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.
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ $(a)(-1)+(a)(1)=(a)(-1+1)=(a)(0)$. Prove that $(a)(0)=0$ using $(a)(0+0)=(a)(0)$. $\endgroup$– André NicolasCommented Apr 15, 2015 at 16:07
-
$\begingroup$ Do you know Peano's Axioms? $\endgroup$– Chan HuntCommented Apr 15, 2015 at 17:00
-
1$\begingroup$ @ForrestChanningHunter The ring axioms are quite different from Peano's. $\endgroup$– ArthurCommented Apr 15, 2015 at 21:17
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Use distribution for all it's worth, together with $1\cdot a = a \cdot 1 = a$: $$ a + (-1) \cdot a \\= 1\cdot a + (-1)\cdot a \\= (1 + (-1)) \cdot a \\= 0\cdot a \\= 0$$ and thus, by the uniqueness of additive inverses (remember, if we forget multiplication, any ring becomes an abelian group), we get that $(-1) \cdot a = -a$. By the same reasoning, we get $$ a + a\cdot (-1) = a(1 + (-1)) = 0 $$
$\begingroup$
$\endgroup$
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 $a\cdot (-1)=(-1)\cdot a=-a$.
