16 December 2009, by Melanie Bayley

What would Lewis Carroll’s Alice’s Adventures in Wonderland be without the Cheshire Cat, the trial, the Duchess’s baby or the Mad Hatter’s tea party? Look at the original story that the author told Alice Liddell and her two sisters one day during a boat trip near Oxford, though, and you’ll find that these famous characters and scenes are missing from the text.

As I embarked on my DPhil investigating Victorian literature, I wanted to know what inspired these later additions. The critical literature focused mainly on Freudian interpretations of the book as a wild descent into the dark world of the subconscious. There was no detailed analysis of the added scenes, but from the mass of literary papers, one stood out: in 1984 Helena Pycior of the University of Wisconsin-Milwaukee had linked the trial of the Knave of Hearts with a Victorian book on algebra. Given the author’s day job, it was somewhat surprising to find few other reviews of his work from a mathematical perspective. Carroll was a pseudonym: his real name was Charles Dodgson, and he was a mathematician at Christ Church College, Oxford.

The 19th century was a turbulent time for mathematics, with many new and controversial concepts, like imaginary numbers, becoming widely accepted in the mathematical community. Putting Alice’s Adventures in Wonderland in this context, it becomes clear that Dodgson, a stubbornly conservative mathematician, used some of the missing scenes to satirise these radical new ideas.

Even Dodgson’s keenest admirers would admit he was a cautious mathematician who produced little original work. He was, however, a conscientious tutor, and, above everything, he valued the ancient Greek textbook Euclid’s Elements as the epitome of mathematical thinking. Broadly speaking, it covered the geometry of circles, quadrilaterals, parallel lines and some basic trigonometry. But what’s really striking about Elements is its rigorous reasoning: it starts with a few incontrovertible truths, or axioms, and builds up complex arguments through simple, logical steps. Each proposition is stated, proved and finally signed off with QED.

For centuries, this approach had been seen as the pinnacle of mathematical and logical reasoning. Yet to Dodgson’s dismay, contemporary mathematicians weren’t always as rigorous as Euclid. He dismissed their writing as “semi-colloquial” and even “semi-logical”. Worse still for Dodgson, this new mathematics departed from the physical reality that had grounded Euclid’s works.

By now, scholars had started routinely using seemingly nonsensical concepts such as imaginary numbers – the square root of a negative number – which don’t represent physical quantities in the same way that whole numbers or fractions do. No Victorian embraced these new concepts wholeheartedly, and all struggled to find a philosophical framework that would accommodate them. But they gave mathematicians a freedom to explore new ideas, and some were prepared to go along with these strange concepts as long as they were manipulated using a consistent framework of operations. To Dodgson, though, the new mathematics was absurd, and while he accepted it might be interesting to an advanced mathematician, he believed it would be impossible to teach to an undergraduate.

(Ms Sia’s notes – imaginary numbers are also known as unreal numbers!)

Outgunned in the specialist press, Dodgson took his mathematics to his fiction. Using a technique familiar from Euclid’s proofs, reductio ad absurdum, he picked apart the “semi-logic” of the new abstract mathematics, mocking its weakness by taking these premises to their logical conclusions, with mad results. The outcome is Alice’s Adventures in Wonderland.

Algebra and hookahs

Take the chapter “Advice from a caterpillar”, for example. By this point, Alice has fallen down a rabbit hole and eaten a cake that has shrunk her to a height of just 3 inches. Enter the Caterpillar, smoking a hookah pipe, who shows Alice a mushroom that can restore her to her proper size. The snag, of course, is that one side of the mushroom stretches her neck, while another shrinks her torso. She must eat exactly the right balance to regain her proper size and proportions.

While some have argued that this scene, with its hookah and “magic mushroom”, is about drugs, I believe it’s actually about what Dodgson saw as the absurdity of symbolic algebra, which severed the link between algebra, arithmetic and his beloved geometry. Whereas the book’s later chapters contain more specific mathematical analogies, this scene is subtle and playful, setting the tone for the madness that will follow.

