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1 Adam: {holding a book} Hey Jamie! Have you heard about this wacky thing? The BanachTarski theorem?
2 Adam: Apparently you can cut a sphere into five pieces and reassemble them into two complete spheres, each the same size as the original. It's mathematically proven!
3 Jamie: Yeah, it's really counterintuitive. It's surprising that it can't be done with just four pieces.
4 Adam: You're a sick, sick individual.
4 Jamie: I used to be just sick.
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I think I may really have bitten off more than I can chew this time. Let's start and see how far we get.
The BanachTarski theorem is a mathematical theorem that states that a solid sphere can be divided into pieces such that the pieces can be reassembled into two spheres, with each sphere the same size as the original sphere.
This sounds crazy, and for good reason. If you take a rubber ball and cut it up, there's no way you can reassemble the pieces into two balls, each the same size as the original. Assuming you're not doing anything tricky like stretching the pieces.
It's important to point out up front that the BanachTarski theorem only applies to mathematically ideal objects, not to physical objects. The theorem relies on certain properties of real numbers, namely that there are infinitely many of them, and that you can't match up real numbers oneforone with the natural numbers (refer back to the annotation to comic #2292).
In any physical sphere, there are not infinitely many atoms  there are only a finite number that you can pull apart and rearrange. You can't suddenly double the number of atoms  there would be twice as many!
This is not such a problem for the real numbers. How many real numbers are there between 0 and 1? Referring back to #2292 again, there are infinitely many. What's more, there are so many that you can't match them up oneforone with natural numbers. How many real numbers are there between 0 and 2? Again, infinitely many... but:
Consider a function (refer to #1640) that multiplies numbers by 2. If we apply this function to any number x between 0 and 1, we multiply the number by 2, and we end up with a number y between 0 and 2. Furthermore, for every number y between 0 and 2, there is exactly one number x  between 0 and 1  that when multiplied by 2 gives the number y. (Specifically, it is y/2.)
In other words, the real numbers between 0 and 1 can be matched oneforone with the real numbers between 0 and 2. In one sense, there are twice as many numbers between 0 and 2 as there are between 0 and 1 (because 0 to 2 is twice as "long" as 0 to 1); but in another sense there are exactly the same number of numbers between 0 and 2 as there are between 0 and 1.
This is beginning to approach the level of comfort you need with the quirks of infinite collections of numbers to understand the BanachTarski theorem. If you're not quite comfortable with this so far, please go back and reread before continuing on.
Okay, so we can see that applying a function like multiplication can match up numbers within a range of a certain size (or alternatively, points on a number line between two ends) oneforone with numbers within a range of a different size. But multiplication is by its nature something that "stretches" the size of things. To show the BanachTarski theorem, we need to get a similar oneforone matching by using an operation that doesn't "stretch" things.
The operation we are going to use is rotation in three dimensions. If you take an object in three dimensions and rotate it, it doesn't "stretch" or "shrink", right? This is not meant to be a trick question. Take a rubber ball and rotate it a bit. It doesn't get bigger or smaller. Okay, good.
Now, from here on in, I am going to gloss over some of the fiddly details. If you know some advanced mathematics, you'll probably spot the tricky bits I'm carefully not mentioning. The only reason I'm not going to mention them is because I don't want to get bogged down in fiddly details that must be taken care of in a formal proof. This is not a formal proof, and I just want to get the basic flavour across. But to have things open up front  yes, there are some details I'm skipping; they do need to be taken care of if you want to do this properly; but they can be taken care of if you know what you're doing, so they're not really a problem.
All right. Consider a sphere. Let's take a familiar example so I can define some directions: the Earth. Okay, the Earth isn't exactly spherical, but let's ignore that and imagine it's a perfect sphere, with no bumps or mountains or oceans or anything.
We're going to imagine rotating the Earth around a couple of different axes. One convenient one is the actual axis of rotation, passing through the north and south poles. We need a second one, at right angles to this. Let's imagine an axis passing through the equator, puncturing the Earth at the Greenwich meridian (in the Gulf of Guinea on the west side of Africa) and the 180° meridian (in the middle of the Pacific Ocean). Spin the Earth on this axis and the north and south poles go whizzing around the middle.
