memory and answers

Visualisation of the (countable) field of alge...

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I attended a public lecture yesterday by Professor E.Victor Flynn on some fields within Algebraic Geometry…It was incredibly fascinating. Although maybe 20% of it went over my head, it did feel like i’d simply have to jump up to be able to reach it. So when I got home, I decided to google some of the ideas that had been expressed.

I always knew number theory was a diverse field, but I never realised HOW diverse it was. At one point, I had 6 or 7 tabs open, with each one linking to one of the others, sharing ideas and definitions – you could not read any one page completely without knowing content from the others. Most of the time, it was simply a case of definitions – when talking about an algebraic number field, you need to know about field extensions, fields and Rational numbers, and knowing the definitions of these terms allows you to understand the definition of an algebraic number field – and so this bought up some interesting questions for me.

Before I’d started looking into these ideas, I’d done my normal routine of checking 20 or so math based blogs for new content. one blog –  “Godel’s lost letter and P=NP” – spoke about the importance of memorisation. Now, if you’ve read my previous blogs, you may notice that I find memorisation of theorems and definitions to be a complete waste of time.

I think now I am beginning to see the error of my ways…

without knowing definitions, we cannot hope to know other definitions that depend on the earlier ones. If an algorithm F works because algorithm G works, we need to know how G works to show F works.

On analysing these thoughts (and ideas expressed by others), I re-evaluated what I thought and why. I think one of the comments on the other blog found the real issue – ‘rote’ memorisation.

however, my thoughts are still developing on this front, and maybe at another point I shall come back to the idea. I agree – and think I have thought this way for some time, but failed to notice it –   that knowing definitions and theorems etc are incredibly important. I think what I have issue with is how we learn them, and how we learn to apply them.

now, moving on to the questions from the last post:

I still don’t have much of an idea for the second question – in truth I’m a  bit bored of it, so I’m just going to leave it.

but as for the first, this one appeals to me!

so: what do we know?
the number plate is only 4 digits long and contains 2 unique digits, so it’s of the form aabb or abba or abab. As the eldest child is 9 years old, it must be divisible by 9, so


⇒ 9|(a+b) with  0 ≤ a,b ≤ 9


now, what else do we know? 8 children, each with a different age. the eldest is 9, which means that the other 7 children are either 1,2,3,4,5,6,7 or 8. therefore, there is either a child of 4, or a child of 8, which means the number is divisible by 4.

this tells us the number plate was one of:

9900, 1188,7272,2772, 3636,6336 or 5544.

this gives my smith an age of 00, 88, 72, 36 or 44. logic would dictate that 00 is impossible, 88 and 72 highly improbable. we shall include them for now, but we won’t include 00.

now notice that none of these possible numbers are divisible by 5, which means the children’s ages are 1,2,3,4,6,7,8 and 9. so the number must be divisible by 504=(9*8*7).

simple calculation gives:

1188 mod 504=180

7272 mod 504=216

2772 mod 504=252

3636 mod 504=108

6336 mod 504=288

5544 mod 504=0 (504*11=5544)

so the number plate was 5544, the children’s ages were 1,2,3,4,6,7,8,9 and Mr Smith is a (presumably) very tired 44-year-old.


Rambling puzzles…Number plates, shoelaces and you.

Square root of x formula. Symbol of mathematics.

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I recently read 2 rather interesting questions online. The first was this:

Mr Smith has 8 children, and each one is a different age. His entire family is in the car, when his eldest, aged 9, shouts out “Daddy! That number plate has only two numbers, and each number is repeated twice! And the number is divisible by each of our ages!” “You’re Correct!” Mr Smith shouts as he whips out his iPhone to check. “and look, the last two digits are my age!”

