Mathematics Through Problem Solving

by Margaret Taplin

What Is A 'Problem-Solving Approach'?

As the emphasis has shifted from teaching problem solving to teaching via problem solving (Lester, Masingila, Mau, Lambdin, dos Santon and Raymond, 1994), many writers have attempted to clarify what is meant by a problem-solving approach to teaching mathematics. The focus is on teaching mathematical topics through problem-solving contexts and enquiry-oriented environments which are characterised by the teacher 'helping students construct a deep understanding of mathematical ideas and processes by engaging them in doing mathematics: creating, conjecturing, exploring, testing, and verifying' (Lester et al., 1994, p.154). Specific characteristics of a problem-solving approach include:

interactions between students/students and teacher/students (Van Zoest et al., 1994) mathematical dialogue and consensus between students (Van Zoest et al., 1994) teachers providing just enough information to establish background/intent of the problem, and students clarifing, interpreting, and attempting to construct one or more solution processes (Cobb et al., 1991) teachers accepting right/wrong answers in a non-evaluative way (Cobb et al., 1991) teachers guiding, coaching, asking insightful questions and sharing in the process of solving problems (Lester et al., 1994) teachers knowing when it is appropriate to intervene, and when to step back and let the pupils make their own way (Lester et al., 1994) A further characteristic is that a problem-solving approach can be used to encourage students to make generalisations about rules and concepts, a process which is central to mathematics (Evan and Lappin, 1994).

Schoenfeld (in Olkin and Schoenfeld, 1994, p.43) described the way in which the use of problem solving in his teaching has changed since the 1970s: My early problem-solving courses focused on problems amenable to solutions by Polya-type heuristics: draw a diagram, examine special cases or analogies, specialize, generalize, and so on. Over the years the courses evolved to the point where they focused less on heuristics per se and more on introducing students to fundamental ideas: the importance of mathematical reasoning and proof..., for example, and of sustained mathematical investigations (where my problems served as starting points for serious explorations, rather than tasks to be completed).

