Lemma 1   Let $R$ be a ring, and let $a$ be any element of $R$ . There exists a unique element $b$ of $R$ such that $a+b=0$ , i.e. there is a unique additive inverse for $a$ .

Proof. Let $a$ be an element of $R$ . By definition of ring, there exists at least one additive inverse of $a$ , call it $b_1$ , so that $a+b_1=0$ . Now, suppose $b_2$ is another additive inverse of $a$ , i.e. another element of $R$ such that $$a+b_2=0$$ where $0$ is the zero element of $R$ . Let us show that $b_1=b_2$ . Using properties for a ring and the above equations for $b_1$ and $b_2$ yields \begin{eqnarray*} b_1 &=& b_1+0 \quad \text{(definition of zero)}\\ &=& b_1+(a+b_2) \quad (b_2 \text{ is an additive inverse of }a)\\ &=& (b_1+a)+b_2 \quad (\text{associativity in }R)\\ &=& 0+b_2 \quad (b_1\text{ is an additive inverse of }a)\\ &=& b_2 \quad \text{(definition of zero)}. \end{eqnarray*}Therefore, there is a unique additive inverse for $a$ .