Complex complex numbers

  1. Extensions of complex numbers are available for 2^n dimentions.

    For example:
    1. (a+bi+cj+dk) = ((a+bi)+(c+di)j)
    where: i<>j, i^2=j^2=-1, ij=ji=k, ik=ki=-j, jk=kj=-i, k^2=+1.

    Unlike quaternions, these hypercomplex numbers are:
    commutative and associative wrt addition and multiplication, distributive with addition, all multiplicative inverses exist except zero and zero divisors.
    And, the elementary functions, e.g. e^(a+bi+cj+dk) are available.

    2. (a+b(i2)+c(i3)+d(i4)+e(i5)+f(i6)+g(i7)+h(i8)) =
    (a+b(i2)+c(i3)+d(i4)) + (e+f(i2)+g(i3)+h(i4))(i5).

    where: i2<>i3, i2<>i5, i3<>i5, (i2)(i3)=(i4), (i2)(i5) =(i6), (i3)(i5)=(i7), (i4)(i5)=(i8).

    All other product combinations are easily found granting commutativity and associativity. e.g. (i6)(i4)= (i2)(i5)(i2)(i3)= -(i3)(i5)=-(i7).

    Unlike octonions, these hypercomplex numbers are:
    commutative and associative wrt addition and multiplication, distributive with addition, all multiplicative inverses exist except zero and zero divisors.
    And, the elementary functions,
    e.g. e^(a+b(i2)+c(i3)+d(i4)+e(i5)+f(i6)+g(i7)+h(i8)) are available.

    Quaternions and Octonions, etc., are produced within these hypercomplex numbers via special product functions.

    Complex numbers of any dimention can be constructed in this way.

    Whatdoyouthink?
     
  2. jcsd
  3. matt grime

    matt grime 9,396
    Science Advisor
    Homework Helper

    That you've not proven any of the things you say are true. That makes it bad mathematics. The things you are producing are called algebras and they'vebeen studied for many years.

    Oh,and you're also wrong:

    i^2=j^2
    hence

    i^2-j^2=0

    since everything is commutative this implies

    (i-j)(i+j)=0

    since there are no zero divisors it follows that i+j=0, since i=/=j, that is j=-i. so it isn't an extension after all.
     
    Last edited: Mar 26, 2005
  4. The proofs follow from the definitions, shall I do the arithmetic for you as well?
     
  5. matt grime

    matt grime 9,396
    Science Advisor
    Homework Helper

    Please see my above counter example to your claims.

    Please look up "finite dimensional division algebras" to see what you should know already before telling me I'm wrong.

    (As an extra hint, Forbenius proved what you're trying to do is impossible in 1877, I just found out)
     
    Last edited: Mar 26, 2005
  6. matt grime

    matt grime 9,396
    Science Advisor
    Homework Helper

    Ah so you're not claiming these are divison algbras, my mistaek. YOu're just saying they are algebras. SO why aer you claiming these are "like" the complex numbers?

    All you've done is take some ring and add in inverses to the non-zero divisors? So? I was assuming you were doing something that wasn't well known and been aroudn for many years.
     
  7. Of course there are zero divisors!! They are numbers that are specifically defined.

    (i-j) is a zero divisor because there is a non-zero number, (i+j), which when multiplied by (i-j), becomes 0.

    "(As an extra hint, Forbenius proved what you're trying to do is impossible in 1877, I just found out)"

    Nonsense!
     
  8. matt grime

    matt grime 9,396
    Science Advisor
    Homework Helper

    Like I said, I misunderstood and thought you were trying to define a division algebra "like C".

    What you've got is an algebra, two dimensional over C. See the Wedderburn structure theorem for a classification of all semi simple ones. If they are not semisimple have you found the Jacobsen radical of it?

    Associative (and commutative) algebras have been studied for many years and exist in arbitrary dimension.
     
    Last edited: Mar 26, 2005
  9. matt grime

    matt grime 9,396
    Science Advisor
    Homework Helper

    Just forget this (my comments), ok. Sorry if I'm sounding harsh, but there are many algebras out there, and I can't see for the life of me why you think these deserve to be called the "proper" generalizations of complex numbers, and think Hamilton and Cayley were wrong, whatever wrong may mean in this context.
     
