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SHLOMO L I B E S K I N D
A P R O B L E M S O L V I N G A P P R O A C H TO T E A C H I N G
M A T H E M A T I C S
"How does it happen that so many refuse to understand mathematics", asked
Poincar6 (1929), p. 43 and 1969, p. 295) at the beginning of the century.
This question is at least as relevant today as it was then.
Polya (1957, p. 50) points out that:
Teachers and authors of textboods should not forget that the intelligent student and the intelligent reader are not satisfied by verifying that the steps of a reasoning are correct but also want to know the motive and the purpose of the various steps. The introduction of an auxiliary element is a conspicuous step. If a tricky auxiliary line appears abruptly in the figure, without any motivation, and solves the problem surprisingly, intelligent students and readers are disappointed; they feel that they are cheated. Mathematics is interesting in so far as it occupies our reasoning and inventive powers. But there is nothing to learn about reasoning and invention if the motive and purpose of the most conspicuous step remain incomprehensible. To make such steps comprehensible by suitable remarks or by carefully chosen questions and suggestions takes a lot of time and effort; but it may be worth while.
This heuristic approach is advocated by many mathematicians and math-
ematics educators, among them Polya (1954, 1957, 1962-65), Poincar6 (1929
and 1969), Wilder (1967, especially p. 610), Avital and Shettleworth (1968,
pp. 39 and 42), Avital (1973), Henderson (1963, in particular p. 1015), Kinsella
(1970), Wittmann (1971) and the educational psychologists Bruner (1966, in
particular p. 72), Ausubel and Robinson (1969, in particular p. 530).
Avital (1973, p. 144) emphasizes the gain in comprehension and retention
when the teaching is done heuristically:
In exposing a student to logically structured, sequentially ordered material, we can improve the learner's grasp of it if we succeed in arousing his awareness of the need for each step.
In efficient expository teaching of mathematical proof, we have to present the material step by step so that the learner first sees the need for the next step, then becomes aware of how each step produces the next. Only if care is taken to organize and present the material in this way can we hope that the learner will be able to store it and retrieve it when necessary. It should be clear that the greatest facilitation is obtained when a rationale can be given for each step that has to be taken. We shall call such a procedure a naturally meaningful approach.
Unfortunately in textbooks as well as in classrooms the heuristic approach is
rarely used. Mathematics is usually presented as a Finished product. Students
are rarely told how to start a proof and how to proceed from one step to the
Educational Studies in Mathematics 8 (1977) 167-179. All Rights Reserved. Copyright 0 1977 by D. Reidel Publishing Company, Dordreeht-Holland.
168 SHLOMO LIBESKIND
next. Consequently many find it difficult to reproduce proofs that they have learned and find it almost impossible to prove new statements and to solve challenging problems.
This paper will consider an aspect of heuristic teaching which has not received the attention it deserves, namely that of teaching 'theory'. Most
students see mathematics as divided between problems and theory. Theorems are presented and proved, but students are rarely required to know the proofs. Those proofs for which the students are responsible are frequently just mem- orized before a test. A student who does not have a complete understanding of the theory is handicapped in problem solving and in the comprehension of subsequent topics.
It seems, therefore, desirable to introduce theorems in the form of problems and the proofs of these theorems as solutions to problems. This approach to teaching mathematics was successfully used in an experiment in teaching
Number Theory to high school students (Libeskind, 1971).* This paper advo- cates that similar principles be followed in teaching mathematics as often as possible.
Although we will see in this paper that theorems can be treated as problems, a student must be aware of the distinction between problems and theorems. Theorems are distinct from problems in that a problem is designated as a theorem if it is either unusually interesting in itself or valuable time and time again in the solution of other problems.
In the following examples several well known theorems are discussed, first
with a brief description of the traditional treatment followed by an illustration of a heuristic problem solving approach.
