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Session 3

In this third session, we look into mathematical tools from a historical perspective with in mind perhaps a reflection on what this tell us about the nature of mathematical activity.

We well know those Babylonian tablets:

(source)

By collating important results, as well as heuristics in the form of problems and their solutions, these clay tablets can probably count as one of the oldest known mathematical "tool". Interesting to think how the material (clay, wooden stylus) could influence the mathematical work (for the numbers: positional system using only 2 symbols). The methods used to obtain certain results could also be considered as mathematical "tools". Let us therefore offer a little exploration in this direction!

 

A little problem

Here's a somehow very simple problem: 

Find the side of a square whose surface is 200 hops.

Solve this problem without using any algorithm (or any instrument)!

A Babylonian Method

It is obviously a problem of "square root extraction" which can be seen as part of the larger family of problems requiring, as we say today, to solve a second degree equation: ax²+bx+c=0 . The Babylonians  had different techniques to solve this kind of problem. The simplest, which applies here, is to average. It is the same technique of present Heron of Alexendria some 1700 years later:

(source)
 

Allowing yourself to use our symbols as well as your favorite ways to do additions and divisions solve the problem of 200 hops with this formula, where e is an approximation of the root you are looking for!

Reusing the formula with your new approximation, e', you'll get an impressive level of precision rather quickly! 

Use the Babylonian method repeatedly until you have a 5 decimal precision for the root of 200 (√200 = 14.1421356237...)

In India

An interesting elaboration of the Babylonian formula was found in India, in a Bakhshali manuscript dating from circa -400. We actually use the formula twice, but also take into consideration the difference between what our first approximation gives and the number we want to square.  In modern notation, the method looks like this:

Let  n = e² + b, where e is the number we want to square (e.g 28), e is our approximation (e.g. 5) and b is, of course, the difference between n and e² (28-5=3).

               Then e' is given by:

Starting with e = 14, calculate √200 with this formula!

The following fragment shows an excerpt from Bakhshali's illustrating the method applied to the difficult problem 3x2/4 + 3x/4 = 7000, which he obviously knew how to solve...

 (source)

Step by step?

How do these approximation methods compare with an algorithm giving the exact result of a root computation? Of course, such an algorithm exists: it looks a bit like the long division algorithm. There is a trace of this algorithm in another Indian manuscript, dating from the year 500 or so, and signed by the hand of a certain Aryabhata. It is essentially the same algorithm find later in the work of Arab and European mathematicians. It was taught in schools a good part of the 20th century. I illustrate it for you in the fairly simple case of the calculation of √150. I leave it up to you to see why it works (good luck !!).

Lets first observe that 150 is also: ...000150.0000...

What we do then is to split the number in 2-digit packets, starting from the decimal point:

01 50 . 00 00 00 ...

We are now interested in the number packet different from 00 at the highest position: here it is 01.

• We are looking for the root r of the closest perfect square c such as c<=01. Here, it is simply r=1 since 1^2=1<=01
• this r is the first digit of √150
• For the next step, we now compute the difference d between r^2 and our packet: 01-1^2=0

We can then move on to the next packet. Note that with another number (eg, 2842) we could obviously have had a remainder different from 0 (e.g. 3 because we would have 5 as the first digit: 5 * 5 = 25 <28 and 28-25 = 3).

• We put the next packet onto the right of our difference d: 050
• We multiply by 20 the number formed by the digits identified so far (regardless of the decimal point if there is one, see below): 1*20=20
• We now look for the digit r giving us the closed number c such as (20+r)*r<=050. Here, we take r=2 since 22*2=44<50    (and 23*3=69, too large)
• This r is the next digit of √150
• For the next step, we compute the difference d : 50-44=6

Up to here we have: √150=12

We now get to the decimal part, by simply repeating what we did before:

• We put the next packet onto the right of our difference d: 600
• We multiply by 20 the number formed by the digits identified so far: 12*20=240
• We look for the digit r giving us the closed number c such as (240+r)*r<=600. Here: r=2 since 242*2=484<600  (and 243*3=729, too big)
• This r is the next digit of √150
• For the next step, we compute the difference d: 600-484=116

 We now have: √150=12.2

One last digit, just for the fun of it:

• We put the next packet onto the right of our difference d: 11600
• We multiply by 20 the number formed by the digits identified so far: 122*20=2420  (this is where we start ignoring the decimal point)
• We look for the digit r giving us the closed number c such as (2420+r)*r<=11600. It'a a bit more difficult this time, but we can make it: r=4 since 2424*4=9696<11600
• The next digit for √150 is thus 4
• etc.

There is (at least) one "shortcut" we can use to find r relatively easily event with large numbers. For example, to get the next digit we nee to solve : (24480+r)*r<=190400). Note however how r is, of course, near the floor value of the division of our "augmented" d by the product of our digits by 20. Here we have r ≈ [190400/24480] = [7.77...]=7. So if making a division is not a problem, we can actually save ourselves some time/work. We can even get a pretty good guess with simple approximation, e.g. : 19/2.4=7.9166... Here, 7 is in fact the next digit.

√150=12.247...

Now try this algorithm to calculate the digits of √200

By the rules

The often slow and painful side of this algorithm explains that we often prefer approximative methods. In terms of efficiency, it is advantageous to return to tools in the sense in which we generally think about the term. The slide ruler is one of these ingenious instruments! Depending on the slide ruler, it is possible to get trigonometric values for example, but also square or cubic roots, etc. But unlike modern calculators, this requires a little more work. And then of course, there are several variation slide rulers.

(source)

Let us simply say that the general idea is to have a rule graduated so that we have in relation to the numbers 1,2,3,4 ... and their root 1, 1.414, 1.732, 2. The rule is then used on the basis that √ab = √a • √n, taking either a = 10 and a = 100 (or a power of 10 or 100). In this way, one can always "reduce" a number to a value between 1 and 10. With 200, we will have a = 100 and n = 2. With 14.3 we would have a = 10 and n = 10. For larger numbers (or smaller ones, because we can also divide by 10 or 100), we are still always in one case either we have a factor of 100, or we have a factor of 10. And since √10 and √100 have different values, the rule has two graduations.

In fact, the rule generally gives the value of √n for 0>n>10. If the number we are interested in is larger or smaller, we count the digits to know if we will read on the rule of "odd" (for 20, 2000, etc.) or "pair" (for 2, 200 etc. .). It is the user who must then "place the decimal point" according to the order of magnitude. Thus, for √200 or go to the "even" rule, the value of √2, and multiply by 10. The root to be extracted is chosen on the A rule, and the root obtained is Read on the rule C. Thus, on the left one recognizes √2 ≈ 1.41 and on the right √20 ≈ 4.47

Similarly we see on the left √4.5 ≈ 2.12 and on the right √45 ≈ 6.70

 (Note: As commented on the source page of the ruler image, slide rules are very difficult to depict well in discrete pixels because they have scales which vary in size in a non-linear way)

 On certain rulers, it is possible to increase the level of precision, by "amplifying" the rule by using the central part of the ruler. But we could save this for another time!

Expliain and justify how to use the ruler to find √200.

[try one here]

Then again?

Nowadays, all this is evidently of ancient history. If I need to solve such a problem, I can simply ask the question to my computer, my phone or my watch, or use a calculator to get the result myself in an instant.

If one accepts that all these are indeed instances of mathematical activity, how do they compare? What does it mean, very concretely, to do math in each case?