Mathematica 2014

MATHEMATICA 2014

DCM ATHEMATICA 2014

E DITORS Xiaofeng Xu Leslie Leung

S UPERVISOR Dr Purchase

C OVER D ESIGNED B Y Kenza Wilks

A RTICLE C ONTRIBUTORS

Kaichun Liu

Charlie Sparkes

Mitchell Simmonds

Xiaofeng Xu

Leslie Leung

Ayman D’Souza

Faraz Taheri

Mr Ottewill

Harry Goodhew

Dr Purchase

Kit George

Table of contents

What is e?

By Kaichun Liu

A-Level Statistics in Investment Banking and Portfolio Theory

By Charlie Sparkes

Mathemagics

By Mitchell Simmonds

How to trap the prisoners?

By Xiaofeng Xu

An Introduction to Modular Arithmetic and its Applications

By Leslie Leung

Proof by Induction or Contradiction

By Ayman D’Souza

Cavalieri’s Principle

By Faraz Taheri

Where’s the policeman?

By Mr Ottewill

Hilbert's 6 th Problem

By Harry Goodhew

BMO question – two equations, three unknowns

By Kit George

How calculators calculate

By Dr Purchase

New Mathematics for an Old Problem

By Mr Ottewill

What is e? By Kaichun Liu

Before writing something really related to the topic, I would like to draw your attention to π first. What is π? If I ask my little sister who is 12 years old what π is she can definitely tell me that π is a constant and has the value of 3.1415926…… I have no idea how many decimal places she can count to but I am pretty sure it will be enough for calculations. However, what is the definition of π,or what does π actually represent? The question is also easy and I will answer for my little sister – it is a mathematical constant that is the ratio of a circle's circumference to its diameter. All right, what if I replace π with e and ask you the same question? If you can recall its value to the nineth decimal place immediately please text Dr. Purchase and see if he is impressed. However, it is not enough, so far, you have only got 1 mark. WHAT IS THE DEFINATION? WHAT DOES e STAND FOR?!!! To be honest, I myself actually had no idea. Thus I asked help from the mighty Wikipedia. Wikipedia generously provided me with a beautifully organized definition -- The mathematical constant e is the base of the natural logarithm. “ What is natural logarithm then?” The natural logarithm, formerly known as the hyperbolic logarithm, is the logarithm to the base e, where e is an irrational constant approximately equal to 2.718281828459.

Did you really tell me anything?

So what is e? Let us take a look at this example first.

Let’s assume that a cell will split into two in a day, so mathematically, after a day, we will get two new cells. If we need to write a formulae of how many cells we ݕ ൌ 2 ௫ However, according to cell biology, each 12 hours, which the split is halfway through, there are a number of new cells which equals the total number of the original cells have been produced and have started splitting into next generation. In fact, the growth speeds up as time goes—although the rate of splitting (the ratio) remains at 100%, the base number is continuously growing, which makes will get after x days, that will be

the 100% arithmetically bigger that the 100% a moment ago.

Thus a day can be divided into two parts, within each part the total number of cells grows by 50%. ݃ ݎ ݋ ݐݓ ݄ ൌ ሺ1 ൅ 100%2 ሻ ଶ ൌ 2.25 In fact, if there are 100 cells at first, after 24 hours, we can get 225 cells. If we can chage the rule of splitting and make it “producing 1/3 of the cells after 8 hours ” The formulae would be ݃ ݎ ݋ ݐݓ ݄ ൌ ሺ1 ൅ 100%3 ሻ ଷ ൌ 2.37037 … If we make the split carry on forever and continuesly, the formulae would be ݃ ݎ ݋ ݐݓ ݄ ൌ lim ௡՜ஶ ൬1 ൅ 100%݊ ൰ ௡ ൌ 2.718281828 “Coincidently”,the number we get here just equals the “e” When the rate of increasing stays at 100%, within a unit of time the cells can only increase to 271.8% In fact e shows up whenever systems grow exponentially and continuously: population, radioactive decay, interest calculations, and moreover it represents the idea that all continually growing systems are scaled versions of a common rate. e shows up whenever systems grow exponentially and continuously: population, radioactive decay, interest calculations, and more.

A-Level Statistics in Investment Banking and Portfolio Theory By Charlie Sparkes Investment banking is often made out to be like an impossible, unpredictable science that only those audacious enough to wear pin striped suits can ever fathom. However, as you may know, a crucial part of an investment banker’s day to day regime requires a lot of pondering over the magical relationship between risk and return and how to optimize the performance of a portfolio based on past and predicted data. Or, in S1 terms, this is the relationship between expected return (E(R) - return), standard deviation (S - risk) and the correlation coefficients (PMCC – just a coefficient in an equation). So, as this is a Maths essay, here is the inevitable set of data around which my explanations will revolve 1 . % Return Probability Security A Security B 0.25 20 45 0.50 10 25 0.25 0 5 Total = 1.00 Total = 30 Total = 75 S A = 7.07% S B = 14.14% Notation: R = % return, W A + W B = weight of portfolio invested in respective securities = 1 (in total), P = % return on portfolio of A and B. To calculate (a rather generalized) return on investments, using the data we can derive that; E(R P ) = W A E(R A ) + W B E(R B ) – i.e the weighted average return on A and B. So if the investor split his or her investment equally in A and B then, using the data; E(R P ) = 0.5(10) + 0.5(25) = 17.5% expected return on investment in the portfolio. Although I may have made some sweeping assumptions here, portfolio theory (and the law of large numbers) does suggest that for some overcompensating discrepancies that have been made, there will be some other discrepancies to counter these externalities, so the calculated expected returns value isn’t far off E(R A ) = 10% E(R B ) = 25% Var(R A ) = 50% Var(R B ) = 200%

1 An Introduction To The Stock Exchange – Janette Rutterford

the truth if you have a very large portfolio. Risk on the other hand is very hard to quantify since it very much depends on future events that even Nostradamus would struggle to predict, for example, very few people categorically predicted the financial crash of 2007 which was why it hit so many financial institutions, households, businesses (so basically everyone) hard. However, the ability to assess the volatility of listed assets using past data to perhaps try to model future predictions is an extremely important ability, for example, a hedge fund manager called John Paulson made $3.7 billion during the financial crisis 2 as he saw the weakness in the subprime mortgage market first and short sold every appropriate share/collateralized debt obligation (CDO) he could. He’s now worth $11.5 billion 3 . So, now I’ve established being very good at Maths can make you very rich, it’s back to the calculations for risk or, in S1 lingo, the variance. The variance of any security can be written as; ∑ ൫R ୔୧ െ EሺR ୔ ሻ൯ ଶ PሺR ୔୧ ሻ ୧ୀ୬ ୧ୀଵ From this, we can sub in the equation W A A + W B B = P and then expand the brackets to give; V P = S P 2 = W A 2 S A 2 + W B 2 S B 2 + ሺ∑ ൫R ୅୧ െ EሺR ୅ ሻ൯൫R ୆୧ െ EሺR ୆ ሻ൯pሺR ୧ ሻ ୧ୀ୬ ୧ୀଵ ሻ As you may have noticed, the final part of this long equation is actually the equation for the covariance of returns for A and B (COV AB ). The covariance is the absolute value that measures how one variable changes with respect to another, where positive values indicate that as one variable increases, so does the other variable, and of course the opposite applies for a negative value. The product moment correlation coefficient, as seen in S1, is simply a version of the covariance (so we call the covariance the determinant of the PMCC) that is scaled by the product of the standard deviations of A and B to give the PMCC a value that is always within the parameter; െ1 ൑ PMCC ൑ 1 Thus, we can say that; COV ୅୆ ൌ PMCC ୅୆ S ୅ S ୆ = ∑ ൫R ୅୧ െ EሺR ୅ ሻ൯൫R ୆୧ െ EሺR ୆ ሻ൯pሺR ୧ ሻ ୧ୀ୬ ୧ୀଵ And subbing this into the equation for the variance of your portfolio gives; V P = S P 2 = W A 2 S A 2 + W B 2 S B 2 + 2W A W B S A S B PMCC AB This is known in the financial industry as the ‘Portfolio Risk Equation’. The PMCC is now the main player in determining the risk of investing in these two securities, therefore, by using the data from the table on page 1 we can deduce that; V i = S i 2 = ∑ x ୧ ଶ ሺPሺX ൌ x ୧ ሻ െ ሺEሺRሻሻ ଶ = 2W A W B

