Heuristic literacy development and its relation to mathematical achievements of middle school students

Abstract

The relationships between heuristic literacy development and mathematical achievements of middle school students were explored during a 5-month classroom experiment in two 8th grade classes (N = 37). By heuristic literacy we refer to an individual’s capacity to use heuristic vocabulary in problem-solving discourse and to approach scholastic mathematical problems by using a variety of heuristics. During the experiment the heuristic constituent of curriculum-determined topics in algebra and geometry was gradually revealed and promoted by means of incorporating heuristic vocabulary in classroom discourse and seizing opportunities to use the same heuristics in different mathematical contexts. Students’ heuristic literacy development was indicated by means of individual thinking-aloud interviews and their mathematical achievements – by means of the Scholastic Aptitude Test. We found that heuristic literacy development and changes in mathematical achievements are correlated yet distributed unequally among the students. In particular, the same students, who progressed with respect to SAT scores, progressed also with respect to their heuristic literacy. Those students, who were weaker with respect to SAT scores at the beginning of the intervention, demonstrated more significant progress regarding both measures.

Appendices

Appendix A: Protocol coding scheme 

   

Appendix B: Episodes from the intervention

Episode 1: Constructing heuristic vocabulary

At the very beginning of the intervention, the students were engaged in the game “Mouse in the maze.” They were given the following task:

There is a 4× 4 maze with the only exit from square A4. A piece of cheese is located near the exit (Figure 3). Some between-square doors are open and others are not. A mouse, striving for the cheese, passed the following path: down – to the right – to the right – up – to the right – cheese. Where was the mouse at the beginning of its way?

Figure 3

Mouse in the maze.

Most of the students obtained the correct answer (B1) quickly and confidently. From the follow up discussion, it appeared that in answering the question the students used one of two different strategies. The first strategy, in one student’s words, was this: “I drew all the steps from the end. Instead of “right” –“left”, instead of “up”–“down” and so on.” Another student articulated the second strategy: “I started from any point ... from C2... and drew the mouse’s way step by step. When I missed the cheese for two squares, I moved the entire path one square up and one square left. Then I got B1.” The teacher encouraged the students to give names to the above strategies. The students suggested: “From the end to the beginning” or “To change a direction” for the first strategy, and “To guess and fix” or “To think forward” for the second one. The teacher listed on the board all the students’ suggestions and then modified the game.

Now the students were asked to play in “Mouse in the maze” under different constraints. For example, a new path of the mouse was offered only verbally, and the students were asked to find where the mouse was at the beginning of its way without using pencil and paper. As was evident from the classroom discussions following every round of the game, the constraint “no writing” leads students to refining their previous strategies. For example, at some point the students noted that the strategy “From the end to the beginning” is difficult to implement without writing, since its use includes memorizing the whole path of the mouse. They also observed that “Thinking forward” works even without writing, but having a drawing of the maze on the board is beneficial.

At the next stage, the drawing of the maze was erased from the board in order to make the students rely only on the memorized images of the maze. This additional constraint triggered a new strategy, which was eventually called “Reject what is impossible” or “Neutralize.” For instance, one of the students observed that the fragment “up–up–up” in a path of the mouse eliminates rows A, B, and C from the list of possible answers. Besides, he noted that fragments like “to the left–to the right” neutralize each other, which was important in overcoming difficulties related to memorizing of the longer paths under the given constraints. Afterwards, at the same lesson, the teachers asked their students to recall situations when they used strategies called “Thinking forward”, “From the end to the beginning” and “Rejection of possibilities” in mathematical problem solving. The introduced strategies’ names helped students to communicate their ideas in many mathematical contexts.

Episode 2: Utilizing heuristic vocabulary in reflective problem-solving discourse

In this episode, “routine” algebraic tasks are utilized to promote the students’ capability to think forward and to think from the end to the beginning (see Appendix A for descriptions of these heuristics). The lesson presented here was on algebraic transformations of polynomials.