What is Mr Smith’s age?Actually the question wasn’t like that, it gave you choices and you had  to say which of those numbers were not the age of one of his kids, but this question is a bit harder and more interesting.
The second question:there are 30 shoelaces in a closed box with all of their ends sticking out (so 60 ends). You tie one end to another, and keep doing this until each end is connected to one other.
How many ways are there of doing this?
What is the expected numberof loops?This was on the “Mind Your Decisions” Blog, and I gave a comment which got beautifully shot down by a fellow commentator, And I’m still uncertain who was right.
But anywho, have a go at the puzzles and see if you can solve them.
The first question I don’t know if it’s possible to answer without extra info. If it isn’t, try work out what extra info you need.
As for the second, the only decent suggestion I have is you start with a smaller number and see what happens.
I also recently discovered that Cambridge University has a very different teaching approach to mathematics. They arrange for 2 students to meet up with a lecturer to go over mathematics that they’ve been doing. I’m quite jealous of this! And yet I wonder if I’d actually take the opportunity…doubtless there are people somewhere who will hear about Auckland University and be amazed I haven’t taken more opportunities that have been offered to me, yet from my point of view, they don’t seem like options.
One thing I’ve always been told to do is go and speak to my lecturers. (In fact, I’ve told people to do it myself). Yet I never have. I always think “what would I talk to them about?” I have one query for one of my lecturers, but it seems incredibly trivial and it has no point other than to give me something to talk about with him for about 43 seconds (it takes about 13 seconds to say hello, how are you etc normally). And then? Do I just leave or what? This is why I would like the chance to have a supervisor, but the truth is I do have the chance, I’m just not taking it.I was aiming to speak to one of my lecturers this last week, but have been unable to attend uni this last week and a half…so I think I’m going to aim to go and speak to some of them next week. With some good questions. Hopefully.
I’ve been reading a fair amount of mathematics recently,online and print, about mathematics and mathematicians. There are a variety of mathematicians, some are ‘gifted’ (or ‘genius’) and others persevere and work hard at it – and of course there are varying degrees of both in some mathematicians, and some great mathematicians have neither. But no matter what ‘category’the mathematician falls into, one key characteristic is passion, passion for mathematics. Another, a willingness to forge a new path that no ones been down before.
It is this second that strikes me most. Too often we do what we are expected to do, and not what we want to do. The joy is doing what we want. If we feel forced into doing it, there is not much joy there.
Some mathematicians will tell you to memorise important theorems and proofs – that is the way to be a good mathematician. Others will tell you to discover the proofs and theorems yourself – that is the way to be a good mathematician. And yet others will tell you find a good mathematician and be their ‘apprentice’ – that is the way to be a good mathematician.
But I guess it all comes down to us as individuals. What do you want to do? How do you want to do it? Forget about the marks or opinions,what do you want to do?

