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Suppose that $\ast$ distributes over $+$, where $0$ is the additive identity. We can conclude the following.

$a \ast b = a \ast \left( b + 0\right)$

$a \ast b = a \ast b + a \ast 0$

$\therefore a \ast 0 = 0$

In other words, the additive identity and multiplicative 'annihilator' become equivalent under the distributive property. However, the converse is not true. Equating the two $0$'s is a necessary but insufficient condition to establish distributivity.

My question: suppose we know that the following two statements are true for any element $a$.

$a+0=a$

$a \ast 0 = 0$

What other conditions (in addition to the above) are sufficient to imply a distributive property?

Ryan
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    You're probably not going to find any nice answer to this. $a+0=0$ isn't really a condition, since it's just the definition of $0$, and $a0=0$ is very weak. You need very little distributivity to get $a0=0$, for example for each $a$, $a(b+0)=ab + a*0$ needs to hold only for a single $b$. It's hard to imagine any extra converse condition that's not essentially the entire distributive law by itself. – Zavosh Aug 25 '13 at 06:27
  • I suspected that might be the answer. I've been trying to find a solution that might somehow involve the assumption of an inverse for multiplication, but it usually ends up going nowhere. – Ryan Aug 25 '13 at 19:21
  • @Prometheus: Maybe with a little rewording you could post that as an answer, since it looks like that is the case. – Ryan Aug 30 '13 at 21:27

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Listen, the converse of the distributive property does work. It is common in algebra for you to use the converse of the distributive property ex.) 2x + 3x = 5x or x(2 + 3).

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