Let ${\displaystyle M:=\{r,p,s\}}$  , standing for the "rock", "paper" and "scissors" gestures respectively, and consider the binary operation ${\displaystyle \cdot :M\times M\to M}$  derived from the rules of the game as follows:

For all ${\displaystyle x,y\in M}$ :

• If ${\displaystyle x\neq y}$  and ${\displaystyle x}$  beats ${\displaystyle y}$  in the game, then ${\displaystyle x\cdot y=y\cdot x=x}$ 
• ${\displaystyle x\cdot x=x}$      I.e. every ${\displaystyle x}$  is idempotent.

So that for example:

• ${\displaystyle r\cdot p=p\cdot r=p}$    "paper beats rock";
• ${\displaystyle s\cdot s=s}$    "scissors tie with scissors".

This results in the Cayley table:

${\displaystyle {\begin{array}{c|ccc}\cdot &r&p&s\\\hline r&r&p&r\\p&p&p&s\\s&r&s&s\end{array}}}$ 

By definition, the magma ${\displaystyle (M,\cdot )}$  is commutative, but it is also non-associative, as shown by:

${\displaystyle r\cdot (p\cdot s)=r\cdot s=r}$ 

but

${\displaystyle (r\cdot p)\cdot s=p\cdot s=s}$ 

i.e.

${\displaystyle r\cdot (p\cdot s)\neq (r\cdot p)\cdot s}$ 

The "mean" operation ${\displaystyle x\oplus y=(x+y)/2}$  on the rational numbers (or any commutative number system closed under division) is also commutative but not in general associative, e.g.

${\displaystyle -4\oplus (0\oplus +4)=-4\oplus +2=-1}$ 

but

${\displaystyle (-4\oplus 0)\oplus +4=-2\oplus +4=+1}$ 

Generally, the mean operations studied in topology need not be associative.

The construction applied in the previous section to rock-paper-scissors applies readily to variants of the game with other numbers of gestures, as described in the section Variations, as long as there are two players and the conditions are symmetric between them; more abstractly, it may be applied to any trichotomous binary relation (like "beats" in the game). The resulting magma will be associative if the relation is transitive and hence is a (strict) total order; otherwise, if finite, it contains directed cycles (like rock-paper-scissors-rock) and the magma is non-associative. To see the latter, consider combining all the elements in a cycle in reverse order, i.e. so that each element combined beats the previous one; the result is the last element combined, while associativity and commutativity would mean that the result only depended on the set of elements in the cycle.

The bottom row in the Karnaugh diagram above gives more example operations, defined on the integers (or any commutative ring).

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

${\displaystyle (x,y,z)=xr+yp+zs,}$ 

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

${\displaystyle \{(1,0,0),(0,1,0),(0,0,1)\}}$  i.e. ${\displaystyle \{r,p,s\}}$ 

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

The same procedure may be used to derive from any commutative magma ${\displaystyle M}$  a commutative algebra over ${\displaystyle K}$  on ${\displaystyle K^{M}}$ , which will be non-associative if ${\displaystyle M}$  is.