. 2 between there 4, 6. A I'd have to f i n d some sequence or. Just a, t h i s squared minus the o r i g i n a l . Yes, see you'd have a, s t a r t i n g off with the f i r s t n cubed plus n + 1 cubed plus n plus oh, however f a r you want to go plus x or something cubed and that w i l l equal the n + n + 1 onto n + x. And i f you started with one i t ' l l be squared but i f , i f you don't i t w i l l have to be a the n + the n + 1, n + x plus that's n squared. Difference 12, n, 6 squared, 2, and i f 1 - 1 yes plus 0. So i t would be squaring It plus n squared -n. These are a l l b i g brackets and you'd come up with an equation sort of l i k e that. Yes, that's your equation then, and then you could sort of try seeing whether or not i t would work with, instead of 2 + cube root of 3, Does i t equal the square root of 6. Maybe I should t r y that. That's 1, that's 1 a l i t t l e over one, that's a 1.3. See you'd come up with about 3.4 over the square root of 6 equals 1. Square root of 6 i s about a 2.6 or something. No. Which i s too small cause that's r i g h t here. 144 Analysis of Interview 14 Problem 9 Student 4 l Extension " I ' l l t ry adding on a, a 6 to the end of the sequence . . . l e t ' s see i f i t would work f o r , i f you had to s t a r t with one . . . then you could sort of try seeing whether or not i t would work with, instead of square, square roots, a cube root of 1 + cube root of 2 + cube root of 3." 145 Interview 15 Problem 4 Student 3 Heuristic-Successive V a r i a t i o n Student F i r s t measure the sides. This CB i s 8 centimeters. 5 and 7/8. And that's roughly 8 again. Interviewer And why did you do that? Student Well, I have to get two a v e r t i c e s on here and one on each side. That way you know, I don't know, I just thought I'd do that. Interviewer Okay. Student Now I'm supposed to construct a square, I guess i t ' s by t r i a l and error. Interviewer So you're t r y i n g to do what now? Student Well, to see l i k e what number. You know, which way one side would be l i k e . Mostly think i t would be between 3 and 4, or something l i k e that. Now I think I got i t . Like the way I d i d I t , I thought i t would be between l i k e 3 and 4. Then I t r i e d out 3. And I figured and I, you know, made a mark here to show where i t i s . And I went p a r a l l e l to the base l i n e . And i t showed that i t was about 3i=f. So I went up a b i t more. So I make i t about 3x. And \ that came out 3i# So i t came out r i g h t . Interviewer You think that's right? Student Yes, 146 Analysis of Interview 15 Problem 4 Student 3 Successive V a r i a t i o n "which way one side would be l i k e . Mostly I think i t would be between 3 and 4 . . . I thought i t would be between l i k e 3 and 4. Then I t r i e d out 3. And I figured and I, you know, made a mark here to show where i t i s . And I went p a r a l l e l to the base l i n e . And i t showed that i t was about 3 l . So I went up a b i t more. So I make i t about 3 i . And that came out 3 i « " 147 Interview 16 Problem 4 Student 22 Heu r i s t i c : Successive V a r i a t i o n Student I don't l i k e geometry. I can't do i t . Oh, l e t me see. We'll, I r e a l l y wouldn't know where to s t a r t . I guess a l i k e f i r s t of a l l I'd just, I don't know why, but I'd just measure to some extent. And l i k e make i t more c l e a r . And see i f , well, wait a second. I'd better do that over again. And then measure down here to see i f that, l i k e make i t perpendicular. Like a I gather I have to make t h i s one, I don't know why, I'd make t h i s , maybe a l i t t l e b i t bigger, to have the, no. That s t i l l wouldn't work. Interviewer Okay, f i r s t of a l l why did you want to make i t bigger? Student To make t h i s smaller. Interviewer To make the one going up and down smaller? Student Yes. Because I needed i t smaller. Cause l i k e that's up there and that's down there and I need to make i t smaller. But then that would make t h i s bigger, and i t wouldn't work. Interviewer Oh, I see. Yes, Student So a I guess I'd just keep on doing that u n t i l I got them both r i g h t . E i t h e r that or I'd just draw an imaginary one i n my head, and just do that. Interviewer What do you mean by that? Student Well l i k e I'd just, I'd just look at i t and construct a square i n my head. And then do i t from that. That's why I happened to choose that point. Cause l i k e I was looking at a square i n my head, and I saw i t there. Interviewer I see what you mean, yes. So you'd t r y to do that would you? Student