Guest Post: The Higgs Boson: How Sure is Pretty Sure?

You'll have heard, unless you particularly like the conditions under rocks, that the folks at CERN's LHC have found some pretty convincing evidence for the existence of what is known as the Higgs boson. At long last they've got something to work with beyond 'it should be there, and it should be a bit like this'. They're pretty sure that what they've found is the Higgs boson and not just something that happens to act a bit like one should act. But how sure is 'pretty sure'? Friend of the blog Colin Beveridge**** answers that question. Apologies for letting through a small instance of bad language. I thought it worked.

“They’re only pretty sure they’ve found the Higgs boson, aren’t they?”

If I had a billion dollars for every time I’d heard that question, I’d have enough to build a medium-sized hadron collider. Technically, that’s right: they’re only pretty sure. But, as with all things scientific, there’s ‘pretty sure’ and there’s ‘pretty sure.’

Luckily, science has a better way of talking about things than just saying ‘we’re pretty sure’, and it’s — surprise! — to put a number on things. You’ll quite often see something described as ‘a three-sigma event’ or — in the case of the Higgs — ‘a five-sigma discovery’. (There’s a business philosophy that, like most business philosophies, had a year or two in the spotlight and then vanished, called ‘six-sigma’ — it’s the same idea.) The higher the number, the more certain you are about the event⁠*.


So what is it this mysterious sigma represents?

Well, it’s to do with hypothesis testing, and it’s based on the mathematical idea of proof by contradiction. The idea is, if you want to show that something is true, you can prove it by showing that the opposite isn’t true. For instance, if you wanted to prove that there was no biggest number, you’d start by saying “let’s assume the opposite! Assume there is a biggest number.” Being a mathematician, you’d probably call it N, but you don’t have to. “Now, add one to that number — you’ve got a bigger number now, so that assumption about having a biggest number? Load of bollocks⁠**. Therefore, there can’t be a biggest number.”

Got it? To prove something is true, you can assume that it isn’t true, and show that reality is inconsistent with that assumption.

That’s sort of how hypothesis testing works, too, but it’s a little bit trickier. When you’re doing nice proofs about real numbers, all the properties are known and you can say for certain (most of the time) whether something is absolutely, definitely true or not. When you’re dealing with the real world, it’s not so simple. As Einstein said, “As far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality.” You really can’t show that something is definitely untrue in science — only that it’s extremely unlikely.

So that’s what you do! You assume the opposite of what you’re trying to prove (and call it the null hypothesis, if you’re that way inclined). You say what you’re trying to prove (the alternate hypothesis). You work out a threshold of how certain you want to be⁠*** — which is about the same thing as saying how unlikely the null hypothesis needs to be — and then you do the experiment.

If your observations would be absurdly unlikely under the null hypothesis, but quite plausible under the alternate hypothesis, you throw out the null hypothesis as a bad job, and accept the alternate hypothesis. If your observations aren’t unlikely enough, no matter how unlikely they are, you don’t have enough evidence for what you were trying to prove. Sorry.


So, where do the sigmas come in?

They come from the normal distribution, which looks to me like a boa constrictor that’s just eaten an elephant. It’s a shape that comes up repeatedly in science — most measurements are pretty close to the mean, in the middle of the graph, and the further away from the middle you get, the less likely the result. For instance, if you look at the heights of a large group of 25-year old adult males, you’d expect most of them to be about 175cm tall, most people within — I’d guess — 15cm of that, and possibly some extremely tall or extremely short people. The graph of those heights would looks something like a normal distribution.

There are two important bits of information there: the 175cm, which is the mean of the distribution — how tall you expect a random person picked from the crowd to be; and the 15cm, which is the standard deviation. The standard deviation is a bit harder to explain — in very loose terms, it’s the ‘give or take’: if you say “a person from this group will be between 160cm and 190cm tall”, you should be right about 70% of the time. That’s the nature of the normal distribution.