At the beginning of the lesson, the teacher seized the opportunity to think “on line” in front of her students. When Ada checked the homework, she found that many students did not solve the following task:

Factor: $30x^3-15x^2-14x^2+7x-4x+2.$

The students asked Ada to explain it. Ada started to solve the problem on the blackboard talking thoughtfully and trying different approaches. At that moment, the observer was sure that Ada did not know how to solve this problem and was really thinking out loud in front of her class:

Let’s try this: $30x^3-15x^2-14x^2+7x- 4x+2=30x^3-29x^2+ 3x+2$

What’s next? Let’s think forward: I can group the first two... and the second two, but it won’t give me something in common. Will it? Perhaps, we can group 30x³+3xand −29x²+2... no ... the same thing ... Let’s start from the beginning ... Oh, I see! 30 and 15, 14 and 7, 4 and 2... Did you try to group the pairs without doing addition of similar terms?”

From this point, the student independently completed the exercise. Then Dan, one of the students who were noticeably impressed by Ada’s problem-solving skills, asked the teacher:

• Dan: How did you understand what to do?

• Ada: I tried different things; I was trying to think forward, to plan the solution one step ahead of my writing.

Later on, the students practiced how to transform a product of binomials into a canonic polynomial. They solved several tasks like the following one:$(x-1)\cdot (x+2)=x^2+2x-x-2=x^2+x-2$. Then the teacher asked the students to solve a similar task without writing the intermediate steps (e.g.,$(x-2)\cdot (x+5)=x^2+3x-10$). Under the constraint “no writing”, the routine task appeared to be a problem to the students. The challenge was to keep in memory an intermediate result, and then to manipulate it mentally. In order to bypass this difficulty, the students were encouraged to consider the structure of a final answer before diving into technical details and to organize computations smartly so that the resulting polynomial would emerge gradually, step by step. For instance, one could think first of how to obtain an element with x ², then elements with x and then the x-free part of the resulting polynomial. In the above example, there is only one combination leading to x ²,$x\cdot x=x^2$, and one can write x ² as a part of the answer. Now, how can one obtain x? There are only two combinations, 5x−2x, and one can write 3x as a part of the answer. Similarly, only one combination contributes to a x-free part in the resulting polynomial, −2· 5=−10, and thus the answer x ²+3x−10 is found. In the follow up discussion a few students connected the game “Mouse in the maze” (see Episode 1) to the above algebra task. They pointed out that the constraint “no writing” turns even routine problems into interesting ones.

Episode 3: Heuristic similarities among different problems and the smallest possible help

When the students learned Pythagorean Theorem and practiced it in many problems on right-angle triangles, we designed a lesson based on not straightforward implementations of the theorem. The following problem was offered at the beginning of the lesson:

Let MNPK be an isosceles trapezoid (see Figure 4a). Prove that $d\,^2=ab+c^2$

Figure 4

For the problem on isosceles trapezoid.

The students were asked to work in small groups on planning solution to this problem. In 15 min, the teacher collected the students’ suggestions and wrote them on the board. To recap, the following plan was created:

• The equation to be proved reminds the equation of the Pythagorean Theorem (at least, in part of d ²=). The idea appears to insert diagonal MP into a right-angle triangle. It can be done using auxiliary construction:$PH\bot MK$ (see Figure 4b)

• The last step in solution will be applying Pythagorean Theorem to the triangle MPH:$d^2=\hbox{MH}^2+\hbox{PH}^2$. Thus, our next goal is to express MH and PH using a, b, and c.

• On finding MH. MH is a part of MK, greater than “a“ and smaller than “b“. To find MH precisely, it is worthwhile to have “a” and “b” together on the picture. For this reason, it is worth to build the second altitude of the trapezoid:$\hbox{NQ}\bot \hbox{MK}$ (see Figure 4c). Now MH = MQ + QH = MQ + a, and the new goal is to find MQ.

• We still did not use the given that MNPK is an isosceles trapezoid; from here it is possible to prove that MQ = HK. Now the new goal is to show that MQ = HK but this is easy since MQ + HK = b−a, and MQ = MK = (b−a)/2.