1089 (, 495 and 6174) and all that

Take any 3 digit number where the first and last digit areat least 2 apart. (1st number)
Reverse it. (2nd number)
Minus the smaller number from the larger one. (3rdnumber)
Reverse it (4th number)
Add the 3rd and 4th number together.
Your result is 1089.
 This problem i first discovered in the book “1089 and all that”…an awesome little book!
Now, take any 3 digit number, with the only requirementbeing that all 3 numbers are not equal (so the number 001 is allowed)
Now, using these three digits, rearrange the numbers to makethe smallest and biggest possible numbers.
Now subtract the smaller number from the larger one.
Repeat this process. After a few steps you will reach 495.
Now do the same process for 4 digit numbers….you get 6174.
The great thing about these processes is they seem magicalto someone who has never heard of it before. Try the 1089 trick on a 10 yearold, you’ll convince them you’re magical!
These ‘tricks’ are – to me – some of the most interesting ‘tricks’around – well, when it comes to mathematics at least.  So what I thought I’d do in this post isexplore these numbers and why these methods work, and then see what othernumbers we can find etc.
So starting off with 1089:
Lets say your three numbers are x,y,z in that order. Lets alsosay that x is greater than z. so our two numbers are xyz and zyx. Subtracting thesegives:
(x-z) (y-y) (z-x)
Now, as z is less than x, we need to ‘borrow’ a 1 from they, so we actually have
But we also need to ‘borrow’ a 1 from the x
As we’re working mod 10, and if we let x-z =c, then we have
( c-1 ) (9) (10-(x-z))
Which is
( c -1) (9) (10-c)
Now we reverse this number
(10-c) (9) ( c-1)
+             ( c-1 )(9) (10-c)
=             ( 9)   (18) (9)
=             (9+1) (8) (9)
=             1089
This is always true.
What about 495?
Lets take the number x,y,z where x>y>z
Then we have
x    y      z
         z     y      x
= (x-z)(y-y)(z-x)
Again, z<x so need to ‘borrow’ a 1 from the y, whichmeans we’ll need to borrow a 1 from the x. again, let x-z=c
If c were 1, we would have the number 990 and 099 for thenext round. If c were 0, it would mean that x=z and as x<y1
If c >5, then 10-c<c-1, so we will first look at when1<c<6
the biggest number will be
(9) (10-c)  (c-1)
And the smallest
(c-1) (10-c) (9)
Now, subtracting them gives
(9-c+1) (10-c-10+c) (c-1-9)
Again, we have to ‘borrow’ so
(9-c) (9) ( c )
With c being 2,3,4 or 5
If we did the same thing when c>5 , we would end up with( c ) , (9 ) , (11-c)
 With c being, 6,7,8or 9.
If c is 2
We have 792, so
972-279=693, which is what we get if c is 3.
693 gives us
963-369=594, which is what get if c is 4…and  if c is 5, we get the number 495, which isthe magic number!
To confirm, 594 is 954-459=495.
If c is 6
We have 695 which is
396 which is the same as 693
If c is 7
We have 794 which is
974-479=495, which is the magic number
If c is 8
We have 893 which is
983-389=594 the magic number almost
If c is 9
We have 992 which is
992-299=693 which we’ve already done.
Quickly looking at c=1
990-099=891 which is
981-189=792 which we’ve already done.
This shows that it works for all numbers, unless all 3digits are the same.
A similar method can be used to show that for four digits,you get the number 6174, and this number is called the Kaprekar number.
What interests me is: what happens if you have n digits? Whathappens if you use repeated different operations (so using addition andsubtraction, but in different orders…maybe you subtract, subtract, add,subtract, subtract, add…)?
You could classify a sequence of operations as a binary operation.You could do operations on reversed numbers, or operations on biggest numberthat can be made from those digits and the smallest.
Any of these methods could yield some interesting observationsin number theory. Applications of these observations could be used in chaostheory or cryptography….but the point of doing number theory is that there areseldom applications!
In a way, these tricks remind me of collatz conjecture. EvenWikipedia thinks so, as the wiki page on 495 and 6174 both link through tocollatz. Interesting correlations.
I’ll end this post by looking at the Kaprekar routine fornumbers with 2 digits.
xy, with x>y
we have xy-yx=(x-y-1) (10-(x-y))
let x-y=c
we have (c-1) (c)
so we have (c) (c-1)-(c-1)(c)=(c-c+1-1)(c-1-c+10)=9
repeatedly subtracting the smallest possible number from thebiggest possible number from a set of digits for n digits:
digits     result
2              9
3              495
4              6174
5              ?

Do you want a mathematician or a computer?

Recently I have been applying to a couple of jobs for next year, and have beenquite fascinated by the anxiety other people express over the whole applicationprocess. It seems everyone is so intent on being perfect for the job they’reapplying for, that they don’t seem to notice that if they have to ‘change’ whothey are to get the job, the job probably won’t suit them. So they doctor uptheir C.V’s and buy different clothes and attend the interviews readyingthemselves to answer in such a way that their weaknesses become strengths….

The best one I’ve heard from this is: “what is your biggest weakness?”
“I’m dedicated to my work, so I often work straight through a lunch break”.

