What a square

I've been reading a bit about magic squares. That is, a square consisting of an arrangement of numbers, such that the numbers in all rows, all columns, and both diagonals sum to the same constant.

Probably the most impressive of these has to be Durer's Square, which is shown below:

The orientation of the numbers 1-16 in this square show an amazing about of magic-square-like properties. In total, the addition of the numbers in 86 different spatial orientations add up to 34.

Fascinating.

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Maths is beautiful

Don't believe me?

Watch these videos.

Fifteen uncoupled simple pendulums of monotonically increasing lengths dance together to produce visual traveling waves, standing waves, beating, and random motion.

For more see here

This is a wave pendulum designed using billiard balls, this device shows how 15 individual pendulum of different (but relative) lengths that have been adjusted to have successively increasing periods. When all the balls are released at the same time, the different periods cause the pendulums to cycle through all possible relative phase relationships, eventually returning to the beginning arrangement. When it cycles through it's phases, different wave patterns are noticed!

For more see here

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A paradoxical post

I was reading the Wiki list of paradoxes. Some of them are very clever - and really make you think.

Here are a few of note...


Curry's paradox

If this sentence is true, then there is no God.

If the sentence is true, then what it says is true: namely that "if the sentence is true, then there is no God". Therefore, without necessarily believing that there is no God, or that the sentence is true, it seems we should agree that "if the sentence is true, then there is no God". But then this means the sentence is true. So there is no God. [adapted from the link]

Quite brilliant! Of course, anything can be 'proven' by simply inserting it into the latter half of the sentence, so in reality it is meaningless. I can't quite get my head around it but I expect it's simply due to the limitations of language.


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Unexpected hanging paradox

A judge tells a condemned prisoner that he will be hanged at noon on one weekday in the following week but that the execution will be a surprise to the prisoner. He will not know the day of the hanging until the executioner knocks on his cell door at noon that day. Having reflected on his sentence, the prisoner draws the conclusion that he will escape from the hanging. His reasoning is in several parts. He begins by concluding that the "surprise hanging" can't be on a Friday, as if he hasn't been hanged by Thursday, there is only one day left - and so it won't be a surprise if he's hanged on a Friday. Since the judge's sentence stipulated that the hanging would be a surprise to him, he concludes it cannot occur on Friday. He then reasons that the surprise hanging cannot be on Thursday either, because Friday has already been eliminated and if he hasn't been hanged by Wednesday night, the hanging must occur on Thursday, making a Thursday hanging not a surprise either. By similar reasoning he concludes that the hanging can also not occur on Wednesday, Tuesday or Monday. Joyfully he retires to his cell confident that the hanging will not occur at all.

It's a funny one. It's clearly wrong, and yet it seems foolproof. I like the end part:

The next week, the executioner knocks on the prisoner's door at noon on Wednesday — which, despite all the above, will still be an utter surprise to him. Everything the judge said has come true.

So it was a surprise after all!


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Barber paradox

Suppose there is a town with just one male barber; and that every man in the town keeps himself clean-shaven: some by shaving themselves, some by attending the barber. It seems reasonable to imagine that the barber obeys the following rule: He shaves all and only those men in town who do not shave themselves. Under this scenario, we can ask the following question: Does the barber shave himself? Asking this, however, we discover that the situation presented is in fact impossible:

- If the barber does not shave himself, he must abide by the rule and shave himself.

- If he does shave himself, according to the rule he will not shave himself.

This is, of course, an applied version of Russell's paradox. I actually prefer the library book example:

Imagine a library that has catalogs for each section. It has a catalog for the science section, one for the British literature section, one for American literature, one for history and so on. These catalogs are also books in their own right, so they may also be listed in catalogs. Now it also has a master catalog, which lists all books which do not list themselves. Now the question is, does the master catalog list itself? If it does, then on the premise that it lists those books that do not list themselves, it doesn’t list itself. If it doesn’t list itself, then by the same logic, it does.

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Liar paradox

An oldie but a goodie. The simplest version of the paradox is this:

This statement is false.

If the statement is true, everything asserted in it must be true. However, because the statement asserts that it is itself false, it must be false. So the hypothesis that it is true leads to the contradiction that it is false. Yet the sentence cannot be false for that hypothesis also leads to contradiction. If the statement is false, then what it says about itself is not true. Hence, it is true. Under either hypothesis, the statement is both true and false.

A criticism of the liar paradox is that it is self-referencing. However, a variation exists that does not self-reference:


Card paradox

Suppose there is a card with statements printed on both sides:

Front: The sentence on the other side of this card is TRUE.

Back: The sentence on the other side of this card is FALSE.

Trying to assign a truth value to either of them leads to a paradox.

I presented a similar argument before using newspapers to show that logical contradictions exist. Neither of the sentences employs self-reference; however, this type of paradox does employ circular referencing. This criticism has also been overcome by a variation that does not self-reference or circular-reference:


Yablo's paradox

The paradox arises from considering the following infinite set of sentences:

(S1): for all k > 1, Sk is false
(S2): for all k > 2, Sk is false
(S3): for all k > 3, Sk is false
...
...

The set is paradoxical, because it is unsatisfiable (contradictory), but this unsatisfiability defies immediate intuition. Moreover, none of the sentences refers to itself, but only to the subsequent sentences; this leads Yablo to claim that his paradox does not rely on self-reference. As it continues to infinite, it does not employ circular reference.


