Sky Issue 14, Fall 2010

In This Issue: (-1) x (-1)=1 . . . but WHY? Palindrome Dates Primes from Fractions Reasoning from the Specific to the General The Mathematics of Climate Modelling

On the Cover Predicting the evolution of the Earth’s climate system is one of the most daunting mathematical modeling challenges that humanity has ever undertaken. In his article on page 17, Adam Monahan discusses the interweaving of mathematics with modern climate science.

Pi in the Sky is aimed primarily at high school students and teachers, with the main goal of providing a cultural context/ landscape for mathematics. It has a natural extension to junior high school students and undergraduates, and articles may also put curriculum topics in a different perspective.

Editorial Board John Bowman (University of Alberta) Tel: (780) 492-0532, E-mail: [email protected] Murray Bremner (University of Saskatchewan) Tel: (306) 966-6122, E-mail: [email protected] John Campbell (Archbishop MacDonald High, Edmonton) Tel: (780) 441-6000, E-mail: [email protected] Florin Diacu (University of Victoria) Tel: (250) 721-6330, E-mail: [email protected] Sharon Friesen (Galileo Educational Network, Calgary) Tel: (403) 220-8942, E-mail: [email protected] Gordon Hamilton (Masters Academy and College, Calgary) Tel: (403) 242-7034, E-mail: [email protected] Klaus Hoechsmann (University of British Columbia) Tel: (604) 822-3782, E-mail: [email protected] Dragos Hrimiuc (University of Alberta) Tel: (780) 492-3532, E-mail: [email protected] Michael Lamoureux (University of Calgary) Tel: (403) 220-8214, E-mail: [email protected] David Leeming (University of Victoria) Tel: (250) 472-4928, E-mail: [email protected] Mark MacLean (University of British Columbia) Tel: (604) 822-5552, E-mail: [email protected] Patrick Maidorn (University of Regina) Tel: (306) 585-4013, E-mail: [email protected] Wendy Swonnell (Greater Victoria School District) Tel: (250) 477-9706, E-mail: [email protected]

Managing Editor Anthony Quas (University of Victoria) Tel: (250) 721-7463, E-mail: [email protected]

Contact Information Pi in the Sky PIMS University of Victoria Site Office, SSM Building Room A418b PO Box 3060 STN CSC, 3800 Finnerty Road, Victoria, BC, V8W 3R4 T/(250) 472-4271 F/(250) 721-8958 E-mail: [email protected] Pi in the Sky is a publication of the Pacific Institute for the Mathematical Sciences (PIMS). PIMS is supported by the Natural Sciences and Engineering Research Council of Canada, the Province of Alberta, the Province of British Columbia, the Province of Saskatchewan, Simon Fraser University, the University of Alberta, the University of British Columbia, the University of Calgary, the University of Lethbridge, the University of Regina, the University of Victoria and the University of Washington.

Submission Information For details on submitting articles for our next edition of Pi in the Sky, please see: http://www.pims.math.ca/resources/publications/pi-sky

Table of Contents Editorial by Anthony Quas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 (-1) x (-1) = 1 . . . but WHY? by Marie Kim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Palindrome Dates in Four-Digit Years by Aziz S. Inan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Primes from Fractions by Alex P. Lamoureux and Michael P. Lamoureux. . . . . . . . . . . 8 Reasoning from the Specific to the General by Bill Russell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Book Review Pythagorean Crimes Reviewed by Gord Hamilton . . . . . . . . . . . . . . . . . . . . . . . . . . 15 The Mathematics of Climate Modelling by Adam H. Monahan. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Pi in the Sky Math Challenges Solutions to problems published in Issue 13. . . . . . . . . . . . . . . 22 New Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Editorial Anthony Quas MATHEMATICAL Modelling: WHAT, HOW AND WHY?

When light is reflected in water the reflection seems to be stretched out along the line connecting the viewer and the source. Why is this? Why isn't it stretched in the perpendicular direction (horizontally in this picture)?

In my work and outside it too I often hear about Mathematical Modelling. In this editorial I'll say a bit about what it is; how to do it; and why it's important.

What is Mathematical Modelling? First let's try a quick answer: Mathematical Modelling is the process of using mathematics to understand something `in the real world'. For high school students the closest thing to this in the curriculum might be the (often-dreaded) word problems. Here is an example taken from the internet: Mr.S is planting flowers to give his girl friend on Valentine's day, which is half a year away. Currently the flowers are 7 inches tall and will be fully grown once they reach one foot. If the flowers grow at a rate of half an inch per month, then how large will they be on the day? Will the flowers be fully grown? or will he have to find another gift?

In the case of a fast-spreading epidemic involving an unknown disease (e.g. the SARS epidemic that hit China in late 2002 / early 2003 and spread to other countries including Canada) what are the best policies to follow to minimize sickness and death from the disease while avoiding overreacting? Finally one of the most important cases of all: climate modelling. Here you're trying to predict the weather patterns in 10 years; 20 years; 100 years. For more information on this see the article in this issue by Adam Monahan.

The aim here is to take something which is on the surface not a mathematical question, translate it into a mathematical question, solve the mathematical question and translate the answer back into the framework of the original question.

How do you do Mathematical Modelling?

In this case we let x denote the height of the flowers on Valentine's Day and say that x = 7 + t / 2 where t is the number of months until Valentine's Day. Since we're told it's half a year away we have t = 6 so that we can compute x = 10. Since we're told the flowers will be 12 inches when they're fully grown it sounds as though `Mr.S' will have to think of a new present.

Just as in the case of the word problem, a first step is trying to decide which variables to study in the problem. There is an important difference though. The word problem is typically constructed to give you all of the information you need to solve the problem (and no more). Generally the variables are given to you in the statement of the problem itself. When you're doing more advanced modelling you have to decide which variables to include (and often equally importantly which ones to leave out of consideration).

For some more advanced examples, here are some other questions that might be approached using mathematical modelling. 1

where you know the answer? If it's not working you may need to tweak the model to improve it.

This last idea is at first sight quite surprising. Surely the best model is the one that takes everything into account? Actually that's not usually considered to be the case. Rather a good model is usually considered to be the simplest one that explains the behaviour you're trying to study. This is an important idea that is sometimes called `Occam's Razor' (named after a 14th Century monk who first expressed it).

Why does mathematical modelling matter? Mathematical modelling can be great or it can be useless. It all depends on the model chosen. Good mathematical modelling should make some predictions that you can test with existing data (it's important that you haven't already used the data to build the model-otherwise you end up with a circular argument: you build the model to work for a certain set of data and then check that it works with that same data). It should then make some predictions that you can test in the future.

For example in the case of the reflected lights one might try to take account of the positions of all of the water molecules! This would probably be a bad idea because people don't know exactly how liquids work. Also because there are a number of trillion trillion water molecules in the picture it would be impossible to do any computations even if you did know where they all were.

You can use mathematical modelling to test ideas that would be impractical or unethical to do as an actual experiment. For instance in the SARS modelling example you might want to predict what would happen if we took no action and compare it to the effect of (say) shutting down schools and airports for two weeks. Doing an actual experiment would be politically impossible but doing it in the model doesn't impact people in the same way.

In practice what you do is this: pick out some variables that you think could be important (in the water case I used the `slopiness' of the water: how far away from being flat it is as a variable for example) and try to write down some equations relating the observations to the situation you're modelling. This may involve making some guesses as to how things work. Test your model: does it give the results you expected?; can you apply it to make predictions in situations other than the one

In the best case, mathematical modelling tells you how things work and informs decisions. It is an invaluable tool across science and economics.

www.xkcd.com 2

(-1) x (-1) = 1 . . . but WHY? Marie Kim, Cheong Shim International Academy Positive and negative numbers are often explained with the model of `debt and profit.' Profit is positive and debt is negative; adding positive numbers results in profit while adding negative numbers deducts profit - adding to debt. This model however, failed to make people understand the concept of multiplying a negative number by a negative number since it is hard to find such cases in everyday life.

Marie Kim is at Cheong Shim International Academy, currently in 11th grade. She's from South Korea, and plans to go to the U.S. for college and major in biology or neurology.

Negative numbers. Most students find the concept hard to understand and to accept at first. Mathematicians before Descartes refused to accept negative numbers, including the great Pascal himself. Negative numbers actually are believed to have been found in the East the earliest, in China. An ancient Chinese text, written in B.C. 1000, titled \Ku Jang San Sul" - meaning \nine arithmetic formulas"- includes in part a computation of negative numbers.

How is the concept `negative number x negative number' explained, then? There are quite a few illustrations and models that are available to help people understand and accept such a concept. Out of those explanations, here are three easy ones.

1. Pattern 2 x 2 = 4

1 x 2 = 2

0 x 2 = 0

(-1) x 2 = (-2)

(-2) x 2 = (-4)

2 x 1 = 2

1 x 1 = 1

0 x 1 = 0

(-1) x 1 = (-1)

(-2) x 1 = (-2)

2 x 0 = 0

1 x 0 = 0

0 x 0 = 0

(-1) x 0 = 0

(-2) x 0 = 0

Take a look at the pattern above. In each row, you'll be able to find that constant decrease in the number multiplied results in constant change in the product. The same for each column. Applying this same pattern, the next row will be: 2 x (-1) = (-2) 1 x (-1) = (-1)

0 x (-1) = 0

(-1) x (-1) = 1

(-2) x (-1) = 2

2. Number line

This pattern is an initial indication why the statement \negative number multiplied by negative number results in positive number" might be correct.

On the number line, `2 x 3' is thought as `moving by 2 three times in the 2. Number line

same direction as 2 - the positive direction.'

On the number line, `3 x 2' is thought of as `moving by 2 three times in the same direction as 2 - the positive direction.'

3

direction of 2 - the negative direction.' Similarly, `3 x (-2)' is thought of as `moving by 2 three times in the opposite direction of 2 - the negative direction.'

Now, keeping those two models in mind, `(-2) x (-3)' can be thought as `moving by (-2) three times in the opposite direction of (-2)'; so it will be the same as `moving by 2 three times in the positive direction.' This results in `(-2) x (-3) = 6'; negative number times negative number equals positive number which Now, would look keeping like the firstthose numbertwo line. models in mind, `(-2) x (-3)' can be thought

as (-2) three times in the opposite direction of (-2)'; so it will be the same as `m 3. Proof three times in the positive direction.' This results in `(-2) x (-3) = 6'; nega times negative number equals positive would look likeit can thebe first This one may be the most complicated out of the number three. Usingwhich the distribution property, proved that multiplying two negative numbers result in a positive number. Let's put `a' and `b' as two real numbers.

x = ab + (-a)(b) + (-a)(-b)

Let's expand it this way first.

x = ab + (-a) {(b) + (-b)} (factoring `-a' out)

Proof 3.

x = ab + (-a) (0)

x = ab + 0

x = ab

This one may be the most complicated out of the three. Using the distribution proper proved that multiplying two negative numbers result in a positive number. Do it again but this time in a different order.

x = {a + (-a)} b + (-a)(-b) (factoring `b' out)

x = (0)b + (-a)(-b)

x = 0 + (-a)(-b)Let's

x = (-a)(-b)

put `a' and `b' as two real numbers.

So `x = ab and x = (-a)(-b),' which leads us to `ab = (-a)(-b)' x=

ab

+

(-a)(b)

+

(-a)(-b)

Other explanations for `(-) x (-) = (+)' do exist and can be found easily, for example using complex numbers. But, don't just go and look it up; take your time and think of your own way to explain `(-) x (-) = (+).'

*2

4

Palindrome Dates in Four-Digit Years Aziz S. Inan Aziz Inan was born in Turkey, did his PhD at Stanford and now teaches Electrical Engineering at University of Portland. He enjoys posing Recreational Math Puzzles. He has noted that even his name has a puzzling geometric property. Write out AZIZ INAN. Swap A’s and I’s and rotate the consonants by 90 degrees. The two names switch places!

is, \Can some of these eight-digit full dates be palindrome numbers?" (A palindrome number is a number that reads the same forwards or backwards [1-2].)