Yeah right.
Anyways, a Game theorist wrote a lot more on this here :

an awesome blog to read, and the articles make you think.

My favourite part of interviews are the questions that makeyou think. Asking what I would do in a given situation is boring, but asking mehow to solve an abstract idea – that’s fun.
My girlfriend had an interview a few months ago, and one ofthe questions was:

using square tiles that have an area of 1 unit squared, how can you lay thetiles such that the number of tiles used is twice the number of tiles in theperimeter?

the way people approach this is fascinating. Most of us will start writing outequations, so perimeter of a rectangle is 2 x (width+length) and the area iswidth x length, so, using w,l to denote width, length respectively,

which isn’t easily solved. It would probably be best to start throwing integersinto the equation, and see what comes out.
So if we define w as 8, we get 4+l/2=l4=l/2 8=l so a square of 8 by 8 works! Right?
We forgot to notice that thecorner tiles are counted twice. The perimeter calculation should actually be2*(w+l-1) (but the area still stays the same).
4+4(l-1)/w =l

again, try values of w

Which gives a perimeter of 28tiles and an area of 56, which works.
But to do this in an interviewwhere people are stressed out….that’s a bit more tricky. You’ll forget todouble the perimeter. You’ll forget the repeated tiles. And you won’t noticethat if you tile around the equator, you simply need to place 4 rows of tiles,one above the other, to satisfy the requirements…
There are quite a few questionslike this.  
Here’s another one from the blog Ilisted before:

the solution is awesome.

An alternative solution is tocreate a gold cutting device that has 6 knives. Then you just need one cut.
Now, from my limited experience ofinterviews and applications, it seems that most companies claiming to belooking for mathematicians are actually looking for a computer or a softwaredeveloper (which is good news to software developers!). Almost all graduatejobs base part of the application on marks attained at university.  Interestingly, those marks are made up ofexams, tests and assignments. Assignments seldom give more than 20%  towards the final mark. Roughly 40% of theassignments are based on an ability to think like a mathematician.  The rest is rote learning/memorisation or useof a method.  Unfortunately, the examsand tests are also based on this.  This meansthat most courses have at most 10% of the course based on the ability to think,and 90% on being able to memorise proofs and theorems and methods. 
This is all well and good…except for a little thing called the internet. Throughthe use of google, wolframalpha and Wikipedia, that 90% of the course can bedone by someone with no previous knowledge of the course.  I would also guess that they can do it moreaccurately, and faster.
So to base your applicants ontheir university scores seems to me to be a waste of time. If that is all youwanted, you’d be better off getting a software developer or simply using thosethree sites – this way will be much cheaper! If you ever come across a problemyou can’t solve using those sites…well, then there isn’t a high chance yourgraduate with straight A’s could have solved it either…

But enough of that. It’s fairly obvious I have crap marks and think I’m abetter mathematician than those marks show, but that’s my opinion. I likesolving problems, and this is a taster of the next post:

what do the numbers 1089 and 495 have in common?

mathematical english

I read the question in the link above (you may have to click on the title) a couple of months ago, and have been thinking about it on and off for the last little bit. here are my ideas on creating a basis for the english language using a computer: Please note this is just in note form. When it piques my interest again I’ll come back and rewrite this 🙂

Finding the basis of the english language.

2 matrices, 1 with the word and their classification (word in column 1, classification in column 2.) n rows for n words.

Different versions of same word (ie, the animal bear, and can’t bear to see you go)are different words.

Matrix 2 would be n rows and m columns where m is the maximum length of the definition of the word. Each row of matrix 2 (the definition) would be on the same row as matrix 1 (the word).

Create a third matrix, undefined size, called basis.

Run through matrix 2, inserting each word into basis after checking that the word does not already exist in basis. At the end of this loop, basis will be a large matrix of independent terms.

Create another matrix, smaller than basis, called basis2. Run through all the elements of basis, converting the words to a number such that the numbers refer to the row that the word appears in matrix 1. May have some issues here with words with multiple meanings.