Yes. Analysis of Interview 16 Problem 4 Student 22 148 Successive Variation "I guess like f i r s t of a l l I'd just . . . measure to some extent . . . And then measure down here to see i f that, li k e make i t perpendicular . . . I'd a make this, maybe a l i t t l e b it bigger, . . . to make i t smaller . . . just keep on doing that until I got them both right" APPENDIX E INFORMATION ON THE STUDENTS INTERVIEWED Grade Problems Sex School Age Math 11 Grades Plans to take Plans to take math 11 T r i e d Male Female Years Months Easter Christmas Math 12 a f t e r high school Student Yes No Undecided Yes No Undecided 1 1 2 X 1 16 8 A B X X 2 1 3 X 1 16 6 A A X X 3 1 X 1 16 7 A B X X 2 3 X 1 16 5 A B X X 5 2 X 1 16 6 B B X X 6 3 X 1 16 11 B B X X 7 5 8 X 2 17 6 A A X X 8 6 7 X 2 16 7 B B X 9 6 8 X 2 16 8 A A X X 10 7 8 X 3 16 0 A A X X i l 3 9 X 3 15 8 A A X X 12 4 9 X 3 '16 7 A A X X t-1 VO APPENDIX E (continued) Grade Problems Sex School Age Math 11 Grades Plans to take Plans to take math 11 Tried Male Female Years Months Easter Christmas Math 12 after high school Student Yes No Undecided Yes No Undecided 13 5 9 X 3 17 1 A A X X 14 3 5 X 3 18 3 A A X X 15 3 6 X 3 16 7 B B X X 16 1 5 X 4 16 11 A A X X .17 1 6 X 4 16 6 A A X X 18 2 5 X 4 16 2 A A X X 19 2 7 X 4 17 0 A A X X 20 3 7 X 4 16 9 B A X X 21 3 8 X 4 17 0 B A X X 22 4 7 X 6 16 1 A A X X 23 1 8 X 6 17 4 A A X X 24 2 6 X 5 16 8 A A X X 25 4 5 X 5 16 7 A A X X 26 27 1 5 7 6 X X 6 6 16 15 4 9 A A A A X X X X M o APPENDIX E (continued) Grade Problems Sex School Age Math 11 Grades Plans to take Plans to take math 11 Tried Male Female Years Months Easter Christmas Math 12 after high school Student Yes No Undecided Yes No Undecided 28 7 9 X 6 16 1 A A X X 29 4 8 X 6 16 6 A A X X 30 1 9 X 7 16 6 A B X X 31 2 8 X 7 17 6 A A X X 32 4 6 X 7 16 6 A A X X 33 2 9 X 8 16 10 B B X X 34 6 9 X 8 16 8 A A X X 35 1 3 X 8 k7 2 A A X X 36 3 4 X 8 16 11 A A X X 37 8 9 X 9 16 5 A A X X 38 5 9 X 9 16 2 B A X X 39 5 7 X 9 16 8 B B X X 40 4 9 X 9 17 3 B B X X 41 3 9 X 8 17 4 B B X X 42 2 3 X 8 17 0 A A X X i-J APPENDIX F THE STUDENTS' USE OF HEURISTICS Heu r i s t i c s Used Student Problem Cases Deduction Inverse Deduction Invariatlon Analogy Preserva-t i o n of Rules Varia-t i o n Extension Succes-sive V a r i a -t i o n 1 1 2 X X X 2 1 3(a) X X X 3 1 X 4 . 2 3 ( a ) X -5 2 X X 6 3 ( a ) 4 • X M IV APPENDIX F (continued) Student Problem Heurisi t i c s Used Cases Deduction Inverse Deduction Invariation Analogy Preserva-t i o n of Rules Varia-t i o n Extension Succes-sive V a r i a -t i o n 7 5 8 X X 8 6 7 9 6 8 X 10 7 8 X X X 11 3(b) 9(a) X X X 12 k 9(a) X 13 5 9(a) X X — M Vn APPENDIX F (continued) Student Problem Heuristics Used Cases Deduction Inverse Deduction Invariation Analogy Preserva-tion of Rules Varia-tion Extension Succes-sive Varia-tion 14 3(b) 5 X 15 3(b) 6 X 16 1 5 X X 17 1 6 x X 18 2 5 j ! 19 2 7 | 20 3(b) 7 APPENDIX F (continued) He u r i s t i c s Used Student Problem Cases Deduction Inverse Deduction Invari a t i o n Analogy Preserva-t i o n of Rules Varia-t i o n Extension Succes-sive V a r i a -t i o n 21 3(b) 8 . X 22 k 7 X 23 1 8 X • x 2 6 X • X 25 5 X • 26 1 7 X X l 27 5 6 „....-.. •-1 i M Vn Vn APPENDIX P (continued) Student Problem Heurisl bics Used Cases Deduction Inverse Deduction Invariation Analogy Preserva-tion of Rules Varia-tion Extension Succes-sive Varia- , tion 28 7 9(b) x . 29 4 8 X X X 30 1 9(b) X X 31 2 8 X X -32 4 6 X X X 33 2 9(b) X X 34 6 9(b) x M ON APPENDIX P (continued) Student Problem He u r i s t i c s Used Cases Deduction Inverse Deduction Invariation Analogy Preserva-t i o n of Rules Varia-t i o n Extension Succes-sive V a r i a -t i o n 35 1 3(b) X 36 3(b) 4 X 3? 8 9(b) x • X 38 . 5 9(b) X 39 .' 5 7 X 40 \ 4 9(b) X 41 3(b) 9(b) X U\ 42 2 3(b) X APPENDIX G INFORMATION ON THE SCHOOLS SAMPLED School Grades E n r o l l e d Type School Enrollment Math 1 1 A Enrollment Sample S o c i a l Size C l a s s b Mathematics Teacher Male Female 1 9-12 p r i v a t e coeducational 450 43 6 3 X 2 8-12 private boys 350 12 16 4 1 . 3 X 3 8-12 private g i r l s 205 34 6 2 X 4 8-12 private c oeducat i onal 165 29 6 4 X 5 8-12 private g i r l s 250 22 2 1 X 6 1-12 private g i r l s 305 30 6 2 X 7 8-12 public 1050 25 3 3 X 8 8-12 public 1050 33 6 1 X 9 8-12 public 920 31 4 1 X ^ h i s f i g u r e represents the enrollment of the class where the students that were interviewed were enrolled. ^ CO Joseph A. Kahl, The American Class Structure (New Yorki Holt, Rinehart, and Winston, 1957), P. 215-2167 : ~ 

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