On Mathematics Education: Algebra vs. Calculus

This time I am going to take a break from heavy use of $ \mathrm\LaTeX$ like I used to do in my earlier posts. I will focus on the education of mathematics as is being provided to students in India. My sources of information in this regard are:

• memories from my school/college days (which are still relevant in current time)
• interaction with current school/college students
• mathematics books (both Indian and foreign, old and new) available on the market

The subtopic of the post should not be taken too literally, rather it reflects two approaches in the teaching of mathematics: 1) algebraic approach which I don't like and find unsuitable for any serious teaching of mathematics, and 2) the calculus approach which is the way mathematics teaching should be, but is currently not being practiced anywhere as far as the books show.

The words "Algebra" and "Calculus" might not sound quite appropriate in context of mathematics education so I need to provide some justification of these terms and I proceed to do the same in what follows.

The Algebra Approach

The subject of algebra deals with common mathematical operations like addition, subtraction, multiplication and division. To spice things up algebra also adds the operation of exponentiation which is just repeated multiplication but gradually becomes a new operation altogether when we introduce the operation of root extraction. These operations follow certain rules which are:

• easy to understand and appreciate
• easy to keep in mind (memorize)
• easy to apply during problem solving
• quite mechanical after certain time

The above properties are very nice and important and people all over the world try to make rules (in other aspects of our lives) which possess most of these properties (at least there is a drive to make rules with these properties). The reason this is so is that it makes life simpler for people at both ends: rule makers and rule followers. The same principle applies to mathematics education. It makes life quite simple for the teacher as well as the student if the mathematical ideas are presented in the form of rules having above properties. And this is what I would call the "algebraical approach".

I also understand that such an approach towards mathematics education is definitely useful and quite relevant till 10th grade. Most of the topics covered upto 10th grade mathematics consists of algebra and its applications (a notable exception being geometry which students find hard to digest in the absence of operations and rules for them). The focus in this approach is towards enabling the student to become a master of manipulating some mathematical symbols (which may be digits after all) based on straightforward rules. In other words the student is trained to become expert in number crunching without really understanding the essence of these numbers.

This is one of the reasons why average students are so poor at solving word problems in their mathematics exercises whereas they are able to solve other problems without much trouble. They are not equipped to relate the concepts the numbers and other quantities represent. Hence they are not able to extract the numbers from a given word problem and then do operations to get the answer. The same argument does not apply altogether to above average students. However their condition is only slightly better. They are able to solve word problems because of one of the following reasons:

• They are very fast at applying the mechanical rules and hence find more time to try various sample word problems and try to map the exercise word problems with sample word problems and understand the technique. These kinds of students might be in trouble if they get a totally different type of word problem.
• They really know whats going behind the scenes. They are able to map numbers to concepts and back and thereby are able to translate a word problem into a mathematical problem devoid of its real world context. These students turn out to be very brilliant in mathematics inspite of the mechanical approach but unless they are very lucky they have hard time in 12th grade.

Now I had mentioned two areas where the usual algebraical mechanical approach is deficient, namely geometry and word problems. However there is another simpler category of problems which are normally difficult for the those trained via the usual approach: the inequalities. In order to demonstrate the gap between the algebraic manipulations and handling inequalities I will offer an example. Lets ask a student the following two questions where he has to find the value of $ x$:
1) when $ x + 5 = 3$
2) when $ |x - 5| < 3$

Almost every student from 6th grade to 10th grade will be able to solve 1) instantly and provide the solution $ x = -2$. However very few (like 2%) of them will be able to solve the 2nd problem instantly. Many average or below average will actually not be able to solve them. And those who are above average will proceed in this fashion for the 2nd problem:

If $ x - 5 \geq 0$ i.e. $ x \geq 5$ then $ |x - 5| = x - 5$ and so $ |x - 5| < 3$ implies that $ x - 5 < 3$ or $ x < 8$. Similarly they get the other part of the solution as $ x > 2$ and finally provide the solution as $ 2 < x < 8$.