2 The Long Short – Michael Lewis 3 http://en.wikipedia.org/wiki/John_Paulson

∑୅ ౟ ୆ ౟ ି ሺ∑ఽ ౟ ሻሺ∑ా ౟ ሻ ౤ √ሺ∑୅ ౟ మ ି ሺ∑ఽ ౟ ሻ మ ౤ ሻሺ∑୆ ౟ మ ି ሺ∑ా ౟ ሻ మ ౤ ሻ =

ସ଴଴ ൫√ଶ଴଴൯ሺ√଼଴଴ሻ = 1

PMCC =

(indicates a perfect positive correlation)

Therefore;

V P

= S P

2 = W A

2 S A

2 + W B

2 S B

2 + 2W A

W B

S A

S B

Simplified;

V P

= S P

2 = (W A

S A

+W B

S B

) 2

So;

S P S B Or to put in it simply, if the returns on A and B are perfectly positively correlated, the risk of the portfolio return (in terms of standard deviation) is just the weighted average of the risks of the constituent assets’ returns. At first impressions then, this may seem to be a positive outcome as the risk is quantifiable and linear and as is the expected returns. Thus, you can model your risk (x-axis) against return (y-axis) as a uniform straight line and make informed decisions on where to invest. However, it is because the two securities have a PMCC of 1 that means investing in these two assets simultaneously brings the same reward as investing all your money in just A or B, hence, we can say that no risk is eliminated and that all this maths has reaped no reward. Or, we could look at the flip side, investment bankers may be interested in buying two stocks that have a PMCC of -1, i.e. as one increases in value, the other decreases in value by the same amount. If we plug PMCC = -1 into the ‘Portfolio Risk Equation’, we get; V P = S P 2 = W A 2 S A 2 + W B 2 S B 2 - 2W A W B S A S B Simplified; V P = S P 2 = (W A S A –W B S B ) 2 So; S P = W A S A – W B S B This equation (that is slightly more complex for non-integer negative values of the PMCC - but nevertheless the theory holds) is very interesting as it implies that risk can be reduced or even completely nullified and the standard deviation would therefore be 0. In fact, for a given set of data (again, let’s use the table on page 1) there is a unique combination of W A : W B that can be found because if the two securities (with a PMCC of -1) ‘are combined in inverse proportion to the ratio of the standard deviations of their respective returns, so the resultant portfolio will yield a constant = W A S A + W B

return and hence be riskless’ 1 . For example, using the data;

W ୅ W ୆

ൌ S ୆ S ୅

So;

W ୅ W ୆

ൌ 14.14 7.07

And;

W B

= 2W A

Since W A + W B =1, then; ଵ ଷ And by plugging this into the equation for returns, we get; W A = ଶ ଷ & W B =

E(R P ଵ ଷ (25) = 15% So, if you have the exact securities A and B as above, and two thirds of your portfolio consists of A and the rest B, then you will get 15% return with 0% risk. Smashing. These equations will actually produce curves on a graph of return against risk, so investors can choose shares with certain levels of risk and return, with the variables being; S, E(R P ), and W A :W B , that they feel will maximise profit (whilst bearing in mind the relative risk associated with the investment). To conclude though, I must mention that the chances that past data will perfectly link to future market fluctuations and hence that the inductive reasoning in the calculations will give you the undisputed values for risk and return is almost as close to nil as you can get. Although a famous stock investor called John Templeton did once say “The four most dangerous words in investment are; ‘this time it’s different’”, suggesting market trends actually mean everything. However, many theories that have been formulated over the years in many different fields of academia have certain annoying parameters that are tricky to get round. Nevertheless, I feel that the underlying principle here that is really interesting is that if an investor can combine any number of securities and combine them in a portfolio, they can expect to get the weighted average of the returns with less than the weighted average of the risk (given the PMCC is less than 1 which reduces the value of the portfolio risk equation and hence brings the value of the standard deviation below that of the returns you get from the same assets). I also think it is worth mentioning that these ideas are not restricted to situations with only two assets. The general equation for return (or expectation in S1) is simple; you just add a summation sign (Σ) and set the limits between 1 and ‘n’ for the % return and portion of portfolio taken by asset ‘i’. The equation for risk is as follows; V ୔ ൌ ෍W ୧ ଶ S ୧ ଶ ୧ୀ୬ ୧ୀଵ ෍෍W ୧ W ୨ S ୧ S ୨ PMCC ୧୨ ୨ୀ୬ ୨ୀ୬ ୧ୀ୬ ୧ୀଵ I leave you with a quote from the late Phillip Fisher, an American stock investor, ) = ଶ ଷ (10) +

that I think perfectly encapsulates the justification for my essay; “The stock market is filled with people who know the price of everything but the value of nothing”

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Division by 11

Here are four ways for different types of numbers:

1. If the sum of every other digit, starting with the first, is equal to the sum of every other digit starting with the second, then the number is evenly divisible by 11. Try 13057. 1+0+7 = 3+5, therefore it should divide evenly by 11. And indeed it does: 13057 / 11 = 1187. 2. If all the digits are the same and there's an even number of digits, then the number is evenly divisible by 11. 333,333 - Yes. 3,333,333 - No. 3. If the number is a 3 digit number with different digits, add the two outside digits. If the difference between the sum and the middle digit is 11 then 11 divides evenly into the 3 digit number. If the sum is the same as the middle digit, then 11 will also divide evenly into the number. Try 484. 4 + 4 = 8 which equals the middle digit so 11 divides into 484 evenly. How about 913? 9 + 3 = 12 and 12 - 1 = 11 so 913 is evenly divisible by 11. 4. If the digits are different, count them from the right and then add the numbers in the odd positions and the even positions. Subtract the smaller number from the larger. If the difference is evenly divisible by 11, so is your original number. Take the number 181,907. The numbers 8,9, and 7 are in the odd positions. They sum to 24. The numbers 1,1, and 0 are in the even positions. They sum to 2. Subtract 2 from 24 to get 22. 22 divides by 11 into 2, so 181,907 is evenly divisible by 11.