Sigma is just the number of standard deviations an observation is away from the mean. So, if you had a person in the group who was exactly 175cm tall, that would be a zero-sigma event. Someone 160cm tall is 15cm away from the mean, or one standard deviations, making him a 1-sigma event. Someone 205cm tall — a big lad — would be 30cm, or two standard deviations from the mean, making him a 2-sigma event.

The thing about the normal distribution is, it tails off very quickly. A one-sigma event in a given direction shows up about 14% of the time (around one in seven). A two-sigma event, only about 2% (one in 44), and a three-sigma event, barely 0.1% (1 in 740). By the time you get to 4 sigma, you’re talking about one in 32,000, and 5 sigma is one in three and a half million — about the same as winning the lottery if you buy four tickets.


So, the answer is ‘yes, they’re only pretty sure’, but in this context, it means that the alternative is preposterously unlikely. Not impossible, of course, they could just have got lucky and found something Higgs-shaped exactly where they happened to be looking for it — and if that’s the case, we’ll find out more about it in the next few months.

* I’ll get into the details of what the sigma means later.
** Probably not the technical language you’d use in a formal proof.
*** I’m getting to the sigmas, promise!
**** Colin's a mathematician with his finger under many belts, if you'll excuse the mixed metaphor: he's a maths tutor, a maths blogger, a maths writer, and a folk guitarist too.

RL Maths: The Great Pizza Ripoff

I went to a well-known pizza restaurant chain for dinner last night. I won't tell you which one, but I'll refer to it as 'Pizza Shed' and you can read from that what you will. When we were ordering, my girlfriend and I briefly wondered whether we should get a regular pizza each, or a large pizza with 'half-and-half' toppings. We went for the latter, not really thinking too much about it (we wanted a stuffed crust anyway).

However, when the bill came we were surprised to note that our half-and-half large pizza had been charged as two regular ones. This got us thinking: is one large pizza really worth the same as two regular ones? The good thing is that we don't even need to know the price to work this out. We do need to know the sizes, though: a regular pizza of the type we ordered is 11" across, and a large is 14".

At first glance, you might say "the large pizza isn't twice the size of the regular one, so no, it's not worth it!" but be careful: you're making a mistake that loads of people make when considering the mathematics of pizza ordering*.

Consider the hastily produced graphic to the left**. It shows two small pizzas, side by side, plonked on top of a large pizza. The important bit is that the two small pizzas together have the same diameter as the large pizza, but when plonked on top there's still a large expanse of pizza that isn't hidden. This means that twice the diameter means much more than twice the pizza.

So we've got to be a bit more cleverer about this. When talking about how much pizza you get, it makes more sense to talk about the area of a pizza than its diameter (which is the standard quoted measurement for pizzas worldwide). If you've got a maths GCSE (or O-Level) you'll have at least a dim recollection that the area of a circle and its radius (which is half the diameter) are linked. Specifically, the formula is:

A = π x r²

Where A stands for 'area' and r stands for 'radius'. And π, far from being something scary, is just a symbol that means "a bit more than 3"***.

So, for our pizzas, we can find the regular one's area by working out π x 5.5², and finding the answer to π x 7² will give us the area of a large pizza.

A good idea in maths is to estimate things before working them out exactly. This gives you a feeling for what kind of number you're expecting the actual answer to be, so you can tell if you've messed something up (or hit the wrong button on the calculator if you've got fat fingers like me). In maths 'estimating' is not just guessing, but working out a simpler version of the problem. I did this, last night, in a minute or so whilst driving home (that's how easy it is****).

So the regular pizza has an estimated area of "about 3 multiplied by about 30*****", which is about 90 square inches, and the large pizza has an estimated area of "about 3 multiplied by about 50******", which is about 150 square inches. It looks like two regular pizzas gives you more total pizza than one large one!

What's the solution, exactly?

• π x 5.5²  = 95.03 square inches (to 2dp).
• π x 7² = 153.94 square inches (to 2dp).

