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Piin the

SkyIssue 14, Fall 2010

I n T h I s I s s u e :(-1) x (-1)=1 . . . but WHY?Palindrome DatesPrimes from FractionsReasoning from the Specific to the GeneralThe Mathematics of Climate Modelling

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On the Cover Predicting the evolution of the Earths climate system is one of themost daunting mathematical modeling challenges that humanity has ever undertaken. In his article on page 17, Adam Monahandiscusses the interweaving of mathematics with modern climatescience.

Editorial BoardJohn Bowman (University of Alberta)Tel: (780) 492-0532, E-mail: bowman@math.ualberta.caMurray Bremner (University of Saskatchewan)Tel: (306) 966-6122, E-mail: bremner@math.usask.caJohn Campbell (Archbishop MacDonald High, Edmonton)Tel: (780) 441-6000, E-mail: jcdotcalm@shaw.caFlorin Diacu (University of Victoria)Tel: (250) 721-6330, E-mail: diacu@math.uvic.caSharon Friesen (Galileo Educational Network, Calgary)Tel: (403) 220-8942, E-mail: sfriesen@ucalgary.caGordon Hamilton (Masters Academy and College, Calgary)Tel: (403) 242-7034, E-mail: gamesbygord@gmail.comKlaus Hoechsmann (University of British Columbia)Tel: (604) 822-3782, E-mail: hoek@pims.math.caDragos Hrimiuc (University of Alberta)Tel: (780) 492-3532, E-mail: dhrimiuc@ualberta.caMichael Lamoureux (University of Calgary)Tel: (403) 220-8214, E-mail: mikel@math.ucalgary.caDavid Leeming (University of Victoria )Tel: (250) 472-4928, E-mail: leemingd@uvic.caMark MacLean (University of British Columbia)Tel: (604) 822-5552, E-mail: maclean@math.ubc.caPatrick Maidorn (University of Regina)Tel: (306) 585-4013, E-mail: Patrick.Maidorn@uregina.caWendy Swonnell (Greater Victoria School District)Tel: (250) 477-9706, E-mail: wswonnell@shaw.ca

Managing Editor Anthony Quas (University of Victoria)Tel: (250) 721-7463, E-mail: aquas@uvic.ca

Contact InformationPi in the Sky PIMS University of Victoria Site Of ce, SSM Building Room A418bPO Box 3060 STN CSC, 3800 Finnerty Road, Victoria, BC, V8W 3R4T/(250) 472-4271 F/(250) 721-8958 E-mail: pims@math.uvic.ca

Pi in the Sky is a publication of the Pacifc Institute or the MathematicalSciences (PIMS). PIMS is supported by the Natural Sciences and EngineeringResearch Council of Canada, the Province of Alberta, the Province of BritishColumbia, the Province of Saskatchewan, Simon Fraser University, the University of Alberta, the University of British Columbia, the University of Calgary, theUniversity of Lethbridge, the University of Regina, the University of Victoria andthe University of Washington.

Pi in the Sky is aimed primarily at high school students andteachers, 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 alsoput curriculum topics in a different perspective.

Submission InformationFor details on submitting articles for our next edition of Pi in theSky , please see:

http://www.pims.math.ca/resources/publications/pi-sky

Table of Contents

Editorialby Anthony Quas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

(-1) x (-1) = 1 . . . but WHY?by Marie Kim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Palindrome Dates in Four-Digit Yearsby Aziz S. Inan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Primes rom Fractionsby Alex P. Lamoureux and Michael P. Lamoureux. . . . . . . . . . . 8

Reasoning rom the Specifc to the Generalby Bill Russell . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Book ReviewPythagorean CrimesReviewed by Gord Hamilton . . . . . . . . . . . . . . . . . . . . . . . . . . 15

The Mathematics o Climate Modellingby Adam H. Monahan. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Pi in the Sky Math Challenges

Solutions to problems published in Issue 13. . . . . . . . . . . . . . . 22New Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

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E it rialAnthony Quas

MATHEMATICAL ModELLIn :

WHAT, HoW And WHY?In my work and outside it too I often hear aboutMathematical Modelling. In this editorial I'll saya bit about what it is; how to do it; and why it'simportant.

What is Mathematical M elli g?First let's try a quick answer: MathematicalModelling is the process of using mathematicsto understand something `in the real world'. For

high school students the closest thing to this inthe curriculum might be the (often-dreaded) wordproblems. Here is an example taken from theinternet:

Mr.S is planting flowers to give his girlfriend on Valentine's day, which is half a year away. Currently the flowers are 7inches tall and will be fully grown oncethey reach one foot. If the flowers growat a rate of half an inch per month, then

how large will they be on the day? Will theflowers be fully grown? or will he have tofind another gift?

The aim here is to take something which is onthe surface not a mathematical question, translateit into a mathematical question, solve themathematical question and translate the answerback into the framework of the original question.

In this case we let x denote the height of the flowers

on Valentine's Day and say that x = 7 + t / 2where t is the number of months until Valentine'sDay. Since we're told it's half a year away we havet = 6 so that we can compute x = 10. Since we'retold the flowers will be 12 inches when they'refully grown it sounds as though `Mr.S' will haveto think of a new present.

For some more advanced examples, here are someother questions that might be approached usingmathematical modelling.

When lightis reflected

in water ther e f l e c t i o nseems to bestretched outalong the lineconnecting theviewer andthe source.Why is this?Why isn't itstretched in theperpendiculard i r e c t i o n(horizontally inthis picture)?

In the case of a fast-spreading epidemic involvingan unknown disease (e.g. the SARS epidemic thathit China in late 2002 / early 2003 and spreadto other countries including Canada) what arethe best policies to follow to minimize sicknessand death from the disease while avoiding over-reacting?

Finally one of the most important cases of all:climate modelling. Here you're trying to predictthe weather patterns in 10 years; 20 years; 100years. For more information on this see the articlein this issue by Adam Monahan.

H w y u Mathematical M elli g?Just as in the case of the word problem, a first

step is trying to decide which variables to studyin the problem. There is an important differencethough. The word problem is typically constructedto give you all of the information you need tosolve the problem (and no more). Generally thevariables are given to you in the statement of theproblem itself. When you're doing more advancedmodelling you have to decide which variables toinclude (and often equally importantly which onesto leave out of consideration).

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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 theconcept hard to understand and to accept at first.Mathematicians before Descartes refused to acceptnegative numbers, including the great Pascalhimself. Negative numbers actually are believedto have been found in the East the earliest, inChina. An ancient Chinese text, written inB.C. 1000, titled \Ku Jang San Sul" - meaning\nine arithmetic formulas"- includes in part acomputation of negative numbers.

(-1) x (-1) = 1 . . . but WHY?Marie Kim, Cheong Shim International Academy

1. Patter2 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 = 0Take a look at the pattern above. In each row, you'll be able to find that constant decrease in thenumber multiplied results in constant change in the product. The same for each column. Applying thissame 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

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

2. number li e

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

Positive and negative numbers are often explainedwith the model of `debt and profit.' Profit ispositive and debt is negative; adding positivenumbers results in profit while adding negativenumbers deducts profit - adding to debt. Thismodel however, failed to make people understandthe concept of multiplying a negative number by anegative number since it is hard to find such casesin everyday life.

