Re: @ Timothy Golden and ' friends ' ( everyone actually )

On Jul 21, 2:20 pm, master1729 <tommy1...@xxxxxxxxx> wrote:
lwalke wrote :
And so I, like tommy1729, also wait for Tim Golden,
the inventor of polysigned numbers, to give some
suggestions as to how to obtain the quaternions by
using polysigned construction.
2) you mean the polysigned analogue of quaternions i assume , for I) quaternions are not commutative.
I assume you did , but i want to avoid confusion.

This is what I mean: We already know that Golden has
generalized the two-signed reals by coming up with
his polysigned numbers. And he has already told us
how the three-signed numbers P3 are isomorphic to
the complex numbers. So we have an alternate way to
formulate C, merely by coming up with a third sign
"*" and extending the rules of addition as well as
multiplication to include the third sign.

So if there's another way to formulate C using the
polysigned construction, maybe there's another way
to formulate the quaternions as well. Especially
since tommy1729 started talking about noncommutative
polysigned numbers, maybe there's a way to make a
ring of polysigned numbers be isomorphic to the
quaternions the same way that the three-signed
numbers P3 are isomorphic to C.

As Leland McInnis wrote in his last post, there is a
concept of a group ring. As the Wikipedia link in
that post suggests, a group ring is, in some way,
the combination of a group and a ring.

Now Golden's polysigned numbers are in fact group
rings, where the "group" part of the "group" ring is
a cyclic group, Z/nZ (which is the additive group of
integers mod n). The "group" part of the group ring
controls the rules of the signs -- since in P3, for
example, the rule for combining (multiplying) two
signs is identical to the addition table mod 3:

*-+
-+*
+*-

The "ring" part of the group "ring" is officially
the ring R of real numbers. But in order to keep
the coefficients unsigned (nonnegative), we mod out
the term:

*1-1+1 = 0

Without this rule, we'd have to have negative
coefficients for the signs, which removes the
elegance of having unsigned coefficients with the
group elements "*", "-", and "+" being the signs.

Now tommy1729 asked for a noncommutative ring of
polysigned numbers -- and by noncommutative, he
means the multiplication is noncommutative -- and
we recall that the rules of multiplication are
controlled by the signs. So if we change the group
of signs to a noncommutative (i.e., non-Abelian)
group, then the resulting group ring will also be
noncommutative, as just tommy1729 desires.

So all we need now is a non-Abelian group. Now the
simplest non-Abelian group is called D3, also known
as the dihedral group of six elements. And so that
is the resulting group ring will end up having six
different signs.

3) I guess if these numbers exist at all their " dimension " will be quite big.

Well, the dimension of the six-signed numbers will
be at most six. If we want to produce a ring that
is isomorphic to the quaternions, as I have stated
above, then we will want to mod out two different
elements in order to push the dimension to four.

4) dont give up on groups to fast , i suggest making less " sum = 0 " rules , which should create also less " zero-divisors " as a consequence...

It's true that the fewer "sum=0" rules we have,
the less likely it'll be that we'll discover a
zero divisor. But without "sum=0" rules, we'll be
forced to have negative coefficients, which isn't
in the spirit of what Golden intended when he came
up with the polysigned numbers. And we want to be
able to mod out two different elements so that we
can reduce the dimension to four, the dimension of
the quaternions.

I'd still like to mod out:

*1-1+1 = 0

since the three signs "*", "-", and "+" act like
the three signs of P3. These three signs give us
two dimensions after modding out *1-1+1, and their
span should be isomorphic to the two-dimensional
subring of the quaternions generated by (1,i),
which is isomorphic to the complex field C.

But the rules of a group ring alone give us:

(*1-2^1<2)(*1-2^1<2) = *6-6+6^6<6>6

and so we cannot mod out:

^1<1>1 = 0

without *1-2^1<2 being nilpotent. We have already
modded out *1-1+1, so this gives us:

(*1-2^1<2)(*1-2^1<2) == ^6<6>6 (mod *1-1+1)

Hmmm. I've been trying to find a suitable element
other than ^1<1>1 to mod out, but I've been
having trouble finding something. It might have
something to do with the fact that for _real
numbers_, it appears that we have the inequality

x^2+y^2+z^2 >= xy+xz+yz,

with equality only when x=y=z. This means that
nothing other than ^1<1>1 can ever be modded out.

5) maybe rules similar to ab + ba + ac + ca + bc + cb = 0 make more sense then a + b + c = 0 considering the context ... just guessing ...

What are a,b, and c? Are they supposed to be signs,
such as "-", "+", and "*" (or to be more precise,
-1, +1, and *1)? If so, then replace a+b+c with
ab+ba+ac+ca+bc+cb doesn't help us, since, for
example, ab = (-1)(+1) = *1 = c. Nice try though.

I probably just have to admit that the group ring
in which D3 is the group and R is the ring simply
doesn't have a subring that's isomorphic to the
division ring of quaternions.

1) zero divisors cannot be avoided , no problem , they occured for polysigned as well.

