by ayushkhaitan3437

I worry about “small” things.

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.