Search
You can find the results of your search below.
Fulltext results:
- Exercise 2.8 (Solutions) @fsc-part1-ptb:sol:ch02
- $G$ is group so $a^{-1}\in G$ such that \begin{align}&{{a}^{-1}}*\left( a*x \right)={{a}^{-1}}*b \\ \R... ={{a}^{-1}}*b\,\,\,\, \text{by identity law.}\end{align} And for \begin{align} & x*a=b \\ \Rightarrow \,\,\,\,& \left( x*a \right)*{{a}^{-1}}=b*{{a}^{-1}} \te... =b*{{a}^{-1}} \,\,\,\,\text{by identity law.}\end{align} </panel> <panel> **Question # 7** Show that th
- Exercise 1.1 (Solutions) @fsc-part1-ptb:sol:ch01
- rac{b}{c} = \frac{a+b}{c}$$. **Solution** \begin{align} L.H.S &= \frac{a}{c}+\frac{b}{c}\\ &= a \times... ibutive property)}\\ &= \frac{a+b}{c}=R.H.S \end{align} **Question 4(ii)** Prove the following rules o... ac{c}{d}=\frac{ad+bc}{bd}$$. **Solution** \begin{align} L.H.S &= \frac{a}{b}+\frac{c}{d}\\ &= \frac{a}... s \frac{1}{bd} \\ &= \frac{ad+bc}{bd}=R.H.S \end{align} </panel> <panel> **Question 5** Prove that $
- Exercise 1.2 (Solutions) @fsc-part1-ptb:sol:ch01
- )** Simplify: ${-i}^{19}$ **Solutions** \begin{align} {-i}^{19}& =[(-1)(i)] ^{19}=(-1)^{19}\cdot i^{1... 2)^{9} \cdot i \\ & =-(-1)^{9} i=-(-1)i =i. \end{align} **Question 4(iv)** Simplify: $\displaystyle {{(-1)}^{-\frac{21}{2}}}$ **Solution** \begin{align} (-1)^{-\frac{21}{2}}&=\frac{1}{(-1)^\frac{21}{2... ac{i}{i}\\ &=\frac{i}{i^2}=\frac{i}{-1}=-i. \end{align} </panel> <panel> **Question 5(i)** Write $\sq