The way to get an instant answer for the 2nd problem is to reason in the following manner. Since $ |x - 5|$ represents the distance/gap between $ x$ and $ 5$ and this gap should not be more than $ 3$, we need to search two numbers which are at the gap of $ 3$ from $ 5$ and clearly these are $ 5 + 3$ and $ 5 - 3$ one on each side of $ 5$ on the number line and hence $ x$ should be in between these two boundary values to keep the gap from $ 5$ less than $ 3$. The instant answer is available as $ 2 < x < 8$.

The above argument which is slightly geometrical in nature represents a deep understanding of the way numbers work and underlies our many unjustified assumptions about the very concept of a number and its relation to the real world. One of these assumptions is that numbers can represent magnitude of various real world quantities and like the quantities which they represent, the numbers themselves can be compared. This very understanding of comparison of numbers brings me to the "calculus approach" towards mathematics education.

The Calculus Approach

The subject of calculus deals with brand new concepts alien to the world of algebra, but however uses geometry for purposes of illustration. The foundations of calculus lie in the theory of real numbers and it allows us to discuss the concepts of continuity, limits, derivative and integral. The basis of real numbers itself is non-algebraic in nature and it depends ultimately upon order relations (inequalities like $ <$ and $ >$). Rational numbers too have the concept of order relations but unfortunately there is something missing which is filled in by the real numbers to a make a complete ordered system.

It is very difficult to formulate any mechanical rules for calculus although there has been a series of books which are trying to achieve the same and thereby blinding the students forever. I will mention an example of a rule (actually thumb rule) which instead of simplifying things actually tends to complicate the matters and causes even more confusion. In the introductory courses of calculus the student is expected to calculate certain limits by observing the following rule:

To calculate $ \lim_{x \to a}f(x)/g(x)$ when $ f(a) = 0 = g(a)$, first simplify the expression in such a form that putting $ x = a$ does not lead to anything like $ 0/0$ and then put $ x = a$.

This is one of the worst ways to teach limits. It provides a totally different impression of calculus to the students. The student is struggling at this point to understand why he is justified to put $ x = a$ at the end of simplification and not before the simplification. To this question no satisfactory answer is provided. In fact this was the case with calculus as a subject in 18th and 19th century. There were simply no answers to such questions. Mathematicians themselves knew that there were pitfalls like these in calculus and they were smart enough to avoid them while doing their researches. But in general no one was really satisfied with the state of affairs as they were at that time. There was a lot of criticism for this subject as a whole but at the same time no one could challenge its usefulness in the processes of physics as amply demonstrated by Newton.

Things were in quandary until the 20th century when many mathematicians found the answers to such questions and put the subject of calculus on solid foundations. This was the turning point in history of mathematics when a proper theory of real numbers was formulated and the concepts of calculus were presented without any ambiguities. This was an invaluable achievement, but was not something very hard to fathom. It was just that people in past centuries had not paid enough attention to detail and tried to ignore such issues. In fact the theory of real numbers is actually carried out by Eudoxus in his theory of proportions but no one paid any attention to it.

Sorry to digress here in a bit of history, but the point I wanted to emphasize was that concepts of calculus were once without any foundation, but later on put on solid foundations and that these were not that difficult. But it required a departure from the traditional algebraical approach based on operations and rules followed by them. Now you had to really understand the theoretical details as well as the practical aspects if you had to work your way through calculus.

Calculus Approach: Example 1

So the term "calculus approach" to mathematics education means an approach which focuses on the understanding of the concepts behind various symbols and processes used in mathematics. Its not so much about learning the rules of manipulation, but rather learning why the rules work. As an example take the case of an important result from 6th/7th grade which is as follows:
The product of HCF (also called GCD) and LCM of two numbers is the same as the product of the numbers themselves.

Now this is a very well known fact and students in 6th grade have to use this fact while solving many problems. But unfortunately the students are not given any clue as to why this works. Most books contain the infamous line "the proof of this result is beyond the scope of this book/syllabus". This is like telling that its a dead end and there is no need to proceed further on this topic. What we require at 6th grade is not a formal proof containing some symbols, but rather we need to demonstrate the same proof with examples and convince the student that the particular numbers were used for illustration and the same could be demonstrated by any other set of two numbers.