Of course, the classroom discussion of the problem was not as straightforward as the resulting plan. For example, the students had difficult time thinking of how to find MQ. At that point, the teacher suggested the students to read again the problem’s formulation. This was enough in order to trigger the understanding that the problem’s given that the trapezoid is isosceles is still unused, and in turn, that MQ and HK can be found simultaneously. When the plan was understood, the students were asked to complete the solution independently. This work took the rest of the first half of a 90-min lesson, and after a 5-min break the class was continued. The students were given the following problem:

The sides of a triangle are 9, 10, and 17. Find the altitude to the side 17.

At the beginning of small-group discussions the latter problem appeared to be unrelated to the former one. The deep-level connections between the problems became visible when the students drew an appropriate picture (Figure 5) and obtained two right-angle triangles. The students observed:

• As in the first problem, there are two right-angle triangles with the given hypotenuses.

• It is enough to find MH or HK, and then to apply the Pythagorean Theorem to a right-angle triangle in order to solve the problem.

• The sum MH + HK is given, but not MH and HK.

• The given triangle is not isosceles, and thus similarity with the previous problem is limited$(\hbox{MH}\ne \hbox{HK})$.

Figure 5

For the problem on the altitude of the triangle.

Based on these observations, two students were able to complete the solution, whereas the others needed some assistance. After a brief consultation with the two students who solved the problem on their own, the teacher offered the following question:

• Liora: You have two right-angle triangles including the unknown PH, and the idea is to apply the Pythagorean Theorem to one of them. Which right-angle triangle do you choose?

From the whole-class discussion, it became clear that some of the students would prefer MHP and the others would prefer PHK (e.g., for the reason that “9 is a nice number” or “10 is a nice number”), but most of the students could not make a choice in favor of one of the triangles. They also noted that in the previous problem about a trapezoid the choice of a triangle was clear because of the equation that had to be proved. At this point, the following idea was expressed by one of the students: “Let’s not make a choice, let’s work with two triangles simultaneously.” It led the students to using the Pythagorean Theorem twice, and, in turn, to a system of two equations with two unknowns (PH = h, MH = x, KH = 17−x; ${x}^2+{h}^2=10^2$ and $(17-x)^2+h^2=9^2)$, which completed the planning stage. Implementation of the plan evoked additional difficulties to those less proficient in algebraic transformations. The students, who solved the problem first, were asked to help the teacher and to advise their peers with the smallest possible help. The students, who solved the problems with assistance from their peers, also became advisers, and formulated their own heuristic intimations, depending on needs of the classmates they were helping.

Appendix C: Interviewees

Appendix D: Analysis of sample interviews

In the presented episodes, Alon and Dalit solve Problem 3N (see Figure 5). In the excerpts, content units are separated by numbers of different heuristic categories in the protocol coding scheme (see Appendix A). The codes are denoted as follows:

• (3a): This symbol means that the content unit before it is coded as 3a. In the protocol coding scheme, No. 3a is related to “Recalling related problems” (see Appendix A for description of this category).

• (13 + 4): This symbol means that the content unit before it is simultaneously coded as 13 and 4. In the protocol coding scheme, No. 13 is related to “Entering and concentrating” and No. 4 to “Selecting representation” (see Appendix A for descriptions of these categories).

  Alon solves Problem 3N

Two excerpts are of comparable lengths and reflect the same mathematical approach. Namely, the students tried to solve the problem by translating the word problem into algebraic expression(s). In heuristic terms, both Alon and Dalit used the following heuristics: “Planning” (1); “Self-evaluating” (2); “Activating a previous experience” (3) and “Creating a model” (4). The difference in two solutions is particularly apparent when both students achieved the expression 90x+9y−99 and tried to use it. At that point, Alon utilized two new heuristics “Exploring a particular datum” (6) and “Finding what is easy to find” (8), whereas Dalit did not.

We point out that the analysis undertaken essentially reduces the complexity of the protocols. Indeed, the analysis is not sensitive to order in which heuristics were called into play or to numbers of content units to which the fragments were segmented. It is only sensitive to the overall numbers of different heuristics indicated in the solutions. Let us also note that integrative Indices of Heuristic Literacy (IHL) are computed for each student individually, that is, direct comparison between Alon and Dalit’s solutions of Problem 3N is not a necessary part of the analytical procedure. It is presented here anyway to clarify segmenting and coding the protocols.