How is the concept `negative numberx

negativenumber' explained, then? There are quite a fewillustrations and models that are available to helppeople understand and accept such a concept. Outof those explanations, here are three easy ones.

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

Now, keeping those two models in mind, (-2) x (-3)' can be thought as moving by (-2) three times in theopposite 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 whichwould look like the first number line.

3. Pr f

This one may be the most complicated out of the three. Using the distribution property, it can be

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

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) x = (-a)(-b)

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

Other explanations for `(-) x (-) = (+)' do exist and can be found easily, f r example usi g c mplex

umbers. But, don't just go and look it up; take your time and think of your own way to explain`(-) x (-) = (+).'

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Y4Y3Y2Y1Y1Y2Y3Y4,palindrome dates canonly occur in yearsending with a digit Y 4 less than four (since daynumber cannot exceed31). In addition, thehundreds digit Y 2 of the year number of thepalindrome date musteither be zero or one (since month number cannotexceed 12). In the special case when the thousandsdigit Y 1 of the year satisfies Y 1 > 2, Y 2 must equalzero. In other words, palindrome dates in thesecond and third millenniums (years 1001 to 3000)can only occur in the first two centuries of eachmillennium. On the other hand, palindrome datesthat fall between fourth and tenth millenniums

(years 3001 to 10000) can only occur in the firstcentury of each millennium. Also, between years1000 to 10000, palindrome dates in each centuryall fall on the same month with month numberY2Y1. Table 1 provides all possible combinationsof palindrome dates in the DDMMY 1Y2Y3Y4 dateformat categorized in terms of different valuesand ranges of digits Y 4, Y 2, Y 3 and Y 1, where N is the total number of palindrome dates in eachcategory. Based on Table 1, in the DDMMYYYYdate format, a total of 335 palindrome dates existamong all the four-digit years.

In the MMDDY 1Y2Y3Y4 date format, palindromedates represented by Y 4Y3Y2Y1Y1Y2Y3Y4 canonly occur in years ending with digit Y 4 equalto either zero or one since the month numbercannot exceed 12. In the case when Y 4 = 1, thetenth digit Y 3 of the year number cannot exceedtwo. In addition, the second digit Y 2 of the yearnumber of the palindrome date must be less thanfour since the day number cannot exceed 31.Furthermore, if Y 1 > 1, then, Y 2 < 3. That is, forthe MMDDY 1Y2Y3Y4 date format, palindromedates in the second millennium (years 1001 to2000) can occur only in the first four centuries of each millennium. On the other hand, palindromedates between the third and tenth millenniums(2001 to 10000) only occur in the first threecenturies of each millennium. Also, betweenyears 1000 and 10000, palindrome dates in eachcentury all fall on the same day of the month,

with day number equal to Y 2Y1. Table 2 providesall possible combinations of palindrome dates inthe MMDDY 1Y2Y3Y4 date format categorized interms of different values and ranges of digits Y 4Y2, Y 3 and Y 1, where N is the total number of palindrome dates in each category.

According to Table 2, there are a total of 331palindrome dates in the MMDDYYYY dateformat involving all the four-digit years.

Note that even if some palindrome date numbersrepresented by Y 4Y3Y2Y1Y1Y2Y3Y4 are validdate numbers in each date format, unless theday and the month numbers are the same, theycorrespond to different actual dates in each dateformat (e.g., palindrome date number 10022001as mentioned earlier). There are also palindromedate numbers which are only valid dates in onedate format but not the other.

For example, palindrome date number 21022012is a valid date number in the DDMMYYYYdate format and represents 21 February 2012.However, 21022012 is not a valid date number inthe MMDDYYYY format since its month number21 exceeds 12.

Pali r me ates i thesec mille ium

In the DDMMYYYY date format, a total of 61 palindrome dates occurred in the secondmillennium (years 1001 to 2000), 31 in the 11thcentury (all in the month of January) and 30 inthe 12th century (all in November), with the lastone being 29 November 1192 (29111192). No otherpalindrome dates occurred between the 13th and20th centuries, more than 800 years.

In the MMDDYYYY date format, a total of 43palindrome dates occurred during the second

Table 2.Palindrome date combinations in the MMDDY 1Y2Y3Y4 date format. Note that the total

number of palindrome dates in each category adds up to 331.

M=Y 4 M=Y 3 D=Y 2 D=Y 1 Y 1 Y 2 Y 3 Y 4 N

0 1 to 9 0 to 2 1 to 9 1 to 9 0 to 2 1 to 9 0 2430 1,3,5,7,8 3 1 1 3 1,3,5,7,8 0 51 0 to 2 0 to 2 1 to 9 1 to 9 0 to 2 0 to 2 1 811 0,2 3 1 1 3 02 1 2

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millennium, split as 12, 12, 12 and 7 among the11th, 12th, 13th and 14th centuries respectively.The 11th century palindrome dates all occurredon the first day of the month, the 12th centuryones all on the 11th day, 13th century ones all onthe 21st day of the month, and the 14th centuryones all on the 31st day of the month. Thelast palindrome date of the second millenniumoccurred on August 31, 1380 (08311380) and noother palindrome dates existed between the 15thand 20th centuries, over 600 years.

Also, interestingly enough, among all the palin-drome date numbers in the second millenniumcommon to both date formats, only two corre-spond to the same actual dates: 1 January 1010(01011010) and 11 November 1111 (11111111).

Pali r me ates i the 21st ce turyThe 21st century has 29 palindrome dates in theDDMMYYYY versus only 12 in the MMDDYYYYdate formats. The first palindrome date in thiscentury in each date format was 10022001, in2001. The second palindrome date of this centuryin the DDMMYYYY date format was 20022002representing 20 February 2002. The thirdpalindrome date in the DDMMYYYY date format,which also happens to be the second palindromedate in this century in the MMDDYYYY format,is 01022010. This date number represents 1February 2010 in the DDMMYYYY date formatversus January 2, 2010 in the MMDDYYYYformat. The first twelve palindrome dates of thiscentury in both date formats are provided in Table3. Note that in this century, palindrome dates inthe DDMMYYYY date format all occur in themonth of February. On the other hand, in theMMDDYYYY 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 29February 2092 (29022092) in the DDMMYYYYdate format versus September 2, 2090 (09022090)in the MMDDYYYY format. Also, there is onecommon palindrome date to occur in this century,02022020, which corresponds to the same actualdate in each format, that being 2 February 2020.

Pali r me ates after the 21st ce turyAfter this century, in the DDMMYYYY dateformat there will be 31 more palindrome dates,

all in the month of December, to occur duringthe 22nd century and no more after that until theend of the third millennium. In the MMDDYYYYformat, 12 more palindrome dates (all being on the12th day of the month) are to occur in the 22ndcentury followed by 12 more (all on day 22 of themonth) in the 23rd, and no more afterwards untilyear 3001. In addition, there is a second palindromedate to occur in this millennium, 12122121, whichis not only common to both date formats but alsorepresents the same actual day in each format,which is 12 December 2121. In the DDMMYYYYdate format, starting with the fourth millenniumthere will be 31 more palindrome dates (all in themonth 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 inthe 61st, 31 (all in July) to occur in the 71st, 31

(all in August) to occur in the 81st, and 30 (allin September) to occur in the 91st centuries, thelast one being 29 September 9092 (29099092). Noother palindrome dates will exist in all the other

Table 3.The first twelve palindrome dates of the 21st century ineach date format.

