Question

To prove Algebraic proof of property \(2\) as \({\bf{a}} + ({\bf{b}} + {\bf{c}}) = ({\bf{a}} + {\bf{b}}) + {\bf{c}}\).

Short answer

The algebraic proof of property \(2\) for the case of \(n = 2\) is proved.

Step-by-step solution

Concept of Associative law

In mathematics, an associative law is one of two laws relating to the addition and multiplication of numbers, expressed symbolically as:\(\left( {{\bf{a}}{\rm{ }} + {\rm{ }}{\bf{b}}{\rm{ }} + {\rm{ }}{\bf{c}}} \right){\rm{ }} = {\rm{ }}\left( {{\bf{a}}{\rm{ }} + {\rm{ }}{\bf{b}}} \right){\rm{ }} + {\rm{ }}\left( {{\bf{a}}{\rm{ }} + {\rm{ }}{\bf{b}}} \right){\rm{ }} + {\rm{ }}{\bf{c}}\)

Given:

The value of\(n\)as\(2\).

Formula used:

Consider two two-dimensional vectors such as \(a = \left\langle {{a_1},{a_2}} \right\rangle \) and \({\rm{b}} = \left\langle {{b_1},{b_2}} \right\rangle \).

Step 2: Proof of Associative law of Vectors

Sum of vectors:

The vector sum of two vectors\((a + b)\)is,

\(\begin{aligned}{l}a + b &= \left\langle {{a_1},{a_2}} \right\rangle + \left\langle {{b_1},{b_2}} \right\rangle \\ &= \left\langle {{a_1} + {b_1},{a_2} + {b_2}} \right\rangle \end{aligned}\)

The property\(2\)is basically an associative law.

write the expression for property\(2\)

\({\bf{a}} + ({\bf{b}} + {\bf{c}}) = ({\bf{a}} + {\bf{b}}) + {\bf{c}}(1)\)

Here,

\({\bf{a}},{\bf{b}}\), and\({\bf{c}}\)are vectors.

The value of\(n\)is\(2\), hence vectors are two-dimensional. Consider the vectors as\({\bf{a}} = \left\langle {{a_1},{a_2}} \right\rangle ,{\bf{b}} = \left\langle {{b_1},{b_2}} \right\rangle \), and\({\rm{c}} = \left\langle {{c_1},{c_2}} \right\rangle \).

Substitute\(\left\langle {{a_1},{a_2}} \right\rangle \)for\({\bf{a}},\left\langle {{b_1},{b_2}} \right\rangle \)for\({\bf{b}}\), and\(\left\langle {{c_1},{c_2}} \right\rangle \)for\({\bf{c}}\)in equation (1),

\(\begin{aligned}{l}\left\langle {{a_1},{a_2}} \right\rangle + \left( {\left\langle {{b_1},{b_2}} \right\rangle + \left\langle {c1,{c_2}} \right\rangle } \right) &= \left( {\left\langle {{a_1},{a_2}} \right\rangle + \left\langle {{b_1},{b_2}} \right\rangle } \right) + \left\langle {{c_1},{c_2}} \right\rangle \\\left\langle {{a_1},{a_2}} \right\rangle + \left\langle {{b_1} + {c_1},{b_2} + {c_2}} \right\rangle &= \left\langle {{a_1} + {b_1},{a_2} + {b_2}} \right\rangle + \left\langle {{c_1},{c_2}} \right\rangle \\\left\langle {{a_1} + {b_1} + {c_1},{a_2} + {b_2} + {c_2}} \right\rangle &= \left\langle {{a_1} + {b_1} + {c_1},{a_2} + {b_2} + {c_2}} \right\rangle \end{aligned}\)

Therefore, LHS is equal to RHS for\(n = 2\)in equation (1) and hence, the associative law is proved.

Thus, the algebraic proof of property\(\;2\)for the case of \(n = 2\) is proved.