Mathemagics By Mitchell Simmonds

After performing magic for many years, you learn that magic isn’t simply about sleight of hand; it also about probability and maths. Whilst many will be familiar with the standard: ‘Think of a number. Double it. Add 10. Halve it. Take away your original number. Your answer is five,’ there is more to Mathemagics than that. In fact, some of the best card tricks invented revolve around Mathematics. The best way to explain how theory can be used in magic is to reveal a trick itself. Again, many will be familiar with the 21 card trick in which three rows of seven cards are laid out and a volunteer picks a card. The volunteer tells the magician what row their card is in and this is repeated twice further. After this, the magician can name the card. In my opinion this trick lacks imagination. It is dull, boring and a waste of the audience’s time. Magic is all about amazing your spectators, and giving them that ‘Wow!’ moment. The only way to do this is to make it appear that maths is not part of the magic trick, even though it may be the very basis of the trick. Without further ado I will explain, Fitch Cheney’s Five-Card Twist, which I feel is a perfect example of Mathemagics. The magician exits the room leaving his assistant with the audience. She gives a full deck of ordinary playing cards to an audience member and asks him/her to shuffle it and then pick out any five cards. The assistant then takes the cards, looks at them and places one card face down and the four others face up. The magician is then allowed back in; he takes a look at the cards and in an instant names the card that is face down! How is it done? The explanation is simple; the assistant uses a pre-arranged code with the magician to tell him what the card is. As there are five cards and 4 suits, at least two of the cards have to be of the same suit. The assistant therefore chooses one of these cards to be the hidden one and places the others face up. For example, if there are two hearts, the assistant hides one of these hearts and the other heart is the first card on the left. Therefore, when the magician sees the first card, he instantly knows the suit. Now, the clever part - how does the magician use the other three cards to determine the value of the card? There are 13 values in a deck of cards. The order goes as follows: A,2,3,4,5,6,7,8,9,10,J,Q,K. You have to imagine these numbers as a clock. So after King comes Ace, for example. It is a rule, that if you pick any two cards, their values can be, at most, six positions apart. So, for example, if you pick 2 and 9, there is 9,10,J,Q,K,A,2- six positions apart. The lowest card must be revealed, when the assistant turns over one card, so that the card that is hidden is either 1,2,3,4,5 or 6 values above it. The magician must then deduce what value it is from the other three cards and count that number from the value of the first card to gain the value of the hidden card…simple!

Ace is low and King is high and the suits also go as follows: Clubs Hearts Spades Diamonds Low---------------------------------------------High

So the Ace of Clubs is the lowest card and the King of Diamonds is the highest card. The assistant must find the Lowest, Middle and Highest ranking cards- let’s call these L, M and H. The order of these cards determines how many numbers to add the lowest value card. So, LMH=1 LHM=2 MLH=3 MHL=4 HLM=5 HML=6 The example therefore is as follows: The spectator picks out 5 cards. The assistant puts the 3 Clubs on the far left. The other cards are arranged as the 7 of Hearts, the 3 of Spades and the Ace of Diamonds. After carefully thinking, the magician says that the card faced down is the 8 of Clubs! Everyone is amazed! You may have to read it through a couple of times, to understand the trick, but I hope that this article has changed your mind on Mathemagics and that you can see quite how complex maths tricks can be.

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Division by 6

The number has to be even. If it's not, forget it. Otherwise, add up the digits and see if the sum is evenly divisible by 3. It it is, the number is evenly divisible by 6. Try 108,273,288. The digits sum to 39 which divides evenly into 13 by 3, so the number is evenly divisible by 6. If you want, you can keep adding numbers until only one digit remains and do the same thing. So in this case, 3 + 9 = 12 and 1 + 2 = 3, and 3 is evenly divisible by 3!

How to trap the prisoners?

By Xiaofeng Xu

This is a story I have recently read from a book which draws my great interest. It is said that once there was a cave an emperor or used to trap the prisoners. The prisoners tried to find a way to escape but the guards were somehow always one step ahead of them. The prisoners, paranoid as ever, tried to seek out the traitor who had leaked out their escape plans, only to find out that it was due to the

shape of the cave. How does this happen? Because the shape of the cave is an ellipse, the prisoners are just at one focus of the ellipse and the guards are located at the other focus of the ellipse. Therefore, the sound of the prisoners will just be heard by the guards after reflection.

Can this be proved by vigorous mathematics proof ? The answer is ‘yes’. The rule I get from the story is that after reflection, the ray passing through one focus of an ellipse will then pass through the other focus of the ellipse. This can be shown by the diagram below. Proof:

Firstly, after transferring or rotating, any ellip coordinate system with its two foci on the x

se can be shown in the rectangular -axis. Therefore, by proving that the

x y

2 A B + = ( A is greater than B ) obeys the 2 1

ellipse with the function

rule I have mentioned, I can say that this is right for any single ellipse. Here is the graph for the model. The line coming from the focus F the other focus F 2 . The normal line is drawn perpendicular to the tangent through the point of contact. The incident angle is marked as l is the tangent to the ellipse. The ray is 1 . At the point of contact T it reflects and passes through α , and the reflective angle as

β .

(Picture drawn by Geometer’s

Sketchpad)

My way to prove β is equal to α is to prove that the complementary angles of

β and α are equal. So two perpendiculars are drawn from the foci to the line l with the feet G and H respectively. The problem then obviously turns into proving that the two triangles are similar. Here come the details.

x y

2 A B + = ( A > B ), I can derive the coordinates of the two 2 1

From the equation

2 2 2 C A B = − , so the coordinates of foci F 1 and F 2 are ( C , 0) and

foci. Let’s say

( C − , 0) respectively.

0 y = , the tangents(on the left and right) are parallel with the

Apparently, when

y-axis. The ray will just reflect back in the same way as it comes. So it will pass through both of the points.

When y ≠ , we can get the derivative function of the ellipse by implicit 0

derivation(1).

2 dy B x dx A y = −

We assume the point of contact T have the coordinate ( , ) a b .

2 B a A b

= −

The gradient of the tangent:

gradient

So the line l , which is the tangent, has the equation 2

2 2

0 B ax A by A b B a + − − =

The distance between the right focus F 1 and point T: F 1 T =

( ) a C b − +

Similarly, the distance between the left focus F 2 and point T: F 2 T =

( ) a C b + +

Because we have the equation of the tangent and the coordinates of the two foci, we can calculate the distance from the foci to the line.

0 A x B y C + + = :

This is the formula. The distance from a point (x 1

y 1 ) to a line

0 1 0 1 + + + 2 2 A B A x B y C

(2)

So the distance from the right focus to the line l : F 1 G= 2

2 2 − − B aC A b B a 2 2

2 2 2 2 ( ) ( ) B a A b +

The distance from the left focus to the line l : F 2 H= 2

2 2 − − − B aC A b B a 2 2

2 2 2 2 ( ) ( ) B a A b +

2 2 2 + + B aC A b B a 2 2

2 2 2 2 ( ) ( ) B a A b + Because point T(a , b) is on the ellipse.

a b

2 B B a A b A B b B a + = ⇒ = − 2 2 2 2 2

1 + = ⇒

A B

2 2 2 − − B aC A b B a 2 2

2 2 2 − − 2 2 ( ) ( ) B a A b FG F H B aC A b B a B aC A b B a + = = − − − + + 2 2 1 2 2 2 2 2 2 2 2 B aC A b B a

2 2

2 2 ( ) ( ) B a A b + 2 2

a A −

2 2 2 2 B aC A B A C B aC A B − = + 2 2

a A +

What’s more, an interesting thing can be found that the ratio of F 1 T and F 2 T is the same as the ratio above!