Introduction

For example, in the DDMMYYYY date format, the first palindrome date of this (21st) century occurred on 10 February 2001 since this date, expressed as the single date number 10022001, is indeed a palindrome number. Note that palindrome date 10022001 is also the first palindrome date of this century that occurred in the MMDDYYYY date format, but it corresponds to a different actual date, which is October 2, 2001.

The answer is yes, and these special dates are called palindrome dates.

In most of the world's countries, a specific calendar date in a four-digit year is expressed in the format DD/MM/YYYY (or DD.MM.YYYY or DD-MM-YYYY) where the first two digits (DD) are reserved for the day, the next two (MM) for the month and the last four (YYYY) for the year numbers. (The United States is one of the few countries which use the MM/DD/YYYY date format in which the places of the month and the day numbers are switched.) In general, if one removes the separators between the day, the month and the year numbers, a full date number consists of a single eight-digit number sequence given as DDMMYYYY. For example, the birth date of the famous American recreational mathematician Martin Gardner is 21 October 1914 which can be expressed as a single date number as 21101914 in the DDMMYYYY format (or 10211914 in the MMDDYYYY format).

Palindrome Dates Assuming each date in all the four-digit years is assigned a single eightdigit date n u m b e r DDMMYYYY, a question that comes to mind

D=Y4 0 0 1,2 1,2 3 3 3 3

D=Y3 1 to 9 1 to 9 0 to 9 0 to 9 0 1 0 1

M=Y2 0 1 0 1 0 0 1 1

Generally speaking, palindrome dates are very rare and sometimes don't occur for centuries. If any, only a single palindrome date can exist in a given fourdigit year Y1Y2Y3Y4. Also, among all four-digit years, a specific date designated by both month and day numbers as D1D2M1M2 (or M1M2D1D2) can be a palindrome date only once represented by a date number D1D2M1M2M2M1D2D1 (or M1M2D1D2D2D1M2M1). In the DDMMY1Y2Y3Y4 date format, since each single palindrome date number must satisfy

M=Y1 Y1 1 to 9 1 to 9 1,2 1,2 1 to 9 1 to 9 1,2 1,2 1,3 to 9 1,3 to 9 1,3,5,7,8 1,3,5,7,8 1 1 0,2 0,2

Y2 0 1 0 1 0 0 1 1

Y3 1 to 9 1 to 9 0 to 9 0 to 9 0 1 0 1

Y4 0 0 1,2 1,2 3 3 3 3

N 81 18 180 40 8 5 1 2

Table 1 Palindrome date combinations in the DDMMY1Y2Y3Y4 date format. Note that the total number of palindrome dates in each category adds up to 335.

5

one being 29 November 1192 (29111192). No other palindrome dates occurred between the

1001 to 2000) can occur only in the first four centuries of each millennium. On the other

Y4Y3Y2Y1Y1Y2Y3Y4, M=Y4 M=Y3 D=Y2 D=Y1 Y1 Y2 Y3 Y4 N palindrome dates can 0 1 to 9 0 to 2 1 to 9 1 to 9 0 to 2 1 to 9 0 243 only occur2 in years TABLE 0 1,3,5,7,8 3 1 1 3 1,3,5,7,8 0 5 ending with a digit Y4 1 0 to 2 0 to 2 1 to 9 1 to 9 0 to 2 0 to 2 1 81 less than four (since day 1 0,2 3 1 1 3 02 1 2 Table cannot 2. Palindrome combinations in the MMDDY1Y2Y3Y4 date format. Note that the total number exceed date Table 2. number palindrome 31). In ofaddition, thedates in each category adds up to 331. Palindrome date combinations in the MMDDY1Y2Y3Y4 date format. Note that the total hundreds digit Y2 of number of palindrome dates in each category adds up to 331. the year number of the *5 palindrome date must with day number equal to Y2Y1. Table 2 provides either be zero or one (since month number cannot all possible combinations of palindrome dates in exceed 12). In the special case when the thousands the MMDDY1Y2Y3Y4 date format categorized in digit Y1 of the year satisfies Y1 > 2, Y2 must equal terms of different values and ranges of digits Y4, zero. In other words, palindrome dates in the Y2, Y3 and Y1, where N is the total number of second and third millenniums (years 1001 to 3000) palindrome dates in each category. can only occur in the first two centuries of each millennium. On the other hand, palindrome dates According to Table 2, there are a total of 331 that fall between fourth and tenth millenniums palindrome dates in the MMDDYYYY date (years 3001 to 10000) can only occur in the first format involving all the four-digit years. century of each millennium. Also, between years 1000 to 10000, palindrome dates in each century Note that even if some palindrome date numbers all fall on the same month with month number represented by Y4Y3Y2Y1Y1Y2Y3Y4 are valid Y2Y1. Table 1 provides all possible combinations date numbers in each date format, unless the of palindrome dates in the DDMMY1Y2Y3Y4 date day and the month numbers are the same, they format categorized in terms of different values correspond to different actual dates in each date and ranges of digits Y4, Y2, Y3 and Y1, where N format (e.g., palindrome date number 10022001 is the total number of palindrome dates in each as mentioned earlier). There are also palindrome category. Based on Table 1, in the DDMMYYYY date numbers which are only valid dates in one date format, a total of 335 palindrome dates exist date format but not the other. among all the four-digit years. For example, palindrome date number 21022012 In the MMDDY1Y2Y3Y4 date format, palindrome is a valid date number in the DDMMYYYY dates represented by Y4Y3Y2Y1Y1Y2Y3Y4 can date format and represents 21 February 2012. only occur in years ending with digit Y4 equal However, 21022012 is not a valid date number in to either zero or one since the month number the MMDDYYYY format since its month number cannot exceed 12. In the case when Y4 = 1, the 21 exceeds 12. tenth digit Y3 of the year number cannot exceed Palindrome dates in the two. In addition, the second digit Y2 of the year number of the palindrome date must be less than second millennium four since the day number cannot exceed 31. In the DDMMYYYY date format, a total of Furthermore, if Y1 > 1, then, Y2 < 3. That is, for 61 palindrome dates occurred in the second the MMDDY1Y2Y3Y4 date format, palindrome millennium (years 1001 to 2000), 31 in the 11th dates in the second millennium (years 1001 to century (all in the month of January) and 30 in 2000) can occur only in the first four centuries of the 12th century (all in November), with the last each millennium. On the other hand, palindrome one being 29 November 1192 (29111192). No other dates between the third and tenth millenniums palindrome dates occurred between the 13th and (2001 to 10000) only occur in the first three 20th centuries, more than 800 years. centuries of each millennium. Also, between In the MMDDYYYY date format, a total of 43 years 1000 and 10000, palindrome dates in each palindrome dates occurred during the second century all fall on the same day of the month, 6

millennium, split as 12, 12, 12 and 7 among the 11th, 12th, 13th and 14th centuries respectively. The 11th century palindrome dates all occurred on the first day of the month, the 12th century ones all on the 11th day, 13th century ones all on the 21st day of the month, and the 14th century ones all on the 31st day of the month. The last palindrome date of the second millennium occurred on August 31, 1380 (08311380) and no other palindrome dates existed between the 15th and 20th centuries, over 600 years.

all in the month of December, to occur during the 22nd century and no more after that until the end of the third millennium. In the MMDDYYYY format, 12 more palindrome dates (all being on the 12th day of the month) are to occur in the 22nd century followed by 12 more (all on day 22 of the month) in the 23rd, and no more afterwards until year 3001. In addition, there is a second palindrome date to occur in this millennium, 12122121, which is not only common to both date formats but also represents the same actual day in each format, which is 12 December 2121. In the DDMMYYYY date format, starting with the fourth millennium there will be 31 more palindrome dates (all in the month of March) to occur in the 31st century, 30 (all in April) to occur in the 41st, 31 (all in May) to occur in the 51st, 30 (all in June) to occur in the 61st, 31 (all in July) to occur in the 71st, 31 (all in August) to occur in the 81st, and 30 (all in September) to occur in the 91st centuries, the last one being 29 September 9092 (29099092). No other palindrome dates will exist in all the other

Also, interestingly enough, among all the palindrome date numbers in the second millennium common to both date formats, only two correspond to the same actual dates: 1 January 1010 (01011010) and 11 November 1111 (11111111).

Palindrome dates in the 21st century The 21st century has 29 palindrome dates in the DDMMYYYY versus only 12 in the MMDDYYYY date formats. The first palindrome date in this century in each date format was 10022001, in 2001. The second palindrome date of this century in the DDMMYYYY date format was 20022002 representing 20 February 2002. The third palindrome date in the DDMMYYYY date format, which also happens to be the second palindrome date in this century in the MMDDYYYY format, is 01022010. This date number represents 1 February 2010 in the DDMMYYYY date format versus January 2, 2010 in the MMDDYYYY format. The first twelve palindrome dates of this century in both date formats are provided in Table 3. Note that in this century, palindrome dates in the DDMMYYYY date format all occur in the month of February. On the other hand, in the MMDDYYYY date format, palindrome dates of this century all fall on the second day of the month. The last palindrome date of this century will be 29 February 2092 (29022092) in the DDMMYYYY date format versus September 2, 2090 (09022090) in the MMDDYYYY format. Also, there is one common palindrome date to occur in this century, 02022020, which corresponds to the same actual date in each format, that being 2 February 2020.

Number 1 2 3 4 5 6 7 8 9 10 11 12

Palindrome dates after the 21st century After this century, in the DDMMYYYY date format there will be 31 more palindrome dates,

MMDDYYYY 10022001 October 2, 2001 01022010 January 2, 2010 11022011 November 2, 2011 02022020 February 2, 2020 12022021 December 2, 2021 03022030 March 2, 2030 04022040 April 2, 2040 05022050 May 2, 2050 06022060 June 2, 2060 07022070 July 2, 2070 08022080 August 2,2080 09022090 September 2, 2090

DDMMYYYY 10022001 10 February 2001 20022002 20 February 2002 01022010 1 February 2010 11022011 11 February 2011 21022012 21 February 2012 02022020 2 February 2020 12022021 12 February 2021 22022022 22 February 2022 03022030 3 February 2030 13022031 13 February 2031 23022032 23 February 2032 04022040 4 February 2040

Table 3.

7

The first twelve palindrome dates of the 21st century in each date format.

centuries that fall between the fourth and tenth millenniums (year 3001 to 10000).

date in four- digit years in the MMDDYYYY date format will be 09299290, which is September 29, 9290 and after this one, no more palindrome dates will occur until the next one which is October 10, 10101 (101010101) to occur in year 10101.

The number of palindrome dates to occur in the MMDDYYYY date format starting with the fourth millennium up to the tenth is 36 per each millennium, distributed as 12, 12 and 12 between the first three centuries of each millennium. The 12 palindrome dates in each century all fall on the same day of the month, which equals the reverse of the first two digits of the year number. For example, there will be 36 palindrome dates in the 6th millennium and these 36 palindrome dates will be distributed as 12, 12 and 12 between the 51st, 52nd, and 53rd centuries respectively. The 12 in the 51st century will all happen on the fifth day of the month, the 12 in the 52nd on day 15 of the month, and the 12 in the 53rd are all going to be on the 25th day of the month. No further palindrome dates will exist in the other centuries (54th to 60th) of the 6th millennium. The last palindrome

Lastly, between the fourth and tenth millenniums, there is one common palindrome date that exists in each millennium that corresponds to the same actual date in each date format. For example, common palindrome date number 05055050 to occur during the first century of the 6th millennium represents 5 May 5050 in each date format.

References: 1. M. Gardner, The Colossal Book of Mathematics, Chapter 3, W. W. Norton & Company, 2001. 2. J. N. Friend, Numbers: Fun and Facts, Chapter VIII, Charles Scribner's Sons, 1954.

Primes from Fractions Alex P. Lamoureux and Michael P. Lamoureux decimal gives you 1," he points out, \which is not a prime. Next one is 16, also not a prime. And after that is 166, 1666, 16666, none of which are primes. So what are you talking about?"