My estimates were both slightly low, but not by a lot. The same result stands: two regular pizzas would have given us more pizza than one large one, yet they charged us the same amount!

The scoundrels!

* O.k, the number of people who actually stop to consider the mathematics of pizza ordering might be quite small.
** I am available for freelance graphic design jobs for a modest fee.
*** π is, of course, a specific number (3.14159... etc), but to write it down accurately would take more time and energy than is available to us, even if we enlist our children's children's children, so we're lazy and use a symbol instead.
**** If you'd like some tips for being more accurate with your mental maths for not much extra effort, you could do worse than check out my pal Colin's 'Mathematical Ninja' series.
***** 5² is 25, and 6² is 36, so 5.5² is about half way between, which is about 30.
****** 7² is 49, which is about 50.

Codebreakers: Bletchley Park's Lost Heroes

Most people have at least heard of Bletchley Park, the UK's codebreaking centre during World War II. Even if you haven't heard of the place, you've probably at least heard the name of the most famous code to be cracked by them: Enigma.

This BBC program details the history behind an even tougher code that is less well known, but appears to have been just as vital to the war effort as deciphering Enigma-encoded messages. The story is also one of the people behind the scenes, especially two mathematicians whose contributions, although invaluable, have gone largely unrecognised by history.

The Lost Heroes of the title are Bill Tutte and Tommy Flowers, both of whom made significant steps in the race to understand and decode messages sent by Nazi forces: Tutte made great strides in understanding with regards to machine-based ciphers, and Flowers designed and developed the world's first electronic computer, Colossus, in an effort to speed up the process by which messages were decoded.

Timewatch - Code-breakers: Bletchley Park's Lost Heroes: 

https://www.bbc.co.uk/iplayer/episode/b016ltm0/timewatch-codebreakers-bletchley-parks-lost-heroes

What's the Difference Between Weight and Mass?

For the answer to this question, head on over to Jennie's guest post at Blogstronomy!

What's my score?

I hadn't played Yahtzee in years, but the other night I settled down to possibly the best couple of games I've ever played.

For those not in the know, Yahtzee is a game from Hasbro (My copy's from MB Games... that's how long it's been!) that's been around since 1956. Put simply, it's a game in which players roll five dice and are given a score based on what is rolled.

We played two games, and they were probably my best ever scores (though that doesn't say much; I'm not exactly a pro). In my first game, I scored exactly 80% of the maximum possible mark (assuming no bonuses for extra "yahtzee*"s), and in the second I scored 89% of the maximum, rounded to the nearest whole

1. What was my score in the first game?
2. In the second game, what's the highest score I could have got? What's the lowest?

Have a go at the questions before looking at the answers below!

The solutions

I'll remind you here as well that there are usually many, many ways of solving problems, so if you haven't done it the same way as me that doesn't mean that you've done it wrong. In fact, if you do it a different way I'd like to see an explanation in the comments- it might even be a better** way of doing it!

Part 1)

Look at the information I've given you: I got 80% of the total possible score. Assuming you know something about percentages*** then all you need to know is the total possible score. If you don't know the game, here's a scorecard to look at:

You just need to think of the best possible score for each roll, and record it, then add anything up, taking into account any bonuses. The overall total should come to 375. Comment and ask if you're not sure how to get that.

Now all you need to do is work out 80% of 375, then you've got my score for game 1:

One way of doing this is to find 10% (divide by 10) and then multiply by 8 (as 8 lots of 10% is 80%). So:

375 / 10 = 37.5

37.5 * 8 = 300

• My score for game 1 was 300 points exactly!

Part 2)

This one's trickier. The question asks for my highest and lowest possible scores because I didn't get exactly 89% - I rounded the answer. This means that finding 89% of the maximum possible score won't give the score I got, as it could have been a bit smaller or a bit bigger than that. In fact, if you work out 89% of 375...