If the word does not have a definition, keep the word as it is. Store all these new numbers in basis2, removing them from basis. After this, you should have 2 matrices, one with only words, and the other with only numbers, that together refer to the original basis matrix.

Now use the numbers in basis2. Go to the equivalent row of matrix 2. Run the search again, where it runs through the definition and searches through basis to see if there are any words in the definition that are not in basis, but store these words in a new matrix called secondary.

At this point, you should have a core matrix of words in basis, and some secondary words in deleted. If basis is empty, then the english language is circular. If there is at least 1 word in basis, then there is a basis of undefinable words that make up the language. The words in secondary make up a list of words that refer to each other but that can be used in conjunction with each other to make all the words of the english language.

A few potential problems:

getting around multiple definition words. Perhaps using a number system, where a number is stored after a word (so bear becomes bear1 and bear2) could be arranged. Would be best to apply this to a smaller group of definitions so that the potential problems could be spotted and destroyed, and so that the method could be refined.

There is also a chance that, if the english language is circular, there will be a potential for infinite loops to appear. This is just something that will have to be considered in the programming, and avoided as best as possible by using escape routes.

A bigger problem is getting a text based dictionary whose format is easy to convert for a program like matlab (which is all I have) to be able to convert it to matrices. Once that step is overcome, the rest is fairly easy. If anyone does have any ideas on this last bit/ a plaintext dictionary that they think will be suitable, please let me know, as I wouldn’t mind having a crack at writing this program.

If anyone else thinks they would like to use my idea, they may, I just ask that they let me know before they start, and make sure they mention me in the program.

I don’t know if there is any potential for a program like this to be useful. perhaps a canonised set of words that form a basis of english can be compared to a similar set of words in another language and hence aid translation. But I don’t have that knowledge, nor do I know if this has been done before, so I leave it open to whoever wants to use it.

