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How to solve it? 1

NANYANG JUNIOR COLLEGE DEPARTMENT OF MATHEMATICS

To imbue our students with excellent mathematical habits of thought that will enable

them to become critical thinkers, effective problem-solvers and proficient in the use of the language of mathematics and information technology

HOW TO SOLVE IT?

I INTRODUCTION AND RATIONALE The great Hungarian mathematician Paul Halmos once stated:

“the heart of mathematics is its problems.” Indeed, problem-solving lies at the heart of mathematics. Mathematics are discovered and created by way of problem-solving. It is the foremost skill any mathematician, or for that matter, any student of mathematics must possess. Many people believe that mathematical ability is granted only to an elite few while the rest are not blessed with it. According to the popular folklore, you are either a mathematical genius or you simply cannot do mathematics at all. On closer scrutiny, the above viewpoint is apparently oversimplified. Mathematical attitudes and aptitudes do vary across the entire subject. Some people prefer algebra to geometry while others enjoy geometry and dislike algebra. Some people are good in calculus but can be quite hopeless in trigonometry. Some find ‘O’ level additional mathematics easy and ‘A’ level mathematics difficult. Some are attracted to pure mathematics but are put off by applied mathematics. Some can handle computational mathematics fairly well but are completely defeated by “theorem-proving” mathematics. Finally, there are those who can perform routine and standard mathematical calculations with tremendous ease but have absolutely no idea how to begin tackling a non-routine problem couched in a non-standard setting. Many factors affect an individual’s mathematical performance. These factors can be “internal” like a person’s natural ability, intelligence, inclination and personality while the “external” factors can include the environment, upbringing, training as well as experiences with things and people. While we recognise that there are these factors that influence an individual’s ability in doing mathematics, we want to find out if there is anything that one can do to improve performance in mathematics. In short, we ask the simple question: “Is there anything one can do to become better at doing mathematics?” The most common answer to this question is ‘practice’. If you sit down and work through all the problems in the ‘10-year-

Prepared by Leong Chong Ming / Department of Mathematics / Nanyang Junior College

How to solve it? 2

series’ conscientiously you will eventually become good at solving ‘10-year-series’ problems. True, to be good at something and that of course includes mathematics, you need to devote a generous amount of time to practice. Certainly hard work is crucial in acquiring proficiency in mathematical problem-solving. Let me reiterate that I am not dismissing the importance of setting time for practice in order to gain experience and confidence in problem-solving. But is drill and practice the only answer to the above question? Besides drill and practice, are there ways, methods, habits or perhaps even attitudes that one can adopt in order to become a more effective problem-solver? The answer is a definite yes! There are in fact certain mathematical behaviours that we can adopt to help us become better at doing mathematics. While drill and practice are necessary in preparing one to sit for mathematics examinations, they alone are not able to guarantee acceptable results. Firstly, drill and practice alone are not very helpful to someone who has been “programmed” to solve only 10-year-series type of problems if the problems set in the examination are non-routine and unfamiliar. Secondly, they are not sufficient to guarantee success for the individual who wants to pursue higher mathematics in the university which requires a lot more rigour and understanding of concepts and ideas. Proficiency in mathematics requires the following three ingredients: 1. knowledge of the subject matter, 2. awareness of strategies and tactics for solving problems and learning new material, 3. ability to monitor and to control one’s activities when doing mathematics. Most mathematics textbooks are concerned only with item 1. The purpose of our discussion is to focus on items 2 and 3. What I am presenting here are some fundamental rules-of-thumb that might be helpful in the problem-solving process. These rules-of-thumb or heuristics, or ‘thinking techniques’ if you like to call them, are no magical formulae that will guarantee success in any given problem but they are good guidelines that can help you in the thinking process. We shall base our exposition on the work of George Polya (1887 – 1985, Hungarian), one of the leading mathematicians of the early 20th century who had done stupendous work in analysis and probability theory. Polya was also a great teacher who devoted considerable efforts to the study of mathematical behaviours in problem-solving. II POLYA’S MODEL FOR PROBLEM-SOLVING Polya once said:

“Mathematics, taught and learnt appropriately, improves the mind and implants good habits of thought.”


How to solve it? 3

Without doubt, the benefits reaped with appropriate teaching and learning of mathematics are great. Problem-solving, being the heart of mathematics, is one extremely important aspect of mathematics that improves the mental faculty and promotes good thinking habits. Polya’s principal method in problem-solving was introspection. He carefully observed and monitored his own actions as he was solving mathematical problems, noting particularly the steps that enabled him to solve a difficult problem. In his well-known book How To Solve It, Polya systematically spelt out his findings by dividing the problem-solving process into 4 parts: 1. Understand the Problem. 2. Devise a Plan. 3. Carry out the Plan. 4. Look Back. Polya’s problem-solving model can be organised into the following framework: I Understand the Problem

1. Write down clearly what is to be accomplished and identify the unknown(s). 2. Isolate the hypotheses and the data. 3. Restate the problem. 4. Develop a representation of the problem.

a. Draw a figure. b. Introduce suitable notation.

5. Examine special cases. a. Select special values to acquire a “feel” for the problem. b. Consider extreme cases. c. Evaluate integer parameters at n = 1, 2, 3, … and look for a pattern.

6. Simplify the problem. a. Exploit symmetry. b. Choose appropriate units.

II Devise a Plan

1. Consider essentially equivalent problems. a. Replace conditions by equivalent conditions. b. Recombine the elements of the problem in various ways.

c. Introduce auxiliary elements. d. Reformulate the problem.

i. Change perspective or notation. ii. Use induction whenever appropriate. iii. Argue by contradiction or contraposition. iv. Take the problem as solved; i.e., assume you have a solution and

determine its properties.


How to solve it? 4

2. Consider slightly modified problems. a. Aim for subgoals. b. Relax a condition, then restore it. c. Consider case analysis.