375 / 100 * 89 = 333.75

... you don't even get a whole number, so it's actually impossible to get exactly 89% of the maximum score in a game of Yahtzee. So we'd have to make a guess. Luckily we can use the information given to make that guess as accurate as possible- we can find a sensible low- end for our guess, and a sensible high- end, and go somewhere in between. That's what the question means when it asks for the highest and lowest scores that I "could have" got.

To do this we need to look at that percentage first: I said I rounded it to the nearest whole percent, which was 89. So what are the lowest and highest answers I could have got before I rounded it? Any more than 89.5%, and I'd have rounded it up to 90%. Any less than 88.5%, and it'd have gone down to 88%. If we work out the two percentages in bold, we'll have our answers:

375 / 100 * 88.5 = 331.875

375 / 100 * 89.5 = 335.625

They're the highest and lowest scores, but now our common sense should kick in: there's no way of getting a decimal score in Yahtzee! That means the for the lowest score, we've got to pick 331 or 332, and for the highest, we must choose 335 or 336. But which one? Well, the lowest score that would end up being rounded to 89% would be 331.875. Any lower than that and we won't get 89%, so it must be higher than that: we pick 332. Using similar reasoning, but at the other end of the scale, we must pick 335 as the highest score that I could have got.

• My score for game 2 was somewhere between 332 and 335 inclusive.

Is it possible to be any more accurate than this?





* In the game Yahtzee, a 'yahtzee' is rolling five of the same number in one turn, i.e. after three rolls you end up with 5 twos, or 5 sixes, etc.
** What makes one answer 'better' than another, assuming that both are correct? That might be a subject for another post...
** If you don't, ask me for a post ;-)

If you fold a piece of paper 50 times, will it reach the Moon?

This post was prompted by @mikemcsharry 's tweet of a similar nature the other day. But is it true? For the pedants out there, I'm assuming that we're folding our imaginary sheet of paper exactly in half each time.

Stuff we need to know:

• How far away is the Moon?

It's about 384,400 kilometres away, and counting.

• How thick is a sheet of paper?

That all depends on the quality of paper and the manufacturer. Typical office paper (80 "gsm"*) is about 0.1mm thick, I think. As it makes sense to have everything in the same units, 0.1 mm is 0.0000001 kilometres.

Answering the question, part 1: How thick is a piece of paper folded 50 times?
I've seen this question tackled before, and the biggest mistake is to rush straight in and say "if we fold the paper 50 times, the paper will be 50 times thicker".

Think about it like this (feel free to grab a sheet of paper and try it):

• Fold a piece of paper in half once. It's twice as thick as it was, right?
• Fold it in half again. It's now twice as thick as it was last time- if you've got the paper in front of you, you can count the layers: it's four times as thick as the original piece of paper.
• Fold it in half again. if you count the layers, you'll see that it's now eight times thicker than the original.

By now, it's fairly easy to see what's happening: each time you fold, you're doubling the thickness of your lump of paper. That means it's fairly easy to see the long, slow way of calculating the thickness of the paper after 50 folds:

Start off with the thickness of one sheet, and then double it. Double it again. Double once more. Double again. And again. And again. Keep going until you've done it 50 times. If you're into button-mashing** on a calculator, you could type this in:

• 0.0000001 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 112,589,991

Because the original number (0.0000001) was in kilometres, we know that the answer must be. So if we fold a piece of paper 50 times, it would end up being 112,589,991 kilometres thick!

Answering the question, part 2: Would this reach the Moon?

This is the easy bit: we just have to compare the two numbers:

• Distance to Moon: 384,400 km
• Height of paper folded 50 times: 112,589,991 km

The second number is bigger (and, importantly, given in the same units), so we have our answer: yes, it would!

But it'd be boring to stop there...

• 112,589,991 / 384,400 = 292.897999...

That means that our folded paper would reach nearly 300 times further than the Moon! So what's that far out?

