(02/14/2021, 02:24 AM)JmsNxn Wrote: Off topic but I like that you use a day planner/agenda to write your math in, lol.

(06/19/2022, 12:17 AM)tommy1729 Wrote: ok here is a 5 dim candidate

the coefficients are " magnitudes " ; nonnegative reals.


1 + a + b + c + d + e = 0.

so when " reduced " not all magnitudes can be nonzero.

mod ( 1 + a + b + c + d + e ) if you want.
( so the dimension reduces from 6 ( 5 letters and real ) to 5 )


a^2 = b^2  = ... = 1.

a b = c
a c = d
a d = e
a e = b

b c = e
b d = a
b e = d

c d = b
c e = a

d e = c

som and products are commutative.

som and product behave distributive.



regards

tommy1729

(06/18/2022, 11:36 PM)tommy1729 Wrote: The basic ideas are

1) unital and commutative but nonassociative numbers.
2) power-associative numbers so we can use taylor theorems.
3) no nilpotent elements
4) every element has at least 1 square root.
5) the smallest ones
6) no subnumbers only real coefficients. and not iso to an extension of 2 type of numbers ( like complex coefficients or other extensions of smaller dimensions )

then there are 2 cases left

the units sum to 0.

the units are linear independant.

assuming solutions exist ofcourse.  I conjecture yes.

On the other hand I conjecture only a finite amount of them ... probably between 0 and 3.
And all solutions having dimension below 28.

The 8 dimensional number given here has nilpotent elements. So it violates one of the conditions.
They always have a square root though.

I will post a candidate soon.

I was not able to find this relatively simple idea in the books.

I see applications in physics and math as I believe they are the " next quaternion ".

regards

tommy1729

(06/25/2022, 09:13 PM)MphLee Wrote:

(06/18/2022, 11:36 PM)tommy1729 Wrote: The basic ideas are

1) unital and commutative but nonassociative numbers.
2) power-associative numbers so we can use taylor theorems.
3) no nilpotent elements
4) every element has at least 1 square root.
5) the smallest ones
6) no subnumbers only real coefficients. and not iso to an extension of 2 type of numbers ( like complex coefficients or other extensions of smaller dimensions )

then there are 2 cases left

the units sum to 0.

the units are linear independant.

assuming solutions exist ofcourse.  I conjecture yes.

On the other hand I conjecture only a finite amount of them ... probably between 0 and 3.
And all solutions having dimension below 28.

The 8 dimensional number given here has nilpotent elements. So it violates one of the conditions.
They always have a square root though.

I will post a candidate soon.

I was not able to find this relatively simple idea in the books.

I see applications in physics and math as I believe they are the " next quaternion ".

regards

tommy1729

That's more reasonable, but still not clear to me. Let's see if I get it...
Your are trying to axiomatize something, an agebraic structure that has the desired properties...
Those properties you want to enforce on them are motivated by your need to define functions valued in that structure that are enough well behaved for your purposes...
So given the properties, you are trying to pin down how many models those axioms have... how many isomorphic algebras have those properties

To be more precise you are looking for all the \(\mathbb R\)-algebras, i.e. \(\mathbb R\)-vector spaces \(V\) equipped with a bilinear product map \(\cdot:V\otimes V\to V\) that satisfies your list.

This seems a hell of an enterprise... I feel that classically (in the last 50/80 years) problems of this kind are treated via model theoretic methods, if not by universal algebra. And nowadays i feel that the way to examine this should be category-theoretic. For example, recently it was proved that a class of agebras interesting to physicians actually has only three models \(\mathbb R, \mathbb C\) and \(\mathbb H\) and it was proved using category theory if I remember good.

My question now is... why those properties? What do you want to do on those algebras? Extending tetration? Or is it a more vast programme?

(06/19/2022, 12:17 AM)tommy1729 Wrote: ok here is a 5 dim candidate

the coefficients are " magnitudes " ; nonnegative reals.