Alex Lamoureux is a senior at Queen Elizabeth School in Calgary. He enjoys reading, fencing, and video games – in addition to his school work. Michael Lamoureux is a professor of mathematics at the University of Calgary, He does research in analysis and its applications to geophysics and signal processing, and teaches courses from calculus to graduate research seminars.

\No, Dad, you have to round up the numbers. So you start with 1.666 . . . which rounds up to 2, which is a prime. Next is 16.666 . . ., which rounds up to 17, which is also a prime. After that is is 166.666 . . ., which rounds to 167, also prime, and after that is 1667, which I'm pretty sure is a prime too."

Coming home \Hey Dad," announced Alex one day, arriving home from school. \There is something weird about the fraction one-sixth. See, if you write it out in decimal form,

\You're right Alex, all of those are primes. But the next one, 16667, is that prime?" \I don't know, but let's check. Of course two doesn't divide it, since it is not even. Three does not divide it since the digits don't add to a multiple of 3. Five doesn't divide it since it doesn't end in 0 or 5. What about seven?"

1/6 = .16666 . . . and start moving the decimal place, you get primes!" \That's nuts," said his father, Michael. \The first 8

\Yes, what about seven? With our calculator, or even in our head, we can check that 16667/7 = 2381. So that one is not prime." \Still, Dad, this fraction one-sixth is doing pretty good. See, by comparison, if you look at one-half, in decimal it is

3 7 31 127

= = = =

11(binary) 111(binary) 11111(binary) 1111111(binary)

and even the huge prime number 243112609 - 1 = 111 . . . (binary) with 43112609 ones, which was discovered in 2008.

1/2 = .5000 ...,

So perhaps our one-sixth prime generator can also produce big primes for us. How can we tell?

so shifting decimals gives you 5 (a prime), then the numbers 50, 500, 5000, etc., none of which are prime."

Testing for primes

\Same thing with one-third, in decimal

To find out experimentally what is going on, we can use a computer program that can look at a big integer and decide whether it is prime. There are lots of tools out there: MAPLETM, MathematicaTM, GIMPS, among others. Since the second author is a mathematician, he has access to lots of these tools. So we just use any one of them.

1/3 = .3333 . . . , so shifting gives you 3 (a prime), then the numbers 33, 333, 3333, etc., also none of which are prime. The fractions one-fourth, and one-fifth also give only one prime each." \So the one-sixth is pretty special. Why is that, Dad?"

Mathematica is a useful mathematical tool, with plenty of commands to do all sorts of mathematical calculations quickly and painlessly. It has a very simple command, PrimeQ[n], which tests whether an integer n is prime or not. Mathematica is also smart enough to keep track of all the digits of a very long number. Another nice command, FactorInteger[n] will compute the prime factors of n, if we are interested in those.

Why is that? Why indeed? This is a question for a number theorist, and neither of us are such. But it still is interesting to think about. The pattern of the numbers generated by one-sixth are pretty special. Except for the first number 2, the integers we get are of the form of a one, followed by many sixes, and ending in seven. That is, they look like this:

So, for instance, we type in PrimeQ[16667] and discover the number 16667 is not a prime. Type in PrimeQ[166667] and discover 166667 is actually a prime.

1666 ... 6667. So these are always odd numbers, which is good, as they will not be divisible by 2. The digits never add up to a multiple of 3 (since the 1+7 is not a multiple of 3, while the sixes all are), so this number is not divisible by 3. And of course it is not divisible by 5. So at least we have a few factors that \can't" happen.

A little \FOR" loop can be used to test a bunch of numbers: For[n=0, n<101, If[PrimeQ[Round[(1/6)*10^n]], Print[n]]; n++]. We can make the code a bit nicer by typing out both the prime and its index n:

In fact, this pattern of repeating digits might remind us of Mersenne primes, those primes of the form n = 2k - 1. In binary notation, where we use only the digits 0 and 1, these primes are in the form

For[n=0, n<101, If[PrimeQ[x = Round[(1/6)*10^n]], Print[x,n]]; n++]. With this little piece of code, we can find the prime numbers hidden in the decimal expansion of any fraction.

n=111 . . . 111(binary),with exactly k repeats of 1 Of course, not all numbers binary numbers of this form are prime, but many are, including 9

Some fractions We start with the fraction 1/6 = .16666 . . ., and see how many primes appear in the first 100 digits. We test for number of digits n up to 100, and get the following twelve primes: 2, 17, 167, 1667, 166667, 1666666667, 166666666667, 166666666666667, 166666666666666666666666666666667, 1666666666666666666666666666666666666666666666666666667, 16666666666666666666666666666666666666666666666666666667, 1666666666666666666666666666666666666666666666666666666666667,

a ratio of two integers. Everyone knows about the number pi, which is irrational. Its decimal expansion starts out as π = 3.1415926 . . . Here, in the first 100 digits, we see only four primes:

In fact, if we are patient (and have a fast enough computer), we can use Mathematica to find more primes of this form. Up to one-thousand digits, we find primes of the form 1666 . . . 6667 for digit lengths of n = 154, 201, 462, 570, 841, and 848. There may be more!

3, 31, 314159 314159265359

Alex pointed out that one-sixth seems really special. So let's use this numerical test to compare what we get with other fractions.

There are only two primes that appear in the first 100 digits of the expansion.

n=4 n=7 n = 10 n = 16 n = 25

3, n=1 2718281828459045235360287471352662 4977572470936999595749669676277240 76630353547594571, n=85 Wow, that's not a lot of primes!

For the fraction 1/9 = .1111 . . . we only get get three primes in the first one hundred digits: 11, 1111111111111111111, 11111111111111111111111,

n=1 n=2 n=6 n = 12

Another prime that we learn about in calculus is the exponential number e = 2.718 ....

To start, let's try the fraction 1/7 = .142857 . ... We get the following five primes, with digit length less than 100: 1429, 1428571, 1428571429, 1428571428571429, 1428571428571428571428571,

n=1 n=2 n=3 n=4 n=6 n = 10 n = 12 n = 15 n = 33 n = 55 n = 56 n = 61

Conclusion

n=2 n = 19 n = 23

Well, one-sixth is pretty special. We get a dozen primes in its first one hundred digits, which is a lot more than our other test cases! We have no idea why. Michael thinks Alex should experiment with more fractions, and more digits, to see what is going on. Or maybe some of the readers of this article might be inspired to look into why we get primes this way.

So again, it looks like one-sixth is pretty special.

Irrational numbers Not all number are fractions, some of them are irrational - numbers that cannot be written as 10

Reasoning from the Specific to the General Bill Russell concept such as odd x odd = odd. The following is a more advanced example of this principle.

Bill Russell teaches math at James Bowie High School in Austin, Texas, where he's worked for the past 22 years.

Problem A (Take 1)

Solving a math problem that involves numbers can often open the door to discovering a more general mathematical concept. Much of mathematics is about relationships, and once these relationships are recognized, the next logical step is to try to extend them into more profound revelations.

A total of 240 meters of fencing is to be used to enclose a rectangular region and divide it into 15 smaller rectangular regions (see Figure 1). Find the values of x and y that will maximize the total enclosed area. Solution: From the picture, 240 = 4x + 6y, or y = - x + 40. The goal is to maximize the area.

For example, while practicing multiplication tables, an elementary student might observe that the product of two odd numbers always appears to be an odd number. However, since there are infinitely many 2-number combinations of odd numbers, it is not possible to check each pairing and make sure that the product is odd. However, a first-year algebra student has the tools necessary to prove this relationship is always true.

(

)

A(x) = xy = x - x = 40

(2)

We observe that the function A(x) is quadratic and therefore its graph has a vertical line of symmetry midway between its zeros. Since equation (2) above is in factored form, one can easily finds its zeros by setting each factor equal to zero and solving. This reveals that the function has zeros at x = 0 and x = 60. Furthermore, since the leading coefficient of (2) is negative, the function will take on a maximum value at the vertex, which, as symmetry dictates, is located midway between the zeros at x = 30. This gives a corresponding width of y = - (30) + 40 = 20 ft. Thus, the maximum area enclosed is 20 m x 30 m = 600 m2, a correct but less than provocative outcome.

Any odd number can be represented as 2n + 1, where n is some integer. A second odd number, then, can be represented as 2m + 1 for some (possibly different) integer m. The product of these two integers, then, is (2m + 1)(2n + 1)=4mn + 2m + 2n +1 =2(2mn + m + n)+1 This last expression represents one more than an even integer, and is therefore an odd integer, thus proving that the product of 2 odd integers is always odd. This very simple example shows how a specific numerical problem such as 3 x 5 = 15 can be extended to arrive at a more general mathematical Figure 1 Specific example of the fencing problem

11

Of much greater interest is to observe that in this optimum situation, the total horizontal fencing used is 4 x 30 = 120 m, and the total vertical fencing used is 6 x 20 = 120 m! The question that should arise is whether this is a coincidence | an accident of the specific numbers chosen | or whether this is a numerical example of some greater law of rectangular regions. Moreover, if

true, it is necessary to solve the related general problem by replacing the numbers with v mind, the original problem can now be restated. ue, true,it itis isnecessary necessarytotosolve solvethe therelated relatedgeneral generalproblem problembybyreplacing replacingthe thenumbers numberswith withvariables. variables.With Withthis thisinin Problem A (Take 2): A total of T meters of fencing is to be used to enclose a rectangular mind, mind,the the original original problem problem can can now now be be restated. restated. this is a universal truth, then how would one prove a general theorem that would encompass into smaller rectangular regions using m horizontal pieces, each of length x and n vertica prove this? many specific problems, we restated the length y (see Figure 2). Show thatsuch when the enclosed area isdivide maximized roblem ProblemA A(Take (Take2):2):A Atotal totalofofT Tmeters meters ofoffencing fencing is is totobebeused used totoenclose enclose a arectangular rectangular region region and and divideit it that the total am original in more general terms and tried equalseach the total amount of nhorizontal fencing used. nto intosmaller smaller rectangular rectangular regions regions using using mfencing mhorizontal horizontal pieces, pieces, each ofof length length x and xproblem and nvertical vertical pieces, pieces, each eachof of It would be easy enough to repeat theused computations to prove it.the Although our ofproof depended mostly ength lengthy (see yusing (seeFigure Figure Show Showthat that when when the the enclosed areais ismaximized maximized that that the total totalamount amount ofvertical vertical other2).2). numbers, but even ifenclosed similararea results on some simple algebra, we supplemented that T mx encing fencingused usedequals equalsthe thetotal totalamount amountofofhorizontal horizontalfencing fencingused. used. , and the goal is to m In this version of the problem, T = mx + ny or y = were observed the only thing Solution: that would be proven

