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- Exercise 2.8 (Solutions) @fsc-part1-ptb:sol:ch02
- s: (i) Name the identity element if it exists? (ii) What is the inverse of 1? (iii) Is the set $G$, under the given operation a group? Abelian or non-... $. This show that $0$ is the identity element. (ii) Since $1+1=0$ (identity element) so the inverse of $1$ is $1$. (iii) It is clear from table that element of the give
- Exercise 1.2 (Solutions) @fsc-part1-ptb:sol:ch01
- (i^2)^4 \cdot i$ $=1 \cdot i$ =$i$. **Question 4(ii)** Simplify: $i^{14}$ **Solutions** $i^{14}$ $=(i^2)^7 $ $=(-1)^7 $ $=-1$. **Question 4(iii)** Simplify: ${-i}^{19}$ **Solutions** \begin... *Solutions** $\sqrt{-1}b= bi$ ---- **Question 5(ii)** Write $\sqrt{-5}$ in term of $i$ **Solution... \sqrt{-1}\cdot \sqrt{5} =\sqrt{5}i$ **Question 5(iii)** Write $\sqrt{-\frac{16}{25}}$ in term of $i$
- Exercise 1.1 (Solutions) @fsc-part1-ptb:sol:ch01
- closure property w.r.t. '$\times$'. **Question 1(ii)** Is the set $\{1\}$ has closure property w.r.t... closure property w.r.t. '$\times$' **Question 1(iii)** Is the set $\{0,-1\}$ has closure property w... roperty:** Commutative property w.r.t. '+'. ---- (ii) $(a+1)+ \frac{3}{4}= a+(1+\frac{3}{4})$ **Property:** Associative property w.r.t. '+'. ---- (iii) $(\sqrt{3}+\sqrt{5})+\sqrt{7}= \sqrt{3}(\sqrt{5