Schoenfeld also suggested that a good problem should be one which can be extended to lead to mathematical explorations and generalisations. He described three characteristics of mathematical thinking: valuing the processes of mathematization and abstraction and having the predilection to apply them developing competence with the tools of the trade and using those tools in the service of the goal of understanding structure - mathematical sense-making (Schoenfeld, 1994, p.60). As Cobb et al. (1991) suggested, the purpose for engaging in problem solving is not just to solve specific problems, but to 'encourage the interiorization and reorganization of the involved schemes as a result of the activity' (p.187). Not only does this approach develop students' confidence in their own ability to think mathematically (Schifter and Fosnot, 1993), it is a vehicle for students to construct, evaluate and refine their own theories about mathematics and the theories of others (NCTM, 1989). Because it has become so predominant a requirement of teaching, it is important to consider the processes themselves in more detail. The Role of Problem Solving in Teaching Mathematics as a Process Problem solving is an important component of mathematics education because it is the single vehicle which seems to be able to achieve at school level all three of the values of mathematics listed at the outset of this article: functional, logical and aesthetic. Let us consider how problem solving is a useful medium for each of these. It has already been pointed out that mathematics is an essential discipline because of its practical role to the individual and society. Through a problem-solving approach, this aspect of mathematics can be developed. Presenting a problem and developing the skills needed to solve that problem is more motivational than teaching the skills without a context. Such motivation gives problem solving special value as a vehicle for learning new concepts and skills or the reinforcement of skills already acquired (Stanic and Kilpatrick, 1989, NCTM, 1989). Approaching mathematics through problem solving can create a context which simulates real life and therefore justifies the mathematics rather than treating it as an end in itself. The National Council of Teachers of Mathematics (NCTM, 1980) recommended that problem solving be the focus of mathematics teaching because, they say, it encompasses skills and functions which are an important part of everyday life. Furthermore it can help people to adapt to changes and unexpected problems in their careers and other aspects of their lives. More recently the Council endorsed this recommendation (NCTM, 1989) with the statement that problem solving should underly all aspects of mathematics teaching in order to give students experience of the power of mathematics in the world around them. They see problem solving as a vehicle for students to construct, evaluate and refine their own theories about mathematics and the theories of others. According to Resnick (1987) a problem-solving approach contributes to the practical use of mathematics by helping people to develop the facility to be adaptable when, for instance, technology breaks down. It can thus also help people to transfer into new work environments at this time when most are likely to be faced with several career changes during a working lifetime (NCTM, 1989). Resnick expressed the belief that 'school should focus its efforts on preparing people to be good adaptive learners, so that they can perform effectively when situations are unpredictable and task demands change' (p.18). Cockcroft (1982) also advocated problem solving as a means of developing mathematical thinking as a tool for daily living, saying that problem-solving ability lies 'at the heart of mathematics' (p.73) because it is the means by which mathematics can be applied to a variety of unfamiliar situations. Problem solving is, however, more than a vehicle for teaching and reinforcing mathematical knowledge and helping to meet everyday challenges. It is also a skill which can enhance logical reasoning. Individuals can no longer function optimally in society by just knowing the rules to follow to obtain a correct answer. They also need to be able to decide through a process of logical deduction what algorithm, if any, a situation requires, and sometimes need to be able to develop their own rules in a situation where an algorithm cannot be directly applied. For these reasons problem solving can be developed as a valuable skill in itself, a way of thinking (NCTM, 1989), rather than just as the means to an end of finding the correct answer. Many writers have emphasised the importance of problem solving as a means of developing the logical thinking aspect of mathematics. 'If education fails to contribute to the development of the intelligence, it is obviously incomplete. Yet intelligence is essentially the ability to solve problems: everyday problems, personal problems ... '(Polya, 1980, p.1). Modern definitions of intelligence (Gardner, 1985) talk about practical intelligence which enables 'the individual to resolve genuine problems or difficulties that he or she encounters' (p.60) and also encourages the individual to find or create problems 'thereby laying the groundwork for the acquisition of new knowledge' (p.85). As was pointed out earlier, standard mathematics, with the emphasis on the acquisition of knowledge, does not necessarily cater for these needs. Resnick (1987) described the discrepancies which exist between the algorithmic approaches taught in schools and the 'invented' strategies which most people use in the workforce in order to solve practical problems which do not always fit neatly into a taught algorithm. As she says, most people have developed 'rules of thumb' for calculating, for example, quantities, discounts or the amount of change they should give, and these rarely involve standard algorithms. Training in problem-solving techniques equips people more readily with the ability to adapt to such situations. A further reason why a problem-solving approach is valuable is as an aesthetic form. Problem solving allows the student to experience a range of emotions associated with various stages in the solution process. Mathematicians who successfully solve problems say that the experience of having done so contributes to an appreciation for the 'power and beauty of mathematics' (NCTM, 1989, p.77), the "joy of banging your head against a mathematical wall, and then discovering that there might be ways of either going around or over that wall" (Olkin and Schoenfeld, 1994, p.43). They also speak of the willingness or even desire to engage with a task for a length of time which causes the task to cease being a 'puzzle' and allows it to become a problem. However, although it is this engagement which initially motivates the solver to pursue a problem, it is still necessary for certain techniques to be available for the involvement to continue successfully. Hence more needs to be understood about what these techniques are and how they can best be made available. In the past decade it has been suggested that problem-solving techniques can be made available most effectively through making problem solving the focus of the mathematics curriculum. Although mathematical problems have traditionally been a part of the mathematics curriculum, it has been only comparatively recently that problem solving has come to be regarded as an important medium for teaching and learning mathematics (Stanic and Kilpatrick, 1989). In the past problem solving had a place in the mathematics classroom, but it was usually used in a token way as a starting point to obtain a single correct answer, usually by following a single 'correct' procedure. More recently, however, professional organisations such as the National Council of Teachers of Mathematics (NCTM, 1980 and 1989) have recommended that the mathematics curriculum should be organized around problem solving, focusing on: (i) developing skills and the ability to apply these skills to unfamiliar situations (ii) gathering, organising, interpreting and communicating information (iii) formulating key questions, analyzing and conceptualizing problems, defining problems and goals, discovering patterns and similarities, seeking out appropriate data, experimenting, transferring skills and strategies to new situations (iv) developing curiosity, confidence and open-mindedness (NCTM, 1980, pp.2-3). One of the aims of teaching through problem solving is to encourage students to refine and build onto their own processes over a period of time as their experiences allow them to discard some ideas and become aware of further possibilities (Carpenter, 1989). As well as developing knowledge, the students are also developing an understanding of when it is appropriate to use particular strategies. Through using this approach the emphasis is on making the students more responsible for their own learning rather than letting them feel that the algorithms they use are the inventions of some external and unknown 'expert'. There is considerable importance placed on exploratory activities, observation and discovery, and trial and error. Students need to develop their own theories, test them, test the theories of others, discard them if they are not consistent, and try something else (NCTM, 1989). Students can become even more involved in problem solving by formulating and solving their own problems, or by rewriting problems in their own words in order to facilitate understanding. It is of particular importance to note that they are encouraged to discuss the processes which they are undertaking, in order to improve understanding, gain new insights into the problem and communicate their ideas (Thompson, 1985, Stacey and Groves, 1985). Conclusion It has been suggested in this chapter that there are many reasons why a problem-solving approach can contribute significantly to the outcomes of a mathematics education. Not only is it a vehicle for developing logical thinking, it can provide students with a context for learning mathematical knowledge, it can enhance transfer of skills to unfamiliar situations and it is an aesthetic form in itself. A problem-solving approach can provide a vehicle for students to construct their own ideas about mathematics and to take responsibility for their own learning. There is little doubt that the mathematics program can be enhanced by the establishment of an environment in which students are exposed to teaching via problem solving, as opposed to more traditional models of teaching about problem solving. The challenge for teachers, at all levels, is to develop the process of mathematical thinking alongside the knowledge and to seek opportunities to present even routine mathematics tasks in problem-solving contexts.