The first clue may be in the pipe itself: the word “hookah” is, after all, of Arabic origin, like “algebra”, and it is perhaps striking that Augustus De Morgan, the first British mathematician to lay out a consistent set of rules for symbolic algebra, uses the original Arabic translation in Trigonometry and Double Algebra, which was published in 1849. He calls it “al jebr e al mokabala” or “restoration and reduction” – which almost exactly describes Alice’s experience. Restoration was what brought Alice to the mushroom: she was looking for something to eat or drink to “grow to my right size again”, and reduction was what actually happened when she ate some: she shrank so rapidly that her chin hit her foot.

(Ms Sia’s notes – that sounds like factorization and expansion of algebraic expressions!)

Jean-Victor Poncelet, the French mathematician who set out the principle, describes it as follows: “Let a figure be conceived to undergo a certain continuous variation, and let some general property concerning it be granted as true, so long as the variation is confined within certain limits; then the same property will belong to all the successive states of the figure.”

The case of two intersecting circles is perhaps the simplest example to consider. Solve their equations, and you will find that they intersect at two distinct points. According to the principle of continuity, any continuous transformation to these circles – moving their centres away from one another, for example – will preserve the basic property that they intersect at two points. It’s just that when their centres are far enough apart the solution will involve an imaginary number that can’t be understood physically.

Of course, when Poncelet talks of “figures”, he means geometric figures, but Dodgson playfully subjects Poncelet’s “semi-colloquial” argument to strict logical analysis and takes it to its most extreme conclusion. What works for a triangle should also work for a baby; if not, something is wrong with the principle, QED. So Dodgson turns a baby into a pig through the principle of continuity. Importantly, the baby retains most of its original features, as any object going through a continuous transformation must. His limbs are still held out like a starfish, and he has a queer shape, turned-up nose and small eyes. Alice only realises he has changed when his sneezes turn to grunts.

The baby’s discomfort with the whole process, and the Duchess’s unconcealed violence, signpost Dodgson’s virulent mistrust of “modern” projective geometry. Everyone in the pig and pepper scene is bad at doing their job. The Duchess is a bad aristocrat and an appallingly bad mother; the Cook is a bad cook who lets the kitchen fill with smoke, over-seasons the soup and eventually throws out her fire irons, pots and plates.

(Ms Sia’s notes – yep, I know that happens sometimes when we do algebra )

Alice, angry now at the strange turn of events, leaves the Duchess’s house and wanders into the Mad Hatter’s tea party, which explores the work of the Irish mathematician William Rowan Hamilton. Hamilton died in 1865, just afterAlicewas published, but by this time his discovery of quaternions in 1843 was being hailed as an important milestone in abstract algebra, since they allowed rotations to be calculated algebraically.

Just as complex numbers work with two terms, quaternions belong to a number system based on four terms (see “Imaginary mathematics”). Hamilton spent years working with three terms – one for each dimension of space – but could only make them rotate in a plane. When he added the fourth, he got the three-dimensional rotation he was looking for, but he had trouble conceptualising what this extra term meant. Like most Victorians, he assumed this term had to mean something, so in the preface to his Lectures on Quaternions of 1853 he added a footnote: “It seemed (and still seems) to me natural to connect this extra-spatial unit with the conception of time.”

Where geometry allowed the exploration of space, Hamilton believed, algebra allowed the investigation of “pure time”, a rather esoteric concept he had derived from Immanuel Kant that was meant to be a kind of Platonic ideal of time, distinct from the real time we humans experience. Other mathematicians were polite but cautious about this notion, believing pure time was a step too far.

The parallels between Hamilton’s maths and the Hatter’s tea party – or perhaps it should read “t-party” – are uncanny. Alice is now at a table with three strange characters: the Hatter, the March Hare and the Dormouse. The character Time, who has fallen out with the Hatter, is absent, and out of pique he won’t let the Hatter move the clocks past six.

