Ring-like strusture with non associative addition
$begingroup$
There is this structure i found which is the set of continuous maps from [-1, 1]^n into itself, endowed with a "sum" which is the pointwise sum of two functions divided by 2, and a "product" which is the pointwise multiplication of two functions, where the product vector is obtained by component-wise multiplication.
I found that:
ABOUT THE SUM:
sum IS NOT associative
sum IS commutative
a + a = a (sum is idempotent)
for all a there exists a unique b such that a + b = 0
NB: by zero I meant the constant null function, not addition's neutral element (which doesn't exist). This property is meaningful cause it only holds for zero.
ABOUT THE PRODUCT:
prpduct is associative
product is commutative
it's got a unit, the constant (1,1,...,1) function, which is the only invertible element together with the other combinations of 1's and -1's stuffed in a vector
product distributes over addition
Question is: has this got any name? Are there any publications dealing with similar structures? I found some on nonassociative rings (which maybe are called somethig else), i mean ring with non associative multiplication, but I didn't find anything about structures like this.
Post scriptum: there could also be an additional operation: composition.
ABOUT COMPOSITION:
associative
got identity element (x --> x)
right-distributes over both addition and multiplication
abstract-algebra ring-theory associativity
asked Dec 9 '18 at 16:38
ReLonzoReLonzo
13
$endgroup$
add a comment |
$begingroup$
There is this structure i found which is the set of continuous maps from [-1, 1]^n into itself, endowed with a "sum" which is the pointwise sum of two functions divided by 2, and a "product" which is the pointwise multiplication of two functions, where the product vector is obtained by component-wise multiplication.
I found that:
ABOUT THE SUM:
sum IS NOT associative
sum IS commutative
a + a = a (sum is idempotent)
for all a there exists a unique b such that a + b = 0
NB: by zero I meant the constant null function, not addition's neutral element (which doesn't exist). This property is meaningful cause it only holds for zero.
ABOUT THE PRODUCT:
prpduct is associative
product is commutative
it's got a unit, the constant (1,1,...,1) function, which is the only invertible element together with the other combinations of 1's and -1's stuffed in a vector
product distributes over addition
Question is: has this got any name? Are there any publications dealing with similar structures? I found some on nonassociative rings (which maybe are called somethig else), i mean ring with non associative multiplication, but I didn't find anything about structures like this.
Post scriptum: there could also be an additional operation: composition.
ABOUT COMPOSITION:
associative
got identity element (x --> x)
right-distributes over both addition and multiplication
abstract-algebra ring-theory associativity
asked Dec 9 '18 at 16:38
ReLonzoReLonzo
13
$endgroup$
$begingroup$
You haven't really changed the ring. You've just taken the associative operation and willfully made it not associative. What is the reason for doing this?
$endgroup$
– Matt Samuel
Dec 9 '18 at 20:25
$begingroup$
to keep it C([-1,1]^n, [-1, 1]^n)
$endgroup$
– ReLonzo
Dec 9 '18 at 22:09
1
$begingroup$
It's possible you'll get some responses, but there is a better chance if you format your question properly. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Matt Samuel
Dec 10 '18 at 3:08
add a comment |
$begingroup$
There is this structure i found which is the set of continuous maps from [-1, 1]^n into itself, endowed with a "sum" which is the pointwise sum of two functions divided by 2, and a "product" which is the pointwise multiplication of two functions, where the product vector is obtained by component-wise multiplication.
I found that:
ABOUT THE SUM:
sum IS NOT associative
sum IS commutative
a + a = a (sum is idempotent)
for all a there exists a unique b such that a + b = 0
NB: by zero I meant the constant null function, not addition's neutral element (which doesn't exist). This property is meaningful cause it only holds for zero.
ABOUT THE PRODUCT:
prpduct is associative
product is commutative
it's got a unit, the constant (1,1,...,1) function, which is the only invertible element together with the other combinations of 1's and -1's stuffed in a vector
product distributes over addition
Question is: has this got any name? Are there any publications dealing with similar structures? I found some on nonassociative rings (which maybe are called somethig else), i mean ring with non associative multiplication, but I didn't find anything about structures like this.
Post scriptum: there could also be an additional operation: composition.