Number MMDDYYYY DDMMYYYY

1 10022001October 2, 2001

1002200110 February 2001

2 01022010January 2, 2010

2002200220 February 2002

311022011

November 2, 201101022010

1 February 2010

4 02022020February 2, 2020

1102201111 February 2011

5 12022021December 2, 2021

2102201221 February 2012

6 03022030March 2, 2030

020220202 February 2020

7 04022040April 2, 2040

1202202112 February 2021

8 05022050May 2, 2050

2202202222 February 2022

9 06022060June 2, 2060

030220303 February 2030

10 07022070July 2, 2070

1302203113 February 2031

11 08022080August 2,2080

2302203223 February 2032

12 09022090September 2, 2090

040220404 February 2040

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

The number of palindrome dates to occur in theMMDDYYYY date format starting with thefourth millennium up to the tenth is 36 per eachmillennium, distributed as 12, 12 and 12 betweenthe first three centuries of each millennium. The

12 palindrome dates in each century all fall on thesame day of the month, which equals the reverseof the first two digits of the year number. Forexample, there will be 36 palindrome dates in the6th millennium and these 36 palindrome dates willbe distributed as 12, 12 and 12 between the 51st,52nd, and 53rd centuries respectively. The 12 inthe 51st century will all happen on the fifth dayof the month, the 12 in the 52nd on day 15 of themonth, and the 12 in the 53rd are all going to be onthe 25th day of the month. No further palindromedates will exist in the other centuries (54th to60th) of the 6th millennium. The last palindrome

Alex Lamoureux is a senior at Queen ElizabethSchool 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 analy-sis and its applications to geophysics and signal pro-cessing, and teaches courses from calculus to gradu-ate research seminars.

C mi g h me\Hey Dad," announced Alex one day, arrivinghome from school. \There is something weirdabout the fraction one-sixth. See, if you write itout in decimal form,

1/6 = .16666 . . .and start moving the decimal place, you getprimes!"

\That's nuts," said his father, Michael. \The first

Primes fr m Fracti sAlex P. Lamoureux and Michael P. Lamoureux

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

Lastly, between the fourth and tenth millenniums,there is one common palindrome date that exists

in each millennium that corresponds to the sameactual date in each date format. For example,common palindrome date number 05055050 tooccur during the first century of the 6th millenniumrepresents 5 May 5050 in each date format.

Refere ces:

1. M. Gardner, The Colossal Book of Mathematics,Chapter 3, W. W. Norton & Company, 2001.

2. J. N. Friend, Numbers: Fun and Facts, ChapterVIII, Charles Scribner's Sons, 1954.

decimal gives you 1," he points out, \which is nota prime. Next one is 16, also not a prime. Andafter that is 166, 1666, 16666, none of which areprimes. So what are you talking about?"

\No, Dad, you have to round up the numbers. Soyou start with 1.666 . . . which rounds up to 2,which is a prime. Next is 16.666 . . ., which roundsup 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 prettysure is a prime too."

\You're right Alex, all of those are primes. Butthe next one, 16667, is that prime?"

\I don't know, but let's check. Of course twodoesn't divide it, since it is not even. Three doesnot divide it since the digits don't add to a multipleof 3. Five doesn't divide it since it doesn't end in 0or 5. What about seven?"

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\Yes, what about seven? With our calculator, oreven in our head, we can check that 16667/7 =2381. So that one is not prime."

\Still, Dad, this fraction one-sixth is doing prettygood. See, by comparison, if you look at one-half,in decimal it is

1/2 = .5000 ...,

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

\Same thing with one-third, in decimal1/3 = .3333 . . . ,

so shifting gives you 3 (a prime), then the numbers33, 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?"

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

1666 ... 6667.

So these are always odd numbers, which is good,as they will not be divisible by 2. The digits neveradd up to a multiple of 3 (since the 1+7 is nota multiple of 3, while the sixes all are), so thisnumber is not divisible by 3. And of course it isnot divisible by 5. So at least we have a few factorsthat \can't" happen.

In fact, this pattern of repeating digits mightremind us of Mersenne primes, those primes of the form n = 2 k - 1. In binary notation, where weuse only the digits 0 and 1, these primes are inthe formn=111 . . . 111(binary),with exactly k repeats of 1

Of course, not all numbers binary numbers of thisform are prime, but many are, including

3 = 11(binary)7 = 111(binary)

31 = 11111(binary)127 = 1111111(binary)

and even the huge prime number243112609 - 1 = 111 . . . (binary) with 43112609ones, which was discovered in 2008.

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

Testi g f r primesTo find out experimentally what is going on, wecan use a computer program that can look at a biginteger and decide whether it is prime. There arelots of tools out there: MAPLE TM, Mathematica TMGIMPS, among others. Since the second author isa mathematician, he has access to lots of thesetools. So we just use any one of them.

Mathematica is a useful mathematical tool, withplenty of commands to do all sorts of mathematicalcalculations quickly and painlessly. It has a verysimple command, PrimeQ[n], which tests whetheran integer n is prime or not. Mathematica is alsosmart 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.

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

A little \FOR" loop can be used to test a bunchof numbers:

For[n=0, n

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S me fracti s

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, n = 117, n = 2167, n = 3

1667, n = 4166667, n = 61666666667, n = 10166666666667, n = 12166666666666667, n = 15166666666666666666666666666666667, n = 331666666666666666666666666666666666666666666666666666667, n = 5516666666666666666666666666666666666666666666666666666667, n = 561666666666666666666666666666666666666666666666666666666666667, n = 61

a ratio of two integers. Everyone knows aboutthe number pi, which is irrational. Its decimalexpansion starts out as = 3.1415926 . . . Here, inthe first 100 digits, we see only four primes:

3, n = 131, n = 2314159 n = 6314159265359 n = 12

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

There are only two primes that appear in the first100 digits of the expansion.

3, n=12718281828459045235360287471352662497757247093699959574966967627724076630353547594571, n=85

Wow, that's not a lot of primes!

C clusiWell, one-sixth is pretty special. We get a dozenprimes in its first one hundred digits, which is alot more than our other test cases! We have noidea why. Michael thinks Alex should experimentwith more fractions, and more digits, to see whatis going on. Or maybe some of the readers of thisarticle might be inspired to look into why we getprimes this way.

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

Alex pointed out that one-sixth seems reallyspecial. So let's use this numerical test to comparewhat we get with other fractions.

To start, let's try the fraction 1/7 = .142857 . ...We get the following five primes, with digit lengthless than 100:

1429, n = 41428571, n = 71428571429, n = 101428571428571429, n = 161428571428571428571428571, n = 25

For the fraction 1/9 = .1111 . . . we only get getthree primes in the first one hundred digits:

11, n = 21111111111111111111, n = 1911111111111111111111111, n = 23

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

Irrati al umbers

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

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Bill Russell teaches math at James Bowie HighSchool in Austin, Texas, where he's worked for thepast 22 years.