Okay, let's imagine that we can only rotate the Earth around these axes by one very specific angle. We can do it multiple times, but each time we rotate the Earth, it must be by exactly the same angle. The angle we can choose somewhat freely, but it needs to be an irrational number of degrees  that is, it cannot be a rational number of degrees.
[Recall that a rational number is one integer divided by another. For example, 5/8, or 2 (equal to 2/1), or 3 4/9 (equal to 31/9). Rational numbers can be represented by decimal numbers that either terminate after a finite number of decimal places, or recur with the same sequence of digits repeatedly forever. The examples in this paragraph are equal to 0.625, and 2.0, and 3.4444... An irrational number is a number that cannot be expressed as one integer divided by another. An irrational number can also be written as a decimal number, but the digits after the decimal place never terminate, nor do they end up in a repeating pattern. Example irrational numbers are the square root of 2, the number π (= 3.1415926535897932...), and the logarithm of 7.]
If we choose a rational number of degrees, say p/q degrees, then if we apply this rotation q times around the same axis, we end up with an overall rotation of exactly p degrees. Since there are 360° in a full rotation, if we apply the rotation 360q times, then the Earth will be in exactly the same orientation where it started. We don't want this to happen.
By choosing an irrational number for the size of our rotation, we can guarantee that no matter how many times we apply the rotation, we will never end up with the Earth in exactly the same orientation as it started. (This is just another way of stating the property that if you multiply an irrational number by any integer, you never end up with a whole number. This is basically the definition of an irrational number, so we know it's true.) You might get extremely close to the orientation where you started, but you'll never match it exactly.
The only exception to this is if you apply backwards rotations, in the exact reverse sequence, to cancel out the forwards ones. If you rotate forwards 6 times, then rotate backwards 6 times, you do end up back where you started.
The other interesting thing is that when you combine the rotations about the two axes, you still never end up back where you started, no matter how many rotations you do or in what order  except for the case where a reverse sequence of reversed rotations exactly cancels out the forward sequence of original rotations.
All right, we're going to need a little bit of mathematical notation now. Not much, so don't run away.
Let's call the operation "rotate by our chosen angle around the Earth's normal axis in the normal direction" by the letter A. The backwards operation "rotate by our chosen angle around the Earth's normal axis in the reverse direction", we'll call A^{1}.
The superscript 1 is standard mathematical notation for an inverse operation. An inverse operation is one which when combined with the original operation, gets you back where you started. Rotating in one direction, then rotating by the same amount in the opposite direction gets you back where you started, so the two operations are inverses of one another.
We have two more operations to give labels to. "Rotate by our chosen angle around the new axis, clockwise looking at Africa" we'll call B. And the anticlockwise version is the inverse, so we call it B^{1}.
Now we can write down a sequence of rotations by using our four labels. We'll just read them from left to right. So, for example: AB^{1}A^{1}BB means:
In terms of this notation, we can be a bit more precise about the earlier statement that you never end up back where you started, no matter what string of rotations you do. In other words, for example, AAB, ABA, and BAA all produce different overall rotations of the Earth. Also, something like ABA^{1} is not the same as B  you can't cancel inverse rotations if there's something in between. (You can actually see these results for yourself with a sixsided die and defining rotations simply to be 90° for the sake of this limited example. Seriously, grab a die and play around with rotating it around different axes a bit. It won't hurt your understanding of the rest of this discussion.)
Right.
Now, let's consider the collection of all possible rotations of the Earth (subject to our constraints). Let's call this collection S. Since we can only rotate the Earth by our chosen angle, forwards or backwards, about our two axes, this means that every possible rotation can be written as a string of the symbols A, B, A^{1}, and B^{1}. Because we never end up where we started (because the chosen angle is irrational), a string of symbols can be as long as we like, and  assuming we've cancelled out any inverse pairs, to simplify each string to its shortest representation  it will always be a new, unique, rotational orientation. So there are an infinite number of different strings, and an infinite number of different rotational orientations.
Let's split this collection of rotations S up into four smaller sets:
Let's take (set 3), the set of rotations which starts with A^{1}. The first operation in each of these rotations is A^{1}, but we don't say anything at all about what the second operation is. Except we know it can't be A, because then the rotation would start with A^{1}A, and these two operations would cancel and we could have just removed them before listing it. So the second operation in each of the rotations in (set 3) is either B or A^{1} or B^{1}.
