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Non Integer Exponents for Cartesian Products

  1. Jan 25, 2012 #1
    I know the Cartesian product for an algebraic structure: A x B = {(a,b): a ∈ A, b ∈ B}

    Which naturally gives An = {(a1, a2, ... , an): ai ∈ A ∀ i}

    Some of the time, at least we can also have a non integer n.
    For example [A x A x A]2/3 = A x A.

    Is there any way of continuing the exponent n into non integer numbers for any exponent? It might involve creating an algebraic structure very unlike the original A, but which is still isomorphic when raised to an appropriate power, with respect to the Cartesian product.

    I have another question as well:

    Is there a way to define a group, which has as elements all groups, or all algebraic structures of a certain kind? Could this be made into a ring? If it could, perhaps we could create a structure exponential using the two operations. A metric might also be required on the set of all structures to make sure the series converges.

    Anyway, I would be grateful if anyone could tell me if any of this exists already, is complete nonsense, or which area of mathematics it would belong to.
     
  2. jcsd
  3. Jan 25, 2012 #2

    morphism

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    I don't think there's any way to do this without creating some sort of mess. You will already run into a problem if you try to define the square root of a group to be a group. Indeed, there are examples of nonisomorphic groups A and B such that [itex]A^2 \cong B^2[/itex]. This means that "square roots" of isomorphic things won't have to be isomorphic!

    I don't know what kind of structure you would ask the "square root" of a specific algebraic object to be...

    Due to set-theoretic difficulties, the answer to your first question is no for groups (and many other types of algebraic structures). This is because any nonempty set can be given a binary operation that turns it into a group. (This assertion is equivalent to the axiom of choice.) Thus your group of all groups will in particular be a set of all (nonempty) sets, which is problematic.
     
  4. Jan 25, 2012 #3

    chiro

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    Hey Daron and welcome to the forums.

    This is a very interesting question.

    It seems that you could attack this from two ways.

    The first one is cardinality. Lets say we have a set A that has 10 elements. We know that A^2 has 100 element pairs. Now somewhere in-between A and A^2 you would have somewhere between 10 and 100 elements.

    The elements would have to obey a specific permutation law that goes hand in hand with the ordered definition of elements in the set. So for example if my set was {0,1,2,3,4,5,6,7,8,9} then we would expect permutations involving the lower numbers would appear before those involved with higher numbers in terms of the first element of the pair (i.e. pairs (1,x) will come before (9,x)).

    The next thing to address is how one would handle countable and uncountable sets.

    If you had a countable set like my example above, you would need to use a function like say a floor function to get the right number of elements to permute and then use the standard permutation function to get your output set.

    In order to find the number of elements it would have to be a derivative of the log function in a given base since we are dealing with systems involving exponents.

    Basically every order of magnitude will add an extra dimension to the object of the final set element. The easiest way to picture this is to think a normal number in a given base: we can represent 10 and 98 with two digits with 101 needs three digits and each digit corresponds to representing a sub-element as part of a final element in your final set.

    I think I could derive a formula to do this for countable sets in terms of logs, floors and/or ceilings and a standard permutation function to generate the right set that gives the right answers for integer exponents and the "interpolated" versions thereof if you are interested.
     
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