ok.
although minus is not commutative , associative or distributive , its inverse + is !
this § does not have such an inverse. ( as an operator at least )
the issue is people will argue § is + or § is - , compare it to defining a number j , that also has j^2 = -1 but j =/= i or - i.
or k^2 = 1 with k not -1 or +1.
i hope you are familiar with zero-divisors and quaternions.
" tessarines " and analogues ( sometimes other names ) also cross my mind.
usually , these kind , they have matrix representations.
these mainly 18 and 19 th century ideas where popular in their days , and one of the key ideas is to define a " muliplication table for the units " , call it a group or so if you like.
this to stretch the importance of multiplication and the remark that some things CANNOT BE DERIVED , but MUST BE DEFINED.
not that this is strictly necc but i think you should take a look at it if you havent already.
i wrote the above so to make clear the " actual reply " below :
let a , b , c be distinct nonnegative reals.
i refuse to take § as an operator at the moment , but as a sign/number i will try to define it :
here x§y is considered as x + §y and also equal to §y + x
+a-b§c => max(a,b,c) * the sign that matches that max (+,- or §)
the cases when some a,b,c are equal is trivial.
so now we only need a multiplication table.
since we dont want § to be + or - ( see the intro before the " actual reply " , i hope it is clear ) it seems wise to define §1^2 different from -1 , 0 , 1.
so we only are left to define (§1)^2 and -1 * §1.
( well actually there are other solutions than this post , but those are " old ideas " such as group rings etc and not in the spirit of this thread , closer to quaternions and such )
thus we are basicly forced to define §1*§1 = §1
3 problems remain
x+y = -x+y
-1 * §1
(-1+1) * §1
it is a kinda basic thing in math to be ABLE TO DO BOTH THINGS ON BOTH SIDES OF THE EQUATION.
examples
a = b then a^2 = b^2
although we must note
a^2 = b^2 does not imply a=b necc.
so take with a grain of salt.
but for + and - that grain of salt is not needed in standard math.
however with those § and the equation -x+y = x+y we get another situation.
if we cannot freely use substraction , addition on both sides of the equation AND we do not have commutativity , associativity and distributivity on either hand ... well its hard to take algebraic steps in a proof !
im not trying to shoot this idea to the moon , but im pointing out the issues.
on the other hand , the definition of -1 * §1 seems key here.
if we take -1 * §1 to be §1 , then the equation
-x+y = x + y
reduces to
-x = x
if we substract y on both sides , and then
-x = x has the solutions § (c^2) for any real c.
that seems consistant HOWEVER
if we take
-x +y = x+y
and add x on both sides , we get
y = 2x + y
now substract y on both sides
0 = 2x
and now x can only be 0 and not the other solution arrived at earlier !!
HENCE it seems that there is no satisfactionary solution to -1 * §1 UNLESS WE ALSO DEFINE A 4 TH SIGN
&1 = - §1 ( or §-1 )
but that has issues too
3+§4 = §1
§1+&4 = &3
thus (3 + §4) + &4 = &3 however 3+(§4+&4) = 3 + 0 = 3
one CANNOT DEFINE &3 as 3 ( and hence &1 = +1 ) because we get the same with
-3+§4 = §1
§1+&4 = &3
thus (-3 + §4) + &4 = &3 however -3+(§4+&4) = -3 + 0 = -3
and then we must get §3 = -3 , but we just defined &3 = 3 above !!
SINCE we cannot state & as another sign ( &1 =/= -1 , &1 =/= 1 , &1 =/= §1 , &1 =/= 0 ) we remain with the fact that if we accept & we loose associativity for addition !
( recall thus (-3 + §4) + &4 = &3 however -3+(§4+&4) = -3 + 0 = -3 so we lost associativity of addition )
since we have &1 = -1 * §1 and -1 * -1 * §1 = 1*§1 = §1
and §1*§1 = §1
we are forced to have &1^2 = (-1)^2 * (§1)^2 = §1
and &1 * §1 = - §1 = &1 which is ... undesired. ( since §1 =/= 1 )
---
issues/problems or no properties , that seems to be the decision case ...
hence my lack of enthousiasm and the non-introduction in math i think.
good luck anyway.
regards
tommy1729
although minus is not commutative , associative or distributive , its inverse + is !
this § does not have such an inverse. ( as an operator at least )
the issue is people will argue § is + or § is - , compare it to defining a number j , that also has j^2 = -1 but j =/= i or - i.
or k^2 = 1 with k not -1 or +1.
i hope you are familiar with zero-divisors and quaternions.