3. Consider broadly modified problems. a. Construct an analogous problem with fewer variables. b. Generalise the problem. c. Hold all but one variable fixed to determine that variable’s impact. d. Try to exploit any problem that has a similar form, hypothesis, or conclusion.

III Carry out the Plan

1. Check each step. 2. Prove that each step is correct.

IV Look Back 1. Apply these specific tests to your solution:

a. Does it use all the pertinent data? b. Does it conform to reasonable estimates or predictions? c. Does it withstand tests of symmetry, dimension analysis or scaling?

2. Apply these general tests: a. Can the result be obtained differently? b. Can the result be verified in special cases? c. Can the result be reduced to known results? d. Can the result be used to derive other known results?

We shall consider some of the more frequently used strategies and give examples on how these can be employed to tackle problems. II.1 UNDERSTAND THE PROBLEM Understanding a problem is the first important step to solving it. II.1.1 Write down clearly what is to be accomplished and identify the unknown(s) Before embarking on a problem, it is always a good practice to write down explicitly what we are supposed to do in the problem, be it finding an unknown or proving a statement. This will help us to be focused on the task and prevents digression or “beating-about-the-bush”. Also, focusing on the unknowns provides coherence to the problem and directs the efforts of the problem solver towards a specific aim. In other words, if you know and understand what you are looking for, then you will develop a clearer understanding of the problem.


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Example 1 Prove by induction that 1 + 3 + 5 + … + (2n – 1) = n2. When proving the inductive step Pk is true ⇒ Pk+1 is true, write down explicitly what we want to prove. Assuming the induction hypothesis 1 + 3 + 5 + … + (2k – 1) = k2 for some positive integer k, we want to prove that “1 + 3 + 5 + … + (2k – 1) + [2(k + 1) – 1] = (k + 1)2.” Having written this down, we know exactly what to do and then sets out accomplishing our goal. A few careful steps of simple algebra would see us achieving our end. Example 2 Find the centre and radius of the circle having the equation x2 + 4x + y2 – 6y – 3 = 0. We are required to find the centre and radius of the circle. These are therefore the unknowns that we have to find. Having identified the unknowns and pinpointing their meaning, the next question is, “how to find them?”. Now you need to recall and understand the meaning of the centre and radius of a circle. Here it is: If the centre of a circle is (a,b) and its radius is r, then any point (x,y) on the circle is at a distance r from (a,b).

Thus by the distance formula, ( ) ( ) rbyax 22 =−+− which gives the general equation of a circle (x – a)2 + (y – b)2 = r2 (*) where (a,b) is the centre and r the radius of the circle. So to find the centre and radius of the circle, it seems reasonable to transform the given equation into the general form (*) so that the values of a, b and r can be obtained. Performing some elementary algebraic manipulation (basically completing-the-squares) carefully should yield the equation (x + 2)2 + (y – 3)2 = 42. We therefore conclude that the centre and radius of the circle are (-2,3) and 4 respectively. Notice that by first identifying the unknowns and understanding their meaning, we have drawn the problem into sharper focus. Finding the centre and radius of the circle reduces to finding real numbers a, b and r by rewriting the given equation into the form (*). If we succeed, we are done. II.1.2 Isolate the hypotheses and the data Once you have a clear understanding of the problem, focus on what information is given in the problem, the hypothesis and the data. Ask yourself these questions: What is given in the problem? What is the hypothesis? What are the data? What do the terms in the hypothesis mean? Example 3 If a2 < 4b, prove that x2 + ax + b > 0 for all real numbers x. What is given here? a2 < 4b.


How to solve it? 6

How do we make use of this piece of given information to solve our problem? How does it play a pivotal role to guarantee that x2 + ax + b > 0 for all real numbers x? It may not be apparent at first sight but it becomes clear when we complete the squares

for the expression x2 + ax + b to obtain 42

22 abax −+⎟⎠⎞

⎜⎝⎛ + .

Observe that whether the above expression is positive for all x hinges on whether

04

2>−

ab .

It is now clear that since we are given a2 < 4b, we must have 04

2>−

ab .

Consequently x2 + ax + b = 42

22 abax −+⎟⎠⎞

⎜⎝⎛ + > 0 for all real x.

II.1.3 Restate the problem Very often, by restating a given problem, the task at hand becomes clearer and more focused. Example 4 Prove that if f(x) is a continuous function on the interval (a,b) such that f(a) and f(b) have different signs, then f(x) has a zero in the interval (a,b). The information given and the task we are supposed to perform are not very clear at the onset. So we restate the problem as follows: Prove that if f(x) is a continuous function on the interval (a,b) such that either f(a) > 0 and f(b) < 0 or f(a) < 0 and f(b) > 0, then there exists a number r where a < r < b such that f(r) = 0. It now becomes clearer what are given and what we have to do. II.1.4 Develop a representation of the problem By a representation of the problem, we mean a method of describing the problem that

is internally coherent or consistent, corresponds closely to the problem, is closely connected with the knowledge of the problem solver about the items of the

problem and related concepts. The representation may take any form – a diagram, a graph, an equation or even a new idea. Example 5 Prove the triangular inequality: If a, b and c are the lengths of the sides of a triangle, then a + b > c.


How to solve it? 7

The inequality becomes immediately obvious if we draw a diagram. A Fig. 1 c b B a C By using a diagrammatic representation, we see that the inequality makes perfect sense and we get a visual “feeling” of the problem. It is absolutely consistent with our knowledge of and experience with triangles. Example 6 Refer to Example 4. The statement of the problem becomes clear via a graphical representation. (a,f(a)) y = f(x) Fig. 2 a r b x (b,f(b)) We see that the problem which seems rather abstract initially is really common sense. Notice also that by looking at Fig. 2 above, it becomes clear why the condition that f(x) is continuous is necessary, for if f(x) is discontinuous on (a,b), then r may not exist at all. (See Fig. 3 below) (a,f(a)) Fig. 3 a b x (b,f(b)) There is another condition present in the problem which may not be so apparent. Can you pinpoint what it is and why is it important?