EXAMPLE 1. The Euclidean Algorithm The Euclidean Algorithm for finding the greatest common divisor (G.C.D.) of
two integers is usually introduced (in the United States at least) at the univer- sity level in Abstract Algebra and Number Theory courses. Given two positive integers a and b (a > b), the standard procedure for finding the G.C.D. of a and b applies the division algorithm repeatedly and is described as follows:
"We divide a by b and write:
a = q o b + r l , 0 ~ rl < b .
Next we divide b by rl and get
b = q l r l +r2, 0 ~ r 2 ( r 1.
* The first two examples in this article appeared in that study.
A PROBLEM SOLVING APPROACH 169
Next we divide rl by r2 and get
rl = qzr2+ r3, 0 <~ ra < r2.
We continue getting
r2 = qar3 + r4, 0 ~ r4 ( r3.
r~_ 2 = q k = l r ~ _ l + rk, 0 <~ rk < r ~ - l .
r k -1 = qkrk + O .
The process stops when the zero remainder is obtained". Then the following
theorem is stated: "The last positive remainder obtained in the above process
is the G.C.D. of a and b". Many instructors find that their students' retention span when exposed to
the above described Euclidean Algorithm is very short and their retention of
the proof is even shorter. A problem solving approach produced better results.
The following approach was tried with a group of 14-15 year old high
school students in an experiment in teaching Number Theory (Libeskind,
1971). The notion of the greatest common divisor (G.C.D.) of two numbers was
first introduced informally by asking: "What is the largest number that divides
both 6 and 15". Then the students were told that the largest number that divides both a and b is called the greatest common divisor of a and b and will
be denoted by (a, b). It was explained that using this notion one may write: (6, 15) = 3, (6, 18) = 6, etc. Students were told that when the numbers are
small it is easy to guess what their G.C.D. is, but it will be almost impossible
to guess the values of (455,221) or (584, 1606). Thus it would be desirable to
find a systematic procedure for finding the G.C.D. of any two numbers. Sup- pose one wants to compute (455,221). If he could find two small numbers,
say less than 10, whose G.C.D. is the same as (455,221), he could easily fred the G.C.D. It was pointed out to the students that (15, 2 0 ) = 5 and ( 1 5 - 10, 10)= 5, ( 1 5 , 6 ) = 3 and ( 1 5 - - 6 , 6 ) = 3 . They were able to conjecture the
following generalization from these facts:
(a, b) = ( a - b, b).
The students were encouraged to check the conjecture with a few more numeri- cal examples. They were then challenged to prove it. One student observed that the set of all common divisors of 24 and 18 is the same as the set of all com- mon divisors of 24 -- 18 and 18, namely {1,2, 3, 6} and suggested that the
class try to prove this in general. The following proof was found in a joint effort between the students and the teacher.
170 SHLOMO L I B E S K I N D
Proof." Let S be the set o fa l l common divisors of a and b, and T the set
of all common divisors o f a - b and b.
We first show that S C T. Indeed, if d is in S, then dla and dlb. (dla if
a = kd for some integer k.) By a previously proved theorem (if dla and d[b,
then d[a -- b) we have that d]a - b and therefore d is also in T.
Next we show that T C S. If d is in T then dla -- b and dlb. Since dlx and
dLv implies dlx + y , we have dl{a -- b) + b, i.e., dla, and conclude that d is in S. S C T and T C S imply that S = T. Since S = T, the greatest number in the
set S - w h i c h is (a, b ) - e q u a l s the greatest number in the set T which is
(a --b, b). Next the students were asked to use statement (1) as many times as was
necessary to compute (288, 51). After writing:
(288, 51) = (237, 51) = (186, 51) = (135, 5 t ) = (84, 51)
= (33, 51),
most students realized, either by themselves or after being asked for a shortcut,
that, rather than subtracting 51 five times from 288, it would have been
easier to divide 288 by 51 and find the remainder. Eventually students were
able to find the G.C.D. of two given numbers like 975 and 105 using the
following procedure: 9
(975,105) 105/975
= (30 ,105) 945 30
= (30, 15)
= (0, 15)
= 15.
