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Let's discuss Mathematics

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

  1. ParadigmShifter

    ParadigmShifter Random Nonsense Generator

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    Indeed, root 2 and root -1 being the most famous. Both are algebraic of course.

    e and pi are roots of power series but not (rational) polynomials.
     
  2. Ammar

    Ammar King

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    Well, my point was just that your original statement is correct and your edit is not. :)
     
  3. Luckymoose

    Luckymoose The World is Mine

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    Why does 2 + 2 = 4?
     
  4. dutchfire

    dutchfire Deity Retired Moderator

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    2 + 2 = (by the definition of 2) (0++)++ + (0++)++ = (by the definition of addition) (((0++)++)++)++ which is defined as 4.
     
  5. ParadigmShifter

    ParadigmShifter Random Nonsense Generator

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    In modulo 4 arithmetic 2+2 = 0.
    In the ring with only 1 element (the trivial ring), x+x = x and x*x = x so 1 = 0 ;)
     
  6. dutchfire

    dutchfire Deity Retired Moderator

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    Is 0 a natural number? :)
     
  7. ParadigmShifter

    ParadigmShifter Random Nonsense Generator

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    Not in my book. Some people say it is though.
     
  8. dutchfire

    dutchfire Deity Retired Moderator

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    Why do you think it isn't?
     
  9. ParadigmShifter

    ParadigmShifter Random Nonsense Generator

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    Probably because we start at 1 when counting.
    Unless you are a programmer of course ;)

    0 isn't nice to divide by either.
     
  10. dutchfire

    dutchfire Deity Retired Moderator

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    Natural numbers aren't nice to divide anyway :)

    It just seems that a lot of times you use natural numbers (IE. dimension of a vector space, degree of a polynomial), you want to include 0.
     
  11. ParadigmShifter

    ParadigmShifter Random Nonsense Generator

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    Sure they are, 5/3 = 1 remainder 2, lots of people forget that although that's how we generally learn about division. It's more general than fractions, e.g. polynomials can be divided the same way.

    I don't think a 0-dimensional vector space is very useful ;)
    In Topology, any countable set has dimension zero however.

    What is the degree of the zero polynomial?

    Spoiler :

    It is given the formal degree of negative infinity.
     
  12. dutchfire

    dutchfire Deity Retired Moderator

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    But very possible! It's just the hull of the empty set.
     
  13. ParadigmShifter

    ParadigmShifter Random Nonsense Generator

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    True. It's very important for the dimension theorem in vector spaces.
     
  14. Ammar

    Ammar King

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    Not if the topological space is weird enough.
     
  15. ParadigmShifter

    ParadigmShifter Random Nonsense Generator

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    Have you got an example? I'm pretty sure that the rational numbers and the Cantor set are considered dimension zero.

    Are you talking about Hausdorff dimension by any chance?
     
  16. Ammar

    Ammar King

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    Hmm, it's just a thought that I just had. You are right that both rational numbers and the Cantor Set do have dimension zero, but that's because they are in R which has a pretty nice topology.

    For example, consider the space S:={1,2,3} and define a subset of S as open if it contains the 1 or is empty. Yields a valid topology on S if I'm not mistaken. Now consider the covering of S consisting of {1,2} and {1,3}. It's not refinable and the 1 is in both sets, so S has a topological dimension of 1.

    Note that the space isn't very nice, it's not even Hausdorff. I'm pretty sure that your statement holds on nicer spaces (uncertain which exact attributes they needs to have) and that the topological dimension doesn't even make much sense on the sort of space I have given. Still, it should be formally valid. It's some times since I heard topology, so there might be mistakes somewhere.
     
  17. ParadigmShifter

    ParadigmShifter Random Nonsense Generator

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  18. Atticus

    Atticus Deity Retired Moderator

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    The answers on previous page didn't perhaps tell very well what's the point of trancendentals, focusing more on the power series business...

    Consider numbers in general. Children first learn that numbers are 1,2,3,... Then they find out about 0 and negative numbers, -1,-2,-3,... and then the rational numbers, and finally about real numbers like squareroot 2, pi and e and so on, even though they're not very aware what real number really means.

    Now as square root 2 can easily be proven to be irrational and is the first number known to be irrational, question arises: are all irrationals squareroots? Or more generally are all irrationals n-roots? The answer to both these questions is no, and can most easily be seen when considering numbers like sqrt 2+ sqrt 3 (exercise: prove that sqrt n is rational if and only if n is a square of a natural number).

    All of these counter examples are nonweird, because they are roots of polynomials, sqrt 2 +sqrt 3 for example is root of (x^2 -5)^2 -24. While they are perfectly good answers, they aren't entirely satisfying, because we wanted to ask if there's really weird real numbers. So it's natural to reformulate the question: are there any real numbers which aren't roots of polynomials (with rational coefficients)? Numbers which are such roots are called algebraic, and the rest are called trancendental. For all I know, no number was known to be trancendental before late 19th century when pi and e were proven to be ones.

    Now back to the numbers in general, the concept of number has been upgraded by adding something to the previous definition: First you had just the natural numbers, then you add 0 and the negative numbers and get integers, and then you add the fractions and get the rationals. Now tempting expansion would be to add the roots of polynomials, but the existence of trancendentals shows that it's insufficient.

    Trancendetality of pi is also important, because as a corollary a circle can't be squared (with compass and ruler). The problem is pretty much as old as the western civilization.

    I wouldn't be too sure about it, although it isn't used explicitly. Think this standard undergraduate exercise as an example: You have to prove that x in R^n is at the boundary of a nonempty set A if and only if there's a sequence of A's points which converges to x.

    The proof of course is that you take x_k from the intersection of A and B(x;1/k), where k=1,2,3,..., and the sequence (x_k) converges to x. But: Do you use the axiom of choice when picking that x_k? In all honesty, I don't know if it's used. However this example, I think, is enough to show that the matter isn't as easy and clear cut as it is often seen to be.
     
  19. dutchfire

    dutchfire Deity Retired Moderator

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    0 is a natural number :gripe:
     
  20. Olleus

    Olleus Deity

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    To get on a different track, I saw this theorem in a physics textbook and hoped someone might give me an explanation for why it is true.

    Let F(z) be some function for a complex z = x + iy.
    The F(z) = U(z) + iV(z)
    where both U and V are real.

    No problems so far, this is just an obvious property of the complex number F(z).

    However if you then differentiate in the following you (partial differentiation of course, but hard to type that on a computer)

    dU/dx = dV/dy
    and
    dU/dy = -dV/dx


    Anyone care to try to explain why this is to me?
     

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