Hypatia of Alexandria

  (born c. AD 350 - 370; died 415)

About Hypatia

Hypatia was one of the earliest known female mathematicians in the world, known not only for her excellence in mathematics and astronomy, but also for the brutal death she suffered from the hands of the early Christians due to her pagan beliefs and political involvement. She was the daughter of a renowned Greek mathematician Theon Alexandricus. Being born to an educator, Hypathia herself had all the qualities that an ideal teacher should have. She was a great orator and people from all across the globe came to gain from her teachings.

Contributions and Achievements

✦ She edited the book On the Conics of Apollonius where she contributed on the ideas introduced by Apollonius on conic sections. It was through this that the development of concepts such as hyperbolas, parabolas, and ellipses took place later.

✦ Some sources also attribute her to be the inventor of hydrometer; however, it was invented before Hypatia and was already known at the time. Apart from editing the existing works of renowned mathematicians of the time, most of her works are believed to have been in collaboration with her father's work.

Meaning And Examples of Converse Statements

Different types of statements are used in mathematics to convey certain theorems, corollaries, or prove some ideas. One such statement is the converse statement.

Converse vs. Inverse

In inverse statements, the opposite of the original hypothesis and conclusion is written, whereas in a converse statement, only the hypothesis and the conclusion is exchanged. The meaning of the statement does not change in an inverse statement.

A very important type of statement, the converse statement is mostly used in geometrical theorems. Understanding or writing a converse theorem is not very difficult.

In this Buzzle write-up, we discuss the meaning of a converse statement, how it is written, and some examples.

Converse statements are a type of conditional statements. So let us understand in brief what conditional statements are. They are basically if and then statements.

Here's an example: If it rains, then I won't go to school. The first part, i.e., if it rains is called the hypothesis, whereas the latter part then I won't go to school is called conclusion. When one condition is fulfilled, then the other condition can happen. Conditional statements are used in mathematical theorems. Sometimes, there can be two or more conditions as well.

                   What is a Converse Statement?

Converse statement is a statement in which the hypothesis and conclusion is interchanged. For example, statement: If the angle is less than 90º, then it is an acute angle.
Converse: If the angle is acute, it is less than 90º.
Here you can see that the hypothesis of the statement becomes the conclusion in the converse, and the conclusion becomes hypothesis.

The most important factor to be considered is that the converse statement may not be true in all cases.
For e.g.: If you are a girl, then you are a human.
Converse of this statement will be: If you are a human, then you are a girl.
It is obviously not true. 

In a converse statement, the hypothesis is called 'p', and the conclusion is called 'q'. The symbol for a converse statement is as follows:

                                     Statement: If p, then q. (p ➔ q)

                                           Converse: If q, then p. (q ➔ p)

Examples of Converse Statements

When converse is true

1. Statement: If a number ends in 0, then it is a multiple of 10.
    Converse: If a number is a multiple of 10, then it ends in 0.

2. If the number is divisible by 2, then it is an even number.

    Converse: If it is an even number, then it is divisible by 2.

When converse is not true

1. Statement: If a quadrilateral is square, then it is a rectangle.

    Converse: If a quadrilateral is rectangle, then it is a square.

2. Statement: If a triangle is equilateral, then it is isosceles.

    Converse: If a triangle is isosceles, then it is equilateral.

Can you CRACK THE LOGIC...............????????????

                    If  1 1 1 1 = R

                         2 2 2 2 = T

                         3 3 3 3 = E

                         4 4 4 4 = N

                   

                   Then

                         5 5 5 5 = ?

Challenge to all Master Brains!!!

Answer:

                     1+1+1+1 = 4 four R.