How to solve it? 8

Therefore a good graphical representation of the problem helps us to understand the problem better and appreciate the importance of the conditions present in the problem. Example 7 Solve the system of linear equations: x + y – z = 0 3x + 2z = 9 -2x + 2y – 3z = -7 Observe that we can represent the problem with a new idea – matrices. The above problem can be put into the form:

. ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−

7-90

322

203111

zyx

By using a matrix representation, the solution can be obtained by simply computing the inverse of the matrix as follows:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−

−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−

7-90

322203111

1

zyx

= . ⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

321

II.1.5 Examine special cases Often we can understand a problem better by considering special cases. For example, if the problem involves a positive integer n, then consider the cases when n = 1, 2, 3; if there is a real number x ranging over the interval -1 ≤ x ≤ 1, then consider the cases when x takes the extreme values x = ±1 or x = 0; if we have an arbitrary triangle, then investigate the cases when it is equilateral or right-angled. By considering special cases, we can at least acquire some experience with the problem and develop a feel for various aspects of the problem. This may help us to obtain an idea of how to go about tackling the problem. Example 8

Find a formula in terms of n for ( ) 11

1∑=

+

n

rrr

.


How to solve it? 9

To simplify our working, we introduce the notation ‘S(n)’ where S(n) = ( ) 11

1∑=

+

n

rrr

.

We consider the special cases when n = 1, 2, 3.

S(1) = ( ) 11

1

1∑=

+r

rr = ( )111

1+

= 21 =

111+

.

Similarly, S(2) = ( ) ( )1221

1111

++

+ =

32 =

122+

.

S(3) = ( ) ( ) ( )1331

1221

1111

++

++

+ =

43 =

133+

.

By looking at the pattern, we are led to conjecture that S(n) = 1+n

n for all positive

integers n. It now remains to prove our conjecture by induction. Example 9 12 people are seated at a circular table. How many different handshakes are there if any person is not allowed to shake hand with two of his adjacent neighbours? The number ‘12’ is too big. So we consider smaller numbers. We start with 3. According to our rule, there will be no handshake since they are all adjacent to one another. Next we try 4. Let A, B, C and D denote the 4 people. A Fig. 4 D B C Observe that A can only shake hand with C, B can only shake hand with D and so on. We can easily enumerate all the cases as follows: A → C, B → D, C → A, D → B. Are there 4 different handshakes?


How to solve it? 10

But notice that A → C and C → A are in fact the same. Likewise B → D and D → B

are essentially the same. So we have counted twice and there are actually 2

14× = 2

different handshakes. We move on to 5. Let A, B, C, D and E denote the 5 people. A can only shake hand with C and D, B can only shake hand with D and E and so on. A Fig. 5 E B D C Again we enumerate all the cases as follows: C A D D B E A C E A D B B E C Are there 10 different handshakes? No, we have committed double-counting again.

There are actually 2

25× = 5 different handshakes.

What about 6? Each person can now shake hand with 3 non-adjacent people. Using the

same argument, we should have 2


Now we can easily tackle the case when there are 12 people.

There should be 2



How to solve it? 11

We can now even generalise the problem to any n number of people without giving us any stress. How many different handshakes are there?

We can write down with ease ( )2

3−nn different handshakes.

In the above problem, we have chosen special cases where the numbers are small to help us gain a better understanding of the problem. The method we have used for solving the problem for smaller numbers is repeated for the larger numbers and even for any general n. A simple strategy indeed! Example 10 Prove the following theorem concerning triangles:

The centroid of a triangle is located 32 of the way from the vertex along a median.

Restating the problem, we have to show that if M is the midpoint of the side BC of the triangle ABC, then the centroid G of the triangle must lie on the median AM such that

AG = 32 AM. (See Fig. 6 below)

A Fig. 6 G • B M C Let us consider the special case when the triangle ABC is equilateral as shown in Fig. 7. A Fig. 7 O N G • B M C Since M is the midpoint of BC and AB = AC, AM is perpendicular to BC. Similarly, if N is the midpoint of AC, then BN is perpendicular to AC. If we let the sides of the triangle have a common length of 2a, then we can easily show

that AM = a tan 60° = 3a and AG = 30 cos

a = 3

2a = ( )332 a =

32 AM.


How to solve it? 12

In a similar fashion, BG = 32 BN and CG =

32 CO.

We now have a “feel” that the theorem is correct, at least for the case when the triangle is equilateral. To tackle the more general case, we will need to adopt a different approach in order to prove the theorem for any arbitrary triangle. See Example 13. II.1.6 Simplify the problem Simplifying a problem usually involves introducing some mathematical ideas like the coordinate system, equations and graphs or even a notation to make the problem more familiar and easier to handle. Example 11 The region in Fig. 8 is bounded by a parabolic segment and a straight horizontal line segment PQ that is perpendicular to the axis of the parabola. Given that the distance from the vertex of the parabola to PQ is 4 units and PQ = 4 units, find the dimension of the largest rectangle that can be inscribed in the region. Fig. 8 4 units P Q 4 units To tackle the problem, we look for a suitable representation in order to simplify our problem and “repackage” it in a form we are familiar with. Perhaps the easiest way to do this is to introduce an appropriate Cartesian coordinate system. It is quite natural to set up the coordinate system in such a way that the x-axis coincides with the line segment PQ and the y-axis passes through the vertex of the parabola and is perpendicular to PQ as shown in Fig. 9. y (x,4 – x2) Fig. 9 y = 4 – x2 P(-2,0) -x O x Q(2,0) x


How to solve it? 13

We now have something familiar – a parabola with equation y = 4 – x2 with P and Q having coordinates (-2,0) and (2,0) respectively. We readily see that any inscribed rectangle will have a base of length 2x (0 < x < 2) and a height 4 – x2 (0 < x < 2). Notice also that 0 < 4 – x2 < 4 for 0 < x < 2. The area of the rectangle, being a function of x, is therefore

A(x) = 2x (4 – x2). To find x that corresponds to the maximum value of A(x) is simply a matter of elementary differentiation.