algebra with words that explainn to the reader is that the rule holds for those specific numbers T T mxmx what we were doing. In the end, we were rewarded T T= =mx mx+ +nyny y= =is orory it , and , andthe thegoal goalis istotomaximize maximize olution: Solution: InInthis thisversion version ofofthe the problem, problem, chosen. To prove that it always holds true, n nwith a remarkably simple powerful result T mxyet necessary to solve the related general problem by ( ) A x = x . that precludes the need forn repeating those same replacing the numbers with variables. With this in calculations in the future. For example, we now T T mx mx ) =xrestated. A(Ax()x=be x mind, the original problem can now . . know that if 300 meters of fencing is(3)(3) available, m nn We observe that A(x ) is a then quadratic function and that the leading coefficient the optimum enclosure uses 150 metersof is neg n Problem A fencing in each direction. m m before that the function takes on a maximum value at the vertex. as This (Take 2) ) isa aquadratic is isnegative, We Weobserve observethat thatA(Ax()x is quadraticfunction function and andthat thatthe theleading leading coefficient coefficient negative, indicating indicating as function has zeros n n T T Problem A, these proving case the wastotal hori x =A total = atat . Thus, , andofthe vertex lies In midway between zerosthe at xgeneral Tthe meters efore beforethat thatthe thefunction functiontakes takesonona amaximum maximum valueatat the vertex. vertex.This This function function has has zeros zeros at at and and x x = = 0 0 mvalue 2 m of fencing is to be T Trelatively easy because we were able to use the TT Tthese T zeros exact same methodology for the general to x= = , and , andthe thevertex vertexlies liesmidway midwaybetween between these atatx x= = . .Thus, Thus, the the total total horizontal horizontal fencing fencing used usedis isproblem enclose mused , precisely one-half of the total fencing, thereby completing this proof. = zeros mm 2 m 2 m that we used in the specific numerical problem. The 2 a2m rectangular T T T T only difference was that in the general problem we region and divide mm = , precisely , precisely one-half one-halfofofthe thetotal total fencing, fencing, thereby thereby completing completingthis thisproof. proof. Figure = 2 used variables in the rathera than numbers. Let's take a moment and review what we just did.formula After solving routine specific numeri 2 m 2 m 2 2 it into smaller General case of the fencing observed an interesting result. Attempting to prove a general theorem that Although this is sometimes possible, solvingwould the encom rectangular regions problem specific problems, we restated the original problem in more general and tried to pr et's Let'stake take a amoment moment and and review review what what we we just justdid. did. After After solving solving a aroutine routine specific specific numerical numerical problem, problem, we we terms general problem will sometimes require visualizing elated general problem by replacing the numbers variables. With this in using mwith horizontal proof depended mostly on some simple algebra, we supplemented that algebra wor bserved observed an an interesting interesting result. result. Attempting Attempting to to prove prove a a general general theorem theorem that that would would encompass encompass many many such such w be restated. a completely different approach to the the numbers problem,with aswith true, itnreader isvertical necessary towith solve the related general problem by replacing variabl he related general problem by replacing the numbers variables. With this in pieces, each of length x and pieces, each what we were doing. In the end, we were rewarded with a remarkably simple yet pecific specificproblems, problems,we werestated restatedthe theoriginal originalproblem problemininmore moregeneral general terms terms and and tried triedtoto prove proveit.it.Although Althoughour our the next example illustrates. mind, the original problem can now be restated. nroof now be restated. of length y (see Figure 2). Show that when the precludes the need for repeating those same calculations in the future. For example, we proofdepended depended mostly mostly onon some some simple algebra, algebra, we wesupplemented supplemented that algebra algebra with withwords words that that explaintotothe the meters ofsolve fencing isrelated to be used tosimple enclose aby rectangular region andthat divide it ssary to the general problem replacing the numbers with variables. With this inexplain meters of fencing is available, then the optimum enclosure uses 150 meters of fencing in enclosed area isrestated. maximized that the amount eader reader what we wewere were doing. doing. InInthe end, end, we were were rewarded rewarded with withaeach aremarkably remarkably simple simple yet yet powerful result resultthat that sing mwhat horizontal pieces, each ofthe length x we and n total vertical ofseveral timesis inpowerful the secondary nal problem can now be Problem A (Take 2): Apieces, totalregion of T At meters of fencing to be used to enclosecurriculum, a rectangular regio frecludes T meters of fencing is to be used to enclose a rectangular and divide it the need needfor for repeating repeating those thosesame same calculations calculations ininthe future. For Forexample, example, we wenow know know that that if if300 300the areas ofthe vertical fencing equals thethe total amount ofthe tprecludes when the enclosed area is used maximized that total amount of future. vertical students are exposed tonow aeach unit on finding into smaller regions using m case horizontal pieces, ofbecause length xwe andwere n vertical ns using m horizontal pieces, then each ofthe length xrectangular and n vertical pieces, each of In Problem A, proving the general was relatively easy able topiece use meters meters of of fencing fencing is is available, available, then the optimum optimum enclosure enclosure uses uses 150 150 meters meters of of fencing fencing in in each each direction. direction. unt2): of horizontal fencing used. fencing Ahorizontal total T meters ofused. fencing is to be that used to total enclose a rectangular region and divide it is of triangles. The following a problem y (see Figure 2). Show that the enclosed area is maximized thatthat the they total amount wke that when theofenclosed area is length maximized the amount of when vertical

methodology for the general problem that we used in the specific numerical problem. Th ctangular regions using m horizontal pieces, each of length x and nmight vertical pieces, each of used. fencing used equals the total amount of encounter horizontal fencing amount of horizontal fencing used. such aexact unit. that in the general problem we used variables inuse the formula rather nigure InProblem Problem A,A,proving proving the thegeneral general case case was was relatively relatively easy easy because because we we were were able able tovertical toin use the the exact same same than numbers. Altho T mx 2).Solution: Show thatIn when the enclosed area is maximized that the total amount of this version of the problem, possible, solving the general problem will sometimes require visualizing T = mx + ny y = or , and the goal is to maximize blem, methodology methodology forfor the thegeneral general problem problem that thatwe we used usedininthe thespecific specificnumerical numericalproblem. problem.The Theonly onlydifference differencewas was a completely diff quals the total amount of horizontal fencing used. n T mx T mx problem, as the next example illustrates. Problem hat ininthe the general problem used variables variables in inthe the formula formula rather rather than thannumbers. numbers. Although this thisis issometimes sometimes T =B mx(Take +Although ny y= or 1) , and the goal is to maxim Solution: Inthe this version the problem, Tgeneral = + ny ,, and and the goal to T mx = mx +problem ny or ywe =used or we goal is toisof maximize ethat problem, n ossible, possible,solving solvingthe thegeneral generalproblem problem will sometimes sometimes require require visualizing visualizing a a completely completely different different approach approach totothe the n will � � T mx Find the area of a triangle with vertices (2,on 1),finding th maximize example mx T At in the secondary curriculum, students are exposed to aatunit =illustrates. Texample Tseveral −=mx times, and mx + ny y or the goal is to maximize is version of the problem, roblem, problem, asA as the next next illustrates. (x ) = x . (3) (3) A(x) = xThe following and (4, 6). n is a problem (8, that3) they might encounter n n T mx T mxin such a unit. ( ) Aon xfinding = x the ) A(inxin =the xsecondary . ofoftriangles. .curriculum, (3) tAtseveral several times times the secondary curriculum, students students are are exposed exposed to to a a unit unit on finding theareas areas triangles. We observe thatn A(x) is aProblem quadratic function Solution: Most of my students approach this(4, 6). n vertices B (Take 1): Find the area of a triangle with at (2, 1), (8, 3) and T mx encounter he Thefollowing followingis isa aproblem problemthat thatthey theymight might encounter ininsuch sucha aunit. unit. m ( ) A x = x . (3) problem tic function that leadingcoefficient coefficient as by using Heron's Formula, which states negative,indicating andand that thetheleading n n isisnegative, Solution: that the area of6). a triangle with Heron's sides ofFormula, lengths m Most of my students approach this problem by using which s m roblem ProblemB B(Take (Take1):1):Find Findthe thearea area ofof a atriangle triangle with atat(2,(2,1),1),(8, (8, 3)3)and and(4, (4, 6). (xisvertices ) negative, is axquadratic function is negative, We observe that Avertices with adratic function that the leading coefficient indicating as and that the leading coefficient indicating before that the function takes on a maximum valueand at as the vertex. This function has zeros at and at = 0 a triangle with b, andc cisis A = s (s a )(s b )(s c ),, where n s sis the se a, ba, , and where n sides of lengths m T olution: Solution: Most Most of of my my students students approach approach this this problem problem by by using using Heron's Heron's Formula, Formula, which which states states that that the the area area of of x =is is a quadratic function and that the leading coefficient isused negative, indicating as A x )maximum before that the function takes on a 0maximum at the This functiondistance has zeros at x on a(maximum value at This function has zeros at and atis tovalue the semiperimeter of vertex. triangle. value the This function has xat=vertex. yat between these zeros atthe .vertex. Thus, the total horizontal fencing triangle. The distance formula is used find that a, b,the and c are 5, The 40 , and 29 . Sub n 2m ( ( )( )( )( )( ) ) , where , where s is s is the the semiperimeter semiperimeter of of the the A A = = s s s a a s s b b s s c c atriangle trianglewith withsides sidesofoflengths lengthsa, a,b,b,and and c is c is Ta, b, and TT formula is used to find c total are radicals 5, Heron's Formula (and has using a calculator, since thexthat calculations with = and the lies midway between . Thus, thethose horizontag function takesat onxa = 0 maximum vertex. Thishorizontal function zeros at these and atat xis= 0 zeros zeros andatatxvalue ,,the and the vertex lies =x = at.into dway between these zeros Thus, thevertex total fencing used m 2 m 2 m the value of the area as exactly 13. , and . Substituting these values into , and , and . . Substituting Substituting these these values values iangle. triangle. The The distance distance formula formula is is used used to to find find that that a, a, b, b, and and c are c are 5, 5, 40 40 29 29 f the total fencing, thereby completing this proof. T xsince = the einto vertex liesFormula midway between these .. calculations Thus, the total horizontal fencing used is a midway between these zeros Thus, T atatTsince zeros nto Heron's Heron's Formula (and (andusing using a acalculator, the calculations with with those those radicals radicals get get pretty pretty messy) messy) gives givessince the Heron's Formula (and using calculator, 2mthe mcalculator, =thispoint, , proof. precisely one-half of the total fencing, thereby completing thisthough proof. two of the s alf of the total fencing, thereby completing At this observant student should marvel at the result. Even � � he thevalue valueofofthe thearea areaasasexactly exactly13. 13. 2m 2 calculations with those radicals get pretty messy) T these numbers are plugged into a formula that requires T are irrational and taking a hat we just did. solving afencing routinethereby specific numerical problem, we the value of the area as exactly 13. =this m values the totalAfter horizontal used iscompleting , proof. precisely one-half of the total fencing, gives 2m 2many answer came out aresult. nice neat integer. This raises the question of whether or not this will tempting to prove a general theorem that would encompass such tAtthis thispoint, point, the theobservant observant student student should should marvel marvel at at the the result. Even Even though though two two of of the the sides sides had had lengths lengths that that Let's take aspecific moment and review what we just did. After solving a routine specific numerical pro ew whatprecisely we just did. After solving a routine numerical problem, wecoordinates you only use for the ofasquare the vertices. Again, youshould could try repe ofnumbers the total fencing, thereby original problem inone-half more general terms and tried tointeger prove it.values Although our re are irrational irrational values values and andthese these numbers are are plugged plugged into intoaresult. aformula formula that that requires requires taking taking square root, root,the the At this point, the observant student observed an interesting Attempting prove a ageneral theorem that would encompass . Attempting to prove a general theorem that would encompass many such to imple algebra, we supplemented thatAfter algebra with words that explain to the nswer answer came came out out a what anice nice neat neat integer. integer. This This raises raises the question question of ofwhether whether orat or not not this this will always always bebe the the case case if if tried completing this proof. oment and review just did. solving athe routine specific numerical problem, we marvel the result. Even though two of the specific problems, we restated original problem inwill more general terms and to prove it. d the original problem inwe more general terms and tried to prove it.the Although our eou end, we were rewarded with aprove remarkably simple yet powerful result that you only only use use integer integer values values forfor the coordinates coordinates oftheorem ofthe the vertices. vertices. Again, Again, you you could couldtry try repeating repeating the thecalculations calculations teresting result. Attempting tothe athat general thatsome would encompass many such are proof depended mostly on simple algebra, we supplemented that algebra with words tha me simple algebra, we supplemented algebra with words that explain to the sides had lengths that irrational values and Let's take moment and review what wewe just did. hose same calculations in the future. example, now know that 300 ms, restated thearewarded original problem inFor more general terms and tried toif prove it. Although our reader what we were doing. In the end, we were rewarded with a remarkably simple yet powe n thewe end, we were with a remarkably simple yet powerful result that these numbers are plugged into a formula that nd the optimum enclosure uses 150 meters of fencing in each direction. After solving a routine specific numerical problem, mostly on some simple algebra, we supplemented that algebra with words that explain to the precludes the need for repeating those same calculations in the future. For example, ing those same calculations in the future. For example, we now know that iftaking 300 a square root, the answer camewe now requires ethen werethe doing. In the end, we were rewarded with a remarkably simple yet powerful result that weoptimum observed an interesting result. Attempting to meters of fencing is available, then the optimum enclosure uses 150 meters of fencing in each enclosure uses 150 meters of fencing in each direction. case for was relativelythose easy same because we were in able use the exact same we now know that if 300 need repeating calculations thetofuture. For example, em that we usedthen in the numerical problem. The only difference was 12exact ng is available, thespecific optimum uses 150 fencing in each direction. Inenclosure Problem A, proving the case was relatively easy because we were able to use the e neral case was relatively easy because we were able tometers usegeneral theof same ed variables in the formula rather than numbers. Although this is sometimes methodology the general problem that wewas used in the specific numerical problem. The only problem that we used in the specific numerical for problem. The only difference