Let's rotate the Earth by operation A. Okay, done that? Right, now let's look at our sets of rotations again. All of the rotations are now effectively the same rotations, but with an extra A placed in front of them.
Consider (set 3) again. All of the rotations in (set 3) are just the same as they were before, but now with an A placed in front of them.
But the rotations in (set 3) all started with an A^{1} before. If they now have an extra A in front of them, then now they all start with AA^{1}. These two intital operations cancel out, and we can just remove them. If we remove them, then the first operation of a rotation in (set 3) is now the second operation of what it used to be. But the second operation of the rotations in (set 3) is either B or A^{1} or B^{1}. So (set 3), after rotating the Earth by A, now consists of all the rotations in sets 2 and 3 and 4!
In othe words, simply by rotating the Earth by the operation A, we've converted the set of rotations (set 3) into the set of rotations (set 2) plus (set 3) plus (set 4).
Let's stop to think about what this means for the entire collection of every possible rotation  the collection S.
We started by splitting S up into 4 pieces: (set 1), (set 2), (set 3), and (set 4). So we could say:
S = (set 1) + (set 2) + (set 3) + (set 4)Okay so far. But we've just shown that if we rotate the Earth by the operation A, we turn set 3 into (set 2) plus (set 3) plus (set 4). So we could also write:
S = (set 1) + (set 3 after rotating the Earth by A)We don't need (set 2) or (set 4) any more! And, interestingly enough, we can go through exactly the same argument, except rotating the Earth by B rather than A, to show that if we rotate the Earth by the operation B, we turn (set 4) into (set 1) plus (set 3) plus (set 4). So we can also write:
S = (set 2) + (set 4 after rotating the Earth by B)So, what have we shown? If we take the collection S of all possible rotations of the Earth (under our constraints), we can split it up into four smaller sets. By rotating one of these sets by A, and another one by B, we can combine two of the smaller set to generate S, and we can combine the other two of the smaller sets to generate another complete copy of S.
Fine so far, but we're just talking about a collection of rotations, not the actual bits of the Earth itself. But we've done most of the work. Let's now consider actual points on the Earth's surface. By "point" I don't mean something like "London", or even "Piccadilly Circus"  I mean an idealised zerodimensional mathematical point, like 34.432913...° N, 127.348762...° W. Each possible distinct coordinate of latitude and longitude is such a point, and the latitude and longitude may be any real numbers within their respective ranges (i.e. have an infinite number of decimal places, either recurring or nonrecurring).
Imagine one point on the Earth's surface  it doesn't much matter where. Now, when you rotate the point by A (rotating just the point around the centre of the Earth this time, not rotating the entire Earth), you end up at a new point on the surface of the Earth. Similarly if you rotate by B, or A^{1}, or AB^{1}A^{1}BB, or any of our entire collection of rotations S. Imagine all of the possible points on the Earth's surface you can reach by applying all of the possible rotations in S to your starting point. This is an infinitely large collection of points, spread very densely all over the Earth's surface  in every square millimetre of the Earth, there will be infinitely many of these points.
But this is not necessarily all of the points on the Earth's surface. There will also be infinitely many points you can't reach from your initial starting point by applying any of the rotations in S. So pick another point. And imagine the collection of all the points you can reach from that point by applying rotations from S. This is another infinite collection of points on the Earth's surface. You guessed it, there are still points left over. Pick one of those and generate another collection of points by using the rotations in S.
In fact, you need to select an infinite number of starting points to make sure you can reach every point on the Earth by using the rotations in the collection S. But that's okay. In effect, we've sliced the Earth's surface up into an infinite collection of sets of points, each "slice" of points being those that can be moved from one to another using a rotation from S  and you can't move a point from one of the slices to a point in another slice by using any of the rotations in S.
One more step: Make a new set of points, call it M. The set M is made up of exactly one point from each of the slices we've just defined. It doesn't matter which point we take from each slice, as long as we take one point, and only one point, from each slice. (We can do this by using the axiom of choice, assuming you believe it  but that's a story for another annotation.) This set of points M is extremely interesting.
Think of any point whatsoever on the Earth's surface. There is exactly one point in the set M from which you can apply one of the rotations from S, to reach that point. This is true because of how we've defined the set M. Go back and read it again carefully if you need to. So, we can reach every point on Earth by starting at one specific point in M and applying one particular rotation from S, and we can't reach it by starting at any other point, or by using any other rotation from S.