1 FT F T a C b − + = + + 2 2 ( ) a C b ( )

a Ca C b − + +

2 2

a Ca C b + + +

Because point T(a , b) is on the ellipse.

a b

2 B B a A b A B b B a + = ⇒ = − 2 2 2 2 2

1 + = ⇒

A B

So,

2 B a Ca C B a A B a Ca C B a A − + + − + + + − 2 2 2 2 2 2 2 2 2 2

1 2 FT F T

2 2 A A B − A B −

2 a Ca C B − + +

2 a Ca C B + + +

2 a Ca A − +

2 a Ca A + +

a A −

(

)

a A +

(

)

a A −

a A +

1 2 FT FG F T F H = 1 2

and

FGT F HT ∠ = ∠ =

So, Triangle F 1 GT is similar to Triangle F 2 HT. (3)

⇒ ∠ = ∠ ⇒ =

1 2 FTG F TH β α

All considered, the proposition is now proved!

This proof shows a property of an ellipse, which is of great use in real life. We can find it in modern medical equipments, where people use it to concentrate high energy to break the stone in human bodies. It also has connection with sonar system used to locate submarines. Exciting, isn’t it? Finally, ellipses have a lot of properties. Other properties can all be proved mathematically. This is, in my opinion, why mathematics is of great fun and challenge. You may find the other properties through various ways and have a try to prove it on your own! Thank you!

(1) http://en.wikipedia.org/wiki/Implicit_function (2) http://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line (3) http://mathforum.org/library/drmath/view/57674.html (prove triangles similar

An Introduction to Modular Arithmetic and its Applications By Leslie Leung Modular arithmetic can be defined as a system of arithmetic for integers. 4 It is a mathematical tool that is widely used in daily life as well as solving mathematical problems.

Explanation of notation: a ؠ b ሺmod nሻ

ሺread as ′

a is congruent to b modulo n ′ ሻ

ܽ and

ܾ is a multiple of

݊ , or in other

It means that the difference between

ܽ and

ܾ give the same remainder when divided by ݊ .

words, when both

The most common use of modular arithmetic is perhaps the 12-hour clock. A day has 24 hours but a cycle of the clock lasts only 12 hours. Therefore, when the 21 ؠ 9 ሺmod 12ሻ . ࡿ࢚ࢇ࢚࢘ࢋ࢘ ࡽ࢛ࢋ࢙࢚࢏࢕࢔: ࢃࢎࢇ࢚ ࢚࢏࢓ࢋ ࢊ࢕ࢋ࢙ ࢚ࢎࢋ ૚૛ ࢎ࢕࢛࢘ ࢉ࢒࢕ࢉ࢑ ࢙ࢎ࢕࢝ ૚૙૙ ࢎ࢕࢛࢙࢘ ࢇࢌ࢚ࢋ࢘ ࢔࢕࢕࢔? Answer: 100 ؠ 4 ሺmod 12ሻ , so the clock shows 4 o'clock. Before moving on to explore how modular arithmetic can be used to solve other problems, a few 'laws of modular arithmetic' has to be explained. 1) Addition/Subtraction time is 21:00, the clock shows 9 o'clock because

If a ؠ c ሺmod nሻ, b ؠ d ሺmod nሻ , then ሺa െ bሻ ؠ ሺc െ dሻ ሺmod nሻ 2) Multiplication Also,

ሺa ൅ bሻ ؠ ሺc ൅ dሻ ሺmod nሻ

If a ؠ c ሺmod nሻ, b ؠ d ሺmod nሻ, then ab ؠ cd ሺmod nሻ Proof:

c ൌ a ൅ pn, d ൌ b ൅ qn , where

p is an integer. cd ൌ ሺa ൅ pnሻሺb ൅ qnሻ ൌ ܉܊ ൅ aqn ൅ bpn ൅ pqn ଶ

Let

4 http://en.wikipedia.org/wiki/Modular_arithmetic

ܾܽ . Therefore, using

The only term in the expression that is not a multiple of n is

cd ൌ ab ሺmod nሻ

the addition law,

3) Power

If a ؠ b ሺmod nሻ, then a ୩ ؠ b ୩ ሺmod nሻ Proof:

Let b ൌ a ൅ pn ՜ b ୩ ൌ ሺa ൅ pnሻ ୩ =

܉ ܓ ൅ ൫ ୩ ଵ ൯ a ୩ିଵ ሺpnሻ ଵ +

൫ ୩ ଶ ൯ a ୩ିଶ ሺpnሻ ଶ +......+

ሺpnሻ ଶ୩

From the binomial expansion, the only term not divisible by ݊ is ܽ ௞ . ࡼ࢘࢕࢈࢒ࢋ࢓ ૚: ࢃࢎ࢟ ࢏࢙ ࢇ࢔ ࢏࢔࢚ࢋࢍࢋ࢘ ࢊ࢏࢜࢏࢙࢏࢈࢒ࢋ ࢈࢟ ૜ ࢏ࢌ ࢇ࢔ࢊ ࢕࢔࢒࢟ ࢏ࢌ ࢏࢚࢙ ࢊ࢏ࢍ࢏࢚ ࢙࢛࢓ ࢏࢙ ࢊ࢏࢜࢏࢙࢏࢈࢒ࢋ ࢈࢟ ૜? A quick check on a few examples can be shown to be consistent with this rule. ૜૞ૠ ൌ 3x119 Digit sum check: 3 ൅ 5 ൅ 7 ൌ 15 ൌ 3x5 ૜૛ ൌ 10x3. . .2 Digit sum check: 3 ൅ 2 ൌ 5 ൌ 3x1 ൅ 2 This fact amazed me when I first encountered it, but I never understood why it works until I read about modular arithmetic. Proof: Notice that each integer can be broken down into a sum of integers that are multiples of powers of 10. ૜૞ૠ ൌ 300 ൅ 50 ൅ 7 ૜૛ ൌ 30 ൅ 2 We can express them individually as multiples of powers of 10. 300 ൌ 3x10 ଶ 50 ൌ 5x10 7 ൌ 7 Notice that 10 ؠ 1ሺmod 3ሻ . Using the power law, 10 ୩ ؠ 1 ୩ ሺmod 3ሻ , thus 10 ୩ ؠ

1ሺmod 3ሻ for all integers k . Now, using the multiplication law, we have: 300 ൌ 3x10 ଶ ؠ ሺ3ሻx1 ؠ 3ሺmod 3ሻ 50 ൌ 5x10 ؠ ሺ5ሻx1 ؠ 5 ሺmod 3ሻ 7 ൌ 7x1 ؠ ሺ7ሻx1 ؠ 7 ሺmod 3ሻ