6

A1

5

2 look like Figure 3. We want to tryAto visualize the out a nice neat integer. This raises the question 4 area using integers rather than radicals. After a of whether or not this will always be the case if 3 A bit of thought, you might envision the triangle you only use integer values for the coordinates of 2 the vertices. Again, you could try repeating the enclosed inside of a 6 x 5 rectangle, as shown A3 1 deleted for calculations using different points. Eventually, in Figure 4 (the gridlines have been clarity.) It you is now clear area Athe you would find that sometimesusing the different area is points. not is simply Eventually, would findthat that the sometimes area is not an 1 2 3 4 5 6 7 8 9 1 30 A A A , where each of the A ( j = 1, 2, 2 3 3) is changed to (7, 3),j the area becomes 1 multipleof (for (8, of an integer but rather is a multiple (forexample, if 1the point 3) is the area of a right triangle, Figure 2 3 Figure 4 which is half the Specific example of the product its that legs.More trying several youoffind the area always seems example, if the point (8, 3) is changed to (7,sets 3), of points, general picture of to be half of an inte triangle problem actually proved anything yet, you are now ready to state a hypothesis and try to prov the triangle problem the area becomes 10.5.) However, after trying example, then,towe haveFinished? My s further, stateour the numerical property that we will need prove. several sets of points, you find that see theif you areacan For problem appears below. Keep in1 mind that 1 there are 1 other correct ways of stating it. always seems to be half of an integer. Although For our numerical example, then, we have A = 30 (6 2 ) (3 4 ) (5 2 ) = 13 , as previously found you haven't actually proved anything yet,B you are2): Prove that 2if all of the2 vertices2of a triangle have integer coordi Problem (Take as why previously Moreover, though, it should though, it triangle's should the areafound. must always be half of an integer. Armed wit now ready to Moreover, state a hypothesis and trynow to be prove area isapparent always an integer. now be apparent why the area must always be new any insight, we aresee ready to proceed with our proof. it. Before reading further, if you can state of anthis integer. ArmedFormula, with this insight, We could tryhalf to prove using Heron's butnew it should quickly bec the property that we will need toSolution: prove. Finished? A proof is a connected series of statements intended to establish a proposition. As before, we ares (going we are ready to proceed with our proof. be messy. It would be necessary to show that the radicand sa My statement of the general approach problem would appears have to use words in these statements, and we want to choose those words carefully so that our proof s below. Keep in mind that there are other correct perfect square, and since we would have radical expressions for s, a, b, and c, this A proof a connected series intended what we intend for it to say. With this in mind, let's is carefully analyze eachofofstatements these statements one at a lo t Consequently, we arewith motivated to find a different approach to the probl ways of stating Weit.must first get rid nightmare. of all numbers and replace them variables as we in our to establish a proposition. As did before, wefirst areproof, goingidentifyin

as the given information. This is easy enough. to have to use in numerical these statements, Start by drawing the original triangle (thewords one with vertices) onand graph pape we want to choose those words carefully so that Figure 3. We want to try to visualize the area using integers rather than radicals. A (a,ab )triangle and Z (e,our the says coordinates of the vertices triangle XYZ, where , Y (c, d ), have f ) be 1: Let Xof Prove that if Statement all of the vertices proof whatofwe for itofto With might envision the triangle enclosed inside a 6intend 5 rectangle, assay. shown in Figure 4 a , b , c , d , e and f are integers. integer coordinates, then twice the triangle's area It isthis mind, carefully analyze 30 each A1 ofA2these A3 , where deleted for clarity.) nowinclear thatlet's the area A is simply is always an integer. statements one at a time. We must first get rid of the area of a right triangle, which is half the product of its legs. Next we have to describe the rectangle enclosing the triangle, and this is actually a bit tricky. For all numbers and replace them with variables as we now, le the following: Solution: Weclaim could try to prove this using 6

Problem B (Take 2)

Heron's Formula, but it should quickly become 5 clear that this approach would be messy. It would be necessary to show that the radicand 4 s (s - a)(s - b)(s - c) is always a perfect square, 3 and since we would have radical expressions for s, 2 a, b, and c, this looks like an algebraic nightmare. 1 Consequently, we are motivated to find a different approach to the problem. 1

did in our first proof, identifying this as the given information. This is easy enough. A1

Statement 1

A2

A

Let X (a,b ), Y (c,d ) and Z (e,f ) be the vertices of triangle XYZ, where a, b, c, d, e, and f are integers.

A3

Next we4 have to describe the rectangle enclosing 1 2 3 5 6 7 8 9 the triangle, and this is actually a bit tricky. For 3 now, Figure let's claim the following: Figure 4

Start by drawing the original triangle (the one with numerical vertices) on graph paper. It should Specific example of the

triangle problem Statement

2

More general picture o the triangle problem

A rectangle with its to1 the 1 sides parallel 1

) (3so = 30 can(6be 2drawn 4 )that (X, 5 2 ) = 13 , For our numerical example, then, coordinate we have A axes

2 2 2 Y, and Z all lie on the rectangle and at Moreover, though, it should now be apparent why the area must always be half of an leastwith oneour of proof. the vertices of the triangle new insight, we are ready to proceed coincides with a vertex of the rectangle.

Figure 3

A proof is a connected series of statements intended to establish a proposition. As This isand actually a pretty strong (and not carefull have to use words in these statements, we want to choose those words statement andeach would what we intend for it to say. With completely this in mind, true!) let's carefully analyze of these s We must first get rid of all numbers replace them with variables as we did in our beand rather difficult to justify rigorously, as the given information. This is easy enough. but it seems obvious from our existing Figure 4

Specific example of the triangle problem

More general picture (c, d ),proceed and Z (e,with the coordinates the vertices of t X (a, bof), Ythe f ) be the Statement 1: Let assumptionof that it triangle problem

drawing, so for now we're going to

a, b, c, d , e and f are integers. 13

Next we have to describe the rectangle enclosing the triangle, and this is actually a b

2: A rectangle its sides to the coordinate axes can be drawn X, Y, Statement 2: A with rectangle withparallel its sides parallel to the coordinate axes canso bethat drawn soand thatZX,allY,lieand Z all lie exercise caution when making this claim. is true. Bear in mind that in a truly thorough tangle and at least one of the vertices of the triangle coincides with a vertex of the rectangle. on the rectangle and at least one of the vertices of the triangle coincides with a vertex of the rectangle.

proof, this statement would require considerable

Onebeof my more cynical college professors once ually a pretty strong (andstrong not completely true!) statement and would to justify justification. This is actually a pretty (and not completely true!) statement andrather woulddifficult be rather difficult to justify me the \without loss of , rigorously, but it seems from our existing drawing, for nowso we're goingwe're to proceed the words butobvious it seems obvious from our existingsodrawing, for informed now goingthat towith proceed with the nassumption that it Clearly, is true. Bear in mind that in a truly thorough proof, this statement would require considerable generality" are almost always followed by a areBear many different draw proof, that itthere is true. in mind that inways a trulytothorough this statement would require considerable n. statement that results in a loss of generality. the triangle and the rectangle based only on justification. While he likely overstated the extent of its

Statement 2. We can only deal with one such

ere are many ways to draw the triangle and the rectangle based onlybased on Statement 2. We Clearly, there different are many ways draw the and the rectangle on Statement misuse, his overallonly message to use 2. thisWestatement orientation at adifferent time, so weto make the triangle following eal with one such orientation at a time, so we make the following our next statement: can only deal with one such orientation at a time, so we make the following our next statement:

cautiously is worth remembering. For example, in our corner first problem you doand not want to claim, 3: Assume3:without loss of generality that vertexthat X is vertex in the lower of the rectangle Statement Assume without loss of generality X is inleft the lower left corner of the rectangle and Statement 3 \Assume WLOG that and m es on the vertical side to its right. 4, then represents this particular vertex Y lies on the vertical side toFigure its right. Figure 4, then represents this configuration. particular configuration. = 5 and n = 3." Assume without loss of generality that vertex X is There can be little doubt that this condition ,Effectively, we are claiming the proof the same regardless of limits which triangle vertex coincides with the generality of vertex the proof. By contrast, in welower arethat claiming thatproceeds the proof proceeds thevertex same regardless of which triangle coincides with in the left corner ofconditions the rectangle and angle vertex, as long as all of the of Statement 2 are met. The disclaimer in bold (often which rectangle vertex, as long as all of the conditions of Statement 2current are met.problem, The disclaimer in bold (often our we could list each possible Y lies thetime-saver, vertical side right.exercise Figure caution 4, ed "WLOG") is"WLOG") aon great but to oneits should when making thismaking claim. this claim. abbreviated is a great time-saver, but one should exercise caution when configuration implied by Statement 2 and work then represents this particular configuration. out the area for each, but are such an exercise would moreofcynical college professors once informed that theme words loss of generality" One my more cynical college professors onceme informed that "without the words "without loss of generality" are Effectively, we are claiming that the proof be pointless and repetitive. Hence,the in extent this case, the ways followed a statement that resultsthat in aresults loss ofingenerality. While he likely the extent almost alwaysbyfollowed by a statement a loss of generality. Whileoverstated he likely overstated proceeds the same of which triangle se, hismisuse, overall message to useregardless this to statement cautiously is worth remembering. Forseems example, our use of \WLOG" justified. of its his overall message use this statement cautiously is worth remembering. Forinexample, in our and ." There can be little doubt that em you do not want to claim, "Assume WLOG that vertex coincides with which rectangle vertex, as m = 5 n = 3 first problem you do not want to claim, "Assume WLOG that m = 5 and n = 3 ." There can be little doubt that The restweofcould the proof proceeds much as it would if ion the generality of the proof. By of contrast, our 2current problem, long as all the of the conditions Statement are thislimits condition limits generality of the proof. Byincontrast, in our current problem,list weeach couldpossible list each possible numbers were involved. ion implied by Statement 2 and work out the area for each, but such an exercise would be pointless met. The disclaimer in bold (often abbreviated configuration implied by Statement 2 and work out the area for each, but such an exercise would be pointless tive. Hence, in this case, the of the "WLOG" seems and repetitive. Hence, thisuse case, use of "WLOG" seems justified. \WLOG") is aingreat time-saver, but onejustified. should our next statement:

fThe the rest proof wouldas if numbers involved. Statement 4 as it much ofproceeds the proof much proceeds it would ifwere numbers were involved.

dimensions the rectangle, then, f - b). Also, (then, )a. )Also, 4: TheThe dimensions of theofrectangle, then, are c are a )are (c( - f (cab) x ( ( f b ). Also, Statement 4: The dimensions of the rectangle, A1 =

1 1 (e Aa1)(=f 1(be), aA)(2 f=1b()c, Ae2)(=f 1 (dc), eand (c Aa3)(=d 1 (bc). a )( A3d=), and d b ). )( f 2 2 2 2 2 2

(4)

(4)

(4)

(5)

(5)

(5)

(6)

(6)

(6)

Thus, Thus,

1 A = (c a )(Af =(bc)a )([(fe ba))(f1[(be)+a(c)(f e)(bf)+ (dc)+e()( c a )( f dd)+ (bc)] a )(d b )] 2 2 (c 2aA)(=f 2(bc)a[()(e f a )(b )f[(be)+a(c)(f e)(bf)+ (dc)+e()(c f a )(dd)+ (bc)]. a )(d b )]. and and 2 A = 2and

eed interpret algebraic (6) to explain why it establishes our proposition. We to now need tothis interpret thisstatement algebraic (6) to explain why establishes proposition. our proposition. We now need to interpret thisstatement algebraic statement (6) toitexplain whyour it establishes

, b,Since c, d , e and 5: Since a5: and the setand of integers is integers closed under the operations of a5, b,fcare , d , eintegers and f are integers the set of is closed under the operations of Statement Statement ubtraction, and multiplication, it follows that 2A must anmust integer. addition, subtraction, and multiplication, it follows thatbe2A be an integer.