• Travelling inwards from Earth to the Sun, we'd cross the orbit of Venus just after climbing a bit more than a third of our stack of paper.
• About 80% of the way up our stack, we'd cross Mercury's orbit.
• We'd be about 3/4 of the way to the Sun when we were standing on top of our paper tower.
• Travelling outwards from the Earth away from the Sun, we'd cross Mars's orbit a bit more than a tenth of the way up the stack.
• We wouldn't quite reach the main asteroid belt, and we'd need more than 400 more of these paper towers to reach the next planet, Jupiter.

Why don't we do it, then?!

Give it a try! If you get past 8 folds (remember, you're folding the paper exactly in half each time!), I'll happily send you a Mars bar. Any more than 10 and I could probably be persuaded to give you my car.****

* Why they can't use standard notation, I don't know. "gsm" stands for 'grams per square metre', which would properly be written as "g/m²"
** I'm not into button mashing. There has to be an easier way - we can use powers:

• 2 x 2 means the same as 2².
• 2 x 2 x 2 means that same as 2³.
• 2 x 2 x 2 x 2 means that same as 2⁴.

See the pattern? So,

• 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 means the same as 2⁵⁰.

This means that we can write our calculation much more easily and accurately*** as:

• 0.0000001 x 2⁵⁰ 

*** We're typing less, so there's less chance of' 'user error'.
**** As a disclaimer, you're not getting my car.

Your birth year + your age = 111. How?

There's a meme flying round the internet at the moment that tells you to...

... add the last two digits of your birth year to the age you will turn on your birthday this year (2011, for any time travellers). It tells you the answer will be 111.

For me, born in 1982 and turning 29 in October, this would be 82 + 29 = 111. Try it with your own details...

It works!

But how?
To complete this task, you need to know two pieces of information: the last two digits of your birth year, and the age you will turn this year.

Your birth year is easy- just take the two digits off the end: for me, it's 82.

You probably know your own age, but if we're going to figure out how this works, we need to think about it in a different way. To find out your age knowing only the year you were born in and what year it is now, you could subtract your birth year from this year. Using my info, that'd be...

 2011 - 1982 = 29

... which I can confirm is correct!

The thing is, we're only supposed to be using the last two digits of the year, so lets try...

 11 - 82

That gives us a strange value: -71. But consider that the leading two digits of the years we're using are different: 20 and 19 respectively, and remember that this is because we're in different centuries: a century is 100 years, so add this on and what do we get? 29!

So to find out the age this year of anyone who was born during the previous century using only the last two digits of the year, we could do the following:

 Age = 11 - the last two digits of their birth year + 100

This looks a bit clumsy, so I'm going to use the letter 'x' to represent 'the last two digits of their birth year':

 Age = 11 - x + 100

Lets clean it up further: We start off with 11, take something from it, and then add on 100. Why not deal with those two numbers at the same time? If I start with 11 and then later add 100, I may as well start off with 111 in the first place:

 Age = 111 - x

Right, that's the slightly complicated bit sorted. Lets put it all together:

The info we need:

• Last two digits of birth year: we're calling this x.
• Age this year: we're saying this is 111 - x

What we have to do:

• birth year + age;
• Using the notation we've defined above, that's: x + 111 - x

But wait... we're starting off with whatever x is, then adding on 111, then taking x off again. Whatever x turns out to be, it doesn't really matter because we're just getting rid of it, leaving just...

111

... all by itself!

An important point...
... just pointed out to me by @justfin is that, due to the fact that two-digit years go in 100-year cycles, anyone who's over 100 years old this year will find that they get an answer of 211 instead of 111!

How does binary work?

Like the decimal system that we all use on a daily basis, binary notation is just another way of writing down numbers. To get the idea of how it works across, I'll start off by briefly explaining how the decimal number system works: 

How to read decimal numbers

31

The number above is 'thirty-one'. You know that because you've been brought up to count using the decimal number system. We, as a developed society, use it to a large extent not only because there are some nice, easy to remember patterns that make themselves present with a decimal number system, but because that's what our parents used, and their parents, and their parents...