357 ؠ 0 ሺmod 3ሻ if and only if the digit sum is divisible

Using the addition law,

by 3. Something for you to try: Can you find any patterns within numbers divisible by 9 or 11 using modular arithmetic? ࡼ࢘࢕࢈࢒ࢋ࢓ ૛: ࡿࢎ࢕࢝ ࢚ࢎࢇ࢚, ࢌ࢕࢘ ࢋ࢜ࢋ࢘࢟ ࢖࢕࢙࢏࢚࢏࢜ࢋ ࢏࢔࢚ࢋࢍࢋ࢘ ࢔, ૚૛૚ ࢔ െ ૛૞ ࢔ ൅ ૚ૢ૙૙ ࢔ െ ሺെ૝ሻ ࢔ ࢏࢙ ࢇ࢒࢝ࢇ࢙࢟ ࢊ࢏࢜࢏࢙࢈࢒ࢋ ࢈࢟ ૛૙૙૙. ሺ࡮࢘࢏࢚࢏࢙ࢎ ࡹࢇ࢚ࢎࢋ࢓ࢇ࢚࢏ࢉࢇ࢒ ࡻ࢒࢟࢓࢖࢏ࢇࢊ ૛૙૙૙ሻ This problem may look a bit daunting at first sight. However, modular arithmetic can simplify the problem rather nicely. Taking modulo 2000 directly would lead you to nowhere, but we can break down 2000 after which modular arithmetic may be useful. We first express 2000 ൌ 2 ସ x 5 ଷ If the expression is divisible by 16 and 125 , it is divisible by 2000 because 16 and 125 are coprime. This naturally leads us to take modulo 2000 in its prime factors. Solution:

16 and modulo 125. Using the power law,

121 ୬ ؠ 9 ୬ ሺmod 16ሻ 25 ୬ ؠ 9 ୬ ሺmod 16ሻ 1900 ୬ ؠ 12 ୬ ሺmod 16ሻ ሺെ4ሻ ୬ ؠ 12 ୬ ሺmod 16ሻ 9 ୬ െ 9 ୬ ൅ 12 ୬ െ 12 ୬ ؠ 0 ሺmod 16ሻ

The whole expression:

121 ୬ ؠ 121 ୬ ሺmod 125ሻ 25 ୬ ؠ 25 ୬ ሺmod 125ሻ

Similarly,

19 1900 ୬ ؠ 25 ୬ ሺmod 125ሻ ሺെ4ሻ ୬ ؠ 121 ୬ ሺmod 125ሻ

The whole expression: 121 ୬ െ 25 ୬ ൅ 25 ୬ െ 121 ୬ ؠ 0 ሺmod 125ሻ We can now reach the conclusion required that the expression is divisible by 2000 for every positive integer n.

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Division by 3

Add the number's digits. If the sum is evenly divisible by 3, then so is the number. So, will 3 divide evenly into 2,169,252? Yes it will, because the sum of the digits is 27, and 27 is divisible by 3. If you want, you can keep adding numbers until one digit remains. For example, keep going with 27. 2 + 7 = 9, which is also evenly divisible by 3.

Proof by Induction or Contradiction By Ayman D’Souza

Very often in Mathematics we find ourselves needing to prove something: a formula, a fact, a theorem. But what do we do when we can no longer use a 3 ଶ as 9x= 3*3*x? Very often, we can use the method of Proof by Induction, or Proof by Contradiction. In this article I will explain both methods by employing the use of examples. Proof by Induction ଵ ଶ n (n +1) If we already know that to find the value of this some we should take the average simple proof, e.g. 9x must be divisible by We are given the arithmetic progression: 1 + 2 + 3 + ..... + n = sense. But how could prove it by induction? Let us assume that the arithmetic progression is true for n = k, rewriting the formula as: ଵ ଶ k (k + 1 ) If it is true for all values, we can fill the formula in with k + 1 as well as k, reading: {Formula A}: 1 + 2 + 3 + ..... + k = ( ଵ ଶ k + 1 ) (k + 1 ) = ଵ ଶ (k + 1 ) (k + 2 ) We can manipulate the right hand side of the formula to resemble the left hand side ଵ ଶ (k + 1 ) not affecting the overall value. Since we have proved that the original statement to prove works for n + 1 (remember n represents k ), it logically works for any integer equal to or larger than one, and so we have proved the formula to work for any value of n . Proof by Contradiction Often, is something is untrue, we can prove this by contradicting it. For example, let ௔ ௕ is cancelled down as far as possible) which we can contradict to prove the square root of two is an irrational number, similarly to pi. if we halve (k + 2 ), and double us take the example √2 ൌ ௔ ௕ (assuming that and multiply it by the number of terms, the formula ଵ ଶ n (n +1) makes perfect {Formula B}: 1 + 2 + 3 + ..... + k + (k + 1 ) = ଵ ଶ (k + 1 ) ((k + 1 ) + 1 ) Since we already know that 1 + 2 + 3 + ..... + k = ଵ ଶ k (k + 1 ) {Formula A} , we can rewrite the {Formula B} above as: ଵ ଶ k (k + 1 ) + (k + 1 ) = ଵ ଶ (k + 1 ) (k + 2 )

√2 ൌ ܽ ܾ 2 = ௔ మ ௕ మ ܾ ଶ = ܽ ଶ

As ܽ ଶ is double another number (a and b are integers), it must be even. Therefore, let us call it 2n. We now have 2 ܾ ଶ = ܽ ଶ , or : 2 ܾ ଶ = ሺ2݊ሻ ଶ 2 ܾ ଶ = 4݊ ଶ ܾ ଶ = 2݊ ଶ This tells us that ܾ ଶ is an even number, as it is another number doubled, and even squares are always the product of the same even number. As both a and b are even, ௔ ௕ can be further cancelled, but this is a contradiction with the original √2 is irrational. In conclusion, both methods are very powerful for complicated or simple proofs, and the logic behind them cannot be questioned or doubted, making them extremely useful for proving more difficult formulae and theorem. the fraction statement. Thus, by disproving this statement, we can prove that

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Division by 4

If the number's last 2 digits are 00 or if they form a 2-digit number evenly divisible by 4, then number itself is divisible by 4. How about 56,789,000,000? Last 2 digits are 00, so it's divisible by 4. Try 786,565,544. Last 2 digits, 44, are divisible by 4 so, yes, the whole number is divisible by 4.

Cavalieri’s Principle By Faraz Taheri

Abstract Cavalieri’s Principle, named after Bonaventura Cavalieri, is a geometrical method to find the area or volume of an object on a plane or in a three-space respectively.

1 Introduction Cavalieri’s Principle, aka method of indivisibles, presumes lines are made of an infinite number of points, planes are made of an infinite number of lines and volumes are made up of an infinite number of planes. And therefore it aims to find the sum of infinite lengths or areas. It basically uses the concept of integration without using the integral notation. 1.1 Two-dimensional space Suppose the baselines of two regions on a plane are placed on a single line. If every line drawn parallel to the baselines, makes 2 line segments of equal length, the areas of those regions are equal.

1.2 Three-dimensional space Suppose the bases of two objects (with equal base area and height) in a space are placed on a single plane. If every plane parallel to the bases, makes 2 regions of equal area, the volumes of those objects are equal.

1.2.1 How to use the principle The following example illustrates how to use the principle: Prove that any two pyramids with equal height and base area are of equal volume.

