Let us take a ring $R(+,*)$. It is easy to prove that that $0*a=0$. The standard proof works this way: $0*a=(0+0)*a=0*a+0*a$. Adding $-(0*a)$ on both sides, we get $0=0*a$.

This fact could be used to prove that $(-a)*b=-(a*b)$. How? $(-a)*b+a*b=(-a+a)*b=0*b=0$. Hence, as $(-a)*b$ is the additive inverse of $a*b$, and as every element has a unique additive inverse, we have $(-a)*b=-(a*b)$.

Although proving the above is a fairly uninvolved exercise, it was never intuitive to me, unless I started to imagine $+$ as arithmetic addition and $*$ as arithmetic multiplication. This blog post is an attempt to make the aforementioned facts more easy to visualize, and perhaps intuitive.

What is $0*a$? To answer this, we need to understand what $0$ is in the context of distributivity of addition over multiplication. It is $(-a+a)$. It is $(-2a+2a)$. It is the combination of matter and anti-matter. It takes away as much as it gives. This I feel is the easiest way to think about it in the given context. I would love to elaborate on why this is the best way some other time.

Now what is $(-a)*b$? It is $(a+-2a)*b$. It is $(2a+-3a)*b$. If $(2a+-3a)*b=(a+-2a)*b$, then $a*b+(-a)*b=0$. This proves that $(-a)*b=-(a*b)$, as the additive inverse of every element is unique.

Using this fact, we can now say that

$0*a=(c+-c)*a=c*a+(-c)*a=0$.

All this because I find it easier to visualize $0$ as $(-a+a)$ for any element $a$ rather than $0+0$. And I’m sure you’d find it easier too.