Example of a commutative non-associative magma

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In mathematics, it can be shown that there exist magmas that are commutative but not associative. A simple example of such a magma is given by considering the children's game of rock, paper, scissors.

[edit] A commutative non-associative magma

Let $M : = {r, p, s}$ and consider the binary operation $\cdot : M \times M \to M$ defined, loosely inspired by the rock-paper-scissors game, as follows:

$r \cdot p = p \cdot r = p$ "paper beats rock";

$p \cdot s = s \cdot p = s$ "scissors beat paper";

$r \cdot s = s \cdot r = r$ "rock beats scissors";

$r \cdot r = r$ "rock ties with rock";

$p \cdot p = p$ "paper ties with paper";

$s \cdot s = s$ "scissors tie with scissors".

By definition, the magma $(M, \cdot)$ is commutative, but it is also non-associative, as the following shows:

$r \cdot (p \cdot s) = r \cdot s = r$

but

$(r \cdot p) \cdot s = p \cdot s = s.$

[edit] A commutative non-associative algebra

Using the above example, one can construct a commutative non-associative algebra over a field $K$ : take $A$ to be the three-dimensional vector space over $K$ whose elements are written in the form

(x, y, z) = x r + y p + z s

,

for $x, y, z \in K$ . Vector addition and scalar multiplication are defined component-wise, and vectors are multiplied using the above rules for multiplying the elements $r, p$ and $s$ . The set

{(1,0,0),(0,1,0),(0,0,1)}

i.e.

{r, p, s}

forms a basis for the algebra $A$ . As before, vector multiplication in $A$ is commutative, but not associative.

Example of a commutative non-associative magma

From Wikipedia, the free encyclopedia

[edit] A commutative non-associative magma

[edit] A commutative non-associative algebra

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