looking back in problem solving

LOOKING BACK IN PROBLEM SOLVING Jinfa Cai and Michael Brook propose three ways of encouraging students to ‘look back’ after completing the solution to a problem.

Often after students solve a problem they believe they have accomplished their mission and stop further exploration. The purpose of this article is to discuss ways to encourage students to ‘look back’ so as to maximise their learning opportunities. The idea of using ‘looking back’ in mathematical problem solving is not new. According to Polya, by ‘looking back’ at a completed solution, by reconsidering and re-examining the result and the path that led to it, students can consolidate their knowledge and develop their ability to solve problems (Polya, 1945/1971). Polya suggested asking such questions as ‘can you check the result?’, ‘can you check the argument?’ and ‘can you use the result and/or method for other problems?’. In this article, we extend Polya’s thoughts on looking back and propose three approaches to encourage students to look back: • generating, analysing, and comparing alternative solutions, • posing new problems, and • making generalizations. Our perspective is that looking back goes beyond checking answers and checking the methods used to obtain answers. Instead, the purpose of looking back is to maximise learning opportunities in problem solving. It helps students acquire mathematical knowledge as well as develops their ability to solve problems. Thus, looking back is consistent with the general goals of mathematical problem solving. It is not only a process that allows students to experience the power of mathematics but is also an approach that provides a context for students to learn and understand mathematics.

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Generating, analysing and comparing alternative solutions Generating and analysing alternative solutions can create new learning opportunities for students. Figure 1 shows the ‘two job’ problem (Heid, 1995), which involves two linear relationships. By solving this type of problem, students are able to learn fundamental features of linearity by comparing the two linear relationships. Students that we have observed generated a number of solutions. The teacher asked the students to present each solution and discuss it with the whole class. The solutions shown in figure 1 all highlight how the total amount of earnings for each offer is related to the payment for each hour and the expense required for taking the offer. However, the total amount of earnings for each offer is represented differently in each of these solutions. Through looking back, students were able to generate alternative solution strategies. Most importantly, through looking back, students were guided to analyse and compare these different solutions. An analysis of these solutions can quickly reveal the advantages of asking students to explore different ways of solving the problem. Through comparing and reflecting on these solutions, a many-sided view of linear relationships was fostered in the students’ thinking. This constituted new mathematical knowledge for the students. In particular, students learned how the rate of payment and initial cost affected the linear relation in each offer. Students also better understood the concepts of slope (rate of change) and intercept (initial state) in a linear relation.

MATHEMATICS TEACHING INCORPORATING MICROMATH 196 / MAY 2006

Looking back to generate alternative solutions can be accomplished by asking the class to find multiple solutions to a problem. Regardless of whether there is individual or group effort, it is important for teachers to guide students to reflect and compare various solutions because the comparison helps students recognise similarities and differences between solutions and enhances students’ understanding. Figure 2: the triangle-shaped odd number pattern problem

1 3

5

7 13 21

9 15

23

11 17

25 ......

19 27

29

What is the sum of the numbers on nth row?

Posing alternative problems Posing alternative problems not only increases students’ understanding but also enhances their problem-solving skills. Let us look at the triangleshaped odd number pattern problem (figure 2). In this problem, students are asked to find the sum for the nth row. In order to find the sum for the nth row, students can examine the sums for the first few rows: 1 for the first row, 8 for the second row, 27 for the third row and 64 for the fourth row. This reveals a pattern that the sum for the nth row is n3. The triangle-shaped odd number pattern problem is a mathematically rich one, which can provide students with opportunities to explore mathematics by suggesting other questions related to the pattern. Our experience shows that high school students are capable of asking mathematically significant questions related to the pattern. Here are several examples of the questions posed by students: • How many numbers are there in the 10th row? • How many numbers are there in the nth row? • What is the first number in the nth row? • What is the last number in the nth row? • Is the middle number in each row always a perfect square? • What is the sum of the numbers in the first n rows? It is not completely clear how students pose alternative problems, but research suggests that

