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- Exercise 1.1 (Solutions) @fsc-part1-ptb:sol:ch01
- \frac{1}{c}\\ &= (a+b) \times \frac{1}{c}\quad \text{(Right distributive property)}\\ &= \frac{a+b}{c... \dfrac{1}{4} \times (4\times 1+4 \times 4x)\quad \text{(multiplicative identity)}$ $=\dfrac{1}{4} \times 4 \times (1+ 4x)\quad \text{(distributive property)}$ $= 1 \times (1+4x)\quad \text{(multiplicative inverse)}$ $= 1+4x \quad \text{(
- Exercise 2.8 (Solutions) @fsc-part1-ptb:sol:ch02
- eft( {{a}^{-1}}*a \right)*x={{a}^{-1}}*b\,\,\,\, \text{by associative law}\\ \Rightarrow \,\,\,\,& e*x={{a}^{-1}}*b \,\,\,\,\text{by inverse law.}\\ \Rightarrow \,\,\,\,& x={{a}^{-1}}*b\,\,\,\, \text{by identity law.}\end{align} And for \begin{ali... \,\,& \left( x*a \right)*{{a}^{-1}}=b*{{a}^{-1}} \text{ for } a \in G, {{a}^{-1}}\in G \\ \Rightarrow \,
- Exercise 1.2 (Solutions) @fsc-part1-ptb:sol:ch01
- -iy$, where $x,y\in \mathbb{R}$ \begin{align} \text{Sum} &=z+\overline{z}\\ &=x+iy+x-iy \\ &=2x \in \mathbb{R}, \text{ as } x,y \in \mathbb{R}. \end{align} Now \begin{align} \text{Product}&=z\cdot \overline{z} \\ &=\left( x+iy ... }^{2}} \\ &={{x}^{2}}+{{y}^{2}} \in \mathbb{R}, \text{ as } x, y \in \mathbb{R}. \end{align} Hence, we