The proof of the above simple fact about HCF/LCM can be demonstrated very easily by the help of some particular numbers. Lets choose for example the numbers as 360 and 735. We can then ask students to do a prime factorization of both the numbers (the students at 6th grade are easily convinced that every number can be factored into primes and that this factorization is unique apart from order of factors and so the unique factorization theorem can be assumed here without proof) and get the following: $$360 = 2^{3}\cdot 3^{2}\cdot 5,\,\,\, 735 = 3\cdot 5\cdot 7^{2}$$ Also noting that an exponent of $ 0$ represents $ 1$ and an exponent $ 1$ represents the base number, we can write the above factorizations to include all the prime numbers contained in both the factorizations: \begin{align} 360 &= 2^{3}\cdot 3^{2}\cdot 5^{1} \cdot 7^{0}\notag\\ 735 &= 2^{0}\cdot 3^{1}\cdot 5^{1}\cdot 7^{2}\notag \end{align} The student is clearly aware of finding the HCF and LCM when the factorizations of both the numbers are available. He simply has to pick the lower exponents in each prime factor to get HCF and higher exponents of each prime factor to get the LCM and therefore he can easily say that \begin{align} \text{HCF}(360, 735) &= 2^{0}\cdot 3^{1}\cdot 5^{1}\cdot 7^{0}\notag\\ \text{LCM}(360, 735) &= 2^{3}\cdot 3^{2}\cdot 5^{1} \cdot 7^{2}\notag \end{align} Now when you multiply the two given numbers the exponents get added up for each prime factor and the same happens when you multiply the HCF and LCF (only sometimes the exponents get added up in reverse order, but that does not matter) and hence the product of HCF and LCM is the same as the product of the numbers. The basic idea is that if we have two numbers (may be same or distinct) and we take the greater one of the two and smaller one of the two and add these we get the sum of the given numbers. This above fact in italics is just plain common sense and I have emphasized it only to show that this simple fact forms the basis of the proof of the result about HCF and LCM mentioned above.

The above demonstration also shows why the same result can not be extended for three numbers. Its simply because if we take the greatest of three numbers and least of three numbers and add them up the sum will never be the same as the sum of all three numbers (again common sense, just adding two numbers out of three will never match the sum of three numbers).

However I see that at best the teachers illustrate about this result in the following manner. They just say that the HCF of 360 and 735 is 15 and ask students to verify this. Similarly the LCM of 360 and 735 is 17640 and then the teacher asks the students to verify the equality $ 360 \cdot 735 = 15 \cdot 17640$ by plain old multiplication.

This is a very good example of unnecessarily hiding the concepts from students and asking them to memorize the stuff mechanically. Moreover this was one of the simplest examples where the concepts could have been easily provided.

Calculus Approach: Example 2

A slightly harder example from 8th grade is the Heron's formula for area of a triangle where the students are provided a magical (but easy to memorize) formula which can be used to calculate the area of the triangle given its sides. This formula is actually dependent on the most famous theorem of geometry namely the Pythagoras Theorem (its really a surprise that breaking the tradition textbooks actually prove this theorem in 10th grade and don't hide it considering the fact that textbooks never prove the Heron's formula). It has never been clear to me why the Heron's formula is left out without a proof. The magic behind this beautiful result definitely needs to be explained.

Some books on trigonometry (at 11th grade) do provide a proof of Heron's formula which is unnecessarily based on trigonometry and is presented very late in this subject and by this time the student would have definitely lost the charm of Heron's formula when there are many many many (!) trigonometric identities to memorize and even more problems to solve.

The way to proceed here is to tell the students that the Heron's formula can be derived using Pythagoras theorem in a very simple way. To begin with one needs to calculate the altitude of the triangle given its sides. Let's then suppose that we have a triangle ABC with sides $ a, b, c$ and then consider the following figure:

