Does there exist a nonassociative ring with a cyclic additive

>non-associative ring
>[math]a + b \neq b + a[/math]
Is that even a ring then? Help a brainlet like me understand pls.

>>8327908

You're conflating commutativity and associativity for starters - the property you've stated is the former.

By convention, "ring" often implies commutativity and associativity of both addition and multiplication, but some authors don't require the associativity of multiplication. Thus, a "nonassiciative ring" is a structure that obeys all the ring axioms *except* associativity of multiplication.

For clarification, I am requiring the existence of a multiplicative identity (this axiom is relaxed sometimes, too).

>>8327908
>>a+b≠b+a
False

Anyone?

>>8327957
>false
check out this brainlet

>>8327908
that's non-commutativity not non-associativity nigger

>>8327898
If you multiplication is exponentiation, then you have a non associative ring.

(({0,1,2,3,4}, + mod 5), ^ mod 5) fits your definition right ?

(2^3)^2 = (8 mod 5)^2 = 9 mod 5 = 4
2^(3^2) = 2^(9 mod 5) = 16 mod 5 = 1

Are you explicitly asking it to not be associative? In your second post you said that you just aren't imposing associativity, but that's an obviously boring question.

>>8328587
Sorry that's not distributive.

>>8328587
There is no two-sided identity for ^. Plus, you would have to define 0^0 = 0, which is pretty gay already.
>>8328599
Pretty sure it is distributive?

OP, your question is trivial, at least when you modify it like >>8328590 says.

Write * for the ring multiplication and m $ n for (n + n + ... + n) [m times]. Then a*n = a*(n $ 1) = n $ a, by distributivity. Therefore the only operation that distributes over addition in a cyclic group is the normal multiplication.

OP, no such structure exists (if we assume strict non-associativity). Here is a proof.

There must be a,b,c with abc not equal to a(bc). By the additive group being cyclic, abc is 1+1+...+1 (n times) for some n, and a(bc) is 1+1+...+1 (m times). By inequality and uniqueness of addition, we must have that m≠n. Without loss of generality, assume m>n. Then, a(bc)-abc=m-n, with the RHS using the shorthand that n and m are sums of that many 1's. By distributivity, we get that a(bc-bc)=m-n=a(0)=0, but this implies m=n, a contradiction.

>>8328634
Well done!