## The Fifth Operation of Mathematics

Algebra is often called the “arithmetic of seven operations” thus stressing the fact that the four commonly known mathematical equations are supplemented by three new ones; raising to a power and its two inverse operations.

We start our talk with the fifth operation — raising a number to a power.

Do the practical affairs of everyday life have any need for  this operation? It turns out they do. We often encounter such situations. First recall the many cases of computing areas and volumes that ordinarily call for squaring and cubing figures. Then, the force of universal gravitation, electrostatic and magnetic interactions, light and sound that diminish in force in proportion to  the second power of  the distance. The periods of revolution of  the planets about the sun (and of satellites about planets) are connected with the distances from the centre of revolution by a power relationship: the squares of the periods of revolution are in the same ration as the cubes of the distances.

One  should not think that real life involves only second and third powers with higher powers found only in the problems of algebra books. Engineers dealing with strength-of-materials calculations constantly find themselves up against fourth powers and even (when computing the diameter of a steam pipeline) sixth powers. A water engineer studying the force with which running water entrains rocks also has to deal with a sixth-power relationship. If the rate of  the current in one river is four times that in another, the fast river is capable of rolling rocks that are $4^{6}$ or 4096 times heavier than those carried by the slower river. Even higher powers are encountered when we study the relationship between the brightness of an incandescent body (say, the filament in an electric bulb) and its temperature. The total brightness (luminance) in the case of white heat increases with the twelfth power of  the temperature; in the case of red heat, it increases with the thirtieth power of the temperature( we refer here to absolute temperature, that is we reckon from $-273$ degree Celsius. This means that a body heated say from 2000 degrees to 4000 degrees absolute temperature, that is by a factor of two, becomes brighter by a factor of $2^{12}$, which is more than 4000 times greater.

How much does the earth’s atmosphere weigh?

Here is a convincing way to simplify practical calculations by means of exponential notation; let us determine how many times the mass of the earth is greater than the mass of the earth’s atmosphere.

As we know, each square centimeter of  the earth’s surface supports a column of air equal to one kilogram. The atmospheric shell of  the earth is, as it were, made up entirely of such columns of air — as many as there are square centimeters on the earth’s surface. That is how many kilograms the atmosphere of our planet weighs. From a reference book, we find the earth’s surface to be equal to 510 million, or $51.10^{7}$ square kilometers.

Now figure out how many square centimeters there are in a square kilometre. A linear kilometer contains  1000 meters with 100 centimetres in each, which means it is equal to $10^{5}$ cm, and a square kilometer contain ${10^{5}}^2=10^(10)$ square centimeters. And, so the earth’s surface works out to $51.10^(7).10^{10}=51.10{17}$ square centimeters. And that also is the weight of the earth’s atmosphere in kilograms. Converting to (metric) tons, we get $51.10^{17}:1000=51.10^{17}:10^{3}=51.10^{14}$

Now the mass of the earth is taken to be $6.10^{21}$ tons.

To determine how  much heavier the globe is than its atmosphere, we perform the following division: $6.10^{21}:51.10^{14}=10^{6}$ approximately which means that the mass of the atmosphere is roughly one millionth of that of the earth.

The Changing Weather

Let us describe the weather using only one characteristic — cloudy or not cloudy. Days will be described as clear or overcast. Do you think there will be many weeks with different changes of weather under this condition?

There would appear to be very few: a month or two will pass and all combinations of clear and overcast days in the week will have been exhausted. Then, one of the earlier combinations will inevitably recur. But, let us calculate exactly how many distinct combinations are possible under these conditions. This is a problem that unexpectedly leads us to the fifth mathematical operation.

The problem is: in how many different ways can clear and overcast days alternate in one week?

Solution: The first day of the week is either clear or overcast: that gives us two combinations.

Over a two-day period, we have the following possible alterations of clear and overcast days:

clear and clear;

clear and overcast;

overcast and clear;

overcast and overcast.

Thus, in two days we have $2^{2}$ distinct alternations. In a three-day period, each of the four combinations of the first two days combines with two combinations of the third day; there will be $2^{2}.2=2^{3}$ alternations in all.

In  four days, the number of alternations will reach $2^{3}.2=2^{4}$. In five days, there will be $2^{5}$ alternations, in six days $2^{6}$, in seven days (one week), $2^{7}=128$ distinct alternations.

Hence, there are 128 weeks with a distinct sequence of clear and overcast days. A total of 128.7=896 days will  pass before one of the earlier combinations is repeated. A repetition can of course occur before that, but 896 days is the period after which a recurrence is inevitable. And, conversely, two years and more (2 years and 166 days) may pass during which the weather in one week will not be like that of any other week.

The Number of All Possible Chess Games

Let us make an approximate calculation of the total number of different chess games that can ever be played on a chess board. We will be content with a rough estimate of this. The Belgian mathematician M. Kraichik makes the following calculation in his book entitled The Mathematics of  Games and Mathematical Diversions.

In the first move, white has a choice of 20 moves (10 moves for  the eight pawns, each of which can move one or two squares and two moves each of  the two knights). For each of the white, black can correspond with one of  the same 20 moves, Combining each move of white with each move of black, we have 20.20=400 different games after the first move of each side.

“After the first move, the number of possible new moves increases. For example, if white made the first move P-K4, then it has a choice of  29 moves the next time. The number of possible moves continues to increase. For instance, the queen alone standing on Q5 has a choice of 27 moves (on the assumption that all squares that it can move to are vacant). To simplify matters, let us assume the following average numbers:

20 possible moves for both sides in the first five moves;

30 possible moves for both sides in all subsequent moves.

“Also, we take the average number of moves in a single game to be 40”.

Then, we get the following expression for the number of possible games: $(20.20)^{5}.(30.30)^{35}$

To get an approximate idea of this figure, let us make a few simplifying transformations: $(20.20)^{5}. (30.30)^{35}=20^{10}. 30^{70}=2^{10}.3^{70}.10^{80}$

Replace $2^{10}=1024$ by approximately 1000, or $10^{3}$ and express $3^{70}$ as $3^{70}=3^{68}.3^{2}\approx 10(3^{4})^{17} \approx 10.80^{17} = 10.8^{17}.10^{17}=2^{51}.10^{18} = 2.(2^{10})^{5}.10^{18} \approx 2.10^{15}.10^{18}=2.10^{33}$

And, so we obtain $(20.20)^{5}. (30.30)^{35} \approx 10^{3}.2.10^{33}.10^{80} = 2.10^{116}$ .

This number leaves far behind the legendary number of grains of wheat asked as payment for the invention of chess.Note : that number was a mere $2^{64} - 1 \approx 18.10^{18}$. If the whole population of the world played chess round the clock making one move every second, then it would take no less than $10^{100}$ centuries to exhaust all the games in this marathon of chess!!

More later…

-Nalin

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