3 30/105
105
15
In addition to being able to use the algorithm, students were also able to
reproduce the proof o f statement (1) as well as to generalize statement (1) to
the statement (a, b) = (a - qb, b) where q is any integer and prove this in a
way similar to the proof of statement (1). Most students were also able to solve
the following problem: "Consider (14a + 3, 21a + 4). Find the value of this
G.C.D. for a few different values of a. Make a conjecture. Prove it".
A similar procedure was tried with college students. The college students
were very appreciative of the approach, claiming that it was the first time
that they understood the Euclidean Algorithm.
One can take the principal of the problem solving approach one step further.*
* This extension of the first described procedure for discovering the Euclidean Algorithm was suggested to me by an anonymous reviewer, to whom I am grateful.
A P R O B L E M S O L V I N G A P P R O A C H 171
Notice that the proper ty (a, b) = (a -- b, b) was provided to the students by
the instructor. At the risk of complicating the task and therefore loosing
some students, the instructor could have taken a more exploratory approach
by offering a proper ty like (a, b ) = c-----5 ( a - b, b ) = c and then asking the
students to find similar consequences from the fact (a, b) = c. Students would
likely come up with the observation that not only does (a - - b, b) = c but also
that (a + b, b) = c , (a, a - b ) = c and (a, a + b ) = c .
The students could then be asked which of these possibilities actually aid
the computat ion of the G.C.D. of a and b. The students would most likely
reject the possibility (a + b, b ) = c on the grounds that the numbers would
increase and hence will make the finding o f the G.C.D. harder rather than
easier.
The students would discover that in order to compute the G.C.D. the most
efficient of the above properties is the proper ty (a, b) ---- c => (a - - b, b) = c.
This a t tempt at giving students experience in determining for themselves
most of the procedures in solving a problem is the major principle which
underlines this article. Even when an instructor is lecturing to large numbers
of students or when time considerations make students part icipation in the
solution process impractical, the instructor should always make as explicit as
possible the heuristic considerations he might have followed in solving the
problem.
EXAMPLE 2. The Order of an Integer Modulo m
The following theorem* is taken from a text on Number Theory by Niven and
Zuckerman (1960, p. 48):
Theorem 2.24. If a belongs to the exponent h modulo m, then a le belongs to the exponent hi(h, k) modulo m.
The book gives the following definition and theorem prior to the proof of
Theorem 2.24.
Definition 2.7. Let m denote a positive integer and a any integer such that (a, m) = 1. Let h denote the smallest positive integer such that a h ~- 1 (rood m). We say tha t a belongs to the exponent h modulo m. (h is also called the order of a modulo m.)
Theorem 2.23. If a belongs to the exponent h modulo m then hl~o(m). Furthermore ad _--a k (mod m) if and only ffhl(/" --k) .
* The theorem (and the following discussion) can be stated in analogous group theor- etical terms:
Let G be a cyclic group with h elements generated by a. Let b ~ G and b = a k. Then b generates a cyclic subgroup of G containing hi(h, k) dements.
172 SHLOMO L I B E S K I N D
Proof of Theorem 2.24. According to Theorem 2.23, (ak) J ~ 1 (mod m) if and only if hlk]. But hlk] if and only if h/(h, k) I k/(h, k) ] and hence if and only if hi(h, k) I]. Therefore the least positive integer] such that (ak) j ----- 1 (mod m) is] = h/(h, k).
The texts proof does not explain why each step is taken. A student reading
the proof may persuade himself that each step is correct and consequently
that the theorem is correct; however, that may be of limited value to him if
he is called upon to prove this or a similar theorem.
An alternative approach is to state the theorem in the form of a problem.
PROBLEM. If a belongs to the exponent h modulo m, then to what exponent
does a k belong?
