Making Your Own Sense

Reflections on maths, learning and maths learning support, by David K Butler

Tag: proofs

  • Archimedes’s Integrals

    One of my staff (thanks Fergus) told me ages ago about Archimedes’ proof that the volume of a sphere is 4/3 π R3 (where R is the radius of the sphere). It is a very very cool proof and it’s high time I shared it! One of the reasons it is so cool is that it uses the concept that a volume can be produced by stacking up a whole lot of thin slices. This is the idea behind integration, and Archimedes used this idea thousands of years before Newton or Leibniz.

    So here is a modernised version of the proof:

    We’re going to assume we already know a few things:

    1. How to calculate the area of a circle.
    2. Pythagoras’ theorem.
    3. The fact that all the points on a sphere are the same distance from the centre (the radius).
    4. The fact that a cone has a third the volume of a cylinder with the same base and height.
    5. How to calclate the volume of a cylinder.

    Ok, now let’s get a sphere of radius R and slice it in half to make a hemisphere. Next we’ll get a cylinder of radius R and height R, and we’ll scoop a cone out of it to make a sort of bowl (see the pictures).

    A diagram of a hemisphere and a cylinder the same height as the hemisphere with a cone scooped out of it.

    The volume left for our cylinder-minus-cone bowl is two thirds of the volume of the cylinder (since the cone is a third the volume of the cylinder). We’re going to show that the hemisphere has the same volume.

    We are going to do this by working our way up from the bottom and slicing both shapes into very thin slices. If we can show that at every possible height the slice for the bowl has the same area as the slice for the hemisphere, then it must be true that the two shapes have the same volume.

    A hemisphere and a cyclinder with a cone scooped out. Each has a bright green shape shown where both have been sliced at

    So suppose we are at height y above the base:

    Consider the bowl: when we slice the bowl, we will get a circle of radius R with a hole in the middle. The edge of this hole is on the outside surface of the cone. Now a line in the curved surface of the cone has slope 1 (since it goes across R and up R), so since we’ve gone up y, we must also have gone across y to get to the edge of the cone. That is, the radius of the circular hole is y. Hence the area of the slice at height y is π R2 – π y2.

    Consider the hemisphere: when we slice the hemisphere, we will get a circle – let’s say the radius of this circle is x. Now a point on the edge of this circle is a distance x from the vertical centre line of the hemisphere, and it’s a distance y above the base, and it’s a distance R direct from the big centre of the hemisphere in the middle of the base. Using Pythagoras’ theorem, we get that x2 = R2 – y2. But the area of our circular slice is π x2, which is equal to π (R2 – y2) = π R2 – π y2.

    A hemisphere and a cyclinder with a cone scooped out. Each has a bright green shape shown where both have been sliced at. The height of the slice is labelled y, the radius of the sphere/cylinder is labelled R, and the radius of the slice of the hemisphere is labelled x.

    So the areas of the slices at every height are the same! So when we stack up those areas to produce a volume, the two volumes must be the same!

    Since the bowl has two thirds the volume of the cylinder, this means our hemisphere’s volume is also two thirds the volume of the cylinder! Therefore a whole sphere must be four thirds the volume of the cylinder.

    Now the volume of the cylinder is the base area times the height, which is π R2 × R. So therefore the volume of the sphere must be 4/3 π R3.

    How cool is that?!

    These comments were left on the original blog post:

    David Robers 24 July 2014:
    Fact 4 is not _hugely_ obvious… One could look up a proof online, but how did the Greeks do it?

    David Butler 25 July 2014:
    Can’t find information about how they did the volume of a cone. However my instinct is that it is by analogy to the volume of a pyramid. If you take a cube and construct inside it a pyramid with base the same base as the cube and vertex at one of the top vertices of the cube, then you can observe that three such pyramids can fit neatly into the box, so a square pyramid has a third the volume of a cube. By stretching you can see that a recangular-based prism has a third the volume of a prism with the same base and height. By analogy the volume of the cone ought to work

  • The sizes of infinity

    Last week a student visited the Drop-In Centre to talk about the different sizes of infinity. His lecturer had been talking about the sizes of sets and had made an off-hand comment that there were different sizes of infinite sets, and he wanted to know what the hell that meant.

    So I explained it. It’s not the simplest explanation, because you have to define some new ideas in order for it to make sense. But it is one of my favourite things about the families of number that some are the same size and some are different sizes, even though they are all infinite.

    So I want to explain it again here…

    The key to the whole thing is to realise that when talking about infinite sets, that counting is not really a useful way to find the sizes of sets. When you have a lovely finite set, such as the letters in the word “sluiced”, you can simply count the members of the set and you know how many there are (7 in this case). But the process of counting works because at some point you stop, and the number you’re up to is the number of objects there are.

    But what if you can’t stop? If your set has infinitely many objects in it, you’ll never get to a final number where you can stop and say “this is how many”. We need a new way to talk about the sizes of sets, and the way that we use is in some sense even more fundamental than counting.

    If you had two piles of things and were asked which pile has more, you’d probably count both and whichever had the bigger number would be bigger. But there’s another way: you could pair them off, one object in each pile, and if one pile runs out before the other one does, then you know that the other pile must be bigger. And if you are able to pair off everything in both piles, you know they must be the SAME size. THAT is the solution to our problem.

