On the existence of hyperrings associated to arithmetic functions

A brief survey on algebraic hyperstructures: Theory and applications

“…In [17], Asghari and Davvaz introduced a hyperoperation associated to the set of all arithmetic functions and analyzed the properties of this hyperoperation. In [6], Al Tahan and Davvaz defined a new hyperoperation associated to the set G of all arithmetic functions. Here, we review some definitions and results from [6].…”

“…In [6], Al Tahan and Davvaz defined a new hyperoperation associated to the set G of all arithmetic functions. Here, we review some definitions and results from [6]. An arithmetic function is a function in which its domain of definition is the set of natural numbers and its codomain is the set of complex numbers.…”

Hyperhomographies on Krasner Hyperfields

“…Consider f (x) = x 2 over the finite field F = Z 5 . Then we have a parabola in F and the Cayley table of its points Q f (F) = {O, (1, 1), (2,4), (3,4), (4, 1)} where, O = (0, 0) is the identity element of the group, is as follows.…”

“…The main idea consists in substituting the field with a hyperfield, in particular with the associated quotient Krasner hyperfield. The power of this algebraic hyperstructure has been already used in solving different problems in affine algebraic schemes [2], theory of arithmetic functions [3], tropical geometry [4], algebraic geometry [5], etc. The quotient Krasner hyperfield is practically the quotientF = F/G of a classical field F by any normal subgroup G of the multiplicative part (F \ {0}, •).…”

“…2),(4,4) (3,2),(4,4) O (2, 3) (3, 2),(4,4) (1, 1), (4, 4) O (1, 1), (2, 3) (3, 2) (3, 2), (4, 4) O (1, 1), (4, 4) (2, 3), (1, 1) (4, 4) (4, 4) O (1, 1), (2, 3) (1, 1), (2, 3) (3In this table, we have O (x, y) = (x, y) = (x, y) O, for all (x, y) ∈ H 0,0 (F) and (x, y) (x , y ) = {(x + x , (x + x ) −1 ), (y + y , (y + y ) −1 )}, for all (x, y), (x , y ) ∈ H 0,0 (F) { O}.The hyperoperation is not associative, but only weak associative, as we can notice here below:O = [[(1, 1) (1, 1)] (3, 2)] = [(1, 1) [(1,1) (3, 2)]] = { O, (3, 2), (4, 4)}.…”