truthful guests of patterns

Let me introduce to some logic games/ideas…
There is an island with only two tribes on it. Tribe number 1 is the “truth tellers”, tribe number 2 is the “liars”. The truth tellers can only tell the truth, the liars can only lie. You’re visiting the island and come across three of the inhabitants. You ask the first what tribe he is from, but he responds in a voice so low and guttural you cannot make out what he said. You ask the second one what the first one said and he states “He said he is a truth teller”. the final inhabitant looks solemnly at the second, then, turning to you says “he lied to you”.
Which tribe was the third speaker from?
I suggest you think about this first before reading on, it’s quite a fun intellectual exercise to do, and requires nothing more than 5 mins of contemplation.
The trick with this question is to look first not at the third speaker, but one of the other two. Now, we don’t know what the first speaker said, so maybe we should see if we can find out what he said.
Imagine the first speaker was a truth teller. He would then tell us he is a Truth teller. But if he was a liar, he’d still tell us he was a truth teller! This tells us that the second speaker is a truth teller, which tells us that the last speaker is a liar.
A slightly more complicated version of this story: the truth tellers love to eat outsiders (ie: you) and the liars love to help outsiders get off the island safely. Both tribes respect the other, and so if you are with a liar, the truth tellers will not try and abduct you and if you are with a truth teller the liars will not try and save you. Inhabitants will never take an outsider, the outsider must always choose to go with them, regardless of the tribe. There are no differences between the tribes other than their honesty and what they do with outsiders. The only way to get off the island is with the liars, and if an outsider doesn’t choose a tribe s/he will starve to death…
So. Imagine you’re on the island and you run into 3 inhabitants. In as few questions as possible, how do you determine who you go with?
I like the idea of asking them questions about themselves (it’s more personal that way…and requires some well thought out questions). I’ll give them names for illustration: Andy, Bob and Carlise. Then you could ask Bob “if I asked Andy which tribe he was from, what would he say?”, asking Carlise the same question but about Bob, then asking Andy about Carlise. Whichever tribe they mention is the tribe they’re from. A simpler way is to ask them whether it’s raining or not!
There are many variations on these – I think there is a book on them, but I can’t seem to find it. Some of the variations involve different conditions for the inhabitants (they all tell the truth on certain days of the week) language issues, etc.
Now, as stated, these are logic games/puzzles. Some people may not realise it ties into mathematics – but it does! All these games can be broken down into symbols and letters and linked together, giving you a set of tools to solve the problems with. If you want to know more about mathematical logic, study it, look it up online, or ask me!
I would now like to go on a small rant if I may about something that bugs me about people who don’t know much about maths…why do people always expect that because you study maths you are going to be a whiz at mental arithmetic? There are people who are amazing at mental arithmetic, and there are others who quite enjoy working out the patterns of arithmetic ( indeed, I’m one of the latter. For example, take any three digit number where the 1st and 3rd digit differ by at least 2. Reverse the number. Subtract the smaller number from the bigger one. Take your new number, and reverse it. Add those two numbers together. And your answer is 1089. Always. And those patterns are very interesting to analyse as they make cool math tricks) but not every mathematician likes or is good at simple calculations – it’s why we have calculators, although it is definitely better for your brain if you spend some time calculating it yourself! In truth, expecting all mathematicians to be able to calculate quickly and easily in their head is like expecting an engineer to build a boeing 747 from a junk site, or a chef to make a wedding cake with ease, or an artist to take stunning photos, all musicians to be able to sing opera. Whilst there are some that can, it does not mean that all of them can.
Moving on 🙂
Using the logic we played with before, we can attack almost any problem and solve it. However, it is probably best to sort out a method of approach first!
First, find the simplest version of the problem.
See how it behaves.
Slowly complicate the problem toward the original.
look for patterns.
prove the patterns.
(if you read the blog about Fibonacci’s rabbits, it ties into the idea of simplifying the problem first).
there are many mini steps that can fit in there, and some steps that could be there but aren’t, and some that are in front of others that should be behind them….but this is just a general guide line.
So now for the next problem.
Mr and Mrs Senzigic were having a party. They invited 4 other couples around, and had a very good time. At about midnight, Mr Senzigic gathered everyone around and asked them all how many hands they had shaken. To his amazement, everyone had shaken a different amount of hands (even himself), ie, one had shaken 0, one 1, one 2…all the way to 8. Nobody had shook the same person twice, and nobody had shaken their spouses hand. How many hands did Mrs Senzigic shake?
On first look, this problem seems impossible. How, without knowing more detail, can you even begin to hope to solve to this??? The only thing you can deduce is that Mrs Senzigic shook the same number of hands as someone else.
So lets take it down to a simple state of affairs, with only two couples, the Hosts, and The Guests. HM will refer to Mr Host, HF will be Mrs Host, GM Mr Guest, GF Mrs Guest.
Let’s imagine that HM – being the nice man he is – shakes every hands except his wife and his own. This means that his wife has shook none, but both of the guests have shook one. if one of the guests shakes hands again, they will be on two shakes, but they can only get there if they shake HF, which would mean she has shaken two hands as well. This would cause a problem, and you would get a similar situation if you started by assuming HF had shaken everybody’s hand. The only option left the is that one of the Guests shakes two hands, so lets assume GF shakes both HM and HF hands, then they each have one, GF has two, and GM has none.
if you look at the same situation with 2 Guest couples, you find that Mr Guest 1 would shake 4 hands, his wife none, Mrs Guest 2 three hands, Mr Guest 2 one hand, and the Hosts would shake the 2 hands each.
The proof for this is very simple. Start with the last couple to arrive. They walk in, walk around the room greeting everybody, one of them shaking everybody’s hand (8 shakes), the other shaking none(0 shakes). The same for the couple before them, they would have walked in and one of them would have shaken everybody’s hand who was there (6 shakes for now) and their partner shaking none. When the couple arrived after them, the earlier couple would have both been shaken by the shaker in the new couple, meaning that the earlier couple would now have shook 7 and 1 hands respectively. The same pattern continues for the first two couples to arrive. Because of this, it ends up that every couple shakes a total of 8 hands. S0 8 and 0, 7 and 1, 6 and 2, 5 and 3…4 and 4?! but that would mean that there was a repeat in the numbers!!
and then you remember that that Mrs Senzigic shook the same amount of hands as one other person…therefore she must be in the couple of four and four, therefore Mrs Senziga shook four hands.
Hope you followed that, if not, let me know :).
Now for the final problem – and this was in an assignment for a fairly difficult paper that I’m not doing – yet.
Imagine a square, 1×1 units.