                     2+2+2+2 = 8 eight T.

                     3+3+3+3 = 12 twelve E.

                     4+4+4+4 = 16 sixteen N.

                     5+5+5+5 = 20 twenty Y.

Using Compatible Numbers for Quick And Easy Arithmetic Calculations

Using Compatible Numbers for Quick And Easy Arithmetic Calculations

A compatible number is close to an actual number in a calculation and is used to estimate the final result. This Buzzle write-up will explain the concept of using compatible numbers for quick and easy arithmetic calculations.


Please Remember

Compatible numbers may help to find an estimated answer if you are trying to find out the percentage of an entity. You can convert the percent to a fraction, and apply the rounding method to figure out an estimate.

To know what are compatible numbers, first we need to understand the meaning of 'compatible'. The word indicates a friendly relation. This means that compatible numbers are those that are well-matched with one another and are useful for the estimation of a sum, difference, quotient, or product. They are widely used in mental computations for conveniently deducing the actual answer. You may be aware that numbers are classified into whole numbers, integers, decimals, fractions, etc.

Working with decimals can be a little time-consuming, hence, using compatible numbers can help you select a margin within which you can identify your answer. Some examples are given below so that you can understand this concept better.

                                 
                                                         The Theory

They are used particularly for quicker mental calculation.
They are close to the original numbers, but are nice, round numbers, and can be conveniently used for calculation.
For instance, in a calculation involving 48 and 18, we can choose compatible numbers 50 and 20 for convenience.
This methodology can be used to estimate different results, even in problems involving percentages and fractions.
To simplify the concept, simply remember that this estimation is a form of rounding.
All you need to do is round off these numbers to the nearest whole numbers and find the solution quickly.

 Compatible Numbers Examples

For Addition

★ When you need to find the sum of 100 + 50, you can answer '150' immediately. However, finding the sum of 347 + 419 takes a while doesn't it? Here is where you use compatible numbers. When such a calculation needs to be done, use the rounding theory to round off both the numbers to the nearest compatible number.

In this case, it would be beneficial to round off 347 to either 345 or 350, and 419 to 420. If you choose 345 and 420, you are actually decreasing one number by two and increasing the other by one. Therefore, you will need to increase one in the final result. Conversely, if you choose 350 and 420, you are increasing both factors by 3 and 1 respectively. Therefore, you will need to decrease the final result by 4.

Once you obtain the factors 345 and 420 or 350 and 420, round them again. Take two groups - in the first case, group 1 will be the addition of 400 and 300, which is 700, and group 2 will be the addition of 45 and 20, which is 65. Add them both, and you will obtain 765. But since we have used compatible numbers, we have to add 1 to the last answer. Thus, the actual answer will be 766, while the estimated answer is 765.

Similarly, in the second case, add 300 and 400, you will obtain 700. Add 50 and 20, which equals 70. Add 700 and 70 and you get 770. But here, you have to reduce 4 from the final answer. Thus, the answer if 766.

★ Consider a bigger number, say 1497 + 423 + 122 +768. In such a case, try to remember compatible pairs. What are compatible pairs? They are the ones that come together for easier, obvious simplification. For instance, you know that (7,3) will make 10 and give you a zero at the unit's place. Similarly, (3,2) will give you 5.

In the above case too, you can use this theory. As you can see, two numbers have a 7 and 3 in their unit place, while two others have 2 and 8. Group them together. Take 1497 + 423 in group 1 and 122 + 768 in group 2.

In group 1, you can round off 1497 to either 1400 or 1500 and 423 to 400 or 420. Ideally, you should choose the latter in both factors, for the rule states that you need to choose the closest number. Now, if you have chosen 1400 and 400, first add them together. You get 1800. Now, take the 97 and 23 that have been cast aside. You know that 7 and 3 make a compatible pair, so adding 97 and 23 will give you 120. Add this to 1800, thus you will obtain 1920.

If you choose 1500 and 420, add them, and you will obtain 1920 directly. Why do you obtain the actual answer in the estimated answer here? It is because you have chosen to add 3 and subtract 3 from both factors to get to the nearest compatible number. The +3 and -3 get canceled. Keep this aside, proceed to group 2.

In group 2, you have 122 + 768. Like case 1, choose compatible numbers 120 and 770, or 100 and 700. If you choose the former, adding 120 and 770 will give you 890. And, since you have subtracted and added 2 in both factors, you will obtain the correct answer in the estimated answer. If you choose the latter, adding 100 and 700 will give you 800. Add 22 and 68 now. You can round this as well, as 20 and 70. Applying the same theory, you get 90. Add this to 800, you get 890, which is the actual answer.

