Question

Prove parts 3, 7, 8, and 9 of Theorem 2. 7.

Short answer

Parts 3, 7, 8, and 9 of Theorem 2.7 are proved.

Step-by-step solution

Obtaining solution for part 3

$\begin{array}{rcl}\left[a\right]\oplus \left[b\right]& =& \left[a+b\right]\\ & =& \left[b+a\right]\\ & =& \left[b\right]\oplus \left[a\right]\end{array}$

Obtaining solution for part 7

$\begin{array}{rcl}\left[a\right]⊝\left(\left[b\right]⊝\left[c\right]\right)& =& \left[a\right]⊝\left[b·c\right]\\ & =& \left[a·\left(b·c\right)\right]\\ & =& \left[\left(a·b\right)·c\right]\\ & =& \left[a·b\right]⊝\left[c\right]\\ & =& \left(\left[a\right]⊝\left[b\right]⊝\left[c\right]\right)\end{array}$

Obtaining solution for part 8

$\begin{array}{rcl}\left[a\right]⊝\left(\left[b\right]\oplus \left[c\right]\right)& =& \left[a\right]⊝\left[b+c\right]\\ & =& \left[a·\left(b+c\right)\right]\\ & =& \left[a·b+a·c\right]\\ & =& \left[a·b\right]\oplus \left[a·c\right]\\ & =& \left[a\right]⊝\left[b\right]\oplus \left[a\right]⊝\left[c\right]\end{array}$

Also,

$\begin{array}{rcl}\left(\left[a\right]\oplus \left[b\right]\right)⊝\left[c\right]& =& \left[a+b\right]⊝\left[c\right]\\ & =& \left[\left(a+b\right)·c\right]\\ & =& \left[a·c+b·c\right]\\ & =& \left[a·c\right]\oplus \left[b·c\right]\\ & =& \left[a\right]⊝\left[c\right]\oplus \left[b\right]⊝\left[c\right]\end{array}$

Obtaining solution for part 9

$\begin{array}{rcl}\left[a\right]⊝\left[b\right]& =& \left[a·b\right]\\ & =& \left[b·a\right]\\ & =& \left[b\right]⊝\left[a\right]\end{array}$