" tessarines " and analogues ( sometimes other names ) also cross my mind.
usually , these kind , they have matrix representations.
these mainly 18 and 19 th century ideas where popular in their days , and one of the key ideas is to define a " muliplication table for the units " , call it a group or so if you like.
this to stretch the importance of multiplication and the remark that some things CANNOT BE DERIVED , but MUST BE DEFINED.
not that this is strictly necc but i think you should take a look at it if you havent already.
i wrote the above so to make clear the " actual reply " below :
let a , b , c be distinct nonnegative reals.
i refuse to take § as an operator at the moment , but as a sign/number i will try to define it :
here x§y is considered as x + §y and also equal to §y + x
+a-b§c => max(a,b,c) * the sign that matches that max (+,- or §)
the cases when some a,b,c are equal is trivial.
so now we only need a multiplication table.
since we dont want § to be + or - ( see the intro before the " actual reply " , i hope it is clear ) it seems wise to define §1^2 different from -1 , 0 , 1.
so we only are left to define (§1)^2 and -1 * §1.
( well actually there are other solutions than this post , but those are " old ideas " such as group rings etc and not in the spirit of this thread , closer to quaternions and such )
thus we are basicly forced to define §1*§1 = §1
3 problems remain
x+y = -x+y
-1 * §1
(-1+1) * §1
it is a kinda basic thing in math to be ABLE TO DO BOTH THINGS ON BOTH SIDES OF THE EQUATION.
examples
a = b then a^2 = b^2
although we must note
a^2 = b^2 does not imply a=b necc.
so take with a grain of salt.
but for + and - that grain of salt is not needed in standard math.
however with those § and the equation -x+y = x+y we get another situation.
if we cannot freely use substraction , addition on both sides of the equation AND we do not have commutativity , associativity and distributivity on either hand ... well its hard to take algebraic steps in a proof !
im not trying to shoot this idea to the moon , but im pointing out the issues.
on the other hand , the definition of -1 * §1 seems key here.
if we take -1 * §1 to be §1 , then the equation
-x+y = x + y
reduces to
-x = x
if we substract y on both sides , and then
-x = x has the solutions § (c^2) for any real c.
that seems consistant HOWEVER
if we take
-x +y = x+y
and add x on both sides , we get
y = 2x + y
now substract y on both sides
0 = 2x
and now x can only be 0 and not the other solution arrived at earlier !!
HENCE it seems that there is no satisfactionary solution to -1 * §1 UNLESS WE ALSO DEFINE A 4 TH SIGN
&1 = - §1 ( or §-1 )
but that has issues too
3+§4 = §1
§1+&4 = &3
thus (3 + §4) + &4 = &3 however 3+(§4+&4) = 3 + 0 = 3
one CANNOT DEFINE &3 as 3 ( and hence &1 = +1 ) because we get the same with
-3+§4 = §1
§1+&4 = &3
thus (-3 + §4) + &4 = &3 however -3+(§4+&4) = -3 + 0 = -3
and then we must get §3 = -3 , but we just defined &3 = 3 above !!
SINCE we cannot state & as another sign ( &1 =/= -1 , &1 =/= 1 , &1 =/= §1 , &1 =/= 0 ) we remain with the fact that if we accept & we loose associativity for addition !
( recall thus (-3 + §4) + &4 = &3 however -3+(§4+&4) = -3 + 0 = -3 so we lost associativity of addition )
since we have &1 = -1 * §1 and -1 * -1 * §1 = 1*§1 = §1
and §1*§1 = §1
we are forced to have &1^2 = (-1)^2 * (§1)^2 = §1
and &1 * §1 = - §1 = &1 which is ... undesired. ( since §1 =/= 1 )
---
issues/problems or no properties , that seems to be the decision case ...
hence my lack of enthousiasm and the non-introduction in math i think.
good luck anyway.
regards
tommy1729