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I have a problem with this definition of anticommutative operators. I found the following: An operation $\circledast$ is called anticommutative if it satisfies the following:

(i) There is a right identity element $r:=r_x$, that is, $\exists r \in \mathbb{X}: x \circledast r = x, x \in \mathbb{X}$.

(ii) $x \circledast y = r \Leftrightarrow (x\circledast y)\circledast (y \circledast x) = r \Leftrightarrow x = y $ for all $x,y \in X$.

Now, my problem is, if you consider a set $\mathbb{X}$ with more than one element, and you choose $x \neq r, y=r$, the following happens:

None of the equivalent equations are true (because of $x \neq y$), so equation 2 says: $(x\circledast r)\circledast (r \circledast x) \neq r$ which is $x \circledast x \neq r$, but this is a contradiction to the equivalence of equation 1 ($x \circledast y = r$) and equation 3 ($x = y $ for all $x,y \in X$).

Is the definition correct?

Thx in advance.

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It appears to me you are assuming the operation is associative. Is that correct? –  RghtHndSd Feb 7 at 18:24
    
Oh, i guess I was just confused with the composition of functions (which is always associative). This might just be the problem :D –  Greg P. Feb 7 at 18:28

1 Answer 1

up vote 1 down vote accepted

In saying $(x \circledast r) \circledast (r \circledast x) \neq r$ is equivalent to $x \circledast x \neq r$, you are assuming the operator is associative:

$$(x \circledast r) \circledast (r \circledast x) = x \circledast (r \circledast x) = (x \circledast r) \circledast x = x \circledast x$$

Take as an example $x \circledast y = x-y$ on the (say) real numbers. Then the axioms are satisfied with $r = 0$.

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