Yes, I know, but I was hoping that since P2 and
P3 avoid zero divisors, I could at least avoid
them for PD3 (i.e., the polysigned numbers based
on the dihedral group D3), but at this point, this
appears to be impossible.On Jul 21, 2:20 pm, master1729
<tommy1...@xxxxxxxxx> wrote:
lwalke wrote :
And so I, like tommy1729, also wait for Tim Golden,
the inventor of polysigned numbers, to give some
suggestions as to how to obtain the quaternions by
using polysigned construction.
2) you mean the polysigned analogue of quaternions i assume , for I) quaternions are not commutative.
I assume you did , but i want to avoid confusion.

This is what I mean: We already know that Golden has
generalized the two-signed reals by coming up with
his polysigned numbers. And he has already told us
how the three-signed numbers P3 are isomorphic to
the complex numbers. So we have an alternate way to
formulate C, merely by coming up with a third sign
"*" and extending the rules of addition as well as
multiplication to include the third sign.

So if there's another way to formulate C using the
polysigned construction, maybe there's another way
to formulate the quaternions as well. Especially
since tommy1729 started talking about noncommutative
polysigned numbers, maybe there's a way to make a
ring of polysigned numbers be isomorphic to the
quaternions the same way that the three-signed
numbers P3 are isomorphic to C.

As Leland McInnis wrote in his last post, there is a
concept of a group ring. As the Wikipedia link in
that post suggests, a group ring is, in some way,
the combination of a group and a ring.

Now Golden's polysigned numbers are in fact group
rings, where the "group" part of the "group" ring is
a cyclic group, Z/nZ (which is the additive group of
integers mod n). The "group" part of the group ring
controls the rules of the signs -- since in P3, for
example, the rule for combining (multiplying) two
signs is identical to the addition table mod 3:

*-+
-+*
+*-

The "ring" part of the group "ring" is officially
the ring R of real numbers. But in order to keep
the coefficients unsigned (nonnegative), we mod out
the term:

*1-1+1 = 0

Without this rule, we'd have to have negative
coefficients for the signs, which removes the
elegance of having unsigned coefficients with the
group elements "*", "-", and "+" being the signs.

Now tommy1729 asked for a noncommutative ring of
polysigned numbers -- and by noncommutative, he
means the multiplication is noncommutative -- and
we recall that the rules of multiplication are
controlled by the signs. So if we change the group
of signs to a noncommutative (i.e., non-Abelian)
group, then the resulting group ring will also be
noncommutative, as just tommy1729 desires.

So all we need now is a non-Abelian group. Now the
simplest non-Abelian group is called D3, also known
as the dihedral group of six elements. And so that
is the resulting group ring will end up having six
different signs.

3) I guess if these numbers exist at all their " dimension " will be quite big.

Well, the dimension of the six-signed numbers will
be at most six. If we want to produce a ring that
is isomorphic to the quaternions, as I have stated
above, then we will want to mod out two different
elements in order to push the dimension to four.

4) dont give up on groups to fast , i suggest making less " sum = 0 " rules , which should create also less " zero-divisors " as a consequence...

It's true that the fewer "sum=0" rules we have,
the less likely it'll be that we'll discover a
zero divisor. But without "sum=0" rules, we'll be
forced to have negative coefficients, which isn't
in the spirit of what Golden intended when he came
up with the polysigned numbers. And we want to be
able to mod out two different elements so that we
can reduce the dimension to four, the dimension of
the quaternions.

I'd still like to mod out:

*1-1+1 = 0

since the three signs "*", "-", and "+" act like
the three signs of P3. These three signs give us
two dimensions after modding out *1-1+1, and their
span should be isomorphic to the two-dimensional
subring of the quaternions generated by (1,i),
which is isomorphic to the complex field C.

But the rules of a group ring alone give us:

(*1-2^1<2)(*1-2^1<2) = *6-6+6^6<6>6

and so we cannot mod out:

^1<1>1 = 0

without *1-2^1<2 being nilpotent. We have already
modded out *1-1+1, so this gives us:

(*1-2^1<2)(*1-2^1<2) == ^6<6>6 (mod *1-1+1)

Hmmm. I've been trying to find a suitable element
other than ^1<1>1 to mod out, but I've been
having trouble finding something. It might have
something to do with the fact that for _real
numbers_, it appears that we have the inequality

x^2+y^2+z^2 >= xy+xz+yz,

with equality only when x=y=z. This means that
nothing other than ^1<1>1 can ever be modded out.

5) maybe rules similar to ab + ba + ac + ca + bc + cb = 0 make more sense then a + b + c = 0 considering the context ... just guessing ...

What are a,b, and c? Are they supposed to be signs,
such as "-", "+", and "*" (or to be more precise,
-1, +1, and *1)? If so, then replace a+b+c with
ab+ba+ac+ca+bc+cb doesn't help us, since, for
example, ab = (-1)(+1) = *1 = c. Nice try though.

I probably just have to admit that the group ring
in which D3 is the group and R is the ring simply
doesn't have a subring that's isomorphic to the
division ring of quaternions.

1) zero divisors cannot be avoided , no problem , they occured for polysigned as well.

Yes, I know, but I was hoping that since P2 and
P3 avoid zero divisors, I could at least avoid
them for PD3 (i.e., the polysigned numbers based
on the dihedral group D3), but at this point, this
appears to be impossible.
.