# Quaternions and hypercomplex numbers are incompatible

1. Feb 28, 2005

### Owen Holden

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.

Note that these hypercomplex numbers are commutative and have elementary functions.

We can extend this idea to hypercomplex numbers to any dimension.

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)

(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.

Any opinions?

Owen

Last edited: Feb 28, 2005
2. Feb 28, 2005

### Hurkyl

Staff Emeritus
http://mathworld.wolfram.com/HypercomplexNumber.html

Anyways, I don't follow your objection to considering the quaternions "entities". Clearly, models of the quaternions exist in other happy domains. Even you speak about a particular model! I suspect you are ascribing a fairly unusual definition to the term "entity".

3. Feb 28, 2005

### matt grime

All you've done, Owen, is define a *different* 4-dimensional Real Algebra from the one that Hamilton considered. The quartenions have the benefit of being naturally isomorphic to a 2-dimensional complex algebra $$\mathbb{C}[j]$$

1. Prove your algebra is also a division ring (as the quarternions are) - that is, as it is commutative, show it is actually a field.

2. Realize that your opinion of what things *ought* to be is no more important that anyone elses. Hamilton provided an example of a division ring that extends C, that was all - in order to do so he had to drop commutativity, but that isn't a big deal.

4. Feb 28, 2005

### jcsd

Davenport's commuataive hypercomplex algebra was proabably investigated by Hamilton anyway, though he would of prefered the quartenions as an extension of the complex numbers as they form a divison ring like rational numbers, real numbers and complex numbers whereas Davenport's algebra does not.

Mathematics is an abstartc subject so sattements like "there are no such things as octonions" don't have anything to do with maths. Even if you are a Platonists you'd proabably prefer the quaretnions to other 4-D real algebras as they have many more obvious physical applications.