    We DEFINE that two sets are the same size when you can pair them off exactly. That is, there is some way to attach to each object in Set 1 a different object in Set 2, and in this way to cover all the objects in Set 2. If this is impossible, then the two sets are different sizes.

    So let’s look at the five families of number: the natural numbers, the integers, the rational numbers, the real numbers and the complex numbers – N, Z, Q, R and C. (Actually I’m going to ignore C, sorry.)

    The natural numbers, being the set of numbers {1, 2, 3, 4, …}, is what most of us think of when we think of an infinite set. We start counting how many objects it has, but we can never stop because the numbers don’t stop. So we have infinitely many. But it’s quite close to the finite sets, because we can at least try to count them. They call the size of N the countable infinity.

    Now let’s compare the sizes of N and Z. If they were the same size, it would be possible to find a way to line them up next to each other. That is, you could find one integer to call 1, one to call 2, one to call 3, and so on in such a way that you cover all the integers.

    Well, look at this:
    1 ↔ 0
    2 ↔ 1
    3 ↔ -1
    4 ↔ 2
    5 ↔ -2
    6 ↔ 3
    7 ↔ -3
    etc

    We have given every integer a different natural number and covered all the integers and all the natural numbers, so by our definition, this means that N and Z are the same size.

    Cool huh?

    What about N and Q? Well, if they were the same size, we’d be able to line them up next to each other like with N and Z. This time the trick takes two steps.

    First we’ll consider just the positive rational numbers. I need a way of organising them into a list so that I can line them up next to the natural numbers and cover all the rational numbers. Well, the denominator and the numerator of a fraction must add to something, so why don’t I organise them into the ones that add up to 1, the ones that add up to 2, the ones that add up to 3 etc. Within each of them, I can organise them in order of the numerator. So if I do this I’ll get a list like this:

    1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, etc

    Of course, I’ve included some of them twice (since for example 2/2 = 1/1 and 2/4 = 1/2), so I’ll just remove those ones:

    1/1, 1/2, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, etc

    And now I have a neat list of all the rational numbers with a clearly-defined way of getting to each of them. Now I’ll line them up with the natural numbers:

    1 ↔ 1/1
    2 ↔ 1/2
    3 ↔ 1/3
    4 ↔ 3/1
    etc

    In this way I can cover all of the rational numbers exactly once, so by our definition, this means that N and the positive numbers in Q are the same size.

    But now we can do the same trick we did with Z. We’ve got a list of positive rational numbers, and a list of negative ones (and 0 as well) and we’ll just start with zero and flip-flop between the positive list and the negative list and we’ll be good:

    1 ↔ 0
    2 ↔ 1/1
    3 ↔ -1/1
    4 ↔ 1/2
    5 ↔ -1/2
    etc
    So N and Q are the same size.

    But what about N and R? I’d need to think of a way of putting R into a decent order so I can line it up next to N. I can’t think of a good way, but just because I can’t, it doesn’t mean it’s impossible – I’ll only know for sure it’s impossible if I prove it. And the best way to prove that something is impossible is to pretend you’ve done it and try to show that this concept is stupid – that is, a proof by contradiction.

    It’s way too hard to do this with all the real numbers, so I’ll just look at the ones between 0 and 1. All these numbers (except 1) will be “0.something”, and the “something” is a decimal expansion which may stop or may go on forever. For those that don’t go on forever, we’ll just say that it is infinite, but all the digits at the end are zeros. So we can represent each of the numbers from 0 to 1 as “0.something”, where the something is an infinite string of digits. (For the pedants, I’m ignoring here the possibility of having a number ending in an infinite string of 9’s.)

    Suppose we managed to line them up next to the Natural Numbers like this:

    1 ↔ real number
    2 ↔ real number
    3 ↔ real number
    etc

    All these real numbers have a decimal expansion, so let’s write down what that might look like:

    1 ↔ 0 . d11 d12 d13 d14 …
    2 ↔ 0 . d21 d22 d23 d24 …
    3 ↔ 0 . d31 d32 d33 d34 …
    etc

    Supposedly, we have covered all the numbers from 0 to 1 in this way. I’m going to show that in fact we haven’t covered all of them, by making a new number that’s not in the list we have.

    Look at the first digit in the first number: d11. Let’s pick some digit different to d11 and call it a11. (We should probably have a rule that says how to pick this different digit – let’s say that if d11 isn’t zero then we’ll pick a11 = 0, but if d11 is zero we’ll pick a11 =1.)

    Now look at the second digit in the second number: d22. Again we’ll pick a22 so that it’s different to d22. And we’ll continue this process for d33, d44, and so on.

    Consider the number 0.a11 a22 a33 …

    This number is not equal to the first number in our list, because its first digit is different. It’s not equal to the second number in our list, because its second digit is different. It’s not equal to the third number in our list, because the third digit is different. And so on. So our new number can’t be equal to any of the numbers in the list! So we didn’t cover all of the real numbers from 0 to 1 after all. But even if we include this number in the list, the same argument would show that there would always be at least one more, so the set of real numbers from 0 to 1 is not a countable infinity. Since the Real Numbers are bigger than just this little section, they can’t be a countable infinity of numbers either.

    So in a very real sense, the Real Numbers are bigger than the Rational Numbers, which means we have at least two different sizes of infinity!