 Divide each of these sides into three new sides.
 On the middle piece of each sides, draw a new square with three new sides.

Repeat this pattern for each side.

What is the total area of this shape?
Have fun! Email me if you want the solution.

golden beauty

So….another blog!

I don’t whether today to write about logic, codes, symmetry or the Fibonacci sequence. I think, seeing as I put it off last time, I should do symmetry, but I know we will be doing the Fibonacci sequence in maths tomorrow, and I’m quite keen on seeing if I can pre-empt my lecturer….therefore, I think I’ll start with the Fibonacci sequence, then move onto symmetry….

So. In the late 12th century, a dude named leonardo of pisa or Fibonacci for short (i can’t remember why) decided he wanted to know how many rabbits he would have if he started with a certain amount.

I’m not entirely certain why he wanted to do it, however, in terms of the mathematical bunny leaps that have come from it, I’m fairly glad that he did!

As with most mathematicians, he decided to skirt certain issues and start at a very simple base to get a rough idea of the pattern that ran through the bunny population. He therefore started off with a couple of assumptions:

Firstly, rabbits don’t die.

Secondly, every female has only one baby every month.

Thirdly, except for the first couple, there are always more females than males.

So he started with one rabbit.(female)

And got bored waiting for it do something.

So he got another one. (male)

after one month, the two rabbits had another rabbit. (female)

Now the one male rabbit was happy. So he did what most males love, and the following month, there were two new babies, one male, one female.

all three females gave birth the following month, resulting in 8 rabbits. There are now five females, three males, so the next month there were 13 rabbits, 8 female, 5 male, etc….

that’s a pretty rough outline of the story, and one that isn’t completely true. The most important part of this story is the pattern: 1 rabbit, 1 rabbit, 2 rabbits, 3 rabbits, 5 rabbits, 8 rabbits, 13, 21, 34, 55, 89, 144….

So how do you get this pattern?

Take any number in the pattern, add the previous number to it, and you will get the following number. So to get 13, you add 8 and 5. there are formulas for working out how many ‘rabbits’ there are after so many ‘months’, but I won’t put those into this blog.

Soubtless some of you are going “ok….there’s a pattern….so what?” well, patterns are important things!!! If there’s a pattern, THERE’S A PATTERN, which generally means there is some interesting maths going on somewhere…

so let’s have a closer look at this pattern: 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610….

now have a look at the ratio’s between these numbers: 1/1 =1

2/1 =2

3/2 =1.5

5/3 =1.6666….

8/5 =1.6

13/8 =1.625

21/13= 1.61528462…

etc. the higher the pair of numbers you use, the closer this ratio gets to a very special number, which is called the golden ratio, and is approximately equal to 1.618(rounded to 3 decimal digits)

It is here I must take a breath, as the branches I could take you down are everywhere. The golden ratio is a truly important number, it appears everywhere, in your body, in beauty, in music, nature, shells….almost everything in nature links to this number.