6 Since a, b, c, d, e, and f are integers and the set of 6 integers is closed under the operations of addition, uInare not familiar with the use of the word "closed" in Statement 5, Figure 5 5 case you are not familiar with the use of the word "closed" in Statement 5, 5 ays that any time you add, subtract, or multiply two integers, the subtraction, and multiplication, it follows that 2A this just says that any time you add, subtract, or multiply two integers, the An obtuse triangle 4 4 always be an integer. closure properties justify the justify the must be an result will always be integer. anThese integer. These closure properties that does not conform 3 nconclusion that 2 A is that an integer. 3 2 A is an integer. to the conditions of In case you are not familiar with the use of the statement 2 2 2 ppear that we are finished with this proof, but have tidy up one word \closed" Statement 5, this thiswe just says that It might appear that we in are finished with proof, buttowe have to tidy up one 1 . loose For Statement 2 toyou be true, each vertex oforthe triangle must have must have 1 any For time add, subtract, multiply two end. Statement 2 to be true, each vertex of the triangle e coordinate that is either a maximum or a minimum value within at leastintegers, one coordinate that will is either a maximum or a minimum the result always be an integer. These value within 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 isifright or acute, but if XYZ is if XYZ is e. is always true if XYZ XYZ is right or acute, but theThat triangle. That is always true closure properties justify the conclusion that 2Athe is sides 1 1 sobtuse, only true if one of the sides of the triangle lies on one of it isinteger. only true if one of the sides of the triangle lies on one of theFigure sides 5 -- An obtuse an Figure 5 -- An obtuse

triangle that does not triangle that does not conform toconform the conditions of to the conditions of Statement Statement 2. 2.

It might appear that we are finished with this proof, but we have to tidy up one loose end. For 14

Statement 2 to be true, each vertex of the triangle must have at least one coordinate that is either a maximum or a minimum value within the triangle. That is always true if ΔXYZ is right or acute, but if ΔXYZ is obtuse, it is only true if one of the sides of the triangle lies on one of the sides of the rectangle. Again, this is not an easy statement to justify rigorously, but a couple of quick sketches should convince you that it is true. Thus, way back at the beginning, we need to split this proof into two separate cases.

Statement 0 Case 1 | ΔXYZ does not contain an obtuse angle. Now that the proof for non-obtuse triangles is complete, Statement 6 should begin the proof of the obtuse case. The picture for the obtuse case looks a bit different (see Figure 5), but otherwise the proof should proceed pretty much the same as the one that we just did. Try to prove the

obtuse case on your own. Choose your wording carefully, and always question whether or not your statements say what you intend for them to say. After writing your proof, have a friend read it and give you feedback about how clearly you made your case. The body of knowledge encompassed by mathematics is constantly growing. The impetus for additions to this body of knowledge is curiosity. New relationships are constantly being discovered because someone perceives an interesting result to a problem and starts wondering how and if this result can be generalized or extended. Although the examples in this article take you down mathematical paths that have already been trodden many times, they hopefully show you how such paths are found in the first place and have helped you develop some of the thinking skills required to experience mathematical adventures of your own. Happy exploring!

century. Here the book delivers with aplomb:

�This was the first time in the history of mathematics that someone had been bold enough to prove the existence of a mathematical solution without illustrating how it was to be constructed. Gord Hamilton The article Hilbert published in 1888 created an For example, to introduce a discussion of prime In the Autumn of 2010, uproar. The numbers, ultraconservative who selfthe author Kronecker, sacrificed authentic Dr. Hamilton opened doubted even the existence of irrational reflections in order to teach thenumbers mathematically www.MathPickle.com to provide practical help unprimed: such as thetosquare root of 2, dismissed the solution K-12 teachers. He is also without further Gordan himself, \I argument. knew that prime numbers were who those that a founding member of the wasofknown his be geniality and his generosityand the cannot divided except by themselves Game Artisans Canada for number one." P. 19 young, talented mathematicians, responded – Canada'stoward premier board game design group. angrily to Hilbert�s paper, saying, �This is the notnarrator ...But within a page, a question from mathematics, but theology.� As for Lindemann, Pythagorean Crimes is written by the Greek ac- shows that he is at a significantly higher level: described his former pupil�s method as Tefcros Michaelides. In a he compact \da \...Isn't that Gauss' conjecture? If I'm not Crimesademic, is written by the Greek unheimlich profane. he Others, however, such as Vinci Code" style, the reader is led helter skel- � mistaken, worked out, but was unable to os Michaelides. In a compact ter through historical, mathematical, and artistic Arthur Cayleyprove, in Cambridge andnumbers Klein are studied how many prime smaller than � style, the reader is led helter events of the early 1900s - all to set up an exciting the proof in detail and, havingP.initially believed it a given number." 20 murder trial. historical, mathematical, and to be impossible, ended up congratulating Hilbert The motivation to expand the readership to the early 1900s � the all book to set up thoroughly an warmly. Hilbert became fanatical By the end of I was satisfied, include thosea that need anproponent explanation of of prime trial. but for the first part of the book, whereof the proofs existence. �Inside this hall,� he would numbers is understandable, but it unfortunately majority of the mathematical history is found, I of the book I was thoroughly often say in his undermines believability the narrator. lectures,the�there is atof least one was distracted: The self-reflections of the narrator, the first part of the book, where student who has more hair on his head than any The believability of the narrator doesn't matter Michael Ingerinos, and the conversations with his f thefriend, mathematical historysometimes is other. We don�t know that is, nor is if you are who reading for person the purpose of getting a Stephanos Kandartzis, felt forced. tracted: The self-reflections of the there any practical way of finding out. But this 15 that he doesn�t exist!�� P.59 el Igerinos, and the conversations doesn�t mean

Book Review

foundations of all these subjects are to be found in mathematics. What you see is an apotheosis of algebra, trigonometry, infinitesimal calculus. It is the tower of wisdom! I'm surprised you don't see it that way."

glimpse of mathematicians at the turn of the last century. Here the book delivers with aplomb: \This was the first time in the history of mathematics that someone had been bold enough to prove the existence of a mathematical solution without illustrating how it was to be constructed. The article Hilbert published in 1888 caused an uproar. The ultraconservative Kronecker, who doubted even the existence of irrational numbers such as the square root of 2, dismissed the solution without further argument. Gordan himself who was known for his geniality and his generosity toward young, talented mathematicians, responded angrily to the paper, saying, \This is not mathematics, but theology." As Lindemann, he described his former pupil' method as unheimlich - profane. Others, however, such as Arthur Cayley in Cambridge and Klein studied the proof in detail and, having initially believed it to be impossible, ended up congratulating Hilbert warmly. Hilbert became a fanatical proponent of proofs of existence. \Inside this hall," he would often say in his lectures, \there is at least one student who has more hair on his head than any other. We don't know who that person is, nor is there any practical way of finding our. But this doesn't mean that he doesn't exist" P. 59

[Michael:] \It's the tower of Hubris," I said, annoyed \The only thing it symbolizes is human arrogance. Mathematics can construct bridges, houses, trains, and ships. It doesn't need such contraptions to justify itself." P. 84 That's an entertaining argument to eavesdrop. I also like that the narrator's observations are sometimes flawed. For example, he totally misses the melancholic mood in Picasso's paintings of harlequins and pierrots - declaring them \cheerful stuff" P. 149. However, despite my enjoyment of the book, I am uneasy about recommending it for grade 12 students. First, the book will not appeal to every top mathematics student because of its impressive breadth of subject matters - the reader should be interested not only in mathematics but also in history and art. Without these interests, a person is liable to get irritated by the interjections - such as the one page summary of the Dreyfus affair in the middle of a description of the delegates at a mathematics conference:

That is the kind of paragraph which made the book such a satisfying read. My reservations about the development of Stefanos and Michael are also offset by insightful psychological sketches of other people:

\French society in 1900 was being torn apart by a scandal revolving around Alfred Dreyfus, a young lieutenant convicted six years earlier of being a spy. After a farcical court-martial, the military had sentenced him to hard labour for life and sent him to Devil's Island, a penal colony off the coast of French Guiana. The royalists, backed by the Catholic Church, seized the opportunity of the conviction to attack the republican constitution. The fact that Dreyfus was a Jew contributed to the rekindling of anti-Semitic feeling among certain sections of French society." P. 20

\... The top brass at [military] headquarters considered me and the three other mathematicians who worked with the French to be geniuses. This didn't stop them from burdening us with all sorts of chores. Whenever they could, however, just so we wouldn't forget our place." P. 155 By the middle of the book Michael and Stefanos were more believable and definitely interestingly opinionated... For example, here is a discussion on the newly constructed Eiffel Tower: [Stefanos:] \This moment symbolizes a new era, the era of technology. What we have before us is a marvel of statics, dynamics, chemistry, electricity - they have all combined to make this the tallest construction in the world. And the

Second, many high schools will not want to expose their students to some of the passages describing a bohemian lifestyle. In a nutshell - I thoroughly recommend the book, but not necessarily for high school student. 16

The Mathematics of Climate Modelling H. Monahan The Mathematics of ClimateAdam Modelling Adam H. Monahan Euros or renminbi. Each of these forms of money represents a different “pool”. You can move money from one pool to another - depositing cash to a bank account, or changing currencies; we can call such transfers “fluxes”. Fluxes don’t change the total amount of money you have (if we neglect service charges) - they just change the form that it’s in. You gain or lose money by earning or spending these processes represent net changes in the total amount of money you have. Earning is a “source” of money, while spending is a “sink”. Different pools may have different sources and sinks - if you buy something with cash, that’s a sink for that pool; if you use your bank card, that’s a sink for your bank account. The amount of money you have in a given pool increases with time if the sum (sources + fluxes - sinks) is positive; it decreases with time if this sum is negative. The process of budgeting involves understanding these processes - how much you have in each pool, what the fluxes are between them, and what the various sources and sinks are. The science (and art) of mechanistic modelling involves representing these processes mathematically. Mechanistic climate modelling is concerned with budgets of physical, chemical, and biological quantities - energy, momentum, water (in various phases), carbon (in various chemical compounds), terrestrial vegetation biomass, and many others. Newton’s Second Law of Motion

Adam Monahan is an associate professor in the School of Earth and Ocean Sciences at the University of Victoria. His research focuses on studying interactions between “large” and “small” scales in the atmosphere and ocean (the “weather-climate connection”).

Introduction Predicting the evolution of the Earth’s climate system is one of the most daunting mathematical modelling challenges that humanity has ever undertaken. The atmosphere and ocean display variability over a bewildering range of space and time scales - from microns to thousands of kilometers, and from seconds to millions of years. The challenge is increased by the fact that the climate system consists not just of the atmosphere and the ocean but also includes the cryosphere (frozen water in land and sea ice), the biosphere (life on Earth), and the geosphere (the solid Earth). While descriptive climate science is a centuries-old discipline, modern climate physics - with a quantitative focus on mechanism - is relatively recent. In fact, the first global atmospheric circulation model was built only in the mid-1950’s (a nice discussion of the history of atmospheric modelling for weather forecasting is given in Harper et al. (2007)). Modern climate physics is a fundamentally mathematical discipline, making use of tools and techniques from across the spectrum of modern applied mathematics. This brief article discusses the interweaving of mathematics with modern climate science.