Setting ( )[ xAdxd ] = 0 gives 8 – 6x2 = 0 ⇒ x =

32 .

It is easy to check that this value of x corresponds to the maximum value of A(x).

The dimension of the rectangle is therefore 3

4 by 38 .

The next two examples illustrate the usefulness of exploiting symmetry. Example 12

Without performing tedious integration, prove that dxxx

x cos sin

sin 2

0∫π

+ =

4π .

Doing the integration by brute force is rather daunting. Is there a simpler and more elegant way to do it? Can we exploit symmetry?

Recall the symmetric relationships xx cos 2

sin =⎟⎠⎞

⎜⎝⎛ −π and xx sin

2cos =⎟

⎠⎞

⎜⎝⎛ −π .

This prompts us to try the substitution x = y−π2

. Effecting the substitution, we have

I = ( )dyyy

y1

2cos

2sin

2

sin0

2

−⎟⎠⎞

⎜⎝⎛ −π

+⎟⎠⎞

⎜⎝⎛ −π

⎟⎠⎞

⎜⎝⎛ −π

∫π

= dyyy

y sin cos

cos2

0∫π

+

= dxxx

x sin cos cos2

0∫π

+ since y is a dummy variable which can be replaced by x.

Therefore 2I = dxxx

x cos sin

sin 2

0∫π

+ + dx

xxx sin cos cos2

0∫π

+


How to solve it? 14

= dxxxxx

cos sin cos sin 2

0∫π

++ = dx 2

0∫π

= 2π .

Hence I = 4π .

Behold the beauty of symmetry! Example 13 Prove the statement in Example 10.

We need to prove that AG = 32 AM, BG =

32 BN and CG =

32 CO.

By symmetry, it suffices to prove that BG = 32 BN.

From the triangle ABC we construct a parallelogram ABCD as shown below. A M’ D Fig. 10 G’ N G B C M Join A to M, the midpoint of BC. Join B to N, the midpoint of AC. Finally, join C to M’, the midpoint of AD. Let G be the intersection of the line segments AM and BN. It is clear by symmetry that AM is parallel to CM’. Observe that triangles CG’B and MGB are similar. Likewise triangles AGD and M’G’D are similar. Consequently, BG = GG’ = G’D since M and M’ are the midpoints of BC and AD respectively. Note also that N is the midpoint of BD since N is the intersection of the diagonals of the parallelogram ABCD.

Hence BG = 31 (BG + GG’ + G’D)

= 31 BD =

31 (2BN) =

32 BN.

II.2 DEVISE A PLAN The heart of the problem-solving process as presented by Polya lies in the second stage – Devising a plan or exploration.


How to solve it? 15

II.2.1 Consider essentially equivalent problems Quite often, a problem presented can be rather complicated or hard to solve. But by considering another problem without losing the essence of the original problem may help us obtain the solution easily. Example 14

Prove that if n > 4, then n

n 44

3+ > 4.

It may be rather daunting to start with the expression n

n 44

3+ and prove directly that

nn 44

3+ > 4.

Instead, we consider an equivalent problem: Prove that n

n 44

3+ – 4 > 0.

Now n

n 44

3+ – 4 =

nnn

416163 2 +−

= ( )( )n

nn4

443 −− > 0

since n > 4 ⇒ n > 0, 3n – 4 > 8 > 0 and n – 4 > 0.

We have therefore established equivalently n

n 44

3+ > 4.

Another way to replace a problem by an essentially equivalent problem is to introduce so-called auxiliary elements. By this we mean any object or feature that is added to the problem which while not changing the problem, enhances or clarifies the relationships among the given elements of the problem. Auxiliary elements are commonly used in algebra and geometry. Example 15

Sketch the graph of the function f(x) = 1

2−xx ( 1≠x ).

We must first obtain the asymptotes of the graph. The line x = 1 is an obvious vertical asymptote. But are there other asymptotes? Lets investigate. Notice that equivalently, f(x) can be written in the form as follows:

f(x) = 1

2−xx

= ( )1

22 1

112 1

1112

1112

12

−+=⎥⎦

⎤⎢⎣⎡

−+=⎥⎦

⎤⎢⎣⎡

−+

−−

=⎥⎦⎤

⎢⎣⎡

−+−

=⎥⎦⎤

⎢⎣⎡

− xxxxx

xx

xx .

It is now clear that y = 2 is a horizontal asymptote.


How to solve it? 16

Notice that in our working, we have deliberately written the “x” in the numerator in the form “(x – 1) + 1” so as to change the feature in order to provide the facility for division with the denominator. We can now proceed with the sketch. Example 16

Define In = . Show that ( ) dxxb

a

n 3 2∫ + ( )∫ +

b

a

ndxxx 3 22 = In+1 – 3In.

Quite naturally, we should consider changing the feature of the integrand so that it contains (3 + x2)n and (3 + x2)n+1. Obviously we will not do anything to (3 + x2)n. So lets see if we can do something to change the feature of x2. We ask how we can create a term “(3 + x2)” so that we can make the integral looks like the form we want. To achieve this, we judiciously write x2 = (3 + x2) – 3.

So = ( )∫ +b

a

ndxxx 3 22 ( )[ ]( ) dxxx

nb

a 3 33 22 +−+∫

= ( ) ( )∫∫ +−++ b

a

nb

a

ndxxdxx 33 3 212

= In+1 – 3In. Example 17 Prove the Pythagoras theorem: If a, b and c are the lengths of the sides of a right-angled triangle with c being the length of the hypotenuse, then a2 + b2 = c2. First we draw a diagram. C Fig. 11 a b A d D e B c Next we drop a perpendicular from C to AB at D. The line CD is called an auxiliary line. Let AD = d and BD = e. Observe that triangles ACD and CBD are similar.