Solut ion. By definition, a belongs to the exponent h means that a h = 1
(mod m), (a, m) = 1 and h is the smallest such integer. We are asked to what
exponent modulo rn does a k belong. If we denote this exponent by x, then
we are looking for the smallest x such that (ak) x ~ 1 (mod m). Our aim is to
find such an x. For that purpose we write the last congruence in the simpler
form, a k x = 1 (mod m). We see now that we have two similar congruences:
a h = 1 (rood m) where h is the smallest such integer and a ~x = 1
(mod m).
Since we want to find x, any information about k x should be of use. From a
previous theorem we know that k x must be a multiple of h, i.e., k x =]h . Since
we want to find x, we write x = ]h/x . Since h and k are given and we want to
find the smallest possible x, we have only to find the smallest integer ] for
which x is itself an integer. (At this point a numerical example hke x = ] .
may be o f value. From this numerical example it should be immediately clear
to the student that one has to reduce first the fraction ]~.) It is obvious that if
we put ] = k we get an integer for x. But is this the smallest possible ] which
will produce an integral value of x? In order to try to find a smaller ], we will
first reduce the fraction h / k as much as possible, i.e., divide both h and k by
their greatest common divisor (h, k). Calling the reduced fraction h'/]', we have:
/h ]h ' h ' k ' x = ~- = --~ where = h/(h, k ) and = k/(h , k) .
It is clear now that the smallest possible value o f ] is k ' (the reason for this is
intuitively clear but requires a proof which can be assigned to the student as
an exercise.) Thus
k 'h ' h x = - - = h ' =
k' (h, k)"
A PROBLEM SOLVING APPROACH 173
What we do when giving only tidy solutions to problems is hide from the
students all of the trials and errors, all of the missteps, that we often go through in attempts to arrive at a solution. By presenting only correct solutions we imply that there is nothing to learn from unsuccessful approaches to
solving a problem. Because it is often not obvious in the course of solving a problem which of
two (or more) next steps (or approaches) will lead to a successful solution,
it is desirable to put ourselves in the position of the students and anticipate what they, who do not know the answer to the problem, will feel are reason-
able ways to proceed. An instructor may collect suggestions from students, add
his own, and explore the ones which - a f t e r examination- seem plausible.
Even if he knows that a certain plausible direction will not lead to a solution,
that direction should still be examined. Because of time considerations, such
complete inquiries into a solution can not always be done. Nonetheless, the
instructor can thoroughly treat some approaches to a problem in class and
leave others for the students to explore as homework. Thus as often as possible
the solution to a problem should be approached more as a trial and rejection procedure than a process by which inevitable sequential steps lead to inevitable
solutions. For example consider the following two problems, one from Number
Theory and the other from School Geometry.
EXAMPLE 3. Fermat's and Euler's Theorems
(i) I f p is a prime and p½a (p does not divide a), then a p-1 _ 1 (rood p)
(Fermat's Theorem).
(ii) I f a and m are relatively prime and ~o(rn) is the Euler ¢ function (which counts the number of integers between 1 and m which are relatively prime to m) then a ~(rn) = 1 (mod m) (Euler's Theorem).
Fermat's Theorem is proved by considering a complete residue system mod m while Euler's Theorem is proved by considering a reduced residue
system mod m. In some textbooks Euler's Theorem is proved first and Fermat's
Theorem is obtained from it as a special case where m is a prime. We propose to prove Fermat's Theorem first and consider Euler's Theorem as a problem to
solve. For completeness and for the purpose of comparison, we include a proof
of Fermat's Theorem. The set B = {1,2, 3 . . . . . p -- 1} is a complete residue system modulo p.