׊ܳ צ ܲ

൬ h ൬ h

൰ 2 ൰ 2

צ P צ P

ื P 1 ื P 2

ื ื S P 2

S P 1 ' S P 1

ۙۖۘ ۖۗ

' ~ P 1

Q Q

஼௔௩௔௟௜௘௥௜ ሳልልልልልሰܸ ଵ

֜ ܵ ௉ భ ′

ൌ ܵ ௉ మ ′

ൌ ܸ ଶ

P 2 '

' ~ P 2 S P 1 =

S 2

Let P and Q be parallel planes. The base of both pyramids are placed on Q. By

ܲ ଵ ′

צ ܲ ଵ and

ܲ ଶ ′

צ ܲ ଶ . Then we know that

ܲ ଵ ′ and

ଵ are similar;

hypothesis,

as ܲ ଶ with the ratio of ܲ ଶ ′ and

ቀ h H

ቁ 2

. According to our assumption, S P 1 = S

. Therefore,

P 2

ൌ S

P 2 ' . These steps could be repeated for every plane parallel to Q. The

P 1 '

Cavalieri’s Principle tells us the volumes of these pyramid equal one another.

2 Initial steps To begin with, we need to define the volume of cube and cuboid. 2.1 Cube

ܽ 3 , where

ܽ is the length of one side.

We postulate the volume of a cube is

2.2 Cuboid

ሺܽ ൈ ܾ ൈ ܿሻ . Then

Suppose three values a, b, c which are the sides of a cuboid

ܽ ൅ ܾ ൅ ܿ . According to our axiom, the

we form a cube with the side length of

volume of this cube would be ሺܽ ൅ ܾ ൅ ܿሻ 3 : ሺܽ ൅ ܾ ൅ ܿሻ 3 ൌ ܽ 3 ൅ ܾ 3 ൅ ܿ 3 ൅ 3ܾܽܿሺܽ ൅ ܾ ൅ ܿሻ ֜ ሺܽ ൅ ܾ ൅ ܿሻ 3 ൅ ܽ 3 ൅ ܾ 3 ൅ ܿ 3 ൌ ሺܽ ൅ ܾሻ 3 ൅ ሺܽ ൅ ܿሻ 3 ൅ ሺܾ ൅ ܿሻ 3 ൅ 6ܾܽܿ ൌ ሺܽ ൅ ܾሻ3 ൅ ሺܽ ൅ ܿሻ3 ൅ ሺܾ ൅ ܿሻ3 ൅ 6ܾܽܿ Then we divide the big cube into 3 cubes which have a 3 , b 3 , c 3 cubes intersecting. Therefore, the remaining pieces are 6 cuboids; the volume of each would be ܽ ൈ ܾ ൈ ܿ . Accordingly, the volume of a cuboid would be : ܪ ݄݁݅݃ ݐ ൈ ܮ ݁݊݃ ݐ ݄ ൈ ܹ݅݀ ݐ ݄ . 3 Implementation In this section we try to work out the volume of some common shapes using the introduced principle. 3.1 Pyramid Let’s define “perfect pyramid”. The base of a perfect pyramid is a rectangle and the height is one of its sides.

As the figures above demonstrate, these three perfect pyramids could be put together to form a cuboid :

The volumes of the pyramids are as follows : ܸ ଵ ൌ 1 3 ൈ ܽ ൈ ܾ ൈ ܿ ܸ ଶ ൌ 1 3 ൈ ܽ ൈ ܾ ൈ ܿ

ܸ ଷ ൌ 1 3 ൈ ܽ ൈ ܾ ൈ ܿ ֜ ܸ ଵ ൌ ܸ ଶ ൌ ܸ ଷ ܸ ௉௬௥௔௠௜ௗ ൌ 1 3 ൈ VCuboid ൌ 1 3 ൈ ݎܣ ݁ܽ ஻௔௦௘

ܸ ் ൌ ܸ ଵ ൅ ܸ ଶ ൅ ܸ ଷ ܸ ଵ ൌ ܸ ଶ ൌ ܸ ଷ

ൠ ֜

ൈ ܪ ݄݁݅݃ ݐ

3.2 Cone Place a cone next to a pyramid with the same base area on a plane.

׊ܳ צ ܲ

൬ h ൬ h

൰ 2 ൰ 2

צ P צ P

ื S 1 ื S 2

1 ื ื ൌ ܵ ଶ 2

S 1 ′ S 1 S 2 ′ S 2

ۙۖۘ ۖۗ

' ~ S

Q Q

஼௔௩௔௟௜௘௥௜ ሳልልልልልሰܸ ଵ ൌ

֜ S 1 '

ൌ S 2 '

ܸ ଶ

' ~ S ܵ ଵ ֜

ܸ ஼௢௡௘ ൌ 1 3 ൈ ݎܣ ݁ܽ ஻௔௦௘ ൈ ܪ ݄݁݅݃ ݐ

3.3 Sphere 3.3.1 Auxiliary Principle

The plane P intersects the center of a sphere with the radius of R. The plane Q is parallel to P with the distance of h. Find the intersection area of Q in terms of R and r.

ݎ ଶ ൌܵ

ܴ ଶ െ ݄ ଶ ൌ ݎߨ ଶ ቅ ֜ ܵ ൌ ߨ ൈ ሺܴ ଶ െ ݄ ଶ ሻ

3.3.2 Hemisphere

The image above illustrates a cylinder containing a cone and a hemisphere with equal radii. The intersecting surface of each plane of the hemisphere equals the excluding part of the cylinder.

ܴ ௉ ൌ ܴ ஼௬௟௜௡ௗ௘௥

െ ݕ

For a random plane passing through the hemisphere, let

Then, the auxiliary principle is taken into account :

ܵ ஼௬௟௜௡ௗ௘௥ ൌ ߨ ൈ ݎ ଷ ܵ ஼௢௡௘ ൌ 1 3 ൈ ݎߨ ଷ ቑ ֜ ܵ ு௘௠௜௦௣௛௘௥௘ ൌ 2 3 ൈ ߨ ൈ ݎ ଷ

3.3.3 Complete Sphere Respecting the previous steps, the volume of a sphere would be :

ܵ ௌ௣௛௘௥௘ ൌ 2 ൈ ܵ ு௘௠௜௦௣௛௘௥௘ ܵ ௌ௣௛௘௥௘ ൌ 2 ൈ 2 3 ൈ ߨ ൈ ݎ ଷ ܵ ௌ௣௛௘௥௘ ൌ 4 3 ൈ ߨ ൈ ݎ ଷ

4 Further steps What other volumes can be found by propagating Cavalieri’s Principle? Is it possible to use this principle to work out the volume of a doughnut? Furthermore, this technique might help you to solve the “The napkin ring problem” and find the volumes of Cycloids.

5 Reference Zill, D. (2009) Calculus: Early Transcendentals, Jones Bartlett Learning.

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Division by 7

Multiply the last digit by 2. Subtract this answer from the remaining digits. Is this number evenly divisible by 7? If it is, then your original number is evenly divisible by 7. Try 364. 4, the last digit, multiplied by 2 = 8. 36, the remaining digits, minus 8 = 28. The last time I checked, 28 is evenly divisible by 7, and thus, so is 364!