Figure 1: the ‘two job’ problem and its multiple solutions

Given the two job offers below, determine the better-paying summer job. Justify your answer. Offer 1: At Timmy’s Tacos you will earn £4.50 an hour. However, you will be required to purchase a uniform for £45. You will be expected to work 20 hours each week. Offer 2: At Kelly’s Car Wash you will earn £3.50 an hour. No special attire is required. You must agree to work 20 hours each week. Solution 1 In a 20 hour week, offer 1 will pay £4.50 × 20 = £90 and offer 2 will pay £3.50 × 20 = £70. Since the difference is £20 per week and the uniform for offer 1 costs £45, it will take £45/£20 = 2.25 weeks to pay for the uniform and break even. If you keep the job for three weeks or more, you should take offer 1. Solution 2 At Timmy’s you make £1 more for each hour of work. After 45 hours of work, you’d make £45 more at Timmy’s than Kelly’s. This extra money would pay for the uniform. From that point on, you’d make £1 more an hour at Timmy’s than Kelly’s. Solution 3 Let x be the number of weeks you intend to work. The total amount for offer 1 is 90x – 45 and the total amount for offer 2 is 70x. If 90x – 45 = 70x, then x = 2.25. So if you work fewer than 3 weeks you should take offer 2, otherwise take offer 1. Solution 4 Let x be the number of weeks you intend to work, y1 be the total amount for offer 1 after working x weeks and y2 be the total amount for offer 2 after working x weeks. Therefore, y1 = 90x – 45 and y2 = 70x. Using a graphical calculator to graph them you will see that they intersect at (2.25, 157.5). From the graph, you will see that if you have the job for three weeks or more you should take offer 1. Solution 5 Construct a table to show the amount of income for offers 1 and 2 for one week, two weeks, three weeks, etc, and then use the information from the table to determine which offer you will take. Solution 6 Let x be the number of weeks you intend to work. The total amount for offer 1 is 4.5 × 20x – 45 and the total amount for offer 2 is 3.5 × 20x. If (90x – 45) < 70x, then x < 2.25. So if you work fewer than 3 weeks you should take offer 2, otherwise take offer 1. students may think about solutions as they pose problems (Cai and Hwang, 2002; Kilpatrick, 1987; Silver, 1994), although they may not yet have complete solutions. Problems are posed based on students’ observation and their understanding of the triangle-shaped odd number pattern. When a student was asked what made him pose ‘Is the middle number in each row always a perfect square?’, he explained:


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‘In the pattern, 9 is the middle number of the third row and it is 32. And 25 is the middle number of the fifth row and 25 is 52. You see, this is the seventh row and 49 is the middle number. In these rows, the middle number is something squared. When I go back and look at the second and fourth rows, there is no middle number. But between 3 and 5 in the second row, you can imagine a middle number, which is 4. You can imagine a middle number in the fourth row, which is 16. Then I sort of guessed . . . the middle number in each row is a perfect square. That’s how I got the question.’ Since these problems came from the students themselves, they were more motivated in exploring them further than they were with those posed by teachers. Students we have observed have also made unexpected discoveries while exploring this problem. For example, two groups of students posed this question: ‘What is the sum of the numbers in the first n rows?’ The first group of students answered the question based on the fact that the sum of the numbers in the first n rows is ‘the sum of the sum’ of the numbers in each of the first n rows. Therefore, this question can be answered by appealing to the original question: the sum of the numbers in the nth row is n3. Therefore, the sum of the numbers in the first n rows should be 13 + 23 + 33 + 43 + . . . + (n –1)3 + n3. The second group used a different approach to answer the question. After making some observations, the students realised that the first row has one odd number (1), the second row has two odd numbers (3 and 5), the third row has three odd numbers (7, 9, and 11), etc. Therefore the nth row should have n odd numbers. Consequently, the sum of the numbers in the first n rows of the pattern should be the sum of the first (1 + 2 + 3 + 4 + . . . + n) odd numbers. Using a ‘square model’, the teacher quickly showed students that the sum of the first m odd numbers is m2; ie, 1 + 3 + 5 + . . . + (2m –1) = m2. And so the sum of the numbers in the first n rows in the pattern should be (1 + 2 + 3 + 4 + . . . + n)2 since the first (1 + 2 + 3 + 4 + ... + n) odd numbers are added. After the two groups of students presented their answers to the class, an unexpected finding emerged. The unexpected finding is that 13 + 23 + 33 + 43 + ... + (n –1)3 + n3 = [n(n + 1)/2]2, since (1 + 2 + 3 + 4 + . . . + n) = n(n + 1)/2. Therefore, in addition to finding the result for the sum of the numbers in the first n rows of the pattern, the exploration of the pattern also provides a tool for students to find the sum of the first n cubes. Without the opportunity for students

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to pose alternative problems, this surprising result would not have been discovered. In fact, this result was unexpected for the teacher as well. It was a real joy to witness students’ excitement in the process. Looking back through posing alternative problems helps students have a better understanding of the pattern, but also provides an environment for students to willingly explore mathematical problems and even make unexpected findings.

Making generalisations Teachers can also maximise students’ learning opportunities by asking students to make generalisations. Generalisation is one of the fundamental mathematical thinking skills. Through generalisation, students can investigate mathematical structures and relationships beyond the concrete and specific. For example, consider the problem in figure 3. Figure 3: the triangle area problem

Triangle ABC has an area of 1 square unit. If side AB is extended to point E such that AB = 1, CA is extended to point D such that