Solving a math problem that involves numbers canoften open the door to discovering a more generalmathematical concept. Much of mathematics isabout relationships, and once these relationshipsare recognized, the next logical step is to try toextend them into more profound revelations.For example, while practicing multiplicationtables, an elementary student might observe thatthe product of two odd numbers always appearsto be an odd number. However, since there areinfinitely many 2-number combinations of oddnumbers, it is not possible to check each pairingand make sure that the product is odd. However,a first-year algebra student has the tools necessaryto prove this relationship is always true.

Any odd number can be represented as 2 n + 1,where n is some integer. A second odd number, then,can be represented as 2 m + 1 for some (possiblydifferent) integer m. The product of these twointegers, then, is (2 m + 1)(2 n + 1)=4 mn + 2 m + 2 n +1

=2(2 mn + m + n)+1This last expression represents one more thanan even integer, and is therefore an odd integer,thus proving that the product of 2 odd integers isalways odd.

This very simple example shows how a specificnumerical problem such as 3 x 5 = 15 can beextended to arrive at a more general mathematical

concept such as odd x odd = odd. The following isa more advanced example of this principle.

Pr blem A (Take 1)A total of 240 meters of fencing is to be used toenclose a rectangular region and divide it into 15smaller rectangular regions (see Figure 1). Findthe values of x and y that will maximize the total

enclosed area.S luti : From the picture, 240 = 4 x + 6 y , ory = - x + 40. The goal is to maximize the area.

A(x ) = xy = x (- x = 40 ) (2)

We observe that the function A( x ) is quadraticand therefore its graph has a vertical line of symmetry midway between its zeros. Sinceequation (2) above is in factored form, one caneasily finds its zeros by setting each factor equalto zero and solving. This reveals that the functionhas zeros at x = 0 and x = 60. Furthermore,since the leading coefficient of (2) is negative, thefunction will take on a maximum value at thevertex, which, as symmetry dictates, is locatedmidway between the zeros at x = 30. This gives acorresponding width of y = - (30) + 40 = 20 ft.Thus, the maximum area enclosed is20 m x 30 m = 600 m 2, a correct but less thanprovocative outcome.

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

Reas i g fr m the Specifict the e eral

Bill Russell

Figure 1Specific example of thefencing problem

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this is a universal truth, then how would oneprove this?

It would be easy enough to repeat the computationsusing other numbers, but even if similar resultswere observed the only thing that would be provenis that the rule holds for those specific numberschosen. To prove that it always holds true, it is

necessary to solve the related general problem byreplacing the numbers with variables. With this inmind, the original problem can now be restated.

Pr blem A(Take 2)A total of T metersof fencing is to beused to enclosea rectangular

region and divideit into smallerrectangular regionsusing m horizontal

pieces, each of length x and n vertical pieces, eachof length y (see Figure 2). Show that when theenclosed area is maximized that the total amountof vertical fencing used equals the total amount of horizontal fencing used.

S luti : In this version of the problem,

T = mx + ny orn

mxT y = , and the goal is tomaximize

A (x ) = xT mx

n (3)

We observe that A(x ) is a quadratic functionand that the leading coefficient

nm is negative,

indicating as before that the function takes on a

maximum value at the vertex. This function has

zeros at x = 0 and at mT

x = , and the vertex lies

midway between these zeros atm

T x

2= . Thus,

the total horizontal fencing used is mT

2 m=

T

2,

precisely one-half of the total fencing, therebycompleting this proof.

Let's take a moment and review what we just did.After solving a routine specific numerical problem,we observed an interesting result. Attempting to

Figure 2General case of the fencingproblem

prove a general theorem that would encompassmany such specific problems, we restated theoriginal problem in more general terms and triedto prove it. Although our proof depended mostlyon some simple algebra, we supplemented thatalgebra with words that explain to the readerwhat we were doing. In the end, we were rewardedwith a remarkably simple yet powerful resultthat precludes the need for repeating those samecalculations in the future. For example, we nowknow that if 300 meters of fencing is available,then the optimum enclosure uses 150 meters of fencing in each direction.

In Problem A, proving the general case wasrelatively easy because we were able to use theexact same methodology for the general problemthat we used in the specific numerical problem. Theonly difference was that in the general problem weused variables in the formula rather than numbers.Although this is sometimes possible, solving thegeneral problem will sometimes require visualizinga completely different approach to the problem, asthe next example illustrates.

At several times in the secondary curriculum,students are exposed to a unit on finding the areasof triangles. The following is a problem that theymight encounter in such a unit.

Pr blem B (Take 1)Find the area of a triangle with vertices at (2, 1),(8, 3) and (4, 6).

S luti : Most of my students approach thisproblem by using Heron's Formula, which statesthat the area of a triangle with sides of lengthsa , b , and c is ( )( )( )csbsassA = , where sis the semiperimeter of the triangle. The distanceformula is used to find that a , b , and c are 5,

40 , and 29 . Substituting these values intoHeron's Formula (and using a calculator, since thecalculations with those radicals get pretty messy)gives the value of the area as exactly 13.

At this point, the observant student shouldmarvel at the result. Even though two of thesides had lengths that are irrational values andthese numbers are plugged into a formula thatrequires taking a square root, the answer came

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out a nice neat integer. This raises the questionof whether or not this will always be the case if you only use integer values for the coordinates of the vertices. Again, you could try repeating thecalculations using different points. Eventually,you would find that sometimes the area is not

an integer but rather is a multiple of 2

1 (for

example, if the point (8, 3) is changed to (7, 3),the area becomes 10.5.) However, after tryingseveral sets of points, you find that the areaalways seems to be half of an integer. Althoughyou haven't actually proved anything yet, you arenow ready to state a hypothesis and try to proveit. Before reading any further, see if you can statethe property that we will need to prove. Finished?My statement of the general problem appearsbelow. Keep in mind that there are other correct

ways of stating it.

Pr blem B (Take 2)Prove that if all of the vertices of a triangle haveinteger coordinates, then twice the triangle's areais always an integer.

S luti : We could try to prove this usingHeron's Formula, but it should quickly becomeclear that this approach would be messy. Itwould be necessary to show that the radicands (s - a )( s - b )(s - c ) is always a perfect square,and since we would have radical expressions for s ,a , b , and c , this looks like an algebraic nightmare.Consequently, we are motivated to find a differentapproach to the problem.

Start by drawing the original triangle (the onewith numerical vertices) on graph paper. It should

look like Figure 3. We want to try to visualize thearea using integers rather than radicals. After abit of thought, you might envision the triangleenclosed inside of a 6 x 5 rectangle, as shownin Figure 4 (the gridlines have been deleted forclarity.) It is now clear that the area A is simply30 - A 1 - A 2 - A 3, where each of the Aj (j = 1, 2,3) is the area of a right triangle, which is half theproduct of its legs.

For our numerical example, then, we have

( ) ( ) ( ) 13252

143

2

126

2

130 ==A ,

as previously found. Moreover, though, it shouldnow be apparent why the area must always behalf of an integer. Armed with this new insight,we are ready to proceed with our proof.

