Let's discuss Mathematics

[IdiotsOpposite]

How do we prove that 0<1?

 

[sanabas]

Thanks Sanabas. I had to reread your explanation a few times, I think I spotted a typo in the numbers, and using numerals for both the tribes and how many times they drew a tech I found confusing, but that helped. Thanks!

Where's the typo? I didn't exactly double-check any of it.

Bah, never mind, just looked myself. There's 6 combinations of 2 techs, not 12. Didn't even notice when I used the correct 6 x 126 when calculating 3 techs. I'm an idiot. I'll go and edit it now.

 

[ParadigmShifter]

How do we prove that 0<1?

http://answers.yahoo.com/question/index?qid=20090208193402AAY9ZWH

That doesn't work if 0 and 1 are treated as complex numbers The best you can do then is |0| < |1|

 

[pboily]

PS, the properties outlined in that proof also hold for the ordered sub-field R in C. So you can also show 0<1 even if you consider them both to also be complex numbers.

 

[ParadigmShifter]

There are infinitely many ordered subfields in C though.

 

[ParadigmShifter]

Show that (the set of) n-tuples (n dimensional vectors), where n is finite, and the tuple elements come from a countable field set, is also countable.

Just realised this is the proof that there are only countably many algebraic numbers

You can do this by defining 2 functions which when combined give a mapping to N which is unique for each tuple.

 

[Atticus]

You mean the set of n-tuples with elements in a countable field is countable? That is: F^n is countable if F is.

 

[ParadigmShifter]

Yes, that is what I mean.

EDIT: F doesn't have to be a field. It could be any countable set.

EDIT2: I suppose I should edit out the word "field" in my post above then

EDIT3: I'm thinking of a number theoretical way to do it. There is probably a method using topology too.

 

[Petek]

It suffices to prove the assertion for n = 2; the general result follows by induction and since there's an obvious bijection between F^n-1 x F and Fⁿ. Since F is countable, there's a bijection between F and N, so we can assume that F = {1, 2, 3, ...}. We list the elements of F x F as follows:

(1,1), (1,2), (2,1), (1,3), (2,2), (3,1), (1,4), (2,3), (3,2), (4,1), (1,5), ...

where we first list all (n,m) such that n + m = 1, then n + m = 2 and so on. Elements that have the same sum are listed in increasing order by the first member of the ordered pair. It is clear that each (n,m) appears in this list exactly once, so we have the required bijection.

Now try this one: Again, let F be a countably infinite set. Define S to be the subset of F x F x F x ... (countably infinitely many products of F) such that (a₁, a₂, a₃, ... ) is an element of S if and only if all but a finite number of the a_i are zero. Show that S is countable.

 

[ParadigmShifter]

Can you do my question without induction? I think my answer may answer your question too

I call shenanigans on "if and only if all but a finite number of the a_i are zero", since we can replace the a_i with ANY (globally) constant value, and swap 0 with the constant, so

 

[Petek]

In your question, is n a fixed natural number, or does it vary over all natural numbers? My solution was for the first interpretation. If the latter, then our questions are probably equivalent. Does your solution involve unique factorization? I don't understand your final comment.

 

[ParadigmShifter]

Yep, my answer answers your question.

Again, we can assume that F maps to { 1, 2, ... }. Extend to { 0, 1, 2, ... } by making sure the a_i's which are a given constant map to 0. Call this representation R_n, say for the nth F in the cartesian product.

Let P_n be a function which returns the nth prime number.

Map each element of F x F x ... to

P_n^R_n

EDIT: And multiply them together

Any constant element in R_n maps to 0 (so P_n^R_n is 1 and doesn't affect the multiplication result) and we can recover the original values of each F in the cartesian product via the fundamental theorem of arithmetic (by factorisation).

EDIT: This is Godel Numbering BTW. I only realised at lunch it can be used to show there are countably many algebraic integers.

http://en.wikipedia.org/wiki/Gödel_numbering

EDIT2: Clarified a bit

EDIT3: LOL, got P and R the wrong way round, spank me. Fixed.

 

[ParadigmShifter]

Hmm there's an easy way to extend that proof so that as long as there are only finitely many of the a_i's which are not drawn from a set of given values (by mapping to { 0, 1, ..., n } and then mapping the countable set to { n+1, n+2, ... }.

Am I missing something there?

EDIT: Yes I am, since in your question there were countably infinite elements in the cartesian product. You need to map the values of the given constant value to 0 so it has no effect on the infinite product.

My maths head hasn't warmed up yet, I'm only starting my second beer.

 

[nc-1701]

Anybody aware of any good FREE computer algebra systems? My main requirement at the moment is that it can solve ODEs, but generally being able to do algebra and calculus symbolically would be important. And if I could do simple programming (logical operators, for do loops, arrays etc.) along with basic number theory stuff divisibility, mod, etc. That would be helpful too, I've been using Maple 11 at school, but I want to be able to work on my laptop, and Wolfram Alpha is great, but not really cutting it for some things.

 

[ParadigmShifter]

Have a look at this:

http://en.wikipedia.org/wiki/Comparison_of_computer_algebra_systems

I think there's a free C++ library as well but the name escapes me.

 

[suiraclaw]

Anybody aware of any good FREE computer algebra systems? My main requirement at the moment is that it can solve ODEs, but generally being able to do algebra and calculus symbolically would be important. And if I could do simple programming (logical operators, for do loops, arrays etc.) along with basic number theory stuff divisibility, mod, etc. That would be helpful too, I've been using Maple 11 at school, but I want to be able to work on my laptop, and Wolfram Alpha is great, but not really cutting it for some things.

http://en.wikipedia.org/wiki/Sage_(mathematics_software)
It's free and I know some mathematic students that use it (despite that they could easily get a free maple license).

 

[SpiritWolf]

First Question: What percentage of people in this thread will choose the same option as you?
Options:
A) 0%
B) 1-25%
C) 26-50%
D) 51%-75%
E) 76-99%
F) 100%

Second question: How would the answer change if given to computers?

Third question: How would your answer change if you posted the poll in CFC-OT?

 

[Souron]

Second question: How would the answer change if given to computers?

Assuming identical, flawless, deterministic algorithims, F.

Flawless Indetermenistic algorithms could result in a wider array of possibilities.

 

[Atticus]

It depends... Does everyone have to answer? Is it better for me if I choose the right one and there are fewer other people that do so, or is it irrelevant? If other people winning isn't away from me, lets just all choose F. Otherwise it won't do.

 

[Mise]

1) If people get money to pick the correct one, I'll pick E and hope that not too many people try and do something stupid. If people get a share of the winnings (share depending on how many other people picked that one) I have no idea what I'd do. Probably go for C, maybe still E. If not, I don't care and won't think about it too hard. I'd probably still pick E though. I don't think it's a tractable problem if the winnings are shared among correct guessers.

2) Depends how the computer is programmed, obviously. If its goal is to guess correctly, and is programmed to iterate over other computers, then it will pick F and everyone will be right. More than 20 or so humans will almost certainly not unanimously pick the same option, though, which is why I'd pick E when faced with humans with an incentive to be correct but not to share winnings.

3) Dunno, not really I guess.