Definition:Commutative B-Algebra

Definition

Let $\struct {X, \circ}$ be a $B$-algebra.

Then $\struct {X, \circ}$ is said to be $0$-commutative (or just commutative) if and only if:

$\forall x, y \in X: x \circ \paren {0 \circ y} = y \circ \paren {0 \circ x}$

Note

Note the independent properties of $\struct {X, \circ}$ being $0$-commutative and $\circ$ being commutative.

To demonstrate consider the $B$-algebra $\struct {\R, -}$ where $-$ denotes real subtraction.

Work In Progress In particular: A page establishing $\R$ with subtraction is a $B$-algebra. Note that it is not only $\R$ that forms a $B$-algebra with subtraction - so does any of the standard number sets, if I'm not mistaken, so that will need to be taken into account. In fact, we now have Group Induces B-Algebra which shows that this is a general rule for ALL groups. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{WIP}} from the code.

$\struct {\R, -}$ is 0-commutative but $-$ is not commutative.

This theorem requires a proof. In particular: Another page establishing the non-equivalence of 0-commutativity and commutativity, rather than hiding it in this "notes" section. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 2002: J. Neggers and Hee Sik Kim: On B-Algebras (Matematički Vesnik Vol. 54: pp. 21 – 29)