BE CA = 1 and BC is extended to point F AD such that BC = 1, what is the area of ⌬ DEF? CF In order to find the area of ⌬ DEF, students worked together to decompose it into smaller triangles (see figure 4). The example is based on the assumption that AB = 1. The teacher led the BE students in looking back, by asking ‘what if the ratio is not equal to 1?. Instead, if AB = x , what is BE y the area of ⌬ DEF?” The students explored a series of related problems that ultimately led to generalisations. For example, one such problem is: What area would result if the ratios given at the beginning of the original problem were all equal to 12 ⎛e.g., AB = 1 ⎞? ⎝ BE 2 ⎠ Students were guided in constructing an argument largely parallel to that of the first case to show that the area of this triangle is 19 square units. Similar results would be obtained for ratios of AB = 1 and AB = 1 the corresponding areas of BE 3 BE 4 ⌬ DEF being 37 and 61 respectively. Students then generalised this problem by letting AB = 1, where BE n n is a positive integer. To make sure that students


Figure 4: Showing that the area of ⌬ DEF = 7 when AB = 1 BE E

Figure 5: Showing that the area of EBC = n

B C

G A

F

D

Draw an auxiliary segment CG perpendicular to AE with G on AB. Draw segment CE to decompose ⌬ EBF into ⌬ EBC and ⌬ ECF. Because ⌬ ABC and ⌬ BCE have the same altitude and AB = 1, the area of ⌬ BCE = the area of ⌬ ABC

BE

= 1. In similar ways, it can be proved that the areas of ⌬ ACF and ⌬ BAD are also 1. Considering ⌬ ECF and ⌬ EBC, CF and BC are congruent and the two triangles have the same altitudes on the base BC. Therefore, the area of ⌬ ECF = ⌬ EBC = 1 square unit. In a similar fashion, the areas of ⌬ BDE and ⌬ DAF are also equal to 1 square unit. In this manner, ⌬ DEF is decomposed into seven sub-triangles, each with area of 1 square unit. Therefore, the area of ⌬ DEF = 7 square units. understood the problem-solving process, it was important to ask students to show the details (figure 5). This example shows the learning opportunities that are created by asking students to make generalisations. In this case, making generalisations not only solidifies the problem-solving skills learned through solving the original problem but is also itself a valuable process to experience. It shows, in the spirit of Polya, how much more learning can be realised from a problem when a ‘looking back’ perspective is adopted. Jinfa Cai is a professor and Michael Brook is an instructor at the Department of Mathematical Sciences at the University of Delaware, USA. [email protected], [email protected]

Acknowledgement The authors would like to thank Jerry Becker for bringing to their attention the ‘triangle area problem’ and Connie Laughlin the ‘triangle shaped odd number pattern problem’. Preparation of this article is supported in part by the USA National Science Foundation (ESI-0454739).

Draw an auxiliary segment CG perpendicular to AE. Draw segment CE to decompose ⌬ EBF into ⌬ EBC and ⌬ ECF. We will show that the area of ⌬ EBC = n. Segment CG is the altitude of both ⌬ ACE and ⌬ ABC. By the formula for the area of a triangle (area = H bh), the area of ⌬ ACE = H(AE)(CG). But since AB = 1n, BE = nAB and BE AE = AB + BE = AB + nAB = (n + 1)AB. Therefore the area of ⌬ ACE = H(n + 1)(AB) (CG). Since the area of ⌬ ABC = H(AB)(CG)=1, the area of ⌬ ACE is (n+1). Since the area of ⌬ EBC = area of ⌬ ACE – area of ⌬ ABC, area of ⌬ EBC = (n+1) – 1= n. The same argument will establish that the area of ⌬ BAD and ⌬ ACF are both equal to n. Therefore, the areas of ⌬ ECF, ⌬ BDE and ⌬ DAF are all n2. Extend segment FB to point Z such that EZ is perpendicular to FB. EZ is the altitude of both ⌬ ECF and ⌬ EBC. Since CF = nBC, the area of ⌬ ECF = n × (area ⌬ EBC) = (n)(n) = n2. In a similar fashion, the areas of ⌬ BDE and ⌬ DAF are also equal to n2. So the total area of ⌬ DEF is 1 + 3n + 3n2. In a similar way, when AB = n = n, where n is BE 1 a positive integer, the area of ⌬ DEF can be 2 shown to be 1 + 3n + n32 = n + 3nn2+ 3 . Likewise, in the case where AB = yx (x and y are BE positive real numbers), the area of ⌬ DEF can be 2 x2 + 3xy + 3y2. shown to be 1 + 3xy + 3y x2 x2 =

References Cai, J. and Hwang, S. (2002) ‘Generalised and Generative Thinking in US and Chinese Students’ Mathematical Problem Solving and Problem Posing’, Journal of Mathematical Behavior, 21 (4), 401-421 Heid, M. (1995) Algebra in a Technological World (Addenda Series Grades 9-12), National Council of Teachers of Mathematics, Reston, VA Kilpatrick, J. (1987) ‘Problem Formulating: Where Do Good Problems Come from?’, Cognitive Science and Mathematics Education, Hillsdale, NJ: Erlbaum, 123-147 Polya, G. (1945/1971) How to Solve It, Princeton, NJ: Princeton University Press Silver, E. A. (1994) ‘On Mathematical Problem Posing’, For the Learning of Mathematics, 14 (1) 19-28


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