Also...

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Fermat's Last Theorem

I'm no mathematician. However there is something that has always intrigued me about the purity of mathematical proof. That is why I tend to read a lot of popular science books on maths. An excellent example is Simon Singh's book detailing the events behind the proof to Fermat's Last Theorem.

We all know Pythagorus' Theorem, it was drilled into us at school:

This equation is true when c is the hypoteneuse of a triangle and a and b are the other two sides. It's a straightforward enough mathematical concept and has been proved many times in completely different ways. Indeed, as a child, I was taught a number of the more easily understandable methods.

Now...replace the number 2 in the above equation with n>2 (that is any number greater than 2) and replace the equals sign with a 'does not equal' sign. This is known as Fermat's Last Theorem and can be summarised as:

If an integer n is greater than 2, then the equation aⁿ + bⁿ = cⁿ has no solutions in non-zero integers a, b, and c.

In principle, this makes the equation no more difficult to understand. Essentially, you can split a square number into a sum of two lower square numbers, but you can't do the same for cube number or higher powers. Easy enough.

Should be simple enough to prove right?

Wrong.


A man called Pierre de Fermat claimed to have a proof in 1637, although bizarrely he didn't record it as he didn't have enough space in the margin of his copy of Arithmetica:

To resolve a cube into the sum of two cubes, a fourth power into two fourth powers, or in general any power higher than the second into two of the same kind, is impossible; of which fact I have found a remarkable proof. The margin is too small to contain it.

There is no doubt that Fermat was a brilliant mathematician in his own right and provided many proofs which were later verified. Thus, the proof he provided that became known as Fermat's Last Theorem was so called as it was the last of Fermat's asserted theorems to remain unproven. This gave it an air of romanticism which attracted many people, all hoping to be the one to finally crack it, but the proof proved elusive.

In the following centuries many people attempted to prove the theorem (technically it was a conjecture, not a theorem) but were unable to do so. Instead they succeeded in proving that there were no solutions for specific integers; Euler provided proof for n=3, Fermat himself for n=4, and so on.

A big step towards a proof came in the 1980s with the realisation that a seemingly unrelated mathematical conjecture, called the Taniyama–Shimura–Weil conjecture, when applied to certain elliptic curves actually implies Fermat's Last Theorem. This is where Andrew Wiles comes in. He had been fascinated by the theorem since childhood and had secretly been attempting his own proof for some time. When he heard about the link with the Taniyama–Shimura–Weil conjecture, he gave up his other commitments and worked on this association for several years in the attic of his house.

I don't want to give away any more of the story as the details can be read in the Simon Singh book Fermat's Last Theorem, also titled as Fermat's Enigma (I recommend it highly), but to summarise, in 1993 Wiles eventually succeeded in proving the theorem, thus fulfilling a life-long dream, although the story didn't end there as there were a few twists and turns after that.

However, the intriguing thing is that in order to prove the theorem, Wiles had to develop several novel mathematical techniques that would be considered as '20th century mathematics' and were thus unavailable to Fermat back in 1637. So the question still remains as to whether Fermat did actually have a proof. It's seems impossible, due to the sophisticated new techniques needed, but it cannot be categorically denied, fueling the intrigue.

Wiki:

Fermat's alleged "marvellous proof"...would have had to be fairly elementary, given the state of the mathematical knowledge at the time, and so could not have been the same as Wiles's. And in fact, most mathematicians and science historians doubt that Fermat had a valid proof of his theorem for all exponents n, as it seems unlikely there is an elementary proof.

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The Monty Hall Problem

This is a strange one. I can't think of another theoretical maths problem quite like it. The answer is so counter-intuitive that generations of gifted mathematicians have got it wrong and undoubtedly generations to come will too. It goes like this:

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 3 and the host opens Door 1, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

So is it better to stick with your original choice or switch to the other unopened door? At first it seems the obvious answer is that it makes no difference what you do. Its a 50/50 choice, right? Well, no, wrong actually. You are always better off switching to the other unopened door. You might not believe it but it's been tested time and again and the conclusion is always the same.

Switch.

I know what you're thinking. I thought the exact same thing. But the fact remains that I was wrong, and so are you. Here's why:

When you first make your choice it's a 1/3 chance that you have picked the door with the car behind it. When Monty opens one of the other doors and offers you the switch, by sticking with your original choice nothing has changed, so your odds are still 1/3. However, if you switch you are effectively getting both the opened door and the other unopened door, meaning yours odds have become 2/3.

Again, I know what you're thinking. "If I stick it's also a 2/3 chance because I now know what's behind one of the doors". But again, you would be wrong. Another way to think of it is to imagine 1,000,000 doors instead of just three. When you choose one door you have a 1/1,000,000 chance of winning. Monty then proceeds to open 999,998 of the other doors, all of which show goats, leaving just one other unopened door. If you now stick, your odds are still 1/1,000,000 because the new information makes no difference to the choice you originally had. If you switch your odds jump to 999,999/1,000,000, because you essentially have been given every door except your original choice. The same rationale holds for 3 doors.

The below figure should explain it:

Hopefully it now makes sense. If not, don't worry. You are in good company. As cognitive psychologist Massimo Piattelli-Palmarini says "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer."

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