One may ask whether adding a constant to each element of the set will pro-
duce again a complete residue system and similarly what happens if each element is multiplied by a constant. It is easy to see that adding a constant
will produce again a complete residue system. When we multiply by a constant (say a), we obtain the set A = {a, 2 a , . . . . (p -- 1)a}. This set will be a complete
174 SHLOMO L I B E S K I N D
residue system if and only if aU the elements are incongruent modulo m. It is
easy to observe and prove that this happens i fp j a. (Indeed k a - - la (mod p) =~
k = / (mod p) if (a, p) = 1 which implies k = l since 1 ~< k ~< p - 1 and 1 ~< l ~<
p -- 1. The condition (a, p) = 1 is equivalent to the condition p I a.) Now
since each set is a complete residue system every element of A is congruent to
some element of the set B and conversely. It follows that the product of the
elements of A is congruent to the product of the element of B. That is:
a " 2 a . . . ( p - - 1 ) a =- l ' 2 . . . ( p - 1 ) ( m o d p ) .
Since each of the numbers 1,2 . . . . . p - 1 is relatively prime to p, cancellation
is legitimate and we obtain
a p - 1 = 1 (modp) ( i fp [a).
The natural question to ask now is whether a similar result can be obtained
modulo m where m is not necessarily a prime. Considering a few special cases,
we see that if p is not a prime Fermat's result does not hold true. We state therefore the following problem:
PROBLEM: Given a and m find x (as a function of m) such that a x = 1
(mod m).
Since in Fermat's Theorem the restriction p l a was necessary, it seems
plausible that we may need a restriction on a in the above problem as well.
Attempting to solve the problem using the proof of Fermat's Theorem,
we define the set B = {1,2, 3 , . . . , m -- 1} which is a complete residue system
modulo m. Next we ask for which a will the set A = {a, 2a, 3 a , . . . , (m -- 1)a}
also be a complete residue system modulo m. For that purpose we find if any
two elements of A are congruent modulo m. If k a - la (rood m) where 1 ~<
k ~< m -- 1 and 1 ~< l ~< m -- 1 then k = l (mod m) and consequently k = l as
long as (a, m ) = 1. Thus A is also a complete residue system modulo m if
(a, m ) = 1. Consequently, as in the proof of Fermat's theorem, we have:
(2) a . 2 a . . . ( m - 1 ) a - 1 . 2 . . . ( m - - 1 ) ( m o d m ) .
Finally it remains to be determined whether the numbers 1, 2 . . . . , m - 1
can be cancelled out. This is possible if and only if each of the numbers
1 , 2 , . . . , m - - 1 is relatively prime to m. A numerical example will help
students to see clearly what happens. Let, for instance, m = 10, then quite a
few of the numbers 1 ,2, 3 . . . . , 9 are not relatively prime to m. (In fact one can easily see and prove that for any composite m there will be some numbers
in B which are not relatively prime to m.) How should we learn from this unsuccessful attempt and remedy the
A PROBLEM S O L V I N G A P P R O A C H 175
situation? We wish we had in (2) only those elements of B which can be
cancelled out, i.e., which are relatively prime to m. This 'wishful thinking'
motivates us to define a subset of B which consists only of those elements
of B which are relatively prime to m. For m = 10, this subset is B ' =
{i, 3, 7, 9}. It seems now that:
(3) a - 3 a ' 7 a ' 9 a - 1 - 3 . 7 " 9 ( m o d 1 0 )
and hence that
a 4 = l ( m o d l 0 ) if (a, 10) = 1.
Statement (3), however, needs to be justified. We need only to show that for
a, such that (a, 1 0 ) = 1 each element of B' is congruent modulo 10 to some
element of A ' = {a, 3a, 7a, 9a} and conversely. (This is a good problem by
itself and can be left as a challenging homework assignment).
In general we define a subset B' of the set B = {1,2, 3 , . . . , m -- 1 } as the
set of all the elements of B which are relatively prime to m. It makes sense
to call B' the reduced residue system modulo m. Let B ' = {rl, r 2 , . . . , rk}.