How does it work?
The number system we're familiar with uses ten 'bases' - the symbols 1, 2, 3, 4, 5, 6, 7, 8, 9 and 0 - displayed in a place value system. That means that where we put them in relation to each other is important. We know that '31' means 'thirty-one' because written like that the 3 represents 'three tens' and the 1 represents 'one unit'. Put together, three tens and one unit make thirty one (of whatever we're counting) altogether.

We can represent ever bigger numbers by adding in more columns to the left- the next one, for example, allows us to say how many hundreds we want (0 - 9), the one after that describes how many thousands there are, and so on for ever and ever and ever, if we wanted to.

That's a crash course in the decimal "base ten" number system- remember that's the one you use every day. On to binary.

How to read binary numbers

11111

The number above is also 'thirty-one', or rather it represents the same amount as the number 31 does in the decimal number system.

How does it work?
The binary system works in a similar way to the decimal one, except it uses only two 'bases': 0 and 1 (i.e. there are no 2s, 3s, 4s, 5s, etc). Instead of each digit telling us how many 'ones', 'tens', 'hundreds', etc are in the number, the columns are labelled in a different way. The first column (starting from the right) is still 'units' (or 'ones'), but the next one to the left now represents how many twos we want in our number. The one after that is how many fours, and then it's how many eights. Can you see the pattern? Each column (as we move to the left) represents double the value of the one before it, so:

11111 means:

One 'sixteen', one 'eight', one 'four', one 'two' and one 'one'. Added together this makes thirty-one.

Another example, on binary day:
Today's date is the tenth of October, 2010, or 10 10 10, and is being called 'binary day' for fairly obvious reasons.

so as a binary number, 101010 means:

One 'thirty-two', no 'sixteen', one 'eight', no 'four', one 'two' and no 'one'. Add the thirty-two, eight and two together and you get:

42

Maths tutoring now available!

Just a quick note to say that I'm now offering private mathematics tuition, primarily to people in the Kettering area of Northamptonshire but if you're further away and still want to hire me get in touch and I'll see what I can do.

I'm offering sessions on anything up to GCSE level at the moment, whether you're at school or a more mature student. Get in touch and we can discuss your needs, whatever they are. My prices start at £25 per hour (but may be more if you want me to travel further).

I have profiles on the following tutoring websites:
Flying Colours Maths | Tutors4me | TutorNet | TutorHunt | Tutor in UK
Feel free to get in touch through any of those, or use the contact me page on this site.

Why me?
I've got experience and understanding: when I first started out in the teaching business I was a teaching assistant in a local primary school. I know my stuff and I got on really well with the kids so I started working with individuals and small groups, sometimes helping those who were struggling to catch up, sometimes taking the more confident kids and pushing them further. I earned myself a good reference from the head teacher and started a PGCE* course, training to teach mathematics at secondary level. I've been teaching maths in a local secondary school ever since.
I've also worked in industry, fast food, a mobile telephones shop and the job centre, so I do have some experience of the real world outside of a classroom!
In terms of understanding, I earned myself a BSc in Mathematics with Astronomy from the University of Leicester. Maths teachers who actually have a maths degree are not as common as you might think.

I know my stuff, and if you're the kind of person who wants to learn and is willing to put the time and effort in, then I am as well.

• Maybe you've had some time off school and want to catch up with what your class has been doing in the mean time: I can help.
• Maybe you're headed for a B in your GCSE but you really want that grade A: I can help.
• Maybe you're not all that keen on maths but you know the value of getting that all important C and want to put that little extra in to get it, if only you can find someone to guide you: I can help.

I don't care if you're eight or eighty: if you want to improve your game in maths: I can help.

Who am I?
Why would you want to hire anybody without knowing whether you were going to get on with them? If I tell you a bit about me then hopefully you'll see that we can get on and have a bit of fun as well as learning some maths.