A proof is a connected series of statements intendedto establish a proposition. As before, we are goingto have to use words in these statements, andwe want to choose those words carefully so thatour proof says what we intend for it to say. Withthis in mind, let's carefully analyze each of thesestatements one at a time. We must first get rid of all numbers and replace them with variables as wedid in our first proof, identifying this as the giveninformation. This is easy enough.

Stateme t 1Let 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.

Next we have to describe the rectangle enclosingthe triangle, and this is actually a bit tricky. Fornow, let's claim the following:

Stateme t 2

A rectangle with its sides parallel to the

coordinate axes can be drawn so that X,Y, and Z all lie on the rectangle and at least one of the vertices of the triangle coincides with a vertex of the rectangle.

This is actually a pretty strong (and notcompletely true!) statement and wouldbe rather difficult to justify rigorously,but it seems obvious from our existingdrawing, so for now we're going toproceed with the assumption that it

Figure 3Specific example of the triangleproblem

Figure 4More general picture of thetriangle problem

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is true. Bear in mind that in a truly thoroughproof, this statement would require considerablejustification.

Clearly, there are many different ways to drawthe triangle and the rectangle based only onStatement 2. We can only deal with one suchorientation at a time, so we make the following

our next statement:

Stateme t 3Assume without loss of generality that vertex X is in the lower left corner of the rectangle and vertex Y lies on the vertical side to its right. Figure 4,then represents this particular configuration.

Effectively, we are claiming that the proof proceeds the same regardless of which trianglevertex coincides with which rectangle vertex, aslong as all of the conditions of Statement 2 aremet. The disclaimer in b l (often abbreviated\WLOG") is a great time-saver, but one should

exercise caution when making this claim.

One of my more cynical college professors onceinformed me that the words \without loss of generality" are almost always followed by astatement that results in a loss of generality.While he likely overstated the extent of itsmisuse, his overall message to use this statement

cautiously is worth remembering. For example,in our first problem you do not want to claim,\Assume WLOG that and m = 5 and n = 3."There can be little doubt that this conditionlimits the generality of the proof. By contrast, inour current problem, we could list each possibleconfiguration implied by Statement 2 and workout the area for each, but such an exercise wouldbe pointless and repetitive. Hence, in this case, theuse of \WLOG" seems justified.

The rest of the proof proceeds much as it would if numbers were involved.

Figure 5An obtuse trianglethat does not conformto the conditions of statement 2

Stateme t 4The dimensions of the rectangle, then, are (c - a ) x (f - b ). Also,

( )( )bf aeA =2

11 , ( )( )d f ecA =

2

12 , and ( )( )bd acA =

2

13 . (4)

Thus,

( )( ) ( )( ) ( )( ) ( )( )[ ]bd acd f ecbf aebf acA ++=2

1

( )( ) ( )( ) ( )( ) ( )( )[ ]bd acd f ecbf aebf acA ++= 22 .(5)

and (6)

We now need to interpret this algebraic statement (6) to explain why it establishes our proposition.

Stateme t 5Since a, b, c, d, e, and f are integers and the set of integers is closed under the operations of addition,subtraction, and multiplication, it follows that 2Amust be an integer.

In case you are not familiar with the use of theword \closed" in Statement 5, this just says thatany time you add, subtract, or multiply twointegers, the result will always be an integer. Theseclosure properties justify the conclusion that 2 A isan integer.

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

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In the Autumn of 2010,Dr. Hamilton

opened www.MathPickle.com toprovide practical help toK-12 teachers. He is alsoa founding member of theGame Artisans of Canada Canada's premier board game design group.

Pythagorean Crimes is written by the Greek ac-ademic, Tefcros Michaelides. In a compact \daVinci Code" style, the reader is led helter skel-ter through historical, mathematical, and artisticevents of the early 1900s - all to set up an excitingmurder trial.

By the end of the book I was thoroughly satisfied,but for the first part of the book, where themajority of the mathematical history is found, Iwas distracted: The self-reflections of the narrator,Michael Ingerinos, and the conversations with hisfriend, Stephanos Kandartzis, sometimes felt forced.

For example, to introduce a discussion of primenumbers, the author sacrificed authentic self-reflections in order to teach the mathematicallyunprimed:

\I knew that prime numbers were those thatcannot be divided except by themselves and thenumber one." P. 19

...But within a page, a question from the narratorshows that he is at a significantly higher level:

\...Isn't that Gauss' conjecture? If I'm not

mistaken, he worked out, but was unable toprove, how many prime numbers are smaller thana given number." P. 20

The motivation to expand the readership toinclude those that need an explanation of primenumbers is understandable, but it unfortunatelyundermines the believability of the narrator.

The believability of the narrator doesn't matterif you are reading for the purpose of getting a

B k ReviewGord Hamilton

Statement 2 to be true, each vertex of the trianglemust have at least one coordinate that is either amaximum or a minimum value within the triangle.That is always true if XYZ is right or acute, butif XYZ is obtuse, it is only true if one of thesides of the triangle lies on one of the sides of therectangle. Again, this is not an easy statement tojustify rigorously, but a couple of quick sketchesshould convince you that it is true. Thus, wayback at the beginning, we need to split this proof into two separate cases.

Stateme t 0

Case 1 | XYZ does not contain an obtuse angle.

Now that the proof for non-obtuse triangles iscomplete, Statement 6 should begin the proof of the obtuse case. The picture for the obtuse case

looks a bit different (see Figure 5), but otherwisethe proof should proceed pretty much the sameas the one that we just did. Try to prove the

obtuse case on your own. Choose your wordingcarefully, and always question whether or notyour statements say what you intend for them tosay. After writing your proof, have a friend readit and give you feedback about how clearly youmade your case.

The body of knowledge encompassed by

mathematics is constantly growing. The impetusfor additions to this body of knowledge is curiosity.New relationships are constantly being discoveredbecause someone perceives an interesting resultto a problem and starts wondering how andif this result can be generalized or extended.Although the examples in this article take youdown mathematical paths that have already beentrodden many times, they hopefully show you howsuch paths are found in the first place and havehelped you develop some of the thinking skillsrequired to experience mathematical adventuresof your own. Happy exploring!

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glimpse of mathematicians at the turn of the lastcentury. Here the book delivers with aplomb:

\This was the first time in the history of mathematicsthat someone had been bold enough to prove the ex-istence of a mathematical solution without illustrat-ing how it was to be constructed. The article Hilbertpublished in 1888 caused an uproar. The ultracon-

servative Kronecker, who doubted even the existenceof irrational numbers such as the square root of 2,dismissed the solution without further argument.Gordan himself who was known for his geniality andhis generosity toward young, talented mathemati-cians, responded angrily to the paper, saying, \Thisis not mathematics, but theology." As Lindemann,he described his former pupil' method as unheimlich- profane. Others, however, such as Arthur Cayley inCambridge and Klein studied the proof in detail and,having initially believed it to be impossible, endedup congratulating Hilbert warmly. Hilbert becamea fanatical proponent of proofs of existence. \In-side this hall," he would often say in his lectures,\there is at least one student who has more hairon his head than any other. We don't know whothat person is, nor is there any practical way of finding our. But this doesn't mean that he doesn'texist" P. 59

That is the kind of paragraph which made the booksuch a satisfying read. My reservations about thedevelopment of Stefanos and Michael are also off-set by insightful psychological sketches of otherpeople:

\... The top brass at [military] headquartersconsidered me and the three other mathematicianswho worked with the French to be geniuses. Thisdidn't stop them from burdening us with all sortsof chores. Whenever they could, however, just sowe wouldn't forget our place." P. 155

By the middle of the book Michael and Stefanoswere more believable and definitely interestinglyopinionated... For example, here is a discussion onthe newly constructed Eiffel Tower:

[Stefanos:] \This moment symbolizes a new era,the era of technology. What we have before usis a marvel of statics, dynamics, chemistry,electricity - they have all combined to make thisthe tallest construction in the world. And the

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

[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 suchcontraptions to justify itself." P. 84

That's an entertaining argument to eavesdrop.