1 + a + b + c + d + e = 0.

so when " reduced " not all magnitudes can be nonzero.

mod ( 1 + a + b + c + d + e ) if you want.
( so the dimension reduces from 6 ( 5 letters and real ) to 5 )


a^2 = b^2  = ... = 1.

a b = c
a c = d
a d = e
a e = b

b c = e
b d = a
b e = d

c d = b
c e = a

d e = c

som and products are commutative.

som and product behave distributive.



regards

tommy1729

(06/25/2022, 10:02 PM)tommy1729 Wrote:

(06/19/2022, 12:17 AM)tommy1729 Wrote: ok here is a 5 dim candidate

the coefficients are " magnitudes " ; nonnegative reals.


1 + a + b + c + d + e = 0.

so when " reduced " not all magnitudes can be nonzero.

mod ( 1 + a + b + c + d + e ) if you want.
( so the dimension reduces from 6 ( 5 letters and real ) to 5 )


a^2 = b^2  = ... = 1.

a b = c
a c = d
a d = e
a e = b

b c = e
b d = a
b e = d

c d = b
c e = a

d e = c

som and products are commutative.

som and product behave distributive.



regards

tommy1729

To solve the sqrt you basically need to solve this set of equation.
See picture.

All variables ( given and solved) need to be non-negative.

The C is not part of input or output But is there to make the reduction ( mod 1 + a + … )

C also nonnegative.

This system was simply obtained by squaring.

regards 

tommy1729

(07/04/2022, 06:05 PM)tommy1729 Wrote:

(06/25/2022, 10:02 PM)tommy1729 Wrote:

(06/19/2022, 12:17 AM)tommy1729 Wrote: ok here is a 5 dim candidate

the coefficients are " magnitudes " ; nonnegative reals.


1 + a + b + c + d + e = 0.

so when " reduced " not all magnitudes can be nonzero.

mod ( 1 + a + b + c + d + e ) if you want.
( so the dimension reduces from 6 ( 5 letters and real ) to 5 )


a^2 = b^2  = ... = 1.

a b = c
a c = d
a d = e
a e = b

b c = e
b d = a
b e = d

c d = b
c e = a

d e = c

som and products are commutative.

som and product behave distributive.



regards

tommy1729

To solve the sqrt you basically need to solve this set of equation.
See picture.

All variables ( given and solved) need to be non-negative.

The C is not part of input or output But is there to make the reduction ( mod 1 + a + … )

C also nonnegative.

This system was simply obtained by squaring.

regards 

tommy1729

ok. it seems the real part of the square of such a number is always the largest.
Therefore we cannot solve for many numbers including the sqrt of the additive inverse of 1 ( the imaginary )

SO im moving on to dimension 6.

Ill keep you all updated

regards

tommy1729

(07/04/2022, 06:05 PM)tommy1729 Wrote:

(06/25/2022, 10:02 PM)tommy1729 Wrote:

(06/19/2022, 12:17 AM)tommy1729 Wrote: ok here is a 5 dim candidate

the coefficients are " magnitudes " ; nonnegative reals.


1 + a + b + c + d + e = 0.

so when " reduced " not all magnitudes can be nonzero.

mod ( 1 + a + b + c + d + e ) if you want.
( so the dimension reduces from 6 ( 5 letters and real ) to 5 )


a^2 = b^2  = ... = 1.

a b = c
a c = d
a d = e
a e = b

b c = e
b d = a
b e = d

c d = b
c e = a

d e = c

som and products are commutative.

som and product behave distributive.



regards

tommy1729

To solve the sqrt you basically need to solve this set of equation.
See picture.

All variables ( given and solved) need to be non-negative.

The C is not part of input or output But is there to make the reduction ( mod 1 + a + … )

C also nonnegative.

This system was simply obtained by squaring.

regards 

tommy1729

ok. it seems the real part of the square of such a number is always the largest.
Therefore we cannot solve for many numbers including the sqrt of the additive inverse of 1 ( the imaginary )

SO im moving on to dimension 6.

Ill keep you all updated

regards

tommy1729