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Exercise 2.8 (Solutions) @fsc-part1-ptb:sol:ch02
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n}&{{a}^{-1}}*\left( a*x \right)={{a}^{-1}}*b \\ \Rightarrow \,\,\,\,& \left( {{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 identit... .}\end{align} And for \begin{align} & x*a=b \\ \Rightarrow \,\,\,\,& \left( x*a \right)*{{a}^{-1}}=b*{{a}^{-
Exercise 1.1 (Solutions) @fsc-part1-ptb:sol:ch01
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d{array} \] As $(-1)+(-1)=-2 \notin \{0,-1\}$. $\Rightarrow \{0,-1\}$ does not satisfy closure property w.r.t... y} \] As $(-1)\times (-1)= 1 \notin \{0,-1\}$. $\Rightarrow \{0,-1\}$ does not have closure property w.r.t. '... ine \end{array} \] As $1+1=2 \notin \{1,-1\}$. $\Rightarrow \{1,-1\}$ does not closure property w.r.t. '+'. ... he entries of the table belongs to $\{1,-1\}$. $\Rightarrow \{1,-1\}$ has closure property w.r.t. '$\times$'.