First and foremost, I'm a geek (but don't change channel just yet...). I'm into most things science related which is where my love of maths comes from: a good understanding of maths can help you to understand and explain pretty much anything else in a bit more depth. I love learning new things and almost equally love passing on the things I've learnt to anyone who wants to learn too. I'm especially interested in astronomy**, which is an incredibly mathematical science, and run a blog in which I answer questions on related issues. Head over to Blogstronomy and have a look!

I don't just do maths and science, though. I also play guitar, sometimes in a band that plays classic rock covers (and the odd slightly more modern number). I have long hair and am fond of leather jackets and a decent Status Quo riff, but I went to see - and thoroughly enjoyed - the Hairspray musical recently, and I'm looking forward to the 25^th anniversary showing of Les Miserables: my tastes are varied.

I'm young (I'll hit 28 next week, but you don't have to get me a card), enthusiastic, flexible and committed to helping you hit the targets that you set for yourself.

If you have any questions about me, maths or what I offer, or would like to enlist my services, please read the new tutoring page on this site, and then get in touch.

I've been given a lot of advice and guidance by Colin over at Flying Colours Maths who has also listed me as part of his expanding empire.




* PGCE stands for Post-Graduate Certificate of Education - it's the most popular route for gaining Qualified Teacher Status in the UK.
** Not to be confused with astrology..,

How many connections?

This post was inspired by @squiggle7's Challenge That 'let's get connected' Post.

In it, @squiggle7 wonders about the number of connections able to be made by people in a 'room', be that virtually or physically.

Let's keep this up to date and use connecting with like-minded individuals on twitter as an example:

Imagine there's only one person on twitter who shares a particular interest of yours- how many connections can be made? By 'connections' I'm meaning mutual followships (to keep it simple), so in this case it's obvious: only one connection can be made.

Spotting the pattern
How about two other like- minded people? That's three altogether. I'll call them A, B & C (because they're cool names, right?). Because we're looking at mutual followships only, we can assume that if A is linked with B, then B is linked with A, so we can have the following links:

• A with B
• A with C
• B with C
• 3 connections altogether.

O.k, now how about four people? A, B, C & D this time:

• A with B
• A with C
• A with D
• B with C
• B with D
• C with D
• 6 connections available to be made.

[I'm being systematic when listing these possibilities- can you see the system I'm using?]

And five people (A, B, C, D & E):

• A with B
• A with C
• A with D
• A with E
• B with C
• B with D
• B with E
• C with D
• C with E
• D with E
• 10 connections possible.

We could carry on, but it's fairly clear by now that it's going to get a lot more complicated a lot more quickly, so can we do anything to simplify things? Can we find a pattern to follow?

People  2 3 4 5
Connections  1 3 6 10

Look at the numbers in that table: as the number of people goes up, the number of connections increases too. But by how much? It's fairly easy to see that as the number of people increases by 1, the number of connections increases by 2, then 3, then 4... so we can be confident that the number of connections for 6 people would be 10 +5= 15, and the number of connections for 7 people would be 15+6= 21 and so on. If we wanted to find out how many connections would be possible for, say, 20 people, we could just carry on the sequence. But what about 100 people? 150 people? 2000? 6000? 132,947 people?

Using algebra
Time to think about it a different way... Imagine there's a large number of people all with the same interest who all connect with each other. How many connections will there be? How many people exactly doesn't matter, so we'll call it "n" (so it can stand for any number).

• Person A can make a connection with everybody (except him/herself), so that's n-1 connections.
• Everyone else can make the same number of connections, so that's n-1 connections, n times: n(n-1)
• But we're forgetting that once a person makes a connection, they're connected both ways, so we've actually counted each connection twice. That's easy to fix, though, we just need to halve what we've got so far: 

And that's that. In simple(ish) terms, this means that to find the number of connections that can be made with any number of people in this situation you have to multiply the number of people by one less than the number of people, and then halve the answer.

So, for 132,947 people, there are (132947 x 132946) / 2 = 8,837,385,931 connections that can be made. That's well over eight billion!