4 ms·
Isn't it something like this? 0=a-a, for all a (where a-a is actually, a+(-a)) so: mul(b,0)=mul(b,(a-a))=mul(b,a)-mul(b,a)=z-z=0
by hhm 18y ago
Isn't it something like this?
0=a-a, for all a
(where a-a is actually, a+(-a))
so:
mul(b,0)=mul(b,(a-a))=mul(b,a)-mul(b,a)=z-z=0
- newt0311 18y agopretty much.