F =

Climate Modelling and Budgets An idea fundamental to climate modelling is that of a “budget”. In everyday life, the budgets we most often think about are budgets of money. The money you have may exist in different forms - in cash or in a bank account; in Canadian dollars or

17

d (mv) dt

is a mathematical description of the momentum budget: accelerations are changes in the “momentum pool” of a physical object resulting from sources and sinks of momentum which we call forces. Newton’s Third Law of Motion - “the force of A on B is equal and opposite to the force of B on A” - is just a statement that forces acting between objects are momentum fluxes that may change the momentum of each object (that is, an individual momentum pool) but don’t change the total amount

ernal energy, etc.). Seen in this way, climate modelling seems aightforward. To build a climate of momentum (the sum over allmodel, pools). allInyou the ed to do is same kind of way, the First Law of Thermodynamics

quantity that we have at some time t, then t budget analysis for the time rate of change of Q talked about earlier be written mathematica Mathematics and can Climate Modelling as the differential equation

is a description of the energy budget - energy is 1. determine what quantities important to not created or destroyed, just are transformed between potential energy, kinetic thepools part(gravitational of the climate system you’reenergy, coninternal energy, etc.). cerned with understanding or predicting, Seen in this way, climate modelling seems straightforward. build a climate model, you 2. decide what the Toimportant pools of all these need to do is

Consider the budget of some climatically significant quantity - fordQ example, the amount of CO2 in the = 1). sources fluxes - sinks atmosphere (Figure If Q is+the amount of this dt quantity that we have at some time t, then the budget analysis for the time rate of change of Q we in tim To make predictions about how Q changes talked about earlier can be written mathematically we need to understand the fluxes, sources, and sin as the differential equation

quantities are,

and represent these mathematically. For exampl what aredQ atmospheric CO2 sources and sinks, a = sources + fluxes - sinks dt what determines how strong these are? An i portant natural source is respiration organis To make predictions about how Q changes in by time, - the of COthe a byproduct extracti we needrelease to understand fluxes, sources, and of sinks 2 as and represent mathematically. For example energy fromthese organic carbon (what we normally c what are atmospheric CO sources and sinks, and 2 “food”). The flip side of this is a major sink what determines how strong these are? An imatmospheric CO2 : photosynthesis, by which CO2 portant natural source is respiration by organisms organic carbon. To model th -transformed the release of into CO2 as a byproduct of extracting processes weweneed to express ra energy from mathematically, organic carbon (what normally call “food”). The flip side of respiration this is a major of of photosynthesis and (insink vegetation, atmospheric CO2 : photosynthesis, CO2 is we a soils, in animals) as functionsby of which the variables transformed into organic carbon. To model these modelling - a wonderful problem of mathematic processes mathematically, we need to express rates biology. of photosynthesis and respiration (in vegetation, in

1. determine what quantities are important to

3. come upthewith descriptions of part mathematical of the climate system you’re concerned with understanding the fluxes between these pools, or ofpredicting, the sources, and of2. the sinks, and decide what the important pools of these quantities are,

4. study the properties of the resulting mathe3. come up to with mathematical descriptions of matical models make predictions about the the fluxes between these pools, of the sources, climate. and of the sinks, and

this is 4. a breeze, Not really - none of study theright? properties of the resulting matheese steps ismatical easy. models As always, devil isabout in the to makethe predictions the tails. All four steps require expertise in physics, climate. emistry, biology, or geology - and (particularly in So this is a breeze, right? Not really - none of ps 3 and mathematics. these4)steps is easy. As always, the devil is in the details. All four steps require expertise in physics, chemistry, biology, or geology - and (particularly in steps 3 and 4) mathematics.

soils, in animals) as functions of the variables we are modelling - a wonderful problem of mathematical biology.

1: Estimate of global carbon cycle pools (or reservoirs) and fluxes. Black numbers are estimates of pre-industrial (e.g. natural) values, gure 1:Figure Estimate of global carbon cycleofpools (or reservoirs) and fluxes.(asBlack are Chapter estimates pre-industrial (e.g. natural) val while red numbers represent estimates changes caused by human activities of thenumbers 1990’s). From 7 ofofSolomon et al. (2007).

le red numbers represent estimates of changes caused by human activities (as of the 1990’s). From Chapter 7 of Solomon et al. (2007).

18

2

(evaporation and precipitation) - that are the domain of the physical disciplines of mechanics and thermodynamics. of energy (for temperatures), budgets of freshwater And whatand determines the concentration (evaporation precipitation) - that are the of do-CO2 main of theWell, physical mechanics and in water? thisdisciplines is a wholeofnew budget problem thermodynamics. - one with all the complications of the budget what determines the concentration of2CO 2 for And atmospheric CO2 . Currents move CO around in water? Well, is a whole new budgetinproblem along with thethis water, photosynthesis the ocean -consumes one withit,all the complications of the budget respiration releases it, CO2 reacts with for atmospheric CO2 . Currents move−CO2 around seawater to form bicarbonate HCO3 and carbonate along 2− with the water, photosynthesis in the ocean CO ions - and all of these processes need to be 3 consumes it, respiration releases it, CO2 reacts with represented mathematically. involves more − seawater to form bicarbonate HCOThis 3 and carbonate 2− budgets, more complications; CO ionsmore - andequations, all of theseand processes need to be 3 as should be clear, even a straightforward quesrepresented mathematically. This involves more tion likemore “howequations, much CO is exchanged between budgets, and complications; 2 more theshould ocean be andclear, atmosphere?” can represent a very as even a straightforward question like problem “how much CO2 is chemistry, exchanged and between involved of physics, biology the ocean and atmosphere?” can represent a very - and of the mathematics needed to model it all. involved problem of physics, matters chemistry,isand Further complicating thebiology fact that -very and of the mathematics needed to model all. total often, we’re interested not just init the Furtherof complicating matters is the fact that or amount some quantity in the atmosphere very often, we’re interested not just in the total ocean, but also in how the abundance varies from amount of some quantity in the atmosphere or place to place. We don’t experience globallyocean, but also in how the abundance varies from averaged temperature: weexperience experiencegloballywhat the place to place. We don’t temperature is right here. Therefore, averaged temperature: we experience whatwetheneed mathematicalis models of how quantities vary in temperature right here. Therefore, we need both space and time -ofthat partial differential mathematical models howis,quantities vary in equations. These local budgets, in which both space and time represent - that is, partial differential the abundance ofrepresent our quantity of interest at every equations. These local budgets, in which the abundance our quantity of interest at every point in space of represents a different pool.

which is the burning of organic carbon stored in geological reservoirs (coal, oil, natural gas). The strength thisother source – thatofis,CO the which Thereofare sources notat the leastwe of 2 -rate burnwhich fossil isfuels is largely by economic the –burning of determined organic carbon stored in geological reservoirs (coal, extraction oil, naturaland gas). The decisions regarding resource energy strength of this source – that is, the rate at which we generation; modelling these emissions represents an burn in fossil fuels – is largelyeconomics. determined by economic exercise mathematical decisions resource extraction and energy And whatregarding about atmospheric CO2 fluxes? Well, generation; modelling these emissions represents an CO2 dissolves in water - in fact, the oceans contain exercise in mathematical economics. more than 50 times the amount of carbon that And what about atmospheric CO2 fluxes? Well, the CO atmosphere does - and CO2 is constantly being 2 dissolves in water - in fact, the oceans contain exchanged between thesethe atmospheric oceanic more than 50 times amount of and carbon that pools 2). This the atmo2 flux the(Figure atmosphere doesCO - and CO2between is constantly being sphere and ocean is very important – butand howoceanic do we exchanged between these atmospheric represent it mathematically? chemistry pools (Figure 2). This CO2 Physical flux between the atmotellssphere us that fluxisshould be proportional andthe ocean very important – but howto dothe we represent mathematically? Physical chemistry air-water COit difference 2 concentration be proportional to the tells us that the flux should CO flux water to air = k [COwater ] − [COair ] 2 2 2 air-water CO2 concentration difference water ] − [COairthe This deceptively simple-looking: flux wateris to air = k [CO ] CO equation 2

2

2

coefficient k bundles together a lot of very compliThis equation is deceptively simple-looking: the cated physics. The air-water interface consists of coefficient k bundles together a lot of very complitwo boundary layers (one in each fluid) which are cated physics. The air-water interface consists of oftentwo turbulent. and fluid) modelling boundary Understanding layers (one in each whichthis are turbulence - the strength and character of which often turbulent. Understanding and modelling this are responsible the exchanges CO2 turbulence - for themediating strength and character ofofwhich (andare other quantities) betweenthe theexchanges fluids - isof “the responsible for mediating CO2 great unsolved problem ofbetween classical (and other quantities) thephysics”. fluids - is This “the problem more budgets - budgets of This mogreat involves unsolved problem of classical physics”. problem more budgets - budgetsbudgets of momentum (forinvolves the winds and the currents),

point in space represents a different pool.

mentum (for the winds and the currents), budgets

Figure 2: Air-sea CO2 fluxes (in moles of carbon per square meter per year) for the year 1995 as computed from a Global Climate Model (left Figure 2: Air-sea CO2 fluxes (in moles of carbon per square meter per year) for the year 1995 as computed from a Global Climate Model (left panel) and as estimated from observations (right panel). Positive values indicate a CO2 flux from the ocean to the atmosphere; negative values

panel) and as estimated from observations (right panel). Positive values indicate a CO2 flux from the ocean to the atmosphere; negative values indicate fluxes from the atmosphere to the ocean. While there is broad agreement between the observed and modelled fluxes, there are many indicate fluxes from the There atmosphere to theofocean. there is broad agreement between observed modelled fluxes, there are many differences in detail. is still plenty work toWhile be done in climate modelling. Adapted fromthe Zahariev et al.and (2008). differences in detail. There is still plenty of work to be done in climate modelling. Adapted from Zahariev et al. (2008).

19

physics. In particular, an emerging school of thought takes advantage of the essentially random character of small-scale turbulence to make use of tools from probability theory and stochastic processes.

For example, the partial differential equation for the temperature field T (x, y, p, t) (in which pressure rather than altitude is used as the vertical coordinate, following common meteorological practice) is

∂T ∂T ∂T ∂T cp ρ +u +v +ω ∂t ∂x ∂y ∂p

− ω = 2 − ∇ · q

In this equation - which is just a mathematical statement of the first law of thermodynamics - (u, v) is the horizontal velocity vector, ω is the vertical velocity (in pressure coordinates), cp is the specific heat capacity of air at constant pressure, ρ is the air density, and 2 is the heating rate associated with viscous dissipation in a turbulent flow. The vector q is the spatial flux of heat energy through the atmosphere due to the emission and absorption of electromagnetic radiation, phase changes of water, chemical reactions, and conduction. Each of u, v, ω, ρ, 2 , and q are themselves (x, y, p, t)dependent fields with their own budgets (expressed as partial differential equations). Clearly, modelling climate variability in both space and time can be very complicated indeed. Once we’ve got these partial differential equations written down, how do we use them to study the climate system? Some things we can learn by looking at the form of the equations. In particular, by considering the relative sizes of different terms we can gain some insight into the relative importance of different processes. We can even carry out formal asymptotic analyses of the equations, to systematically “throw away” less important terms to simplify the models. But even these simplified equations are still very complicated in general, and can’t be solved by hand to make predictions. In this case, all we can do is replace the original equations with discrete approximations appropriate for simulation by computers. That is, we need to make use of numerical analysis. In these approximate models, only space and time scales above a particular threshold are explicitly modelled. However, the smaller scales cannot generally be ignored - the nonlinearity of the original partial differential equations leads to important interactions between large and small scales. The effect of the smaller unresolved scales must be parameterised in terms of the larger resolved scales. There are different approaches to addressing the parameterisation problem, but some of the most promising make use of perspectives from statistical

Such probabilistic perspectives are also useful for appreciating the difference between “weather” and “climate”. Robert Heinlein put it best: “Climate is what you expect, weather is what you get”. This perspective is fundamentally probabilistic: “climate” is the probability distribution of outcomes for many rolls of dice, and “weather” is the outcome of any particular throw. Changes in climate are changes in the weighting of the dice. From a dynamical systems perspective, we can speak of the slowly-evolving “climate attractor” in contrast to the fast “weather trajectory”.