ABOUT COMPOSITION:
associative
got identity element (x --> x)
right-distributes over both addition and multiplication
abstract-algebra ring-theory associativity
asked Dec 9 '18 at 16:38
ReLonzoReLonzo
13
$endgroup$
There is this structure i found which is the set of continuous maps from [-1, 1]^n into itself, endowed with a "sum" which is the pointwise sum of two functions divided by 2, and a "product" which is the pointwise multiplication of two functions, where the product vector is obtained by component-wise multiplication.
I found that:
ABOUT THE SUM:
sum IS NOT associative
sum IS commutative
a + a = a (sum is idempotent)
for all a there exists a unique b such that a + b = 0
NB: by zero I meant the constant null function, not addition's neutral element (which doesn't exist). This property is meaningful cause it only holds for zero.
ABOUT THE PRODUCT:
prpduct is associative
product is commutative
it's got a unit, the constant (1,1,...,1) function, which is the only invertible element together with the other combinations of 1's and -1's stuffed in a vector
product distributes over addition
Question is: has this got any name? Are there any publications dealing with similar structures? I found some on nonassociative rings (which maybe are called somethig else), i mean ring with non associative multiplication, but I didn't find anything about structures like this.
Post scriptum: there could also be an additional operation: composition.
ABOUT COMPOSITION:
associative
got identity element (x --> x)
right-distributes over both addition and multiplication
abstract-algebra ring-theory associativity
abstract-algebra ring-theory associativity
asked Dec 9 '18 at 16:38
ReLonzoReLonzo
13
asked Dec 9 '18 at 16:38
ReLonzoReLonzo
13
edited Dec 9 '18 at 16:55
ReLonzo
asked Dec 9 '18 at 16:38
ReLonzoReLonzo
13
asked Dec 9 '18 at 16:38
ReLonzoReLonzo
13
asked Dec 9 '18 at 16:38
ReLonzoReLonzo
13
13
$begingroup$
You haven't really changed the ring. You've just taken the associative operation and willfully made it not associative. What is the reason for doing this?
$endgroup$
– Matt Samuel
Dec 9 '18 at 20:25
$begingroup$
to keep it C([-1,1]^n, [-1, 1]^n)
$endgroup$
– ReLonzo
Dec 9 '18 at 22:09
1
$begingroup$
It's possible you'll get some responses, but there is a better chance if you format your question properly. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Matt Samuel
Dec 10 '18 at 3:08
add a comment |
$begingroup$
You haven't really changed the ring. You've just taken the associative operation and willfully made it not associative. What is the reason for doing this?
$endgroup$
– Matt Samuel
Dec 9 '18 at 20:25
$begingroup$
to keep it C([-1,1]^n, [-1, 1]^n)
$endgroup$
– ReLonzo
Dec 9 '18 at 22:09
1
$begingroup$
It's possible you'll get some responses, but there is a better chance if you format your question properly. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Matt Samuel
Dec 10 '18 at 3:08
$begingroup$
You haven't really changed the ring. You've just taken the associative operation and willfully made it not associative. What is the reason for doing this?
$endgroup$
– Matt Samuel
Dec 9 '18 at 20:25
$begingroup$
You haven't really changed the ring. You've just taken the associative operation and willfully made it not associative. What is the reason for doing this?
$endgroup$
– Matt Samuel
Dec 9 '18 at 20:25
$begingroup$
to keep it C([-1,1]^n, [-1, 1]^n)
$endgroup$
– ReLonzo
Dec 9 '18 at 22:09
$begingroup$
to keep it C([-1,1]^n, [-1, 1]^n)
$endgroup$
– ReLonzo
Dec 9 '18 at 22:09
1
1
$begingroup$
It's possible you'll get some responses, but there is a better chance if you format your question properly. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Matt Samuel
Dec 10 '18 at 3:08
$begingroup$
It's possible you'll get some responses, but there is a better chance if you format your question properly. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Matt Samuel
Dec 10 '18 at 3:08
add a comment |
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$begingroup$
You haven't really changed the ring. You've just taken the associative operation and willfully made it not associative. What is the reason for doing this?
$endgroup$
– Matt Samuel
Dec 9 '18 at 20:25
$begingroup$
to keep it C([-1,1]^n, [-1, 1]^n)
$endgroup$
– ReLonzo
Dec 9 '18 at 22:09
1
$begingroup$
It's possible you'll get some responses, but there is a better chance if you format your question properly. math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Matt Samuel
Dec 10 '18 at 3:08