Use the vectors \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\), \({\bf{v}} = \left( {{v_1},...,{v_n}} \right)\), and \({\bf{w}} = \left( {{w_1},...,{w_n}} \right)\) to verify the following algebraic properties of \({\mathbb{R}^n}\).

a. \(\left( {{\bf{u}} + {\bf{v}}} \right) + {\bf{w}} = {\bf{u}} + \left( {{\bf{v}} + {\bf{w}}} \right)\)

b. \(c\left( {{\bf{u}} + {\bf{v}}} \right) = c{\bf{u}} + c{\bf{v}}\)

Consider the algebraic property \(\left( {{\bf{u}} + {\bf{v}}} \right) + {\bf{w}} = {\bf{u}} + \left( {{\bf{v}} + {\bf{w}}} \right)\).

Take the left-hand side of the algebraic property as \(\left( {{\bf{u}} + {\bf{v}}} \right) + {\bf{w}}\). Here, \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\), \({\bf{v}} = \left( {{v_1},...,{v_n}} \right)\), and \({\bf{w}} = \left( {{w_1},...,{w_n}} \right)\).

Simplify \(\left( {{\bf{u}} + {\bf{v}}} \right) + {\bf{w}}\)by using \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\), \({\bf{v}} = \left( {{v_1},...,{v_n}} \right)\), and \({\bf{w}} = \left( {{w_1},...,{w_n}} \right)\) as shown below:

\(\begin{aligned}{c}\left( {{\bf{u}} + {\bf{v}}} \right) + {\bf{w}} &= \left[ {\left( {{u_1},...,{u_n}} \right) + \left( {{v_1},...,{v_n}} \right)} \right] + \left( {{u_1},...,{u_n}} \right)\\ &= \left[ {\left( {{u_1} + {v_1}} \right) + ... + \left( {{u_n} + {v_n}} \right)} \right] + \left( {{w_1},...,{w_n}} \right)\\ &= \left[ {\left( {{u_1} + {v_1}} \right) + {w_1}} \right] + ... + \left[ {\left( {{u_n} + {v_n}} \right) + {w_n}} \right]\end{aligned}\)

Show that \(\left( {{\bf{u}} + {\bf{v}}} \right) + {\bf{w}} = {\bf{u}} + \left( {{\bf{v}} + {\bf{w}}} \right)\) arranges the vectors \(\left[ {\left( {{u_1} + {v_1}} \right) + {w_1}} \right] + ... + \left[ {\left( {{u_n} + {v_n}} \right) + {w_n}} \right]\) in such a manner that \({\bf{u}} + \left( {{\bf{v}} + {\bf{w}}} \right)\).

\(\begin{aligned}{c}\left( {{\bf{u}} + {\bf{v}}} \right) + {\bf{w}} &= \left( {{u_1} + {v_1} + {w_1}} \right) + ... + \left( {{u_n} + {v_n} + {w_n}} \right)\\ &= \left( {{u_1},...,{u_n}} \right) + ... + \left[ {{u_n} + \left( {{v_n} + {w_n}} \right)} \right]\\ &= \left( {{u_1},...,{u_n}} \right) + \left[ {\left( {{v_1},...,{v_n}} \right) + \left( {{w_1},...,{w_n}} \right)} \right]\\ &= {\bf{u}} + \left( {{\bf{v}} + {\bf{w}}} \right)\end{aligned}\)

Hence, it is proved that \(\left( {{\bf{u}} + {\bf{v}}} \right) + {\bf{w}} = {\bf{u}} + \left( {{\bf{v}} + {\bf{w}}} \right)\).

Consider the algebraic property \(c\left( {{\bf{u}} + {\bf{v}}} \right) = c{\bf{u}} + c{\bf{v}}\).

Take the left-hand side of the algebraic property as \(c\left( {{\bf{u}} + {\bf{v}}} \right)\). Here, \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\), \({\bf{v}} = \left( {{v_1},...,{v_n}} \right)\), and \(c\) is a constant.

Simplify \(c\left( {{\bf{u}} + {\bf{v}}} \right)\)by using \({\bf{u}} = \left( {{u_1},...,{u_n}} \right)\), and \({\bf{v}} = \left( {{v_1},...,{v_n}} \right)\) is shown below:

\(\begin{aligned}{c}c\left( {{\bf{u}} + {\bf{v}}} \right) &= c\left[ {\left( {{u_1},...,{u_n}} \right) + \left( {{v_1},...,{v_n}} \right)} \right]\\ &= c\left[ {\left( {{u_1} + {v_1}} \right) + ... + \left( {{u_n} + {v_n}} \right)} \right]\\ &= c\left( {{u_1} + {v_1}} \right) + ... + c\left( {{u_n} + {v_n}} \right)\end{aligned}\)

Show that \(c\left( {{\bf{u}} + {\bf{v}}} \right) = c{\bf{u}} + c{\bf{v}}\) arranges the vectors \(c\left( {{u_1} + {v_1}} \right) + ... + c\left( {{u_n} + {v_n}} \right)\) in such a manner that \(c{\bf{u}} + c{\bf{v}}\).

\(\begin{aligned}{c}c\left( {{\bf{u}} + {\bf{v}}} \right) &= c{u_1} + c{v_1} + ... + c{u_n} + c{v_n}\\ &= c\left( {{u_1},...,{u_n}} \right) + c\left( {{v_1},...,{v_n}} \right)\\ &= c{\bf{u}} + c{\bf{v}}\end{aligned}\)

Hence, it is proved that \(c\left( {{\bf{u}} + {\bf{v}}} \right) = c{\bf{u}} + c{\bf{v}}\).