    Last edited: Mar 26, 2005
  10. For hypercomplex numbers (a+bi+cj+dk), k^2=+1.
    For quaternions (a+bi+cj+dk), k^2=-1.

    Which is it? Obviously the k of hypercomplex numbers is different from the k of quaternions.

    Don't you see the inconsistency?

    For me k^2=+1 and kHk=-1, i.e. H is the Hamilton product applied to hypercomplex numbers.

    The same things apply to Cayley numbers and to further extensions, 16-valued etc., of the Hamilton product.
     
  11. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    No. Different systems, different results.
     
  12. matt grime

    matt grime 9,396
    Science Advisor
    Homework Helper

    What? The only inconsistency is that you think those k's ought be the same thing.

    Hamilton produced a division algebra, you haven't. They are different things, that is all, just as there (infintely) many Calyely algebras with generators 1, i,j,k with the expected properties except that k^2=(-)p for some prime p, which are of special intererst in number theory and physics.



    There is also an algebra R[f,g,h] with the relations f^2=g^2=h^2=1, and fg=gf=h, it is the group algebra over R of C_2 x C_2. So what?

    All you're doing is defining some 2-d complex or 4-d real algebra and claiming they are "hypercomplex", as if there was something special about them in relation to the complex numbers.
     



  13. Extending the number system from complex numbers, (a+bi), to 4-D
    hypercomplex numbers, (a+bi+cj+dk), leads to a multiplication
    table such as;

    (A) i^2=j^2=-1, ij=ji=k, k^2=+1, ik=ki=-j, jk=kj=-i.

    Sir W. Hamilton introduced 'quaternions' by presenting the
    multiplication table;

    (B) i^2=j^2=-1, ij=k, ji=-k, k^2=-1, ik=-j, ki=j, jk=i, kj=-i.

    Clearly list (A) is incompatable to list (B).

    Is k^2=-1 or is k^2=+1, it cannot be both. k cannot be the
    same entity in both cases. I believe Hamilton's algebra
    would be consistent with hypercomplex numbers if he had
    introduced a Hamilton (H) product such that;

    iHi=jHj=-1, iHj=k, jHi=-k, kHk=-1, iHk=-j, kHi=j, jHk=i, kHj=-i

    where i,j,k are the same hypercomplex numbers as in (A).

    It was misleading and incorrect for Hamilton to consider that
    quaternions are entities at all. There are no such things as
    quaternions. There is a Hamilton algebra which deals with
    the concepts that Hamilton wanted to deal with but they are using
    hypercomplex numbers in the context of the Hamilton product (H).

    In the 8-D case, (a1+a2i2+a3i3+a4i4+a5i5+a6i6+a7i7+a8i8)
    multiplication leads to the entries;

    (C) (i2)^2=(i3)^2=(i5)^2=-1, (i2)(i3)=i4, (i2)(i5)=i6, (i3)(i5)=i7,
    (i4)(i5)=i8, (i4)^2=+1, (i6)^2=+1, (i7)^2=+1, (i8)^2=-1.

    Sir A.Cayley introduced 'octonions' by presenting a multiplication
    list containing;

    (D) (i2)^2=(i3)^2=(i4)^2=(i5)^2=(i6)^2=(i7)^2=(i8)^2=-1.

    Again (C) and (D) are incompatible. (i6)^2=+1 from list (C),
    contradicts (i6)^2=-1 from list (D). Cayley makes the same
    mistake for 'octonions' that Hamilton made for 'quaternions'

    There are no such things as octonions. There is a Cayley algebra,
    with a Cayley product (Ca), dealing with 8-D hypercomplex numbers
    which expresses what Cayley means.

    (i2)Ca(i2)=(i3)Ca(i3)=(i4)Ca(i4)=(i5)Ca(i5)=(i6)Ca(i6)=
    (i7)Ca(i7)=(i8)Ca(i8)=-1.
     
  14. Hurkyl

    Hurkyl 16,090
    Staff Emeritus
    Science Advisor
    Gold Member

    Thread locked.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?
Similar discussions for: Complex complex numbers
Loading...