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Exercise 1.2 (Solutions) @fsc-part1-ptb:sol:ch01

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iv (3,7)$ **Solutions** $\quad (2,6)\div (3,7)=\dfrac{(2,6)}{\left( 3,7 \right)}$ $=\dfrac{2+6i}{3+7i}=\dfrac{2+6i}{3+7i}\times \dfrac{3-7i}{3-7i}$ $=\dfrac{6-14i+18i-42{{i}^{2}}}{{{\left( 3 \right)}^{2}}-{{\left(

Exercise 1.1 (Solutions) @fsc-part1-ptb:sol:ch01

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step: $$\frac{4+16x}{4}$$ **Solution** $\quad \dfrac{4+16x}{4}$ $=\dfrac{1}{4}\times (4+16x)\quad \because\dfrac{a}{b}=\dfrac{1}{b}\times a$ $=\dfrac{1}{4} \times (4\times 1+4 \times 4x)\quad \text{(multiplicative id

Exercise 2.8 (Solutions) @fsc-part1-ptb:sol:ch02

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ity element. d- For $a\in {{\mathbb{Q}}^{+}}$, $\dfrac{1}{a}\in {{\mathbb{Q}}^{+}}$ such that $a\times \dfrac{1}{a}=\dfrac{1}{a}\times a=1$. Thus inverse of $a$ is $\dfrac{1}{a}$. Hence ${{\mathbb{Q}}^{+}}$ is group under addi