I also like that the narrator's observations aresometimes flawed. For example, he totally missesthe melancholic mood in Picasso's paintings of harlequins and pierrots - declaring them \cheerfulstuff" P. 149.

However, despite my enjoyment of the book, Iam uneasy about recommending it for grade 12students.

First, the book will not appeal to every topmathematics student because of its impressivebreadth of subject matters - the reader shouldbe interested not only in mathematics but also inhistory and art. Without these interests, a personis liable to get irritated by the interjections - suchas the one page summary of the Dreyfus affair inthe middle of a description of the delegates at amathematics conference:

\French society in 1900 was being torn apart by ascandal revolving around Alfred Dreyfus, a younglieutenant convicted six years earlier of being aspy. After a farcical court-martial, the militaryhad sentenced him to hard labour for life and senthim to Devil's Island, a penal colony off the coastof French Guiana. The royalists, backed by theCatholic Church, seized the opportunity of theconviction to attack the republican constitution.The fact that Dreyfus was a Jew contributedto the rekindling of anti-Semitic feeling amongcertain sections of French society." P. 20

Second, many high schools will not want to exposetheir students to some of the passages describing abohemian lifestyle.

In a nutshell - I thoroughly recommend the book,but not necessarily for high school student.

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The Mathematics f Climate M elli gAdam H. Monahan

Introduction

Predicting the evolution of the Earths climatesystem is one of the most daunting mathematicalmodelling challenges that humanity has ever un-dertaken. The atmosphere and ocean display vari-ability over a bewildering range of space and timescales - from microns to thousands of kilometers,and from seconds to millions of years. The challengeis increased by the fact that the climate systemconsists not just of the atmosphere and the oceanbut also includes the cryosphere (frozen water inland and sea ice), the biosphere (life on Earth), and

the geosphere (the solid Earth). While descriptiveclimate science is a centuries-old discipline, modernclimate physics - with a quantitative focus onmechanism - is relatively recent. In fact, the rstglobal atmospheric circulation model was built onlyin the mid-1950s (a nice discussion of the historyof atmospheric modelling for weather forecastingis given in Harper et al. (2007)). Modern climatephysics is a fundamentally mathematical discipline,making use of tools and techniques from across thespectrum of modern applied mathematics. This

brief article discusses the interweaving of mathe-matics with modern climate science.

Climate Modelling and Budgets

An idea fundamental to climate modelling is thatof a budget. In everyday life, the budgets wemost often think about are budgets of money. Themoney you have may exist in different forms - incash or in a bank account; in Canadian dollars or

Euros or renminbi. Each of these forms of moneyrepresents a different pool. You can move moneyfrom one pool to another - depositing cash to a bankaccount, or changing currencies; we can call suchtransfers uxes. Fluxes dont change the totalamount of money you have (if we neglect servicecharges) - they just change the form that its in.You gain or lose money by earning or spending -these processes represent net changes in the totalamount of money you have. Earning is a source of money, while spending is a sink. Different poolsmay have different sources and sinks - if you buysomething with cash, thats a sink for that pool; if you use your bank card, thats a sink for your bankaccount. The amount of money you have in a givenpool increases with time if the sum (sources + uxes- sinks) is positive; it decreases with time if thissum is negative. The process of budgeting involvesunderstanding these processes - how much you havein each pool, what the uxes are between them,and what the various sources and sinks are. Thescience (and art) of mechanistic modelling involvesrepresenting these processes mathematically.

Mechanistic climate modelling is concerned withbudgets of physical, chemical, and biological quanti-ties - energy, momentum, water (in various phases),carbon (in various chemical compounds), terrestrialvegetation biomass, and many others. NewtonsSecond Law of Motion

F =d

dt(mv )

is a mathematical description of the momentumbudget: accelerations are changes in the mo-mentum pool of a physical object resulting fromsources and sinks of momentum which we callforces. Newtons Third Law of Motion - the forceof A on B is equal and opposite to the force of B onA - is just a statement that forces acting betweenobjects are momentum uxes that may change themomentum of each object (that is, an individualmomentum pool) but dont change the total amount

Adam Monahan is an associate professor in theSchool of Earth and Ocean Sciences at the University of Victoria. His research focuses on studying interac-tions between large and small scales in the atmo-sphere and ocean (the weather-climate connection).

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of momentum (the sum over all pools). In thesame kind of way, the First Law of Thermodynamicsis a description of the energy budget - energy isnot created or destroyed, just transformed betweenpools (gravitational potential energy, kinetic energy,internal energy, etc.).

Seen in this way, climate modelling seemsstraightforward. To build a climate model, all youneed to do is

1. determine what quantities are important tothe part of the climate system youre con-cerned with understanding or predicting,

2. decide what the important pools of thesequantities are,

3. come up with mathematical descriptions of the uxes between these pools, of the sources,

and of the sinks, and4. study the properties of the resulting mathe-

matical models to make predictions about theclimate.

So this is a breeze, right? Not really - none of these steps is easy. As always, the devil is in thedetails. All four steps require expertise in physics,chemistry, biology, or geology - and (particularly insteps 3 and 4) mathematics.

Mathematics and Climate Modelling

Consider the budget of some climatically signicantquantity - for example, the amount of CO 2 in theatmosphere (Figure 1). If Q is the amount of thisquantity that we have at some time t , then thebudget analysis for the time rate of change of Q wetalked about earlier can be written mathematicallyas the differential equation

dQdt

= sources + uxes - sinks

To make predictions about how Q changes in time,we need to understand the uxes, sources, and sinksand represent these mathematically. For example -what are atmospheric CO 2 sources and sinks, andwhat determines how strong these are? An im-portant natural source is respiration by organisms- the release of CO2 as a byproduct of extractingenergy from organic carbon (what we normally callfood). The ip side of this is a major sink ofatmospheric CO 2: photosynthesis, by which CO 2 istransformed into organic carbon. To model theseprocesses mathematically, we need to express ratesof photosynthesis and respiration (in vegetation, insoils, in animals) as functions of the variables we aremodelling - a wonderful problem of mathematicalbiology .

Figure 1: Estimate of global carbon cycle pools (or reservoirs) and uxes. Black numbers are estimates of pre-industrial (e.g. natural) values,while red numbers represent estimates of changes caused by human activities (as of the 1990s). From Chapter 7 of Solomon et al. (2007).