Finding the Area of a Triangle

Clearly our objective is to calculate $ h$ from the figure above. The idea is to first calculate $ x$ and then find $ h$ using Pythagoras theorem. Finding $ x$ requires some more manipulation. We only need to see that $$ h^{2} = a^{2} - x^{2} = c^{2} - (b - x)^{2}$$ and from the last equality we can find $ x$ as $$ x = \frac{a^{2} + b^{2} - c^{2}}{2b}$$ and then we can find $ h$ as \begin{align} h &= \sqrt{a^{2} - \left(\frac{a^{2} + b^{2} - c^{2}}{2b}\right)^{2}}\notag\\ &= \frac{\sqrt{(2ab)^{2} - (a^{2} + b^{2} - c^{2})^{2}}}{2b}\notag\\ &= \frac{\sqrt{(2ab + a^{2} + b^{2} - c^{2})(2ab - a^{2} - b^{2} + c^{2})}}{2b}\notag\\ &= \frac{\sqrt{((a + b)^{2} - c^{2})(c^{2} - (a - b)^{2})}}{2b}\notag\\ &= \frac{\sqrt{(a + b + c)(a + b - c)(c + a - b)(b + c - a)}}{2b}\notag \end{align} and if we put $ 2s = a + b + c$ then we get $$ 2(s - a) = b + c - a, 2(s - b) = c + a - b, 2(s - c) = a + b - c$$ so that \begin{align} h &= \frac{\sqrt{2s \cdot 2(s - a) \cdot 2(s - b) \cdot 2(s - c)}}{2b}\notag\\ &= \frac{2}{b}\,\sqrt{s(s - a)(s - b)(s - c)}\notag \end{align} and thereby noting that the area of triangle is $ bh/2$ we have established the Heron's formula.

The above proof is something which can be understood by students of 8th grade but still the student is not provided with the same in his textbook. It is very important to keep in mind that only by demystifying various results and formulas you can excite the students and put them on a sound mathematical foundation. (For another example of demystification see the series of posts on rank of a matrix). When the students are provided some magical formulas without any justification for the same it feels like the student has been tele-ported from one place to another without enjoying the ride between these places.

Common Critique of the Calculus Approach

Many mathematics teachers and even professors (include book authors too!) argue that a systematic study of mathematical topics providing justification for all kinds of formulas is difficult. The reasons offered are that

1. there are some very important results with wide applications in theory and practice without which the student will be handicapped, and at the same time their proofs are quite complicated to present them to students at young age.
2. providing justification at every stage of the mathematical study might not be possible as this will increase the size of textbooks and also put unnecessary burden on those students who are not going to read mathematics as a subject in their later academic career.
3. these are not the part of the syllabus (I think this is the silliest reason).

One very important sentence from preface of "Real Analysis" by Walter Rudin (he is supposed to be best author in "Real Analysis") goes like this: "It is pedagogically unsound to present the theory of real numbers before teaching the concepts of analysis". And he is talking like this in a book meant for undergraduates. Long before I read this preface of his, I had heard of Walter Rudin's fame and his books being the best of the lot. I have to say here that I was very disheartened to read this in the preface itself and never read that book (or any of his books) again. He wants to teach students the same 18th century calculus to make it even more mysterious.

In response to the above critique I have only to say that most of the times the concepts/proofs related to the topics being taught are not difficult at all (as indicated by examples above). Its just a policy to mechanize the whole of education. Also understanding the fact that many students will not pursue mathematics for higher studies, we don't need to keep such material as necessary for exams. But it makes sense to present them in textbooks so that few students who are really going to pursue mathematics will enjoy it and appreciate their subject better. Who knows one of these might turn out to be a Ramanujan in future. Another point which I wish to reiterate here is that the mechanized approach is good for machines because they are designed to function in that way. Humans have different sort of mental facilities which are not suited to a mechanical approach. Such an education will most likely kill the creative faculties of many students.

P.S. If the reader gets a feeling that algebra is a mechanical subject then I offer him my apologies. I have nowhere tried to provide such an impression, but rather told that algebraical systems are made of such laws which can be mechanized easily. But the subject of algebra as such is not mechanical at all and involves a lot of creative effort (one look at this series of posts will convince the reader about the amount of creativity involved in algebra). Its just that it is easier to teach elementary algebra (meaning upto 10th grade) in a mechanical fashion whereas the same is not the case with elementary geometry and calculus. My emphasis in the post was that teachers and textbooks alike are trying to teach the whole of mathematics in this way and are definitely not heading in the right direction.

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