Where’s the policeman? By Mr Ottewill

2 2

y x yx + = is said to represent a policeman on traffic duty. The

The curve

graph looks like the following which illustrates the road junction.

-4

-2

-4

The question is: where is the policeman? See below for the answer.

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Division by 8 If the number's last 3 digits are 000 or if they form a 3-digit number evenly divisible by 8, then the number itself is divisible by 8. How about 56,789,000,000? Last 3 digits are 000, so it's divisible by 8. Try 786,565,120. The last 3 digits, 120, divide by 8 into 15, so yes, the whole number is divisible by 8.

Answer to ‘Where’s the policeman?’

2 2 2 y x yx + = is also true when x = 0 and y = 0, so the point (0, 0) is on the 2

graph as an isolated point – this point represents the policeman.

Hilbert’s 6 th Problem By Harry Goodhew

In 1900 David Hilbert presented 23 problems which he intended to be the focus of mathematical thought for the coming century and beyond. Since then 17 of the problems have been solved although some solutions have not been fully accepted. The remaining 6 are either too vague to be solved or have resisted solution so far, the latter include the Riemann hypothesis and the one which I will be looking at, the 6 th problem, the axiomatic treatment of physics. This problem is somewhat different to the others that Hilbert proposed as it is not a problem in itself but a general task, which he suggested, following his work on the construction of geometric axioms. So far it has not been fully solved despite exhaustive work from many mathematicians including Hilbert himself. However, some fields within physics have succumbed to such axiomatic treatment, including Mechanics and Statistics. In order to advance the discussion of Hilbert’s 6 th problem it is first necessary to define an axiom, it is one of a set of statements, that are assumed to be true, from which all other true statements within a system of logic can be proved. For instance from the axioms of addition: a൅bൌb൅a ܿ݋݉݉ ݐݑ ܽ ݐ ݅ ݒ ݅ ݕݐ [1] a൅ሺb൅cሻൌሺa൅bሻ൅c ܽ ݏݏ ݋ܿ݅ܽ ݐ ݅ ݒ ݅ ݕݐ [2] a൅0ൌa the existence of an identity [3] a൅ሺെaሻൌ0 the existence of an inverse [4] It can be proved that if a൅bൌc൅a, then bൌc ܾ ൌ 0 ൅ ܾ [૜] ൌ ሺെܽ ൅ ܽሻ ൅ ܾ [૝] ൌ ሺെܽሻ ൅ ሺܽ ൅ ܾሻ[૛] ൌ ሺെܽሻ ൅ ሺܿ ൅ ܽሻ ൌ ሺെܽሻ ൅ ሺܽ ൅ ܿሻ [૚] ൌ ሺെܽ ൅ ܽሻ ൅ ܿ [૝] ൌ 0 ൅ ܿ [૛] ൌ ܿ [૜] ■(the bracketed numbers indicate which axiom has been used) The usefulness of the existence of an equivalent set of axioms for physics is clear as it would allow physical facts to be proved logically and provide a framework for testing whether or not a proposition is true, as well as a way to ensure that the system did not rely on any assumptions. Furthermore, the nature of such axioms could reveal facts about the universe itself. Unfortunately, due to our lack of a complete understanding of physics, epitomised by the inconsistency between general relativity and the standard model, a solution to this problem remains elusive. However, the developments that have been made are worth discussing. In the years following his presentation of the 23 problems, Hilbert devoted much of his Similarly all other true statements about field addition can be proved from these axioms.

time to the mathematical treatment of physics, most notably general relativity which has since been axiomatised 5 . Similar attempts have been made with the standard model, though this is far from complete. Other physical theories, with a somewhat longer history have been broken down to this, most basic level, below I will discuss both mechanics and probability calculus as well as their applications in more advanced physics. Probability calculus is again worth defining, as the calculation of the probability of an event. This study that lies on the border between mathematics and physics can be expressed as just three, familiar, axioms: 0 ൑ ܲሺ ܧ ሻ ൑ 1 ܲሺܵሻ ൌ 1 ܲሺ ܧ ଵ ׫ ܧ ଶ ׫ … ሻ ൌ ෍ܲሺ ܧ ௜ ሻ ∞ ௜ୀଵ , ݂݅ ܧ ଵ ܧ , ଶ … ܽ ݎ ݁ ݉ ݑݐݑ ݈݈ܽ ݕ ݁ ݔ ݈ܿ ݏݑ ݅ ݒ ݁ These may seem insufficient, to prove all that you have learnt about probability calculus, however, it has been shown to be consistent and independent, and has managed to prove all statements currently known to be true, although its completeness cannot be proved in accordance with Gödel’s incompleteness theorems. Furthermore, to demonstrate how powerful these seemingly inert statements can be, I will show how they can prove the inclusion, exclusion principle. A/B is deϐined as A ת B ′ visually: ܣ ൌ ሺ /ܣ Bሻ ׫ ሺA ת Bሻ , which are mutually exclusive ܤ ൌ ሺ /ܤ Aሻ ׫ ሺA ת Bሻ , which are also mutually exclusive ܲሺ ܣ ሻ ൌ ܲ൫ሺA/Bሻ ׫ ሺA ת Bሻ൯ ൌ ܲሺA/Bሻ ൅ ܲሺA ת Bሻ – 3rd axiom (if you include an infinite number of empty sets following A and B which all have probability 0, which I have not shown for simplicity) ܲሺ ܤ ሻ ൌ ܲ൫ሺB/Aሻ ׫ ሺA ת Bሻ൯ ൌ ܲሺB/Aሻ ൅ ܲሺA ת Bሻ -3rd axiom ܲሺ ܣ ሻ ൅ ܲሺ ܤ ሻ ൌ ܲሺA/Bሻ ൅ ܲሺA ת Bሻ ൅ ܲሺB/Aሻ ൅ ܲሺA ת Bሻ ൌ ܲሺሺA/Bሻ ׫ ሺA ת Bሻ ׫ ሺB/Aሻሻ ൅ ܲሺ ת ܣ Bሻ – 3rd axiom ൌ ܲሺ ܣ ׫ ܤ ሻ ൅ ܲሺ ܤ ת ܣ ሻ – Which follows from the definition of A/B From here, the inclusion, exclusion principle follows inductively. 6 ■ This system has clearly got a mathematical cleanness, but what about its relevance A/B B/A A ת B