After proving that for a such that (a, m) = 1, A ' = { r l a , r 2 a , • • • , r k a } is also
a reduced residue system modulo m and consequeltly that every element of
B' is congruent modulo m to some element A ' and conversly, we obtain
a r l " a r 2 . . . a r k = r l r 2 • . . r k (modm) ,
and hence that a k = 1 (mod m), where k is the number of elements relatively
prime to m in the set B = {1, 2 . . . . , m -- 1}.
Since k is a function of m Euler introduced the function k = ~o(m) which
is called after him: the Euler ~0 function. Thus we have solved the posed prob-
lem and arrived at Euler's Theorem stated in part (ii) of Example 3.
Notice that in the above approach~ the definition of a reduced system was
obtained naturally arising from a need to modify the method of proof used
in Fermat's Theorem, rather than given without motivation as is the common
practice. The above approach also resembles some of the activities of a working
mathematician when he is looking for generalizations, when he is trying to see
if the same proof carries over to a new problem and if not, why not, and
what can be changed or modified so that it does.
EXAMPLE 4. "The segment between the mid-points of two sides of a triangle
is parallel to the third side and half as long".
This is an important theorem which is usually introduced after the proper-
ties of parallelograms are studied and before the concepts of ratio and pro-
portion are introduced. The theorem can be introduced in a problem form in a
variety of ways. For example:
176 SHLOMO LIBESKIND
PROBLEM: What can be said about the segment between the midpoints of two sides of a triangle in relation to the third side?
Students should be encouraged to explore the relationships and their proofs in some simple special cases like equilateral and right angle isoscles triangles. (In fact students trained to explore special cases should be able to decide by themselves to look for the above mentioned ones). The fact that the midseg- ment is parallel to the third side is observed in any triangle. That the length of the mid segment is half that of the third side can be easily discovered in an equilateral triangle by observing that the triangle MBN (see Figure 1) is isosceles
and hence equilateral (or perhaps simply by observation). Assuming that the
B
,4 C
Fig. 1.
midsegment is parallel to the third side, it is easy to deduce, that in a right isosceles triangle the length of that third side is twice the length of the mid- segment. The above special cases suggest a conjecture that can be stated in the form of a theorem (c.f. Example 4).
We will discuss now ways to prove the conjecture. Focusing on proving that the segments MN and AC in Figure 1 are parallel, we ask students to recall their previous experience in showing that lines or segments are parallel. One common method has been to exhibit two corresponding or alternate angles between the lines and a transversal. Thus one may want to prove that the angles BMN and BAC are congruent. How does one show, however, that the angles are congruent? Again previous experience suggests looking for two congruent triangles which include these angles. If BMN is one such triangle we are looking for another triangle with an angle A which seems to be con- gruent to triangle BMN. Since such a triangle is not found in Figure 1, we try to construct one. One possibility is to draw a line through M parallel to B--C. This line intersects A'-C at D (see Figure 2). We obtain a new triangle (namely triangle AMD) which seems congruent to the triangle MBN.
Another possibility is to find a point D on AC such that M N ~ A D , this seems especially promising, as we want to prove that AC= 2(MN). In either
A PROBLEM SOLVING APPROACH 177
B
D C
Fig. 2.
case, however, it is difficult if not impossible to prove that the triangles are congruent (the details are left out here).
Students may not know now what kind of a new reasonable route to choose. The instructor may suggest (students cannot be expected after all to
discover everything by themselves) to try to show that the segments MN and AC are parallel by showing that they lie on opposite sides of a parallelogram. Thus the objective now is to construct a quadrilateral, which will have as many properties of a parallelogram as possible, and such that MN and ACwill lie on its opposite sides. One possibility of constructing such a quadrilateral is
by choosing the segment MN as one of its sides. Then one could construct
MD parallel to NC (see Figure 2) or find D such that MN ~ DC. This approach will again turn out to be fruitless. Finally the other remaining possibility is to
construct the desired quadrilateral by choosing AC as one of its sides. Since we want the quadrilateral to resemble a parallelogram as much as possible,
after choosing ~ to be one of its sides (we could as well choose ~ it is reasonable to extend ~ and through C construct a line parallel t o ' i n t e r - secting the line M'~ at E (see Figure 3).
