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2010/10/10

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

2010/10/03

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 MathsTutors4me | 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 25th 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..,

2010/08/06

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!

2010/07/24

12 balls: The solution.

At least, I think it's a solution. Please have a good, long look at it, figure out what it's talking about and then tell me if I've got anything wrong.

If you haven't had a go yourself, don't look at the image, just go here for the original problem, and here for a clue. It wouldn't be any fun if you went straight for the solution, would it!


The solution
Warning! This is the solution and contains spoilers!


O.k, that's not very easy to read, is it? Might be a good idea to click on it and view it full-size.

Just in case you're not sure what's going on, here's a walk-through of just one of the possibilities to give you an idea:

  1. First up, you need to name your balls. I've gone for the rather unimaginative (but not too confusing) 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 & 12. You could go for A-L if you wanted to, or you could think of 12 different names, but then you might get too attached.
  2. Weigh* any four balls against any other four. I've gone for 1, 2, 3 & 4 against 5, 6, 7 & 8. Three things can happen here:
    i) The two sets of balls balance;
    ii) The two sets don't balance, with the first set rising and the second falling;
    iii) The two sets don't balance, with the first set falling and the second rising.
    Let's pretend they're balanced:
  3. If the two sets are balanced, you know that balls 1 - 8 are fine, and the odd one out is either 9, 10, 11 or 12. To find out which one, first you need to weigh** two of the suspects against one suspect and one fine ball. In my example, I've gone with 9 & 10 against 11 & 1. Again, three things can happen:
    i) The two sets balance;
    ii) The two sets don't balance, with the first set rising;
    iii) The two sets don't balance, with the first set rising.
    Let's pretend they're unbalanced, and that the first set (9 & 10) rise.
  4. If 9 & 10 rise, then one of two things is true:
    i) Either 9 or 10 is too light;
    ii) 11 is too heavy.
    To work out which it is, weigh*** the balls in the first set against each other (e.g. 9 against 10). Once more, three things can happen, and the solution depends on which:
    i) The two balls are balanced - in which case, it must be that 11 was the odd one out, and it was too heavy;
    ii) The balls are unbalanced, with 9 rising - in which case, it must be that 9 is odd, and it's too light (as 10 has already been in a rising set, so can't be too heavy);
    iii)The balls are unbalanced, with 10 rising - in which case, it must be that 10 is odd, and it's too light (for similar reasons as in the previous case).
I hope that makes some kind of sense. If not, or if you'd like any parts of my flow chart explaining in greater detail, feel free to get in touch (leave a comment, or send me a message on Twitter (@TeaKayB), or something).



* This is the first weigh.
** This is the second weigh.
*** This is the third weigh.

2010/07/12

12 balls: A clue

Carrying on from the 12 pool balls problem posted the other day, here's a clue for anyone who's not too sure where to get started.

The clue comes in the form of a photograph of my whiteboard after a colleague and I had spent part of a free period muddling the problem through. It's not complete and it's not explicit, but then it wouldn't be a clue if it was, would it?

If you want to have a go without seeing the clue first, then CLICK HERE and close your eyes for a few seconds!



The clue
Warning! This is a clue and contains spoilers!
Click to view full-size image!
I'll post the full solution in another few days... subscribe to make sure you catch it!

I've you've got a maths-related question of your own, you could ask it in the comments or use the form at my main blog.

2010/07/11

The 12 pool balls problem

Not my pic... click for originator
O.k, here's one for ya...

You have twelve pool balls and visually they're all exactly identical. One of them, however, has a different mass/weight to the other eleven, but you don't know whether it's heavier or lighter.

You have at your disposal a set of weighing scales so you can weigh the balls against each other, but no weights to go with it. You can only use these scales three times*.

Your job is to find out two things:

  1. Which ball is the odd one out?
  2. Is it heavier or lighter?

The rules
  1. You're only allowed to use the instrumentation and apparatus as described above. No magicking weights or digital scales out of thin air, or inventing science prep rooms nearby.
  2. No cheating. There are solutions available online, but if you look them up you're only cheating yourself. If you look it up and post it here, then you degrade yourself to a point a little beneath most traffic wardens.
  3. There is a solution.
  4. Feel free to comment and ask for clues. My HOD and I have worked out what we think is a pretty nifty solution, so I might post a clue-post in a few days or so if anybody asks for one.
  5. I will post my solution at a future date as yet undecided, then edit this post to link to it. If you're reading this after that's happened, please give it a shot on your own (or with friends/colleagues) first. It's more fun that way!
  6. Feel free to post your own solutions on your own blogs or websites and then let me know so I can link to them.
  7. Subscribe to MathsQS for updates and new posts!


* No, I don't know why. Maybe they're set to explode, or something. Think Die Hard II with the jugs problem.

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