The other day I described the math textbook that was used in the Soviet Union when I attended school there. My question was: how could this textbook teach so much in just 220 pages? My tentative hypothesis was that it eliminated needless “scaffolding” (yes, I recognize that the term is problematic) and went right to the matter. I realized later that this was only a partial explanation.

An acquaintance suggested to me the other day that Russians (and Soviets) had a particular way of presenting mathematics that made it clear to the reader and student. As I started to think about my textbook in this light, I saw that its material was presented not only lucidly, but also artfully. There is a well-considered and well-timed movement from specific problems to larger topics and back.

The textbook’s editor and lead author is the eminent mathematician Andrei Kolmogorov (1903–1987). I suspect that he shaped a good deal of the book and didn’t simply attach his name to it. But since I don’t know that for certain, I’ll leave that aside and look at what’s actually there.

I’ll begin by describing the contents of the first four pages of the book. Then I’ll examine their structure.

The book begins with a discussion of induction. It presents a series of odd numbers (1, 3, 5, 7, … 2*n – *1) and posits that for every natural number *n*, the sum of the first *n* elements of the series is *n*^{2}. That is,

1 = 1 = 1^{2};

1 + 3 = 4 = 2^{2};

1 + 3 + 5 = 9 = 3^{2};

1 + 3 + 5 + 7 = 16 = 4^{2};

1 + 3 + 5 + 7 + 9 = 25 = 5^{2};

and now the hypothesis: “For all natural numbers n, this equation holds true: 1 + 3 + 5 + … + (2*n* – 1) = *n*^{2}.”

All that is on the first page (or half-page) of the text. The book leaves the reader with this concept: having observed a number of examples, you make a hypothesis about the general rule that unites them.

Then, on the next page, the book examines a few more problems in order to shed light on hypothesis-making. It shows, first, how a single counterexample is enough to disprove a hypothesis. For instance, if you have the formula P (*x*) = *x*^{2}; +* x* + 41, and considered the natural numbers 1 through 5 as values of* x*, you might conclude that P (*x*) is always a prime number. However, an obvious counterexample is P (41).

After this, it gives two more problems where you can easily verify the hypothesis.

Now it goes on to explain the principle of mathematical induction. It returns to the original problem and restates it in this manner: If we know that the equation is true for a certain value of *n*, and if we can also demonstrate that where it is true for *n*, it is also true for *n* + 1, then we know that it is true for every natural number* n*.

It then puts this in simple notation. The letter A represents the function. A (5) is true, we know, because 1 + 3 + 5 + 7 + 9 = 5^{2}. We can also see that the truth of A (6) *follows* from that of A (5). (I wont’ go into that; it’s straightforward.) We represent this as follows: A (5) => A (6).

So our goal is to prove that A (*k*) => A (*k* + 1). That is, we want to prove that from the equation

1 + 3 + 5 + … + (2*k* – 1) = *k*^{2}

this equation follows:

1 + 3 + 5 + … + (2*k *+ 1) = (*k* + 1)^{2}

In fact, this is easy to demonstrate, since

1 + 3 + 5 + … + (2*k* – 1) + (2*k* + 1) =

= (1 + 3 + 5 + … + (2*k* – 1)) + (2*k* + 1) =

= *k*^{2} + 2*k* + 1 = (*k* + 1)^{2}.

(In the penultimate equation, just substitute *k*^{2} for everything contined within the outer parentheses. You can do that because we’ve already established that for a cerain value of *k*, the sum of the first *k* elements is *k*^{2})

From here, the textbook goes on to more complex examples.

How is this structured? First we are given a series with a hypothesis about the underlying pattern. The possibility of proving this pattern is dangled before us. We learn a little more about hypotheses themselves and how they might be proved or disproved. We see a few examples. We go on from there to the concept of induction. At this point we return to the original problem and prove, through induction, that it holds for all values of *n*, where* n* is a natural number. What in particular makes this presentation compelling, easy to understand, and at the same time challenging?

For one thing, there’s the lure of something to come. It’s a real lure. We know that a problem has been left hanging and that there’s more to it. Also, each concept is developed and at the same time reduced to the simplest possible notation. The student gets both the concrete examples and the abstractions. The careful reader understands the meaning of A (*k*) => A (*k* + 1). And then the proof of the original problem may jog the mind slightly at first, but then it’s obvious.

Now, some readers may be saying, “I have no idea what you’re talking about.” But that’s the thing. When I first read these pages, I could maybe make a glimmer of a tail’s feather out of them, but that was it. But then I sat down and puzzled over them. Then, when they came clear, they were straightforward and elegant to boot. This applies not only to the first four pages. I found that where this held true through page* n*, it also held true through page (*n* + 1).

Maybe that’s part of the “secret” here, too, if there is one: the understanding that the student must take time with the material, at home, without distraction.

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