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There are other sources of CO 2 - not the least of which is the burning of organic carbon stored ingeological reservoirs (coal, oil, natural gas). Thestrength of this source that is, the rate at which weburn fossil fuels is largely determined by economicdecisions regarding resource extraction and energygeneration; modelling these emissions represents anexercise in mathematical economics .

And what about atmospheric CO 2 uxes? Well,CO2 dissolves in water - in fact, the oceans containmore than 50 times the amount of carbon thatthe atmosphere does - and CO 2 is constantly beingexchanged between these atmospheric and oceanicpools (Figure 2). This CO 2 ux between the atmo-sphere and ocean is very important but how do werepresent it mathematically? Physical chemistrytells us that the ux should be proportional to theair-water CO 2 concentration difference

CO2 ux water to air = k [COwater2 ] [COair2 ]

This equation is deceptively simple-looking: thecoefficient k bundles together a lot of very compli-cated physics. The air-water interface consists of two boundary layers (one in each uid) which areoften turbulent. Understanding and modelling thisturbulence - the strength and character of whichare responsible for mediating the exchanges of CO 2(and other quantities) between the uids - is thegreat unsolved problem of classical physics. Thisproblem involves more budgets - budgets of mo-mentum (for the winds and the currents), budgets

of energy (for temperatures), budgets of freshwater(evaporation and precipitation) - that are the do-main of the physical disciplines of mechanics andthermodynamics .

And what determines the concentration of CO 2in water? Well, this is a whole new budget problem- one with all the complications of the budgetfor atmospheric CO 2. Currents move CO 2 aroundalong with the water, photosynthesis in the oceanconsumes it, respiration releases it, CO 2 reacts withseawater to form bicarbonate HCO 3 and carbonateCO23 ions - and all of these processes need to berepresented mathematically. This involves morebudgets, more equations, and more complications;as should be clear, even a straightforward ques-tion like how much CO 2 is exchanged betweenthe ocean and atmosphere? can represent a veryinvolved problem of physics, chemistry, and biology

- and of the mathematics needed to model it all.Further complicating matters is the fact thatvery often, were interested not just in the totalamount of some quantity in the atmosphere orocean, but also in how the abundance varies fromplace to place. We dont experience globally-averaged temperature: we experience what thetemperature is right here. Therefore, we needmathematical models of how quantities vary inboth space and time - that is, partial differentialequations . These represent local budgets, in whichthe abundance of our quantity of interest at everypoint in space represents a different pool.

Figure 2: Air-sea CO 2 uxes (in moles of carbon per square meter per year) for the year 1995 as computed from a Global Climate Model (leftpanel) and as estimated from observations (right panel). Positive values indicate a CO 2 ux from the ocean to the atmosphere; negative valuesindicate uxes from the atmosphere to the ocean. While there is broad agreement between the observed and modelled uxes, there are manydifferences in detail. There is still plenty of work to be done in climate modelling. Adapted from Zahariev et al. (2008).

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For example, the partial differential equation forthe temperature eld T (x,y,p, t ) (in which pressurerather than altitude is used as the vertical coordi-nate, following common meteorological practice) is

cp T t

+ uT x

+ vT y

+ T p

= 2 q

In this equation - which is just a mathematicalstatement of the rst law of thermodynamics - ( u, v )is the horizontal velocity vector, is the verticalvelocity (in pressure coordinates), cp is the specicheat capacity of air at constant pressure, is theair density, and 2 is the heating rate associatedwith viscous dissipation in a turbulent ow. Thevector q is the spatial ux of heat energy throughthe atmosphere due to the emission and absorp-tion of electromagnetic radiation, phase changes of water, chemical reactions, and conduction. Each

of u,v,,,2

, and q are themselves ( x,y,p, t )-dependent elds with their own budgets (expressedas partial differential equations). Clearly, modellingclimate variability in both space and time can bevery complicated indeed.

Once weve got these partial differential equa-tions written down, how do we use them to studythe climate system? Some things we can learn bylooking at the form of the equations. In particular,by considering the relative sizes of different termswe can gain some insight into the relative impor-tance of different processes. We can even carry outformal asymptotic analyses of the equations, tosystematically throw away less important termsto simplify the models.

But even these simplied equations are still verycomplicated in general, and cant be solved byhand to make predictions. In this case, all we cando is replace the original equations with discreteapproximations appropriate for simulation by com-puters. That is, we need to make use of numerical

analysis . In these approximate models, only spaceand time scales above a particular threshold areexplicitly modelled. However, the smaller scalescannot generally be ignored - the nonlinearity of theoriginal partial differential equations leads to im-portant interactions between large and small scales.The effect of the smaller unresolved scales must beparameterised in terms of the larger resolved scales.There are different approaches to addressing theparameterisation problem, but some of the mostpromising make use of perspectives from statistical

physics . In particular, an emerging school ofthought takes advantage of the essentially randomcharacter of small-scale turbulence to make use of tools from probability theory and stochasticprocesses .

Such probabilistic perspectives are also useful

for appreciating the difference between weatherand climate. Robert Heinlein put it best: Cli-mate is what you expect, weather is what youget. This perspective is fundamentally proba-bilistic: climate is the probability distribution of outcomes for many rolls of dice, and weather isthe outcome of any particular throw. Changes inclimate are changes in the weighting of the dice.From a dynamical systems perspective, we canspeak of the slowly-evolving climate attractor incontrast to the fast weather trajectory.

Of course, all of these mathematical modelswould only be so much science ction without obser-vations to test them and to help estimate parametervalues - so statistical analyses play a fundamentalrole in modern climate science. These analyseshelp us understand our budgets. For example, theincreasing trend in atmospheric CO 2 concentrationis a result of sources being stronger than sinks anduxes (as we burn fossil fuels); statistical analysesof these trends help us understand just how outof balance these budgets are. Statistical analysesalso help us understand how different parts of the climate system are related - for example, howvariability in the Eastern Pacic sea surface temper-ature is related to large scale pressure distributionsin the El Nino - Southern Oscillation phenomenon.Understanding these relationships in observationshelps us both build and assess our mechanisticmodels. Furthermore, all of these mathematical

representations of sources, sinks, and uxes inmechanistic models have various parameters thatneed to be set. Some of these - the speed of light,for example, or the acceleration of gravity - arewell known. Others, such as the dependence of soilrespiration rate on soil temperature and moisture,are not so well constrained. Inverse modellingprovides a systematic framework for estimatingthese parameters so as to bring mechanistic modelsinto closest accord with observations and therebyto make better predictions.

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Conclusions

While the Earths climate system is intimidatinglycomplex, the reality of climate change compels usto build models for understanding the system andpredicting its future behaviour. These models,expressed in the language of mathematics, rangeacross a hierarchy of complexity . At one endof this hierarchy are idealised conceptual modelsthat are useful for developing understanding butcannot be expected to be quantitatively accurate.At the other end of the hierarchy are the fullycomplex Global Climate Models which representthe physics, chemistry, and biology of the climatesystem - but which are so complicated that they canonly be studied numerically using computers. Theworlds most powerful computers require severalmonths of computational time to perform 1000-year

simulations with these complex models; increasesin computer power are generally consumed by in-creases in model complexity or resolution. In fact,models from across the entire range of this hierarchyplay important roles in understanding and predic-tion, and mathematics provides the fundamentalframework by which these models are constructed,studied, and rened.

