(-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{align} & ab + (-a)b \\ &= (a + (-a))b \tag{distributivity} \\ &= 0b \\ &= 0 \tag{$0x = 0 \forall x \in R$} \end{align}

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

\begin{align} & ab + a(-b) \\ &= a(b + (-b)) \tag{distributivity} \\ &= a0 \\ &= 0 \tag{$x0 = 0 \forall x \in R$} \end{align}

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