So cdbbc

dbcea

ac

ea and 22 =⇒==⇒= .

It therefore follows that a2 + b2 = ce + cd = c(e + d) = cc ⋅ = c2. The next example demonstrates the use of an indirect proof called contraposition.


How to solve it? 17

It simply says that given two mathematical statements P and Q, proving “P ⇒ Q” is logically the same as proving “ PQ ⇒ ” where P and Q denote the complement statements of P and Q respectively. Example 18 Let n be an integer. Prove that if n2 is odd, then n is odd. We let the statements P: “n2 is odd” and Q: “n is odd”. Instead of proving P ⇒ Q, we shall prove the equivalent PQ ⇒ . That is, we shall prove that if n is even (n is not odd), then n2 is even (n2 is not odd). If n is even, then n = 2k for some integer k. It follows that n2 = (2k)2 = 4k2 = 2(2k2) which is even since 2k2 is an integer. Consequently, PQ ⇒ . Hence by contraposition, P ⇒ Q. The next example illustrates the use of contradiction. This is another indirect proof where we first of all assume the negation of the desired conclusion and then proceed to show that it leads to a contradiction or an impossible situation, that is, an absurdity. Example 19 Let S denotes the set of all real numbers excluding the number -1. An operation, denoted by ‘*’, is defined on the set S such that for all x, y ∈ S, x * y = xy + x + y. Prove that x * y ∈ S. [ In this case, we say that * is closed on S ] Suppose on the contrary, there exists x, y belonging to S such that x * y ∉ S. Then by definition, x * y = -1. So xy + x + y = -1 xy + x + y + 1 = 0 (x + 1)(y + 1) = 0 ⇒ x = -1 or y = -1. But since both x and y belong to S, x ≠ -1 and y ≠ -1. We now have a contradiction. So the statement “x * y ∉ S” we started with earlier is wrong and therefore we must have x * y ∈ S. II.2.2 Consider slightly modified problems Sometimes it might be useful to consider a slightly modified problem which is supposed to be simpler, in order to tackle the harder problem. We can do this by relaxing a condition and then restoring it, by using analogy or by removing some elements from the problem but still retaining the original structure of the problem. Example 20 Show that if n1, n2, n3 and n4 are real numbers such that 0 < nr < 1 for all r = 1, 2, 3, 4, then (1 – n1)(1 – n2)(1 – n3)(1 – n4) > 1 – n1 – n2 – n3 – n4 . Extend the problem to the general case and prove it.


How to solve it? 18

First we try to understand what we are given. We are given that the real numbers n1, n2, n3 and n4 all lie strictly between 0 and 1. Let us modify the problem slightly by removing some variables from the problem so that we have fewer variables to handle. This will make the problem easier to handle and yet retain the original structure. So we consider (1 – n1)(1 – n2) = 1 – n1 – n2 + n1n2 > 1 – n1 – n2 since n1n2 > 0. We now put back another variable and retain the same structure. Multiplying throughout by (1 – n3) (> 0) gives (1 – n1)(1 – n2)(1 – n3) > (1 – n1 – n2)(1 – n3) = 1 – n1 – n2 – n3 + (n1 + n2)n3 > 1 – n1 – n2 – n3 since (n1 + n2)n3 > 0. Now what about the original problem? We can proceed in exactly the same way to obtain the desired inequality (1 – n1)(1 – n2)(1 – n3)(1 – n4) > 1 – n1 – n2 – n3 – n4 . In fact, we can even extend the problem to the general case. To do this, we shall introduce the notation “ ” which stands for product in exactly the same way as the

sigma notation “ ” stands for sum. The product notation means .

∏

∑ n

n

rr aaaa …21

1 =∏

=With this understanding of the use of the product notation, we shall prove:

where 0 < nr < 1 for all r = 2, 3, 4, … , N. ( ) ∑∏==

−>−N

rr

N

rr nn

22 1 1

Notice that r starts from 2. This is because when r = 1, we actually have an equality. The method we have used to tackle the problem earlier suggests that we can argue by induction. We let Pn denotes the above proposition. P2 has been verified to be true earlier.

Assuming Pk is true for some k > 2, we have . ( ) ∑∏==

−>−k

rr

k

rr nn

22 1 1

We want to prove Pk+1 is also true, that is . ( ) ∑∏+

=

+

=

−>−1

2

1

2 1 1

k

rr

k

rr nn

( ) ( ) ( )∏∏=

+

+

=

−−=−k

rrk

k

rr nnn

21

1

21 1 1

> by the induction hypothesis ( ) ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−− ∑

=+

k

rrk nn

21 1 1

= ∑∑=

++=

+−−k

rrkk

k

rr nnnn

211

21

> since > 0. ∑+

=

−1

21

k

rrn ∑

=+

k

rrk nn

21


How to solve it? 19

II.3 CARRY OUT THE PLAN In carrying out the plan, we must constantly bear in mind the importance of checking each step of our working, making sure that it is error-free and that there is a logical justification. II.4 LOOK BACK It is always a good habit to look back and review what we have done. Polya’s admonition of looking back has three major thrusts. Firstly, he admonishes us to check our work by applying various tests to the solution – use of data, symmetry and alternate derivations. Secondly, he encourages us to sit back and view our work in perspective and to see how it relates to the whole problem. This gives us an overview of all the thinking processes, the interconnectedness amongst the different ideas and concepts as well as the strategies we have used to put the whole puzzle into one seamless piece. Just as an artist enjoys his masterpiece upon completion, so must we enjoy our work. Thirdly, he points us toward future problems by having us to generalise and try to derive new results from the results obtained. In other words, looking back has both a backward and a forward orientation. The questions we should ask ourselves upon the completion of a problem are:

What concepts and ideas have I used? Is there a better way of doing it? Are there alternative approaches? What are the restrictions or conditions that must be satisfied in the problem? Why is the problem asked in that way? Can I relate the problem to other parts of mathematics or to any real-life situations? Can I generalise and extend the problem? Can I obtain new results from what I have found? Does my work point to future possibilities? Does the problem give me a sense of joy and satisfaction?

