Show that $a(-1) = (-1)a = -a $. In a ring $R$ with identity 1, show that $$a(-1) = (-1)a = -a \qquad\forall\, a \in R$$ I have started with $a + (-a) = 0$ but cant proceed from here. 

We have $\space[1+(-1)]=0$.
Therefore $a\cdot [1+(-1)]=a\cdot 0=0$.
By left distributive law,
$$a\cdot 1+a\cdot (-1)=0$$
$$a+a\cdot (-1)=0.$$
Now $-a \in R$.  Adding $-a$ to both sides, we get
$$(-a)+[a+a\cdot (-1)]=(-a)+0,$$
or
$$[(-a)+a]+a\cdot (-1)=-a\quad [\text{associative property}]$$
or
$$0+a\cdot (-1)=-a,$$
or
$$a\cdot (-1)=-a.$$
Similarly we can do $[1+(-1)]\cdot a=0\cdot a=0$.
By right distributive law,
$$1 \cdot a+(-1)\cdot a=0$$
$$a+(-1)\cdot a=0.$$
Now $-a \in R$. Adding $-a$ to both sides and proceeding as before, we get $$(-1)\cdot a=-a.$$
Hence $a\cdot (-1)=(-1)\cdot a=-a$.