Reading this scene with Hamilton’s maths in mind, the members of the Hatter’s tea party represent three terms of a quaternion, in which the all-important fourth term, time, is missing. Without Time, we are told, the characters are stuck at the tea table, constantly moving round to find clean cups and saucers.

Their movement around the table is reminiscent of Hamilton’s early attempts to calculate motion, which was limited to rotatations in a plane before he added time to the mix. Even when Alice joins the party, she can’t stop the Hatter, the Hare and the Dormouse shuffling round the table, because she’s not an extra-spatial unit like Time.

(Ms Sia’s notes – wow! Talking about 4th dimension over here!)

The Hatter’s nonsensical riddle in this scene – “Why is a raven like a writing desk?” – may more specifically target the theory of pure time. In the realm of pure time, Hamilton claimed, cause and effect are no longer linked, and the madness of the Hatter’s unanswerable question may reflect this.

Alice’s ensuing attempt to solve the riddle pokes fun at another aspect of quaternions: their multiplication is non-commutative, meaning that x × y is not the same as y × x. Alice’s answers are equally non-commutative. When the Hare tells her to “say what she means”, she replies that she does, “at least I mean what I say – that’s the same thing”. “Not the same thing a bit!” says the Hatter. “Why, you might just as well say that ‘I see what I eat’ is the same thing as ‘I eat what I see’!”

It’s an idea that must have grated on a conservative mathematician like Dodgson, since non-commutative algebras contradicted the basic laws of arithmetic and opened up a strange new world of mathematics, even more abstract than that of the symbolic algebraists.

When the scene ends, the Hatter and the Hare are trying to put the Dormouse into the teapot. This could be their route to freedom. If they could only lose him, they could exist independently, as a complex number with two terms. Still mad, according to Dodgson, but free from an endless rotation around the table.

And there Dodgson’s satire of his contemporary mathematicians seems to end. What, then, would remain of Alice’s Adventures in Wonderland without these analogies? Nothing but Dodgson’s original nursery tale, Alice’s Adventures Under Ground, charming but short on characteristic nonsense. Dodgson was most witty when he was poking fun at something, and only then when the subject matter got him truly riled. He wrote two uproariously funny pamphlets, fashioned in the style of mathematical proofs, which ridiculed changes at the University of Oxford. In comparison, other stories he wrote besides the Alice books were dull and moralistic.

I would venture that without Dodgson’s fierce satire aimed at his colleagues,Alice’s Adventures in Wonderland would never have become famous, and Lewis Carroll would not be remembered as the unrivalled master of nonsense fiction.

Imaginary mathematics

The real numbers, which include fractions and irrational numbers like π that can nevertheless be represented as a point on a number line, are only one of many number systems.

Complex numbers, for example, consist of two terms – a real component and an “imaginary” component formed of some multiple of the square root of -1, now represented by the symbol i. They are written in the form a + bi.

The Victorian mathematician William Rowan Hamilton took this one step further, adding two more terms to make quaternions, which take the form a + bi + cj + dk and have their own strange rules of arithmetic.

Melanie Bayley is a DPhil candidate at the University of Oxford. Her work was supported by the UK’s Arts and Humanities Research Council.

Hi girls, I’ve come up with 10 more questions on Exchange Rate and 10 more questions on Speed for you to practise. Remember, be PRUDENT with your CALCULATIONS, and show GRACE in your presentation of working. I’m glad to observe SINCERITY and GENEROSITY in terms of effort put in to learn. PERSEVERE, girls, and soon you will be able to reap what you sow.

Hehe.

Here are the questions for you to try out:

Questions on Exchange Rate

1) Convert 5690 Australian dollars (A$5690) to Sing dollars. [S$5490.85]

2) Convert 5324 000 Austrian schiliings to Sing dollars. [S$625 570]

3) Convert S$61243 to New Taiwan dollars (NT$). [NT$113 6233.77 (to 2 d.p.)]