We provide an example on how looking back and checking can help us gain a greater understanding and appreciation of the problem. Example 21 The Mean Value Theorem (MVT) is one of the most famous and important results in calculus. It states: Suppose f(x) is a continuous function on the closed interval [a,b] and that f’(x) exists for all x in the open interval (a,b). Then there exists a number c ∈ (a,b) such that

f’(c) = ( ) ( )ab

afbf−− .

A diagrammatic representation would help us get a better feel of the theorem.


How to solve it? 20

y (b,f(b)) Fig. 12 (a,f(a)) O c x From the diagram, it becomes clear that the theorem is saying that there is a point on the curve corresponding to x = c at which the gradient of the tangent to the curve is the same as the gradient of the chord joining the points (a,f(a)) and (b,f(b)). Even if you have read a general proof of the MVT in a calculus book (or discover one for yourself), it is useful to check and verify the result in simple cases, for instance when f(x) is linear or quadratic. For example, let us assume f(x) = x2. Then for all a, b ∈ R and assuming a < b,

( ) ( )ab

afbf−− =

abab

−− 22

= a + b.

Now f’(x) = 2x for all x ∈ R. Therefore

( ) ( )ab

afbf−− = a + b = ⎟

⎠⎞

⎜⎝⎛ +

22 ba = f’ ⎟

⎠⎞

⎜⎝⎛ +

2ba

and indeed a < 2

ba + < b.

Thus, if f(x) = x2, then the statement of the MVT is satisfied for c = 2

ba + .

By looking back and verifying the result of the MVT in this special case, we develop a more concrete understanding of the general statement. While the checking of the result in the special case is not, strictly speaking, necessary to our logical acceptance of the result, it does give us a strong feel for the statement. To prove the theorem, we require a special case of the MVT called the Rolle’s Theorem (RT) which states: Suppose f(x) is a continuous function on the closed interval [a,b] and that f’(x) exists for all x in the open interval (a,b). Suppose f(a) = f(b), then there exists a number c ∈ (a,b) such that f’(c) = 0. Again we can represent the problem diagrammatically as shown below. y Fig. 13 O a c b x


How to solve it? 21

We see that if f(a) = f(b), then there exists c ∈ (a,b) such that the gradient of the tangent to the curve at the point where x = c is zero. How do we make use of RT to prove MVT? The key is defining a suitable function so that we can use RT. What function is suitable? After some careful ‘investigation’, we are led to consider the function

g(x) = f(x) – ( ) ( ) xab

afbf ⎥⎦⎤

⎢⎣⎡

−− .

Observe that the beauty of this new function g(x) is that it ‘obediently’ satisfies the equation g(a) = g(b). You can easily verify that for yourself. Clearly g(x) is a continuous function on the closed interval [a,b] and that g’(x) exists for all x in the open interval (a,b) since f(x) satisfies all the above by assumption. With all the conditions met, RT is now poised to wield its power. By RT, there exists c ∈ (a,b) such that g’(c) = 0.

Now g’(x) = f’(x) – ( ) ( ) ab

afbf−− .

Therefore g’(c) = 0 ⇒ f’(c) – ( ) ( ) ab

afbf−− = 0

⇒ f’(c) = ( ) ( )ab

afbf−− .

The question you might be dying to ask is, how do you know that you should define g(x)

to be g(x) = f(x) – ( ) ( ) xab

afbf ⎥⎦⎤

⎢⎣⎡

−− ?

The trick is working backwards. Try it for yourself! III MONITORING YOUR OWN THINKING In the preceding section we have discussed the structure of mathematical proofs and the heuristics for solving problems. In the process we have become aware of the general form of the arguments employed and the approaches that can be adopted to solve problems. Nonetheless, all of us, no matter how proficient we are in solving problems, will at some time become ‘stuck’ on a problem. We encounter what appears to be an insurmountable roadblock and we have no idea how to proceed. What do we do? Perhaps the most useful general advice is: REMAIN ACTIVE. Rather than waiting passively for inspiration to strike, engage actively in steps that can help bring about a solution. Specifically, how does this advice translate into action? Here are some suggestions: Read the problem or theorem several times. Rephrase the problem in alternative ways

and use different but equivalent definitions of the terms involved. Play with the problem by drawing diagrams or forming representations. Check out

examples, extreme cases and special cases. Generalise in various ways. Try working backwards.


How to solve it? 22

As a result of this active investigation, try to develop a problem-solving strategy. Execute the chosen strategy and verify the result.

