## Precalculus (6th Edition) Blitzer

Vectors are $u={{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j},v={{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}$, and $w={{a}_{3}}\mathbf{i}+{{b}_{3}}\mathbf{j}$. Now, put these vectors in the expression $\left( u+v \right)+w=u+\left( v+w \right)$. Then, we get \begin{align} & \left( {{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j}+{{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j} \right)+\left( {{a}_{3}}\mathbf{i}+{{b}_{3}}\mathbf{j} \right)=\left( {{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j} \right)+\left( {{a}_{2}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{i}+{{b}_{3}}\mathbf{j} \right) \\ & \left( {{a}_{1}}+{{a}_{2}} \right)\mathbf{i}+\left( {{b}_{1}}+{{b}_{2}} \right)\mathbf{j}+{{a}_{3}}\mathbf{i}+{{b}_{3}}\mathbf{j}={{a}_{1}}\mathbf{i}+{{b}_{1}}\mathbf{j}+\left( {{a}_{2}}+{{a}_{3}} \right)\mathbf{i}+\left( {{b}_{2}}+{{b}_{3}} \right)\mathbf{j} \\ & \left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}} \right)\mathbf{i}+\left( {{b}_{1}}+{{b}_{2}}+{{b}_{3}} \right)\mathbf{j}=\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}} \right)\mathbf{i}+\left( {{b}_{1}}+{{b}_{2}}+{{b}_{3}} \right)\mathbf{j} \\ \end{align} Left- side and right- side are equal. Hence, $\left( u+v \right)+w=u+\left( v+w \right)$.