4) Convert S$469 to Chinese Renminbi (RMB). [RMB2233.33 (to 2 d.p.)]

5) Convert S$999.87 to French Francs. [3603.14F (to 2 d.p.)]

6) Convert 100 111 Canadian dollars (CAD100 111) to Sing dollars. [S$114 927.43 (to 2 d.p.)]

7) Convert 2358 Belgium francs to Sing dollars. [S$94.32]

8 ) Convert S$0.53 to US dollars. [US$0.30 (to 2 d.p.]

9) Convert 12 Philippine peso to Sing dollars. [S$0.40 (to 2 d.p.)]

10) Convert S$83.59 to Swiss francs. [SFR77.47 (to 2 d.p.)]

…

Questions on Speed

Leave your answers in fractions where possible.

1) Express 561 m/s in km/h. [2019 2/3 km/h]

2) Express 3.72 cm/h in m/s. [37/3000 000 m/s]

3) Express 0.163 km/h in cm/min. [271 2/3 cm/min]

4) Express 2511 cm/min in m/s. [837/2000 m/s]

5) Express 1982 km/h in m/min. [33303 1/3 m/min]

6) Express 421.9cm/min in km/h. [0.25314 km/h]

7) Express 621 m/h in km/s. [69/400 000 km/s]

8 ) Express 311.67 km/min in cm/h. [18700 20000 cm/h]

9) Express 2.05 cm/s in km/h. [369/5000 km/h]

10) Express 0.746 km.min in cm/min. [74600 cm/min]

Happy practising!

I’m sorry I haven’t been updating. I guess now that I’ve been going through algebraic equations over and over again (plus the extra worksheets that have been uploaded on Asknlearn), I won’t be going through much here.

Just a point to note about number patterns, though.

Some of you had asked me this: how do you know if the pattern will sure to continue on the same way?

Well the thing that we need to assume is that all patterns that we are/will be dealing with in this unit are all growth patterns, meaning these patterns grow at a constant, predictable.  rate. Thus, this will eliminate the possibility of the patterns going haywire and whichever direction that we cannot predict.

Let’s put our fears for solving such questions aside and start to have some fun!

You are given a sequence of numbers:

1, 4, 7, 10, 13, …

(a) What are the next 2 terms?

Since we know that this is a growth pattern, the next 2 terms will be 16 and 19, as the difference between the term and the term after it is always 3.

(b) What is the nth term?

When we talk about nth term, we are referring to a general rule that determines how the pattern should go. nth term refers to the term at a particular position.

For example in this sequence, the 1st term is 1 and the 2nd term is 4.

How do we go about finding this nth term?

We need to relate the term number and the term itself to come up with the rule.

Did you notice something?

The term is related to its term number by 1 + 3 (term number – 1).

Thus, the nth term = 1 + 3(n – 1) = 1 + 3n – 3 = 3n + 2.

This example is just one of the many patterns that we will encounter, and the general idea of solving the above example applies to other questions most of the time as well. It’s not that bad, is it? Hopefully this will give you some idea as to how to solve other questions too!

Ok! The first post in a longggggg time, and it shall be on algebra, what everyone seems to hate the most. Looking at the problem Paige had, don’t you have a sense of deja vu, like you have had the same problem/feeling before? But her brother had actually clarified her doubts with the algebra problem by relating it to a real-life problem. So actually, algebra is secretly embedded in our everyday lives; we use letters to represent the unknown values of things.

First of all, let’s define a few terms:

Variable: a symbol (normally a letter from the alphabet) used to take on different numerical values, e.g x, y, z, av, bc

This also means that variables have different values in different situations!

