Even stranger ears

****Actually, there is also a 4-component algebra ("quarternions") which does
not obey the commutative property, and an 8-component algebra ("octonions")
which does not obey the commutative or associative properties. But it's been
proved that there are no other possible algebras of n-component continuous
quantities.

Sarn says: Yes, I have heard of both of them, and I know for a fact that
quarternions can be "emulated" with some special types of Hermitean
matrices, which are, by their very nature non-commutative in responce to
multipliaction, but are associative.

I saw these particular beasts on a website, and I cannot for the life of me,
find that darn website again!!!

I asked a Dr of algebra and logic, if we could have "16onions", and
"32-onions", and all the way up to....

"2^n-onions", and he says we could make them, altho they would be logically
inconsistant, and are just plain uninteresting, anyway.

I do wish that I could get more information on perhaps a
"hyper-non-associative law", via which we might get:

a^(b^(c^d)) =, =/=, =v= (a^b)^(c^d) =, =/=, =v=, ((a^b)^c)^d =, =/=, =v=,
d^((a^b)^c) =, =/=, =v=, (c^(a^b))^d

and these symbols are = {"Always equals"}, =/= {"Never equals"}, =v=
{"Sometimes equals"}, and, in this circumstance, the carrot "^" is not
merely just a plain boring old exponent, but what I call a "Wildcard
operator", which can be any algebraic operator.

Is this just a ridiculous pipe dream???

---Sarn.

--- In tuning@egroups.com, Sarn Richard Ursell <thcdelta@i...> wrote:
> ****Actually, there is also a 4-component algebra ("quarternions")
which does
> not obey the commutative property, and an 8-component algebra
("octonions")
> which does not obey the commutative or associative properties. But
it's been
> proved that there are no other possible algebras of n-component
continuous
> quantities.
>
> Sarn says: Yes, I have heard of both of them, and I know for a fact
that
> quarternions can be "emulated" with some special types of Hermitean
> matrices, which are, by their very nature non-commutative in
responce to
> multipliaction, but are associative.
>
> I saw these particular beasts on a website, and I cannot for the
life of me,
> find that darn website again!!!

Tony Smith's is one:
http://www.innerx.net/personal/tsmith/1TSmath.html

> I do wish that I could get more information on perhaps a
> "hyper-non-associative law", via which we might get:
>
> a^(b^(c^d)) =, =/=, =v= (a^b)^(c^d) =, =/=, =v=, ((a^b)^c)^d =,
=/=, =v=,
> d^((a^b)^c) =, =/=, =v=, (c^(a^b))^d
>
> and these symbols are = {"Always equals"}, =/= {"Never equals"}, =v=
> {"Sometimes equals"}, and, in this circumstance, the carrot "^" is
not
> merely just a plain boring old exponent, but what I call a "Wildcard
> operator", which can be any algebraic operator.
>
> Is this just a ridiculous pipe dream???

It can be reduced to the plain old associative law, used twice.