Rules for Multiplication in a Ring
Theorem
Given that $a,\ b,\ c$ is an element of the ring $R$ and $0$ is the identity element for addition, the following properties hold:
- $a0=0a=0$
- $a(-b)=(-a)b=-(ab)$
- $(-a)(-b)=ab$
- $a(b-c)=ac-ac \ \ \And\ \ (b-c)a=ba-ca$ If there exists a multiplicative identity element $1$, then the following properties also hold:
- $(-1)a=-a$
- $(-1)(-1)=1$
Proof
1.
It’s about the property that multiplying any element with the additive identity still results in the additive identity. $$ a0=a(0+0)=a0+a0 $$ Since $a0$ is an element of the ring $R$, it has an additive inverse. Hence, by adding $-a0$ to both sides according to the Cancellation law, $$ a0-a0=a0+a0-a0 \implies 0=a0 $$ Similarly, $$ 0a=(0+0)a=0a+0a \implies 0a-0a=0a+0a-0a \implies 0=0a $$ Therefore, it is $a0=0a=0$.
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2.
Since the ring $R$ follows the distributive law for multiplication, $$ ab+a(-b)=a\left( b + (-b) \right)=a0 $$ According to Theorem 1, since it’s $a0=0$, $$ ab+a(-b)=0 $$ Adding $a(-b)$ to $ab$ yields the identity element $0$, so $a(-b)$ is the additive identity for $ab$. That is, $$ a(-b)=-(ab) $$ And similarly, $$ ab+(-a)b=(a-a)b=0b=0 $$ leads to $(-a)b=-(ab)$. Finally, the following is obtained. $$ a(-b)=(-a)b=-(ab)=-ab $$
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3.
By Theorem 2, $$ (-a)(-b) =-\left( (-a)b\right) =-\left( -(ab) \right) $$ Since the inverse of an inverse is itself, the following is obtained. $$(-a)(-b)=ab$$
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4.
$$ a(b-c)=a(b+(-c))=ab+a(-c) $$ Since it’s $a(-c)=-(ac)=-ac$ by Theorem 2, $$ a(b-c)=ab-ac $$ is obtained. Similarly, the following is obtained. $$ (b-c)a=ba+(-c)a=ba-ca $$
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5.
Since it’s $(-1)a=-(1a)$ by Theorem 2, and $1$ is the multiplicative identity, it is $1a=a$, and therefore, it’s $(-1)a=-a$.
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6.
As in the proof of Theorem 3, by repeatedly applying Theorem 2, the following is obtained. $$ (-1)(-1)=-(1(-1))=-(-(1\cdot 1))=-(-1)=1 $$
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