Constant: a symbol with a fixed numerical value

Term: a single unit containing the product of one or more variables and constants, e.g. 2a, 3b, -5x

Coefficient of a term: the number before the variable e.g. 5 is the coefficient of 5x

Algebraic expression: consists of variables and constants connected by mathematical operations like ‘+’, ‘-‘,
e.g. x + 3, 4 – y, 2a + b

It is important to understand what these words mean, but you won’t be asked to churn out their definitions during a test

The basic notion about algebra is collecting like terms, i.e. you group terms that look exactly alike together. You either add, subtract, multiply or divide like terms together, but for the next 2 examples we’ll deal with only addition and subtraction:

E.g. 1: 5a + 3a +6b – 2a + 4b = ?

a and b are unlike terms, so I can’t just add every single term up just like that. Instead, I need to group them to according their terms:

5a + 3a +6b - 2a + 4b =5a + 3a – 2a +6b + 4b =6a + 10b

Basically what I just did was to group those which look alike together; the “a“s are grouped together, and the “b“s are grouped together. The signs attached to the terms follow them as well, e.g. +4b and + 6b, and 5a, +3a and – 2a. The coefficients of the terms are then added to each other or subtracted from one another

E.g. 2: 2da + 3bc – 3ad + 4cb = ?

This question seems tricky. The terms all don’t seem to look alike… but wait a minute!

Isn’t da = d x a = a x d = ad?

Likewise, bc = cb!

Thus, this question works the same way as E.g. 1:

2da + 3bc – 3ad + 4cb =2da – 3ad + 3bc + 4cb = -ad + 7bc

Note:

1. It is not necessary to write -1ad; -ad is sufficient.

2. You may also write the answer for E.g. 2 as 7bc – ad.

You may try the short exercises below and see how you fare!

1. 6a – 5b +7a – 10b [13a - 15b]

2. 2c + 3c – 5d – 6d + (-c) [4c - 11d]

3. -4sx – 4rt + 4xs – 10tr [-14rt]

4. -9abcd + 8dcab + 3ab – 2bcda + 2ba [5ab - 3abcd]

]]> http://mssia.wordpress.com/2009/07/27/algebra-the-basics/feed/ 3 Miss Sia http://mssia.wordpress.com/2009/05/11/more-practicehelp-on-hcf-lcm-problem-sums/ http://mssia.wordpress.com/2009/05/11/more-practicehelp-on-hcf-lcm-problem-sums/#comments Mon, 11 May 2009 05:44:31 +0000 fireflower http://mssia.wordpress.com/?p=221 ]]>

Problem sums on HCF and LCM can be really tricky as they are not easy to identify. Thus for this post, the main focus is not on going through how to find HCF and LCM (please refer to your notes on those), but more importantly to go through how to determine when to find the HCF and when to find the LCM of the numbers involved in the problem sums.

There will be some problem sums for you to try out at the end of the post, but first let’s take a look at a typical problem involving the HCF.

HCF – A Typical Problem

3 strings of different lengths, 240 cm, 318 cm and 426 cm are to be cut into equal lengths. What is the greatest possible length of each piece?

If you notice, finding the HCF is crucial here because you are trying to find what the 3 numbers have in common, i.e. a common factor. All 3 numbers must be able to be divided by the same number in order for all 3 strings to be cut into equal lengths. HCF is needed here because you are asked to find the greatest possible length.

Therefore,

LCM – A Typical Problem

Two lighthouses flash their lights every 20s and 30s respectively. Given that they flashed together at 7pm, when will they next flash together?

One method to finding the next time the lighthouses flash together is:

20, 40, 60

30, 60, 90

60 is a multiple common to 20 and 30, and thus the lighthouses will flash together in 60s’ time, i.e. at 7:01pm.

This is the same as finding the lowest common multiple, or LCM:

There are other different types of problems involving LCM, but just remember that such questions involve you trying to find a multiple that is common to the numbers involved.

