>>11389872>>11389282Counterexample: take R to be the set of functions Z->Z, with operation + defined as (f+g)(x)=f(x)+g(x) and operation * defined as (f*g)(x)=g(f(x)). Then R has the additive identity x->0 and multiplicative identity x->x. If g is the function x->1, then (-1)*(g)(x)= 1, but (-g)(x)=-1, so (-1)(g) != -g. Even though most of the axioms of a ring are satisfied, even the right distributivity (f+g)(h)=fh + gh.