Now, split up the entire Earth's surface into four sections:
Earth's surface = (section 1) + (section 2) + (section 3) + (section 4){Actually, I'm glossing over something here  in case you've spotted it and are wondering about it. I'll pick it back up at the end*. If you haven't spotted it, don't worry about it just yet.)
But what happens if we rotate the Earth by A? Then the set of rotations (set 3) becomes equal to (set 2) + (set 3) + (set 4), as we showed above. So (section 3) becomes equal to (section 2) + (section 3) + (section 4). And we have:
Earth's surface = (section 1) + (section 3 after rotating the Earth by A)Similarly, we can rotate by B and find:
Earth's surface = (section 2) + (section 4 after rotating the Earth by B)So, we can cut the Earth's surface into four sections. If we rotate one of those sections by a certain amount, we can combine it with another one of the sections to make the entire Earth's surface. And we have two left over sections, and if we rotate one of those by a certain amount we can combine those sections to make another complete copy of the entire Earth's surface.
We're in the home stretch now. The last step is to extend this to the entire volume of the Earth, which is relatively simple. For each section of the Earth's surface, just take the volume of the Earth equal to that section, plus all the space under the points in that section, all the way to the centre of the Earth. We now have four volumes:
Earth = (volume 1) + (volume 2) + (volume 3) + (volume 4)And, well, you guessed it by now:
Earth = (volume 1) + (volume 3 after rotating it by A)and:
Earth = (volume 2) + (volume 4 after rotating it by B)Ta da!
Now, either you're blinking in amazement that you understood everything above and going "wow", or you got a bit lost somewhere (in which case I apologise), or you're thinking there must be something wrong with this because it can't possibly be true.
I should restress that this is a mathematical result, which applies to idealised threedimensional spheres which are infinitely divisible. You cannot really do this with the Earth (technology aspects aside) because the Earth is made of discrete atoms. The sections and volumes discussed above are not simple pieces that can easily be visualised  they are made up of scattered collections of infinite numbers of points. If you could see them, they would look like a fractal pattern of sorts. They're not just orange segments.
One of the things the BanachTarski theorem tells us is that the concept of "volume" of a threedimensional mathematical object is not as welldefined in a strict mathematical sense as it intuitively seems to be in our real threedimensional physical world. This is in fact a wellknown problem in mathematics  it is extremely difficult to define the volume of an arbitrary mathematical object in such a way that is corresponds to our intuition of how "volume" should behave. If you've ever done any university level mathematics involving Lebesgue measure theory, you probably know what I mean. (And if you've done Lebesgue theory and think it's straightforward, then you clearly understand it better than I ever did!)
And, as I stated up front, this is a simplified explanation, in which I have glossed over some of the trickier parts. If you're very clever, you might spot some of the bits I glossed over. But the overall result of Banach and Tarski is sound, and has held up as a mathematical proof for over 80 years. A hole in my simplified argument is not a hole in their rigorous proof.
* One of the points I glossed over, in particular, is the crux of today's comic. Jamie is aware of the BanachTarski theorem, and understands the basic principles of the proof, as outlined above. You'll notice that this annotation implies that you can divide a sphere into just four pieces and reassemble them into two spheres of the same size. The actual BanachTarski result proves that you need five pieces. The extra piece is needed to take care of some "left over" points. These left over points include the centre of the sphere, which can't be included in any of the volumes defined above, and the points along the axes of rotation, which obviously don't move anywhere when you apply some of the rotations. It also includes the points in the set M itself, which you can't get to by starting at a point in M and applying a rotation. The "rotation" you'd need to apply is the "null rotation" consisting of a string of zero of the symbols A, B, A^{1}. While this is a valid operation, it isn't included in the set of rotations begining with either of those symbols, since it has no symbols. So it gets left over at each step of the above discussion. But as explained up front, this detail doesn't break the proof of the BanachTarski theorem  you can take care of it if you go into a lot more detail than I have here.
So if you know a little about the BanachTarski theorem, it's "obvious" that you can double a sphere with just four pieces. It's only if you understand it on a really deep level that you realise you actually need five pieces. The full demonstration of that is left as an exercise for the reader of this comic.
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