Try out the problem sums below and see if you get them right! The starred ones require a little more thinking

1. As a humanitarian effort, food ration is distributed to each refugee in a refugee camp. If a day’s ration is 284 packets of biscuits, 426 packets of instant noodles and 710 bottles of water, how many refugees are there in the camp? [142 refugees]

2. 294 blue balls, 252 pink balls and 210 yellow balls are distributed equally among some students with none left over. What is the biggest possible number of students? [42 students]

3. A group of girls bought 72 rainbow hairbands, 144 brown and black hairbands, and 216 bright-coloured hairbands. What is the largest possible number of girls in the group? [72 girls]

4. A man has a garden measuring 84 m by 56 m. He wants to divide them equally into the minimum number of square plots. What is the length of each square plot? [28 m]

5. Leonard wants to cut identical square as big as he can from a piece of paper 168 mm by 196 mm. What is the length of each square? [28 cm]

6.* 32 girls and 52 boys were on an overseas learning trip, and they were divided into as many groups as possible where the number of groups of girls and the number of groups of boys are the same. How many girls and how many boys are there in each group? [8 girls, 13 boys]

7. A small bus interchange has 2 feeder services that start simultaneously at 9am. Bus number 801 leaves the interchange at 15-min intervals, while bus number 802 leaves at 20-min intervals. On a particular day, how many times did both services leave together from 9 am to 12 noon inclusive? [4 times]

8. Candice, Gerald and Johnny were jumping up a flight of stairs. Candice did 2 steps at a time, Gerald 3 steps at time while Johnny 4 steps at a time. If they started on the bottom step at the same, on which step will all 3 land together the first time? [12th step]

9. Heidi helps out at her mum’s stall every 9 days while her sister every 3 days. When will they be together if they last helped out on June 16 2008? [June 25 2008]

10. A group of students can be further separated into groups of 5, 13 and 17. What is the smallest possible total number of students? [1105 students]

11. Jesslyn goes to the market every 64 days. Christine goes to the same market every 72 days. They met each other one day. How many days later will they meet each other again? [576 days]

12.*Mrs Goh and 3 of her friends went to a supermarket and found that a package of 6 dishcloths cost $10. If they were to share the purchase such that each has the same number of dishcloths, what is the minimum amount each has to pay? [$5]

Hmmm, I don’t think THAT particular square root has any purpose…

Right. Now that you have learnt how to perform prime factorization, we shall transfer this skill and apply it on finding square and cube roots.

The idea is very simple. When a number is squared, you are basically multiplying the number by itself, i.e.

Do take note that the brackets involved in the sum are important, as there is a difference between

and

When you are finding the square root of a number, you are basically splitting the number into 2 equal groups (note: this is NOT the same as dividing the number by 2!).

Follow the step-by-step instruction to finding square roots:

E.g. Find the square root of 36.

Step 1: Perform prime factorization on the number of which the square root needs to be found.

Step 2: Express the number as a product of its prime factors.

Step 3: Split the prime factors into 2 equal groups.

Therefore, the square root of 36 is 6.

This idea is similar for finding cube roots as well, except you need to split the prime factors into 3 equal groups:

Do take note that you can have negative answers for cube roots of numbers, as shown in the example below:

]]> http://mssia.wordpress.com/2009/05/10/square-and-cube-roots/feed/ 0 Miss Sia http://www.math.uvic.ca/~mascu/Resources/Joke/pic/joke27.gif http://mssia.wordpress.com/2009/04/18/more-practice-on-arithmetic-sums/ http://mssia.wordpress.com/2009/04/18/more-practice-on-arithmetic-sums/#comments Sat, 18 Apr 2009 03:41:09 +0000 fireflower http://mssia.wordpress.com/?p=213 ]]>

Well, I’m not going to grade this piece of work, so don’t have to give me the excuse that your dog ate your homework, but it will be good if you have a teeny weeny bit more of practice on arithmetic sums involving fractions and decimals.

I have come up with 20 questions, and you should try practising without using the calculator, otherwise it will be meaningless to do this exercise. Remember to be PRUDENT in your calculations! GRACEFUL presentation is very important for such questions, especially when you can’t use your calculators. Remember, you reap what you sow. Try them and check your answers!

Questions on Fractions & Decimals

Happy practising

I thought this comic is really funny, actually

So, we have started on the new unit on Factors & Multiples. Before we do anything, we need to understand what factors are in the first place.