B
Fig. 3.
E
178 SHLOMO LIBESKIND
Since we want to prove that MN = ½AC another plausible approach is to extent MN and find a point E such that MN~--NE. In either case students should be able to complete the proof.
After the proof is completed it might be instructive to ask why the last
method was productive while the other were not. (That is, in the last method
we obtained extra congruent vertical angles by extending the segment MN.-) It
should also be pointed out that from this method of proof one learns that
sometimes it may be desirable to construct lines and figures which are not inside the original figure.
Finally note that by being flexible enough one may produce a different
proof (without the need to extend lines beyond the triangle) out of the un-
successful early attempts. Trying to show that M N = ½AC (see Figure 2) we
could have asked ourselves then what additional information would enable
us to prove that triangles AMD and MBN are congruent. If we only knew that
M N and AC were parallel this could have been achieved (recall our early
discussion of the problem for a right isosceles triangle). This wishful thinking
suggests changing the hypothesis and proving the following theorem: "A line
through the midpoint of a side of a triangle and parallel to another side bisects the third side. The length of the segment obtained is half of the length of the
side to which it is parallel". Now the proof of the theorem in Example 4
follows by a simple indirect argument.
CONCLUDING REMARKS
The advantage of introducing a theorem in the form of a problem should now
be obvious. Exposed to the 'theorem-proof' approach, students rarely know
how the result was found. If the theorem is forgotten, it will be difficult to
recall it. If the student begins to think of theorems as problems, however, he can recall a theorem by solving it as a problem.
University o f Montana
REFERENCES
Ausubel, David P. and Robinson, Floyd G., School Learning, Holt, Rinehart and Winston, Inc., New York, 1969.
Avital, Shmuel M. and Shettleworth, Sara J., Objectives for Mathematics Learning (Bull. No. 3), Toronto, Ontario, Can., Ontario Institute for Studies in Education, 1968.
Avital, Shmuel M., 'Teaching a Mathematical Proof', Int. £ Math. Educ. ScL TechnoL, Vol. 4, No. 2 (April-June, 1973), pp. 143-147.
A PROBLEM S O L V I N G A P P R O A C H 179
Bruner, Jerome S., Toward a Theory of Instruction, Cambridge: Belknap Press, 1966. Henderson, Kenneth B., 'Research on Teaching Secondary School Mathematics', Hand-
book of Research on Teaching, N. Gagne (Ed.), Chicago: Rand McNally and Company, 1963, 1007-1030.
Kinsella, John J., 'Problem Solving'. The Teaching of Secondary School Mathematics, Thirty-third yearbook, Washington, D.C.: National Council of Teachers of Math- ematics, 1970.
Libeskind, Shlomo, 'A Development of a Unit on Number Theory for Use in High School, Based on a Heuristic Approach', Unpublished Ph.D. dissertation, University of Wisconsin, Madison, 1971.
Niven, Ivan and Zuckerman, Herbert S., An Introduction to the Theory of Numbers, New York: John Wiley and Sons, Inc., 1960.
Poincar6, Henri, The Foundations of Science, New York and Lancaster: The Science Press, 1929.
Poincar~, Henri, 'Mathematical Definitions and Teaching', The Mathematics Teacher, Vol. 62, No. 4 (April, 1969).
Polya, George, Mathematics and Plausible Reasoning, Princeton: Princeton University Press, 1954, 2 vols.
Polya, George, How to Solve It (2rid ed.), New York: Doubleday, 1957. Polya, George, Mathematical Discovery, New York: Wiley, 1962-65, 2 vols. Wittmann, Erich, 'Complementary Attitudes in Problem Solving', Educ. Studies in Math.
4 (1971), pp. 243-251.