20

Of course, all of these mathematical models would only be so much science fiction without observations to test them and to help estimate parameter values - so statistical analyses play a fundamental role in modern climate science. These analyses help us understand our budgets. For example, the increasing trend in atmospheric CO2 concentration is a result of sources being stronger than sinks and fluxes (as we burn fossil fuels); statistical analyses of these trends help us understand just how out of balance these budgets are. Statistical analyses also help us understand how different parts of the climate system are related - for example, how variability in the Eastern Pacific sea surface temperature is related to large scale pressure distributions in the El Ni˜ no - Southern Oscillation phenomenon. Understanding these relationships in observations helps us both build and assess our mechanistic models. Furthermore, all of these mathematical representations of sources, sinks, and fluxes in mechanistic models have various parameters that need to be set. Some of these - the speed of light, for example, or the acceleration of gravity - are well known. Others, such as the dependence of soil respiration rate on soil temperature and moisture, are not so well constrained. Inverse modelling provides a systematic framework for estimating these parameters so as to bring mechanistic models into closest accord with observations – and thereby to make better predictions.

Conclusions

Sarmiento and Gruber (2006) present a detailed discussion of the ocean’s role in the global carbon cycle from a modelling perspective, and an excellent introduction to statistical analyses of climate data is provided in Wilks (2005). Unfortunately out of print, Haltiner and Williams (1980) is a classic introductory text on numerical methods for meteorological modelling.

While the Earth’s climate system is intimidatingly complex, the reality of climate change compels us to build models for understanding the system and predicting its future behaviour. These models, expressed in the language of mathematics, range across a hierarchy of complexity. At one end of this hierarchy are idealised conceptual models that are useful for developing understanding but cannot be expected to be quantitatively accurate. At the other end of the hierarchy are the fully complex Global Climate Models which represent the physics, chemistry, and biology of the climate system - but which are so complicated that they can only be studied numerically using computers. The world’s most powerful computers require several months of computational time to perform 1000-year simulations with these complex models; increases in computer power are generally consumed by increases in model complexity or resolution. In fact, models from across the entire range of this hierarchy play important roles in understanding and prediction, and mathematics provides the fundamental framework by which these models are constructed, studied, and refined.

Acknowledgements The author would like to thank Jim Christian for providing Figure 2, and an anonymous reviewer for very helpful comments on the original version of the manuscript. The author is supported by the Natural Sciences and Engineering Research Council of Canada.

References Colling, A., 2001: Ocean Circulation. ButterworthHeinemann. Haltiner, G. J. and R. T. Williams, 1980: Numerical Prediction and Dynamic Meteorology. Wiley, New york, 477 pp.

Further Reading The climate literature is extensive and can be intimidating to the uninitiated. An comprehensive overview of the state of the art of climate science is given in the most recent report of Working Group I of the Intergovernmental Panel on Climate Change, available as a book (Solomon et al., 2007) or online at http://www.ipcc.ch. A more succinct introduction to climate modelling is presented by Thorpe (2005), while Weaver (2008) provides a non-technical discussion of climate change (with a particular focus on Canada). There too many good books on the physics of the climate to provide an exhaustive list here; an excellent introductory text is Hartmann (1994), while a more quantitatively detailed treatment is provided in Peixoto and Oort (1992). Colling (2001) and Wallace and Hobbs (2006) are good introductions to physical oceanography and meteorology, respectively; deeper treatments of modern geophysical fluid mechanics are given in Holton (2004) and Vallis (2006).

Harper, K., L. W. Uccellini, E. Kalnay, K. Carey, and L. Morone, 2007: 50th anniversary of operational numerical weather prediction. Bull. Amer. Meteor. Soc., 88, 639–650. Hartmann, D. L., 1994: Global Physical Climatology. Academic Press, San Diego, 411 pp. Holton, J. R., 2004: An Introduction to Dynamic Meteorology. Academic Press. Peixoto, J. P. and A. H. Oort, 1992: Physics of Climate. American Institute of Physics, New York. Sarmiento, J. L. and N. Gruber, 2006: Ocean Biogeochemical Dynamics. Princeton University Press.

21

Solomon, S., D. Qin, M. Manning, Z. Chen, M. Marquis, K. Averyt, M. Tignor, and H. Miller, eds., 2007: Climate Change 2007: The Physical Science Basis. Contribution of Working Group

I to the Fourth Assessment Report of the Intergovernmental Panel on Climate Change 2007 . Cambridge University Press.

Weaver, A., 2008: Keeping Our Cool: Canada in a Warming World . Viking Canada.

Thorpe, A. J., 2005: Climate change prediction: A challenging scientific problem. Institute of Physics.

Wilks, D. S., 2005: Statistical Methods in the Atmospheric Sciences. Academic Press.

Vallis, G. K., 2006: Atmospheric and Oceanic Fluid Dynamics: Fundamentals and Large-Scale Circulation. Cambridge University Press. Wallace, J. M. and P. V. Hobbs, 2006: Atmospheric Science: An Introductory Survey. Academic Press.

Zahariev, K., J. R. Christian, and K. L. Denman, 2008: Preindustrial, historical, and fertilization simulations using a global ocean carbon model with new parameterizations of iron limitation, calcification, and N2 fixation. Progress in Oceanography, 77, 56–82.

PiPiininthe Challenges theSky Sky Math Math Challenges

Solutions to the problems published th Solutions to the problems published in the 13 Issue of Pi in the Sky in the 13th issue of Pi in the Sky Problem 1: Find all positive integers n such that log2008 n = log2009 n + log2010 n. Solution: A solution for the above equation is n = 1. Let us prove that there are no other solutions. For any positive integers k > 2, n > 1 we have logk+2 n + logk+1 n − logk n =

1 1 1 + − logn (k + 2) logn (k + 1) logn k

logn (k + 1) (logn k 2 − logn (k + 2)) + logn (k + 2) (logn k 2 − logn (k + 1)) = >0 2 logn (k + 2) logn (k + 1) logn k since k 2 > k + 1. Hence, if we take k = 2008 we get log2009 n + log2010 n > log2008 n.

2

Problem 2: Find the smallest value of the positive integer n such that (x2 + y 2 + z 2 ) ≤ n (x4 + y 4 + z 4 ) for any real numbers x, y, z. Solution: The given inequality can be transformed into an equivalent useful format: 2 2 2 (n − 3) x4 + y 4 + z 4 + x2 − y 2 + y 2 − z 2 + z 2 − x2 ≥ 0.

2 2 2 Since the minimum value of (x2 − y 2 ) + 22 (y 2 − z 2 ) + (z 2 − x2 ) is 0 , we must have 4

4

4

2 2 2 (n − 3) x4 + y 4 + z 4 + x2 − y 2 + y 2 − z 2 + z 2 − x2 ≥ 0. 2

2

2

Since the minimum value of (x2 − y 2 ) + (y 2 − z 2 ) + (z 2 − x2 ) is 0 , we must have (n − 3) (x4 + y 4 + z 4 ) ≥ 0, hence n ≥ 3. An alternative solution, using a geometric argument was given by Carlo Del Noce, Genova, Italy.

Problem 3: Let a be a positive real number. Find f (a) = maxx∈R {a + sin x, a + cos x}. Solution: This problem is trivial. Since maxx∈R {a + sin x, a + cos x} is clear a + 1 then f (a) = a + 1. Problem 4: Prove that the equation x2 − x + 1 = p(x + y) where p is a prime number, has integral solutions (x, y) for infinitely many values of p.

Solution: Let us assume by contradiction that the equation has integral solutions (x, y) only for a finite number of prime numbers, among which the greatest is denoted by P. If we take x = 2 · 3 · 5 · .... · P then x2 − x + 1 = x(x − 1) + 1 is not divisible by any of the prime numbers which are ≤ P. Hence x2 − x + 1 = Qm, where m is an integer and Q is a prime, Q > P. So, if we take y = m − x, then (x, y) would be an integral solution of the equation x2 − x + 1 = Q(x + y), which is a contradiction. This problem was also solved by Carlo Del Noce,Genova, Italy.

Problem 5: Find all functions f : Z −→ Z such that 3f (n) − 2f (n + 1) = n − 1, for every n ∈ Z. (Here Z denotes the set of all integers). Solution: It is clear that f (n) = n + 1 is a solution of the problem. Let us prove that there is no other solution. If we set g(n) = f (n) − n − 1 and using the given equation one obtains 3g(n) = 2g(n + 1) for every integer n. Assume by contradiction that there is an integer m such that g(m) = 0. Then 2g(m) = 3g(m − 1) = 32 g(m − 2) = ... = 3k g(m − k) for any positive integer k. Hence 3k |g(m) for any positive integer k and therefore we must have g(n) = 0 for any integer n. A correct solution was received from Carlo Del Noce,Genova, Italy. 23

= 100◦ . Let D be on the extended Problem 6: In ∆ABC, we have AB = AC and BAC line trough A and C such that C is between A and D and AD = BC. Find DBC. Solution: Ahmet Ardu¸c from Turkey, submitted five solutions to this problem. The following is his first solution. In ∆ABC, we have AB = AC and BAC = 100°. Let D be

on the line through A and C such that C is Let α = m(CBD). Drawextended the equilateral triangle AED. between Awe and and=AD =m( BC.BAE) Find=DBC. = 40◦ , m(ABE) = Since ∆ACB ≡ ∆BAE(SAS) getDAB BE, m(BEA) ◦ i) Dra 100 . As AB = BE and AD = DE, ABED is a deltoid with BD the axis of symmetry. ii) Dra = m(DBA) = 50◦ = 40◦ + α. Thus α = 10◦ . Hence m(DBE)

iii) m(B ⟹m ⟹m iv) ECB

Carlo Del Noce Solution also solved– this 1: problem.

Solution – 3

i) Draw equilateral triangle AED, ii) ACB ≌ BAE (SAS congruency) a. |AB| = |BE| \It is proven that the of birthdays b. celebration m(BAE) = m(BEA) = 40° is healthy. Statistics c. m(ABE) = 100° the most birthdays become show that those people who celebrate iii) Let m(CBD) = α. the oldest." iv) Since |AB| = |BE| and |AD| = |DE| ABED is a deltoid, S. den Hartog, a. PhD Thesis, University of Gronigen b. [BD] is the symmetry axis of the deltoid ABED c. m(DBE) = m(DBA) = 50° = 40° + α ⟹ α = 10°. 24

i) ii) iii) iv)

Dra We Dra EAD a. b. v) Sinc 160 vi) m(E vii) m(C

Math Challenges Problem 1:

Find all the real pairs (x, y) such that log 3 x + logx 3 ≤ 2 cos πy. Problem 2:

Determine all the triples (a, b, c) of integers such that a 3 + b 3 + c 3 = 2011. Problem 3:

In decimal representation the number 22010 has m digits while 52010 has n digits. Find m + n. Problem 4:

Find all the polynomials P(x) with real coefficients such that sin P(x) = P(sin x), for all x ∈ R. Problem 5:

The interior of an equilateral triangle of side length 1 is covered by eight circles of the same radius r. Prove that r ≥ 71– . Problem 6:

Prove that in a convex hexagon of area S there exist three consecutive vertices A, B and C such that Area (ABC ) ≤ S 6 25

Year of Science Science is everywhere and affects everything we do - from driving our cars, to how we communicate and how we learn in schools. Science can improve the quality of our lives at many different levels - from our everyday activities to global issues. You would have to live in an exceptionally remote part of our world not to enjoy at least some offshoot, some benefit of scientific inquiry. But does the average person really think about how science affects our daily lives? In British Columbia and across Canada, scientists and researchers carry out - through chemistry, biology, physics, mathematics and other fields - a wide range of work which is continually leading the way to the development of many new products and practical solutions that are vital to every aspect of our daily lives. Learn more at www.YearOfScienceBC.ca

26