5 An Axiom System for General Relativity Complete with respect to Lorentzian Manifolds 6 http://www.proofwiki.org/wiki/Inclusion-Exclusion_Principle

to the axiomatic treatment of physics? I will discuss here two of the countless important extensionsto probability calculus, the kinetic theory of gases and, as Hilbert called it, “the theory of compensation of errors”, which attempts to calculate the errors associated with the use of measuring equipment in order to reduce it. This study comes from any one of three axioms, the other two being derivable as theorems from the one chosen to be an axiom, these are: “If various values have been obtained from measuring a certain magnitude, the most probableactual value of the magnitude is given by the arithmetical average of the various measurements. … The frequency of error in measuring a given magnitude is given by: න ݁ ି௧ ଶ ఛ ଴ … The most probable value of the variables measured is obtained by minimising the squares of the errors involved in each observation.” 7 The fact that these axioms are interchangeable highlights something important about the nature of axioms, that for a system to be logically coherent the choice of axioms is arbitrary, which seems to undermine their importance. However, it merely serves to demonstrate that the proposed axioms are logically equivalent. Furthermore, in many physical theories those statements chosen to be axiomatic are those that directly follow from experimental data. The theorems are predictions that guide future research. This suggests that in many physical systems this problem does not arise as only one set of axioms will follow directly from the data. However, when there is a choice it may allow a problem to be simplified by choosing an alternative equivalent system. The calculus of probabilities is also relevant to the kinetic theory of gases which again requires just a few additional assumptions; that gases are made up of point particles that collide elastically, that their velocities are independent, and that they are uniformly distributed throughout the volume of space that they occupy. These three “axioms” 8 can be used to derive the average potential energy, kinetic energy and the mean free path of a molecule as well as the second law of thermodynamics, and much more besides. Another of the areas within physics that have been successfully axiomatised is vector addition, vital for much of physics, as well as some areas that are more commonly considered mathematics, these axioms are as follows: ݏݏܣ ݋ܿ݅ܽ ݐ ݅ ݒ ݅ ݕݐ ݋݂ ܽ݀݀݅ ݐ ݅݋݊: ࢛ ൅ ሺ࢜ ൅ ࢝ሻ ൌ ሺ࢛ ൅ ࢜ሻ ൅ ࢝ ܥ ݋݉݉ ݐݑ ܽ ݐ ݅ ݒ ݅ ݕݐ ݋݂ ܽ݀݀݅ ݐ ݅݋݊: ࢛ ൅ ࢜ ൌ ࢜ ൅ ࢛ ܫ ݀݁݊ ݐ ݅ ݕݐ ݈݁݁݉݁݊ ݐ ݋݂ ܽ݀݀݅ ݐ ݅݋݊: ׌ 0 א ܸ| ࢜ ൅ 0 ൌ ࢜׊ ࢜ א ܸ ܫ ݊ ݒ ݁ ݏݎ ݁ ݈݁݁݉݁݊ ݐ ݋݂ ܽ݀݀݅ ݐ ݅݋݊: ׊ ࢜ א ܸ ׌ ሺെ࢜ሻ | ࢜ ൅ ሺെ࢜ሻ ൌ ૙ ܥ ݋݉݌ܽ ݐ ܾ݈݅݅݅ ݕݐ ݋݂ ݏ ݈ܿܽܽ ݎ ݉ ݑ ݈ ݐ ݅݌݈݅ܿܽ ݐ ݅݋݊: ܽሺܾ࢜ሻ ൌ ሺܾܽሻ࢜ 7 http://tau.ac.il/~corry/publications/articles/pdf/hilbert.pdf 8 I use the apostrophes here to indicate that these are not usually viewed as such but function in a similar way to axioms

34 ܫ ݀݁݊ ݐ ݅ ݕݐ ݈݁݁݉݁݊ ݐ ݋݂ ݏ ݈ܿܽܽ ݎ ݉ ݑ ݈ ݐ ݅݌݈݅ܿܽ ݐ ݅݋݊: 1࢜ ൌ ࢜ ܦ ݅ ݎݐݏ ܾ݅ ݐݑ ݅ ݒ ݅ ݕݐ ݋݂ ݏ ݈ܿܽܽ ݎ ݉ ݑ ݈ ݐ ݅݌݈݅ܿܽ ݐ ݅݋݊ ݓ ݅ ݐ ݄ ݎ ݁ ݏ ݌݁ܿ ݐ ݐ ݋ ݒ ݁ܿ ݐ ݋ ݎ ܽ݀݀݅ ݐ ݅݋݊: ܽሺ࢛ ൅ ࢜ሻ ൌ ࢛ܽ ൅ ܽ࢜ ܦ ݅ ݎݐݏ ܾ݅ ݐݑ ݅ ݒ ݅ ݕݐ ݋݂ ݏ ݈ܿܽܽ ݎ ݉ ݑ ݈ ݐ ݅݌݈݅ܿܽ ݐ ݅݋݊ ݓ ݅ ݐ ݄ ݎ ݁ ݏ ݌݁ܿ ݐ ݐ ݋ ݂݈݅݁݀ ܽ݀݀݅ ݐ ݅݋݊: ሺܽ ൅ ܾሻ࢜ ൌ ܽ࢜ ൅ ܾ࢜ 9 Again, from these axioms, it is possible to deduce a vast amount of physical knowledge related solely to vector mathematics and, in conjunction with a few others, can be extended to cover many related studies. For example, the dynamics of point particles can be deduced from newton’s laws of motion which are almost axioms: ܣ ݊ ݋ܾ݆݁ܿ ݓ ݐ ݈݈݅ ܿ݋݊ ݐ ݅݊ ݑ ݁ ܽ ݐ ܿ݋݊ ݐݏ ܽ݊ ݒ ݐ ݈݁݋ܿ݅ ݑ ݕݐ ݈݊݁ ݏݏ ܽܿ ݐ ݁݀ ݋݊ ܾ݁ ܽ ݂݋ ݎ ܿ݁ ࡲ ൌ ݀݉࢜ ݀ ݓ , ݐ ݄݄݅ܿ ݂݋ ݎ ܿ݋݊ ݐݏ ܽ݊ ݐ ݉ܽ ݏݏ ܾ݁ܿ݋݉݁ ݏ ࡲ ൌ ݉ࢇ ݒܧ ݁ ݕݎ ܽܿ ݐ ݅݋݊ ݄ܽ ݏ ܽ݊ ݁ ݑݍ ݈ܽ ܽ݊݀ ݋݌݌݋ ݏ ݅ ݐ ݁ ݎ ݁ܽܿ ݐ ݅݋݊ However, these are not quite axioms as they are not independent as the first one can be derived from the second by setting F =0 you get a =0, which means that the object will continue at constant velocity. Furthermore, the statement that every action has an equal and opposite reaction comes from the law of conservation of momentum, which, along with the other conservation laws is generally assumed to be axiomatic for physics. Therefore, F =m a is the only axiom required, in fact, along with the definitions of time and energy, as the duration of movement, and the integral of force with respect to time, respectively, it is sufficient as the basis of all of classical mechanics. 10 I will now advance the discussion to newer theories, firstly, quantum field theory, our principle theory of particle interactions. This has been described in an axiomatic form by Arthur Wightman.These axioms are of principle concern to both mathematicians and physicists, one of the millennium problems 11 being the realisation of these axioms in the case of Yang-Mills fields. However, they require an understanding of complex mathematics and so stating them will offer very little. Similarly, general relativity requires a reasonable knowledge of vector mathematics, however, it has been summarised by a set of axioms about the equivalence of both inertial and accelerated observers. This seemingly innocuous statement actually has a profound impact on physics at a grand scale as it is possible, with these axioms to show that light has a fixed, finite speed which is the maximum speed at which anything is capable of moving, a fairly unintuitive fact. Similarly, these axioms can prove that mass bends space time in such a way that is seen as gravity to an observer. Unfortunately the mathematics required for this proof was

9 http://en.wikipedia.org/wiki/Vector_space 10 http://www.marinsek.com/files/axiomatic_of_mechanics.pdf 11 Devlin, Keith J. (2003) [2002]. The Millennium Problems: The Seven Greatest Unsolved Mathematical Puzzles of Our Time . New York: Basic Books

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