# Let's discuss Mathematics

Discussion in 'Science & Technology' started by ParadigmShifter, Mar 16, 2009.

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2. ### AtticusDeityRetired Moderator

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Idiotsopposite: What transcendetal numbers do you know? Can you reduce this question to knowledge of them? (Assuming that it is allowed, which I do assume, since otherwise this is quite hard).

EDIT: Oops, didn't notice the page had changed.

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e and pi are the only transcendental numbers most people know about.

Is that a hint, do you know how to show it Atticus? Or just speculative?

EDIT: This looks like the correct theorem, which is an extension of one of the ones I posted links to earlier

http://en.wikipedia.org/wiki/Baker's_theorem

4. ### AtticusDeityRetired Moderator

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It's just speculative, but if Idiotsopposite isn't doing heavy algebra, it's well justified speculation

I'm 97% sure that assuming log m to be algebraic would make e also that.

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I think IdiotsOpposite is still at school. I'd look for a slightly easier problem in that case to investigate

Even the proof that e is transcendental is quite tricky, proof here: http://en.wikipedia.org/wiki/Transcendental_number

EDIT: My analysis textbook has a proof that e is irrational, using just properties of power series, but that is a long way off showing it is transcendental.

6. ### dutchfireDeityRetired Moderator

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I used this method (it's called von Neumann normalization, or something), in a project of mine, building a physical random number generator. I believe I posted about it here...

7. ### dutchfireDeityRetired Moderator

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Hmm, can't find it.

Another interesting one:
100 prisoners are locked in prison. In this prison, there are 100 identical rooms. In every room, there are 100 numbered boxes, each box containing the name of one prisoner in such a way that every prisoners name is in one box* in every room (the rooms are identical, so each name is in the same box in every room).
The prison director proposes the following scheme: "Each prisoner will be led into a separate room and will be allowed to open 50 boxes. If every prisoner opens the box containing his own name, all prisoners will be released."
Initially, the prisoners don't really like this scheme, each individual prisoners has a 50% chance of opening the correct box, so in total they should have a 1/2^100 chance of success. However, luckily, there is a mathematician in the room. He devises a clever scheme that greatly increases the chance of them winning (to more than 30%!) and tells all the prisoners.

Note (Big hint!): His scheme would also work if all other prisoners were robots/computers that could follow a simple algorithm. However, the scheme can't work if the clever mathematician is only allowed to give each prisoner a list of boxnumbers he should open.

*All prisoners have different names

8. ### LulThymeKing

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When you asked this question, I thought you already knew this, so I didn't mention it, to avoid spoiling things, but this trick actually has a name:
Neumann's coin trick
in reference to the first known reference (by von Neumann).

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Hey! I am as great as von Neumann then!

This device will make me famous!

Dutchfire's problem sounds similar to one I heard before about lightbulbs. I'll have a drunken think about it after listening to the Big Black song.

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Dutchfire, do the prisoners get to see the results of each box opening before opening another?

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Well, I know that ln(2) and ln(3) are transcendental, and I believe that a integer multiple of a transcendental number is also transcendental. This means that if the natural log of all primes is transcendental, then the natural log of all composite numbers is transcendental, at least so I believe. (Because, for example, ln(4)=ln(2^2)=2ln(2) ) Although... if a and b are transcendental, is a+b also transcendental? I think that's rather necessary.

And hey, I believe that any good mathematician can prove anything he wants if he has the will (and it's correct.)

12. ### Miseisle of lucy

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I don't even see how that would help?!

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Me neither, but it helps to get all the info beforehand Especially when a bit tipsy.

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Yeah, x=1 is the only rational number such that log(x) is algebraic, but if you know that the proof is trivial

if a and b are transcendental, a+b might not be. b = -a, b = -a + k, where k is algebraic, etc.

EDIT: You are correct that any algebraic rational multiple of a transcendental number is also transcendental, however. That is easy to show.

EDIT2: Might be true for algebraic multiples too, but that would need a proof.

15. ### dutchfireDeityRetired Moderator

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Yes .

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Damn. I was hoping that I could prove numbers like log(6) were transcendental by manner of them being log(2)+log(3).

17. ### dutchfireDeityRetired Moderator

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Gödel disagrees

18. ### Miseisle of lucy

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I'm assuming the solution would have them all open each box with an equal probability, rather than simply opening at random?

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Might be true if the numbers aren't rational multiples of each other. Needs a proof though...

EDIT: But the proof is probably covered by one of the 3 theorems I posted earlier. I think one of those proves the result you are after though.

EDIT2: When I say log I mean log to the base e of course.