Further Reading

The climate literature is extensive and can beintimidating to the uninitiated. An comprehensiveoverview of the state of the art of climate scienceis given in the most recent report of WorkingGroup I of the Intergovernmental Panel on ClimateChange, available as a book (Solomon et al., 2007)or online at http://www.ipcc.ch . A more succinctintroduction to climate modelling is presented byThorpe (2005), while Weaver (2008) provides a

non-technical discussion of climate change (witha particular focus on Canada). There too manygood books on the physics of the climate to providean exhaustive list here; an excellent introductorytext is Hartmann (1994), while a more quantita-tively detailed treatment is provided in Peixotoand Oort (1992). Colling (2001) and Wallace andHobbs (2006) are good introductions to physicaloceanography and meteorology, respectively; deepertreatments of modern geophysical uid mechan-ics are given in Holton (2004) and Vallis (2006).

Sarmiento and Gruber (2006) present a detaileddiscussion of the oceans role in the global carboncycle from a modelling perspective, and an excellentintroduction to statistical analyses of climate datais provided in Wilks (2005). Unfortunately outof print, Haltiner and Williams (1980) is a classicintroductory text on numerical methods for meteo-rological modelling.

Acknowledgements

The author would like to thank Jim Christian forproviding Figure 2, and an anonymous reviewer forvery helpful comments on the original version ofthe manuscript. The author is supported by theNatural Sciences and Engineering Research Councilof Canada.

References

Colling, A., 2001: Ocean Circulation . Butterworth-Heinemann.

Haltiner, G. J. and R. T. Williams, 1980: Numeri-cal Prediction and Dynamic Meteorology . Wiley,New york, 477 pp.

Harper, K., L. W. Uccellini, E. Kalnay, K. Carey,and L. Morone, 2007: 50th anniversary of opera-tional numerical weather prediction. Bull. Amer.Meteor. Soc. , 88, 639650.

Hartmann, D. L., 1994: Global Physical Climatol-ogy . Academic Press, San Diego, 411 pp.

Holton, J. R., 2004: An Introduction to DynamicMeteorology . Academic Press.

Peixoto, J. P. and A. H. Oort, 1992: Physicsof Climate . American Institute of Physics, NewYork.

Sarmiento, J. L. and N. Gruber, 2006: Ocean Biogeochemical Dynamics . Princeton UniversityPress.

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

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

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

Vallis, G. K., 2006: Atmospheric and OceanicFluid Dynamics: Fundamentals and Large-ScaleCirculation . Cambridge University Press.

Wallace, J. M. and P. V. Hobbs, 2006: Atmo-spheric Science: An Introductory Survey . Aca-demic Press.

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

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

Zahariev, K., J. R. Christian, and K. L. Den-man, 2008: Preindustrial, historical, and fertil-ization simulations using a global ocean carbonmodel with new parameterizations of iron limi-tation, calcication, and N 2 xation. Progress in Oceanography , 77, 5682.

Problem 1: Find all positive integers n such that log2008 n = log 2009 n + log 2010 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 + log k +1 n logk n =1

logn (k + 2)+

1logn (k + 1)

1

logn k

=logn (k + 1) (log n k 2 logn (k + 2)) + log n (k + 2) (log n k 2 logn (k + 1))

2logn (k + 2) log n (k + 1) log n k> 0

since k2

> k + 1 . Hence, if we take k = 2008 we get log2009 n + log 2010 n > log2008 n.

Problem 2: Find the smallest value of the positive integer n such that (x 2 + y2 + z 2 )2 n (x 4 + y4 + z 4 ) for any real numbers x,y,z.

Solution: The given inequality can be transformed into an equivalent useful format:

(n 3) x 4 + y4 + z 4 + x 2 y22

+ y2 z 22

+ z 2 x 22

0.

Pi i the Sky Math Challe gesSolutions to the problems published

in the 13th issue of Pi in the Sky

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Since the minimum value of ( x 2 y2 )2 + ( y2 z 2 )2 + ( z 2 x 2 )2 is 0 , we must have(n 3) (x 4 + y4 + z 4 ) 0, hence n 3.

An alternative solution, using a geometric argument was given by Carlo Del Noce, Gen-ova, Italy.

Problem 3: Let a be a positive real number. Find f (a ) = max x R {a + sin x, a + cos x}.

Solution:This problem is trivial. Since max x R {a +sin x, a +cos x} is clear a +1 then f (a ) = a +1 .

Problem 4: Prove that the equation x 2 x + 1 = p(x + y) where p is a prime number, hasintegral solutions (x, y ) for innitely many values of p.

Solution:Let us assume by contradiction that the equation has integral solutions ( x, y ) only for

a nite number of prime numbers, among which the greatest is denoted by P. If we takex = 2 3 5 .... P then x 2 x + 1 = x(x 1) + 1 is not divisible by any of the primenumbers which are P. Hence x 2 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 equationx 2 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 everyn 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 obtains3g(n ) = 2 g(n + 1) for every integer n. Assume by contradiction that there is an integer msuch that g(m ) = 0 . Then

2g(m ) = 3 g(m 1) = 3 2 g(m 2) = ... = 3 k g(m k)

for any positive integer k. Hence 3k |g(m ) for any positive integer k and therefore we musthave g(n ) = 0 for any integer n.

A correct solution was received from Carlo Del Noce,Genova, Italy.

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Problem 6: In ABC, we have AB = AC and BAC = 100 . Let D be on the extendedline trough A and C such that C is between A and D and AD = BC. Find DBC.

Solution:Ahmet Arduc from Turkey, submitted ve solutions to this problem. The following is his

rst solution.

Let = m (CBD ). Draw the equilateral triangle AED.Since ACB BAE (SAS ) we get AB = BE,m (BAE ) = m (BEA ) = 40 , m (ABE ) =100 . As AB = BE and AD = DE, ABED is a deltoid with BD the axis of symmetry.Hence m (DBE ) = m (DBA ) = 50 = 40 + . Thus = 10 .

Carlo Del Noce also solved this problem.

\It is pr ve that the celebrati f birth ays is healthy. Statisticssh w that th se pe ple wh celebrate the m st birth ays bec methe l est." S. en Hartog, Phd Thesis, University of ronigen

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Math Challe gesPRoBLEM 1:

Find all the real pairs ( x, y ) such thatlog3 x + logx 3 2 cos y .

PRoBLEM 2:

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

PRoBLEM 3:

In decimal representation the number 2 2010 has m digits while52010 has n digits. Find m + n.

PRoBLEM 4:

Find all the polynomials P(x) with real coefficients such thatsin P(x) = P (sin x ), for all x R.

PRoBLEM 5:

The interior of an equilateral triangle of side length 1 iscovered by eight circles of the same radius r .Prove that r 17 .

PRoBLEM 6:

Prove that in a convex hexagon of area S there exist threeconsecutive vertices A, B and C such that

Area ( ABC ) S 6

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