(-a)b = a(-b) = -ab

Dependencies:

1. Ring
2. 0x = 0 = x0

Let $a,b$ belong to ring $R$. Then $(-a)b=a(-b)=-ab$.

Proof

$$\begin{array}{rl}& ab+(-a)b\\ \text{(distributivity)}& & =(a+(-a))b& =0b\\ \text{(}0x=0\mathrm{\forall }x\in R\text{)}& & =0\end{array}$$

Therefore, $(-a)b=-(ab)$.

$$\begin{array}{rl}& ab+a(-b)\\ \text{(distributivity)}& & =a(b+(-b))& =a0\\ \text{(}x0=0\mathrm{\forall }x\in R\text{)}& & =0\end{array}$$

Therefore, $a(-b)=-(ab)$.

Dependency for: None

Info:

• Depth: 3
• Number of transitive dependencies: 3

Transitive dependencies:

1. Group
2. Ring
3. 0x = 0 = x0