Factors are numbers making up another number:

A factor is a number that divides a given number exactly into a natural number. So for example for the number 70, it can be:

70 = 1 x 70 = 2 x 35 = 5 x 14 = 7 x 10

Thus, the factors of 70 are 1, 2, 5, 7, 10,14, 35 and 70.

However, what does the comic above mean? Why is there only 3 numbers?

That’s because 2, 5 and 7 are prime factors of 70, i.e. factors that are prime. (Recall the definition of “prime!”)

In order to find the prime factors of any number, we need to proceed with a process called prime factorization. This can be performed by simply dividing the smallest prime number possible over and over again until the number can not be further divided.

Example

Do not give up when you cannot seem to find the next possible number that can be divided. There are really huge prime numbers out there, such as 13, 17, 19, 23, even 113!

]]> http://mssia.wordpress.com/2009/04/16/prime-factorization/feed/ 0 Miss Sia Factoring the Time Blown into prime factors http://mssia.wordpress.com/2009/04/02/arithmetic-laws/ http://mssia.wordpress.com/2009/04/02/arithmetic-laws/#comments Wed, 01 Apr 2009 18:45:16 +0000 fireflower http://mssia.wordpress.com/?p=191 ]]>

I thought this was rather cute

Right. Now that we have learnt about the adding, subtracting, multiplying and dividing of numbers, it’s time to look at some laws that govern the behaviour of arithmetic sums.

There are 5 laws altogether, and these laws may look very trivial and simple, but we need to take note of them and remember their names by hard!

1. Commutative Law of Addition

The word “commutative” means “interchangeable”, so in this law the positions of the terms involved are interchangeable; i.e.

a + b = b + a

where a and b are real numbers.

E.g.

7 + 9 = 16, but 9 + 7 = 16 too.

This shows that switching the positions of 7 and 9 does not affect the answer.

E.g.

(8 + 5) + 9 = 9 + (8 + 5)

(8 + 5) is considered one whole term and 9 is one whole term itself. Thus by switching their positions, you will get the same answer for each side as well (answer is 22).

2. Commutative Law of Multiplication

Just like the 1st law, the terms involved can change their positions too, like this:

a x b = b x a

E.g.

4 x 5 = 20, and so is 5 x 4 = 20.

3. Associate Law of Addition

The word “associative” means “bracketing”, where certain numbers are grouped together in different ways but yet producing the same result; i.e.

(a + b) + c = a + (b + c)

E.g.

(4 + 5) + 3 = 9 + 3 = 12

But 4 + (5 + 3) = 4 + 8 = 12.

Referring to the 2nd example of the 2nd law, you notice that although that example and the example we have here are similar in the sense that brackets are involved, the laws involved are totally different.

4. Associative Law of Multiplication

Similar to the 3rd law, we have:

(a x b) x c = a x (b x c)

E.g.

(3 x 2) x 4 = 3 x (2 x 4)

Work it out to check if this is true!

5. Distributive Property of Multiplication over Addtion

What a horribly long law! But this law will make sense when you take a look at how it’s being carried out:

a (b + c) = = a x (b + c) = a x b + a x c

Basically what happens here is that the number that is being multiplied is distributed to the other numbers involved in addition.

E.g.

3 (5 + 3) = 3 x 8 = 24 = 3 x 5 + 3 x 3

The distributive property can be applied in certain cases to ease our convenience in calculation:

E.g.

51 x 349 + 49 x 349 = 349 x 51 + 349 x 49 = 349 x (51 + 49) = 349 x 100 = 34900

Using the distributive property will be MUCH easier than calculating from left to right straight away.

…

Remember, the commutative and associative properties, although applicable to addition and multiplication, are not applicable to subtraction and division. We have done a worksheet on that, so you should have observed the patterns by now.

]]> http://mssia.wordpress.com/2009/04/02/arithmetic-laws/feed/ 2 Miss Sia