The moral is simple: Keep working and keep trying new approaches. You must, however, bear in mind that not ‘any strategy’ will work. Discretion and judicious judgement must be exercised in order to come up with a strategy that works. While you are working on a problem, it is wise to constantly monitor your own activities and thoughts, and to control your own actions. If your approach leads you to something awfully complicated and messy, stop and check. You might have made a mistake somewhere or it could be your method just simply isn’t going to work. Study the problem again carefully and try a different approach. For instance, suppose you find an approach that seems promising, but that after trying this approach for a while, you are unable to make further progress. Then you should be willing to explore other methods. Suppose you have thought of two possible strategies to solve the problem, evaluate the relative advantages of the two strategies to see which one will require significantly less work and hence is preferable. As these remarks indicate, a diligent and wise problem solver should be aware of more than just the problem-solving techniques. He must constantly monitor his own actions and thoughts. He must be flexible enough to change approaches and be judicious in choosing appropriate strategies. Studies have shown that students who fail to solve problems generally are those who persistently pursue blind alleys, lack focus or simply wander aimlessly, beating-about-the bush. The lack of success boils down to their inability to use efficiently their mathematical resources and their knowledge of heuristics. Some may not even possess any knowledge of the subject matter at all. On the other hand, those who succeed in solving problems generally have good control over their own thought processes. They constantly take stock of their situation and make adjustments based on their findings. If the current method they are pursuing is leading them to nowhere, then that approach is abandoned and another is sought. Successful problem solvers often remind themselves of their goal and relate their progress on the problem to that goal. Throughout their analysis, they ask themselves questions. This self-questioning helps them to monitor their own position and decide on appropriate steps. The questions problem solvers should ask themselves could be: • Can I restate the problem in another way in order to understand it better? • Can I represent the problem pictorially? • Did I misinterpret the question? • Can I introduce auxiliary elements? • I seem to make no headway. Can I try another approach? • Have I used all the information given? Did I miss any? • Can I consider specific cases? • Can I work backwards?


How to solve it? 23

• What am I supposed to do? • What am I doing? • Have I seen something like this before? • Can I relate this to other parts of mathematics I know? • Did I make any careless mistakes? • Did I make a wrong assumption? In the book Thinking Mathematically, J. Mason encourages the readers to develop their own internal monitors and to become their own critics. In this way, students of mathematics are directed to think not only about mathematical facts, but also about their own thought processes. They are encouraged to observe and evaluate their current state of understanding. Such activity engenders problem-solving ability as well as develop mathematical maturity. IV ATTITUDES AND BELIEFS The purpose of this section is to discuss the influence of attitudes and beliefs on mathematical performance. One interesting thing about the discipline of mathematics is what people think of it? There are many people who relish mathematics and thoroughly enjoy doing it, myself included. Then there are those who find mathematics totally distasteful and unappealing. They find no joy and motivation in doing mathematics and would adopt a ‘couldn’t be bothered’ attitude. Such negative attitudes toward mathematics permeate our culture and in turn influence the way students approach mathematics. The effect of these negative societal attitudes is to discourage students from pursuing mathematics for its intrinsic beauty and truth. It conveys the message that doing mathematics is a chore and a perfunctory activity. In addition to societal attitudes toward mathematics, each individual carries his or her own beliefs about mathematics. Personal beliefs and attitudes are important factors that determine mathematical performance. For example, many students regard themselves as unable to do mathematics. This opinion could be due to experiences of failure or could be inherited from people who hold the same belief. Throughout their school years, students learn a number of algorithms or procedures for doing mathematics. They are taught for instance how to add, subtract, multiply, divide, find LCM, HCF and solve simple equations. As a result, students develop the belief that mathematics is all about techniques and procedures. They believe that there is always a specific procedure that will work for every situation. Haven’t we heard of students asking: “Is it always like that?”, “Do I always do this when I see this?”, “Is there a fixed method to solve this problem?”, “Can you tell me one single formula that works all the time?”, “What is the quickest way to solve this problem without doing too much thinking?” The myriads of such similar questions just go on and on….


How to solve it? 24

What comes out of this belief? Firstly, students will reject aspects of mathematics such as proofs that fail to be procedural. If there isn’t a procedure or ‘fixed’ method, then why be bothered with it? Such students can become very frustrated when faced with problems that go beyond being procedural and that require deeper thinking. Often they simply give up completely and take the quickest way out by pleading for the solutions rather than working through the problems diligently. A related belief or perhaps a corollary, is the notion that any mathematical problem that cannot be solved within five or ten minutes is not worth trying. Any student holding this opinion will quit on a problem readily. He or she will neither experiment or “play with” the problem nor carry the problem and mull over it. To the student, giving up is the best policy. Another commonly held belief is that you must be a genius to be good at mathematics. This belief is so widespread that it has become a cultural phenomenon. The result of this belief is this: Since I am not a genius, I am not going to waste my time on something I know I will never be able to succeed at. Any student who embraces this opinion is in effect trying to be absolved of any responsibility with regard to the learning or the doing of mathematics. It is not surprising that the student has little incentive and motivation to perform well in mathematics. So far we have talked about the negative attitudes and beliefs. What about the positive ones? Most mathematicians or students of mathematics know that any interesting problem, especially a difficult one, will require time and effort before succumbing to a solution. Along the way, false starts will occur. There may be long periods in which no progress is made. Mathematicians are not surprised when they are stuck on a problem, neither are they ashamed of getting stuck. They realise that being stuck is not a symptom of stupidity. Despite setbacks and failures, they press on and continue to work actively on the problem until it yields to their persistent efforts. As an example, the famous “Fermat’s Last Theorem” which was unsolved for 358 years finally succumbed to a proof in 1994 by a Princeton professor of mathematics Andrew Wiles who spent 8 long years working tenaciously on the proof! He harnessed very sophisticated mathematics and the works of other mathematicians in order to crack the conundrum and gave it the final blow. The proof is over a hundred pages long, much longer than my presentation here. Another positive attitude amongst all mathematicians is that doing mathematics brings great joy and satisfaction. The uncertainties and complexities of the problem are the things that excite mathematicians all the more to produce a solution rather than discourage them from doing it. Mathematicians are constantly mesmerised and awed by the beauty of mathematical arguments. They are able to discover the intrinsic beauty in mathematics, enjoy and marvel at it just like what an artist would with his own masterpiece. The famous mathematician Sir Michael Atiyah once remarked:

“If you ask me to choose between truth and beauty of mathematics. I would choose beauty.”


