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Exercise 2.8 (Solutions) @fsc-part1-ptb:sol:ch02
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cative inverse of $0$ in not in set $\mathbb{Q}$. Therefore the set of rational number is not a group w.r... $-a\in \mathbb{Q}$ such that $a+(-a)=(-a)+a=0$. Therefore the set of rational number is group under add... 3,...\}$. a- Since sum of integers is an integer therefore for $a,b\in \mathbb{Z}$, $a+b\in \mathbb{Z}$.... e table is symmetric about the principal diagonal therefore $\oplus$ is commutative. Hence $\left\{ E,O
Exercise 1.2 (Solutions) @fsc-part1-ptb:sol:ch01
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ray}{l} \mbox{For each} \; z\in \mathbb{C}, \mbox{there exists}\; -z\in \mathbb{C} \mbox{ such that}\; z+