Curiouser and Curiouser: TimeAuthor(s): Helen MorrisSource: Mathematics in School, Vol. 27, No. 2 (Mar., 1998), pp. 28-29Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30215358 .

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Curiouser and

Curiouser Compiled by Helen Morris

Time

Achilles and the Tortoise In the 5th century BC, the Greek philosopher Zeno proposed a number of paradoxes, one of which relates to Achilles and the tortoise.

Achilles, the legendary athlete, could never overtake the slowest tortoise if the tortoise was given a head start. The argument goes that Achilles must always reach the point from which the tortoise has just departed, by which time the tortoise will have moved on. Where is the flaw in the argument?

Diophantus Diophantus was a Greek mathematician who lived around the 3rd century AD. Some of his books survived and influenced many later mathematicians from Fibonacci (c. 1200), during the early Renaissance, right up to Fermat (c. 1600), who wrote his 'Last Theorem' in the margin of his own copy of

Diophantus' works. We know little more of

Diophantus' life other than the following epitaph.

'This tomb holds Diophantus. Ah, what a marvel!And the tomb tells scientifically the measure of his life. God vouchsafed that he should be a boyfor the sixth part of his life; when a twelfth was added, his cheeks acquired a beard; He kindledfor him the light of marriage after a seventh, and in the fifth year after his marriage He granted him a son. Alas! Late-begotten and miserable child, when he had reached the measure of half his father's life, the chill grave took him. After consoling his grief by the science of numbersfor 4years, Diophan- tus reached the end of his life.'

For how long did Diophantus live?

The Solar Year The Egyptians noticed that the solar year was 365 days, by observing the sun, moon and stars. They observed that the star Sirius appeared in the sky at sunrise on the day of the year when the annual flood of the Nile reached Cairo. But it was i of a day out so that eventually it would not be perfectly synchronised and would only be back on schedule after 365 x 4 = 1460 years. 1460 years was called the Sothic cycle. It is believed

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that the learned Egyptian priests knew that the solar year was 365 a days but deliberately withheld the knowledge from the people. Knowing when the floods were due, the priests could pretend to bring it about with their rites while making the poor farmer pay for the service!

On What Day Were You Born? Let Y = the year you were born, let D = the number of the day of the year, let X =

(Y - 1)/4, ignore the remainder, let S - Y + D + X. Divide S by 7 and note the remainder. This tells you the day on which you were born.

Fri Sat Sun Mon Tues Wed Thurs

0 1 2 3 4 5 6

For example, 9 April 1960:-

Y= 1960, D = 31(Jan) + 29(Feb, leap year) + 31(March) + 9(April) = 100, X = 1959/4 = 489 (rem 3), S = 1960 + 100 + 489 = 2549. Then S/7 = 2549/7 = 364 rem 1 (Saturday).

28 Mathematics in School, March 1998

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Days and Months

Named after Roman gods. March - Mars, god of war April - Aprillus, goddess of spring May - Maia, goddess of the earth June -Juno, goddess of marriage July - Julius Caesar

August - Augustus Caesar

September - 7th month October - 8th month November - 1lth month December - 12th month

January - named after Janus, the god of entrances and beginnings February - Februa, a purification feast

Sunday - Sun's day Monday - Moon's day Tuesday - Tiw's day Wednesday - Woden's day Thursday - Thor's day Friday - Friga's day Saturday - Seater's day

It All Comes to Nought! A very good way of making the point to students of any age is to get them to study cases where a familiar rule or property appears to fail. Two classic cases of fallacious reasoning can lead to a challenging result when put together.

Firstly, using integration by parts, so subtractingfrom both sides gives 0 = 1 (A). Secondly, start with two equal terms, a and b. Giving a and b the value 1 leads to 1 = 2 and consequently n = n + 1 (B). Putting (A) and (B) together gives the curious result that 0 = 1 = 2 = 3 ...

from which we may deduce that all natural numbers (or indeed integers) equal nought! Contributed by John Earle, Exeter

27 Plus Choose any 2-digit number in which the units digit is 3 more than the tens digit, eg 25. Add 27 (25 + 27 = 52). Why has the resulting number the same digits as the original, but reversed? Curious! Contributed by D B Eperson, Canterbury

A Conjuring Trick Problems that involve finding two unknown numbers usually need two simultaneous equations for their solution. But it is possible in some cases to ask for only one piece of information in order to discover two numbers that have been chosen.

Ask a friend to choose two whole numbers less than 10, and then to perform some simple calculations with them. (a) Add them together. (b) Find their difference. (c) Multiply their sum by 10. Then, to this final number, add

their dzifference, and tellyou the answer. From this single piece of information you can tell your friend the numbers they have chosen.

To do this, first add the units figure of the answer to its number of tens, divide by two, and that tells you the larger of the chosen numbers. Second, subtract the units figure of the answerfrom its number of tens, divide by two, and that tells you the smaller number.

For example, if the answer is 153, add 3 to 15 = 18, half of which is 9 (the larger number chosen); subtract 3 from 15 = 12, half of which is 6 (the smaller number chosen).M Contributed by D B Eperson, Canterbury

Solutions 27 Plus Let the first digit of the number be a, and the second digit be a + 3. So 27 + a (a + 3)= (a + 3)a therefore 7 + (a + 3)= a + 10 2+a+1=a+3

A Conjuring Trick If the chosen numbers are x and y, where 10 > x > y, then ten times their sum is 10 (x + y) and their difference is (x - y) < 10. Adding these together, the answer is 10 (x + y) + (x - y), the units digit of which is (x - y). Adding this to

the number of tens, (x + y) + (x - y) = 2x, half of which is the larger number x; subtracting this from the number of tens, (x + y) - (x - y) = 2y, half of which is the smaller number y.

Achilles Euclidean geometry allows for the fact that any line segment contains an infinite number of points, since any line segment can itself be bisected. Following on from this, any interval of time, no matter how small, can be subdivided further. The concept of time regards time as consisting of instants, which follow each other, as do numbers or points on a line. There are an infinite number of instants between any two instants. Since Achilles and the tortoise run the same number of instants, the tortoise must run through as many distinct points as Achilles. But the assertion that a greater distance implies more distinct points is incorrect, as the number of points is the same for Achilles and for the tortoise. Assuming that Achilles gives the tortoise a 100 m start, for example, and has a

speed of 10 m/s, with the tortoise travelling at 1 m/s; in 10 seconds Achilles will have travelled the 100 m to reach the tortoise, but the tortoise will have moved 10 m further forward. In 1 second, Achilles will have moved the required 10 m, but the tortoise will have moved 1 m. But the sum of an infinite number of time intervals can have a finite sum. Achilles will overtake the tortoise! When?

Diophantus Diophantus lived for 84 years.

Let A be the length of his life, then- + - + A + 5 +A + 4 = A

6 12 7 2

Contributions to this column are most welcome and should be sent to: Helen Morris, 57 Woodplumpton Lane, Broughton, Preston, PR3 5JJ

Mathematics in School, March 1998 29

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