How to solve it? 25

Another mathematician has this to say:

“Life is good for two things – discovering mathematics and doing mathematics.” I like to compare doing mathematics with entering a dark tunnel. At first we may be groping around in the dark, unable to find our way. We may knock ourselves against the wall several times and we may get discouraged and even frustrated. The way ahead is strewn with doubts and uncertainties. But when we press on and move forward, undaunted by the circumstances, we begin to find our direction. It seems that our path is becoming clearer as we persevere on. We begin to see light gradually streaming in and we get more and more excited and confident that we are near the other end of the tunnel. Encouraged by the prospect of reaching the end, we press on and suddenly, light shines around us and we are out through the other side of the tunnel. Great joy floods our soul! V PROBLEMS FOR THOUGHT 1. Among all pairs of positive numbers whose product is 10, which pair has the smallest

sum? 2. Find all points at which the tangent line to y = x3 – x2 – 2x + 1 is horizontal. 3. Find the common points of the curves x2 + 2x + y2 = 1 and y = x2 + x + 1. 4. Obtain the area under the curve y = sin x for 0 ≤ x ≤ π.

5. If dn = ( ) ⎥

⎥

⎦

⎤

⎢⎢

⎣

⎡

+n

xdxd

21

1 , show that ( ) ⎥

⎥

⎦

⎤

⎢⎢

⎣

⎡

+

+n

x

xdxd

2

2

1

23 = 3dn-1 – dn.

6. Conjecture and prove a formula for ( )∑= +

N

n nn

1 ! 1. Evaluate ( )∑

∞

= +1 ! 1n nn .

7. Find (a) ( ) ⎥⎥⎦

⎤

⎢⎢⎣

⎡

+−+

∞→ 1212

nnnnlim

n (b)

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛∑= +∞→

n

r r

rn a

alim1 1ln if an → a1 as n → ∞.

8. Prove that if f and g are functions such that f(x) ≥ g(x) for all a < x < b, then

. ( ) ( )∫∫ ≥b

a

b

adxxgdxxf

9. Prove that if 0 ≤ an ≤ bn for n = 1, 2, 3, …, and that ∑ converges, then ∑

also converges.

∞

=1nnb

∞

=1nnb


How to solve it? 26

10. Prove that the diagonals of a parallelogram bisect each other.

11. Prove that ∑∞

= +12 31

n n converges.

12. Prove that the area of the region enclosed by the ellipse 12

2

2

2=+

by

ax is πab.

13. Amongst all rectangular solids of surface area 20 units2, which one has the largest

volume? 14. Notice that 1.2.3.4 = 52 – 1, 2.3.4.5 = 112 – 1. What is the general pattern? Can

you prove it? 15. In any given triangle, prove that (i) the medians, (ii) the altitudes, (iii) the angle

bisectors of the triangle are concurrent. 16. Find a cubic polynomial whose roots are respectively twice as large as the roots of the

cubic polynomial P(x) = x3 – 7x2 + 5x + 2.

17. Let P(x) = ∑ be an arbitrary polynomial. Investigate the relationship

between the roots of P(x) and the polynomial Q(x) = defined by

reversing the order of the coefficients of P(x).

=

n

r

rr xa

0

∑=

−

n

r

rrn xa

0

18. Let a and n be positive integers with n > 1. Prove that if an – 1 is a prime number,

then a = 2. [Hint: consider finite geometric sums] 19. Prove that | x + y | ≤ | x | + | y | for all x, y ∈ R. Deduce that | x – y | ≤ | x | + | y |. 20. Prove that 3 is irrational. If 6 is irrational, prove that 32 + is irrational. 21. Let {fn : n = 0, 1, 2, 3, …} denotes the Fibonacci Sequence where f0 = f1 = 1

and for n > 1, fn = fn-1 + fn-2. Let an = ∑ . Compute an for n ≤ 8. Make a

conjecture about an and prove it. =

n

rrf

0

22. Investigate the problem: Which positive integers can be represented as a sum of any

number of consecutive positive integers? For example, 3 = 2 + 1 can be represented as such a sum but 2 apparently cannot be.


How to solve it? 27

23. Prove that for all positive integers n, n3 – n is divisible by 3. 24. Prove by induction that for all positive integers n, n5 – n is a multiple of 5. 25. Let A and B be two arbitrary sets. Prove or disprove: (a) ( ) . (b) ( ) . ccc BABA ∩=∩ ccc BABA ∩=∪ 26. Let m and n be fixed positive integers. For all a, b ∈ R, define a * b as follows:

a * b = nmnbma

++ . Prove that a * (b * c) = (a * b) * (a * c).

27. Prove that for any given positive integer n, there exists a positive integer k such that

k ! ≤ n ≤ (k + 1) !. 28. Let f be a function defined on the set R+ of positive real numbers such that f(x) > 0

for all x ∈ R+ and ( ) ( ) ( )( ) ( )yfxf

yfxfyxf+⋅

=+ for all x, y ∈ R+.

(a) Express f(2x) and f(3x) in terms of f(x). (b) Conjecture a formula for f(rx) for all r ∈ Q+ and all x ∈ R+. (c) Prove your conjecture.

29. Prove that ∑∏==

≤n

rr

n

rr a

na

11

1 where ar > 0 for all r = 1, 2, 3, … , n.

30. Prove that for each positive integer n, there exists an integer k satisfying the

inequality 3k ≤ n ≤ 3k+1. 31. Let a and b be fixed positive real numbers. Determine the value of

)( nnn b+ . [Hint: Assume n

alim1

∞→n

nclim

1

∞→ = 1 for any fixed positive real number c]

32. Let {fn : n = 0, 1, 2, 3, …} denotes the Fibonacci Sequence defined in problem 21.

Establish the identity 1111

11 1

+−+− ⋅−

⋅≡

⋅ nnnnnn ffffff. Evaluate

∑∞

= +− ⋅1 11

1

n nn ff and ∑

∞

= +− ⋅1 11n nn

nff

f.

33. Evaluate dxxn

tan112

0∫π

+ exactly. [Hint: Use symmetry]


How to solve it? 28


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