Coming to the last part of the problem, you will have to add 1920 from group 1 to 890 of group 2. Again, round off 1920 to 1900 and 890 to 900. The estimated answer will be 2800. The actual answer is 2810 since we have subtracted 20 and added 10.

For Subtraction

★ Consider a problem like 1017 - 421. The first step here involves rounding the numbers. Round 1017 to 1020 and 421 to 420. Now subtract 420 from 1020 and you will obtain 600. This is the estimated answer.

There are several cases in statistics wherein you need only an approximation, and not the exact answer. In this case, the approximation is 600. And remember, we have increased the first factor by 3 and decreased the second factor by 1. So, in the final answer, we have to reduce by 4, thus, the final answer is 596.

★ Consider another example, say 2007 - 512. Round off 2007 to 2000 and 512 to 500. The estimated difference will be 2000 - 500, i.e., 1500. Therefore, the actual answer is somewhere near 1500. In this case, you have subtracted 7 from 2007 and 12 from 512. Subtract 7 from 12, and add the difference to the estimated result. Thus, you will obtain the final answer, i.e., 1495.

For Multiplication

★ Take an example, like 500 X 40. It is rather simple, isn't it? You can quickly multiply 5 and 4 and add 3 zeros to the result. But, what if you had to multiply 72 X 228? For this, round off 72 to 70 and 228 to 230. Now the problem is simpler. Multiply 230 and 70. For this, you can multiply 23 and 7 and add two zeros to the result. You will get the answer, 16100. This is an estimated answer. Here, you will only be able to get somewhere close to the actual product. Finding out the exact answer mentally is slightly tough.

★ Consider another example, 246 X 119. It would be convenient to round off 246 to 250 and 119 to 110. Note that we are increasing 4 in the first factor and decreasing 9 in the second. The estimated product will be 250 110, which equals 27500. If you perform the actual calculation manually or on the calculator, you will get 29274. You might say this as a rather huge difference, but that is how math works. In lengthy calculations, this is tiny difference.

For Division

★ It can get slightly complicated to estimate quotients. Consider an example, like 2400 / 12. It is a simple calculation, you can divide 24 by 12, obtain 2, and add two zeros at the end of 2 for the final answer. But if the problem is slightly complex, you can use compatible numbers to estimate the quotient. For instance, consider 1546 / 117. Round the numbers to 1500 and 100. Note that both, the dividend and divisor are decreased. You will get 15, which means that the answer is somewhere close to 15, it will be lesser than 15 in this case, since we have decreased the numbers.

★ Consider another problem, say 3995 and 9. Round off 3995 to 4000 and 9 to 10. Your estimated quotient is 400. The actual quotient will be more than 400, since we have increased the numbers. Learn to recognize compatible pairs, like (32, 8), (25, 50), (24, 12), (55,11), and more.

For Percentage Problems

★ Remember the commutative property of multiplication while using compatible numbers for solving percent problems. The commutative property states that if x% of y should be the same as y% of x. So here, when you have a problem like 12.5% of 720, you must convert the percent to factors, that act as compatible numbers. In this case, 12.5% means 12.5/100, i.e., 125/1000, which is 1/8. This means that 12.5% of 720 is equal to 1/8th of 720.

★ Similarly, if you have something, like 28% of 50, use the commutative property and the answer will be 50% of 28. 50% indicates ½, the answer will thus be 28/2, i.e., 14.

Important Points to Remember

For quick mental calculations, you should be able to quickly formulate all the required theories so that you get the necessary estimated values.
Calculating the difference between the actual and estimated answer in case of subtraction depends on how much you have added or subtracted during the rounding off.
In multiplication, for rounding off, if one factor is increased, the other should be decreased and vice versa. Failure to do so will give you an incorrect estimation.
Similarly, in division, for rounding off, if one factor is increased, the other should be increased as well, and vice versa. Failure to do so will give you an incorrect estimation.
For solving percentage problems mentally, memorize the common factorizations, like 50% is ½, 25% is ¼, 30% is 3/10, 62.5% is 5/8, etc.


In layman's terms, compatible numbers are nice, friendly numbers that can easily be remembered. That is to say, they are more convenient to use for mental calculations, like 100, 250, 45, 1200, etc. Their basic purpose is to speed up the calculation process without using a calculator. Using them in mental arithmetic will simplify a lot of tedious calculation procedures and will improve your computing speed and thinking power as well.



Read more at Buzzle: http://www.buzzle.com/articles/using-compatible-numbers-for-quick-and-easy-arithmetic-calculations.html