But anyways, let me now tell you about some of the more interesting things about the golden ratio.

firstly, it is easiest to approximate the golden ration by using(sqrt(5)+1)/2. as it is an irrational number, it goes on forever, so we can never get it exactly, which is why we use it’s abbreviated form, 1.618, as this is a much ‘nicer’ number to use when doing calculations. It is also denoted by the Greek letter phi, and often called by such.

as stated above, it appears everywhere…so let’s start with a pineapple (we have to take one to class tomorrow…)

Count the ‘points’ in one clockwise spiral, and the points in one counter clockwise spiral. You will find that the number of points in each spiral will be a Fibonacci number, normally 5,8 or 13-but never 5 and 13, it will always be two consecutive numbers. in other words, the ratio between the spirals approximates the golden ratio.

it’s marvellous fun telling kids to look for a four leaf clover – they’re not very likely to find one. Why? Because four is not a fibonacci number. Seriously. That’s the reason.

now, nature is not saying “oh, four isn’t a Fibonacci number, therefore we can’t have that many leaves/points/ whatever”. people are still searching to find out why this number is so important to nature – for example, why not use a nice number like 1.6 exactly? – and there are some ideas I’ve heard about, such as claiming that the angles that the leaves make to the stem are arranged in the golden proportion to each other, resulting in the fourth leaf being over shadowed by the 6th leaf, therefore causing the 4th to die, resulting in only five. But I did a couple of calculations, and this doesn’t seem to be entirely true, so I will have to keep looking.

Some people claim that the golden ratio was used in the building of the pyramids. However, others say that it is just the fact that the golden ratio is so common that it crops up in measurements, as there is no record of the golden ratio from Egyptian times (it was first mentioned by the Greeks, popularised by Fibonacci). All that I can tell you is that it does seem to crop up almost everywhere, but what I find incredibly interesting is it’s relationship to beauty.

First, let’s start with a rectangle. Draw a rectangle of width 1, and length 1.618.Note that the relationship between the width and length is 1.618. Now, cut a square out of this rectangle with area 1 square unit. You will be left with a rectangle of 1 X 0.618. The relationship between the width and length of this new rectangle, is still 1.618. And you can keep doing this, endlessly, and each time, you will be left with a rectangle whose width and length are in the proportion of 1.618…which is fairly cool!

This rectangle is called the golden rectangle.

Now, I hope you remember some algebra from high-school.

If you want to find a number (lets call it the classic x) such that 1+1/x=x (or x2x-1=0), you find that the answer is phi. This explains why the rectangle cutting works, but I’ll let you try and figure that one out 😛

The rectangle is the first of the geometric shapes that can be drawn in a ‘golden proportion’, but any shape you can think of will have a golden ratio version of it (google it). The golden rectangle, triangle, cross, star, pentagon, spiral etc are generally found to be most ‘pleasing’ to the eye. why? I don’t know. The only common thing between them is this ratio, but you can test it for yourself by drawing a few or googling them and deciding which one you like the most. You may find that you don’t like the ‘golden’ one, but it is a general statement 🙂

Now, the final bit. You need a tape measure.

First, measure your arm, from fingertip to shoulder, then fingertip to elbow. Take the first, divide it by the second….and it’s close to 1.618. Same with your leg to body ratio, hand to elbow, finger to hand….

And then you get to facial features….rather than try to define it for you, go here.

But, whilst general beauty can be ‘created’ using phi, that doesn’t mean that phi is beauty. This is one of the areas that I think maths will never be able to completely explain: that of likes, dislikes, loves, hates, appreciation, ridicule. Whilst we can find (with relative ease) links between things that people like, and therefore create something generic that appeals to most people, we can never find something that anyone will truly find breathtaking. I am a huge fan of the TV series Numb3rs, and on one of the episodes they are dealing with music, and Charlie ‘explains’ that there are some sequences of notes and tones that appeal to everyone, and, using this, we can analyse music and find, with relative accuracy, how well a certain song will do when it’s released. I don’t know how accurate this is, but I do know that we can not quantify something as individual taste. Whilst maths can certainly be used to give us an idea of how people will react to something, we can’t guarantee it’s success. That being said, I do recall a quote that went something like “given all the information, we can predict anything”….