1. Find the reciprocals of:
(i) \( \frac{-1}{5} \)
Answer
\[ \begin{align*} \text{Reciprocal of } \frac{-1}{5} \text{ is } \color{red} \boxed{ -5} \end{align*} \]
(ii) \( 4 \)
Answer
\[ \begin{align*} \text{Reciprocal of } 4 \text{ is } \color{red} \boxed{\frac{1}{4}} \end{align*} \]
(iii) \( \frac{11}{-12} \)
Answer
\[ \begin{align*} \text{Reciprocal of } \frac{11}{-12} \text{ is } \color{red} \boxed{\frac{-12}{11}} \end{align*} \]
(iv) \( \frac{-2}{-19} \)
Answer
\[ \begin{align*} \text{Reciprocal of } \frac{-2}{-19} \text{ is } \color{red} \boxed{ \frac{-19}{-2} } \end{align*} \]
2. Check, if the reciprocal of \( \frac{-2}{3} \) is \( \frac{3}{2} \)
Answer
\[ \begin{aligned} \frac{-\cancelto{1}{2}}{\cancelto{1}{3}} \times \frac{\cancelto{1}{3}}{\cancelto{1}{2}} &= -1 \\ -1 &\neq 1 \\ \\ \text{No, } \frac{-2}{3} & \text{ is not the reciprocal of } \frac{3}{2} \end{aligned} \]
3. Check, if the reciprocal of \( \frac{-1}{5} \) is \( -5 \)
Answer
\[ \begin{aligned} & = \frac{-1}{\cancelto{1}{5}} \times \frac{-\cancelto{1}{5}}{1} \\ & = -1 \times -1 \\ & = 1 \\ \text{Yes, } \frac{-1}{5} & \text{ is the reciprocal of } -5 \end{aligned} \]
4. Verify that \( \color{red}(x - y)^{-1} \neq x^{-1} - y^{-1} \) by taking \( x = \frac{-2}{7} , y= \frac{4}{7} \)
Answer
\[ \begin{aligned} \text{LHS}& = \color{green}(x -y)^{-1} \\ \\ & = \left(\frac{-2}{7} - \frac{4}{7} \right)^{-1}\\ \\ & = \left(\frac{-2 - 4}{7} \right)^{-1}\\ \\ & = \left(\frac{-6}{7} \right)^{-1}\\ \\ & = \boxed {\frac{-7}{6}} \\ \\ \text{RHS} & = \color{green} x^{-1} - y^{-1} \\ \\ & = \left(\frac{-2}{7} \right)^{-1} - \left(\frac{4}{7} \right)^{-1} \\ \\ & = \frac{-7}{2} - \frac{7}{4} \\ \\ & = \frac{-14 - 7}{4} \\ \\ & = \boxed{\frac{-21}{4}} \\ \\ \text{LHS} & \neq \text{RHS} \\ (x -y)^{-1} & \neq x^{-1} - y^{-1}\\ \text{Hence } & \text{ Verified} \end{aligned} \]
5. Verify that \( \color{red}(x + y)^{-1} \neq x^{-1} + y^{-1} \) by taking \( x = \frac{5}{9} \) and \( y = \frac{-4}{3} \)
Answer
\[ \begin{aligned} \text{LHS}& = \color{green}(x + y)^{-1} \\ \\ & = \left(\frac{5}{9} + \frac{-4}{3}\right)^{-1}\\ \\ & = \left(\frac{5-12}{9}\right)^{-1}\\ \\ & = \left(\frac{-7}{9}\right)^{-1}\\ \\ & = \boxed{\frac{-9}{7}} \\ \\ \text{RHS} & = \color{green} x^{-1} + y^{-1} \\ \\ & = \left(\frac{5}{9}\right)^{-1} + \left(\frac{-4}{3}\right)^{-1} \\ \\ & = \frac{9}{5} + \frac{-3}{4} \\ \\ & = \frac{36 -15}{20} \\ \\ & = \boxed{\frac{21}{20}} \\ \\ \text{LHS} & \neq \text{RHS} \\ (x + y)^{-1} & \neq x^{-1} + y^{-1} \\ \text{Hence } & \text{ Verified} \end{aligned} \]
6. Verify that \( \color{red}(x \times y)^{-1} = x^{-1} \times y^{-1} \) by taking \( x = \frac{-2}{3} \) and \( y = \frac{-3}{4} \)
Answer
\[ \begin{aligned} \text{LHS}& = \color{green}(x \times y)^{-1} \\ \\ & = \left(\frac{-\cancelto{1}{2}}{\cancelto{1}{3}} \times \frac{-\cancelto{1}{3}}{\cancelto{2}{4}}\right)^{-1}\\ \\ & = \left(\frac{-1 \times -1}{1 \times 2}\right)^{-1}\\ \\ & = \left(\frac{1}{2}\right)^{-1}\\ \\ & = \frac{2}{1}\\ \\ & = \boxed{2} \\ \\ \text{RHS} & = \color{green}x^{-1} \times y^{-1} \\ \\ & = \left(\frac{-2}{3}\right)^{-1} \times \left(\frac{-3}{4}\right)^{-1} \\ \\ & = \frac{-\cancelto{1}{3}}{\cancelto{1}{2}} \times \frac{-\cancelto{2}{4}}{\cancelto{1}{3}} \\ \\ & = \frac{-1 \times -2}{1 \times 1} \\ \\ & = \frac{2}{1}\\ \\ & = \boxed{2} \\ \\ \text{LHS} & = \text{RHS} \\ (x \times y)^{-1} & = x^{-1} \times y^{-1}\\ \text{Hence } & \text{ Verified} \end{aligned} \]
7. Fill in the blanks
(i) The number \( 0 \) has no reciprocal.
(ii) \( 1 \) and \( -1 \) are their own reciprocals.
(iii) If a is the reciprocal of b, then b is the reciprocal of a
(iv) \( (11 \times 5)^{-1} = (11)^{-1} \times \) \( (5)^{-1} \)
(v) \( \frac{-1}{8} \times \) \( -8 \) = 1
(vi) \( \frac{-3}{16} \) \(\times \left(-5\frac{1}{3}\right) = 1 \)
\[ \begin{aligned} & = -5\frac{1}{3} \\ \\ & = \frac{-16}{3} \\ \\ &\text{Reciprocal of } \frac{-16}{3} \text{ is } \boxed{\frac{-3}{16}} \\ \\ &= \frac{-\cancelto{1}{3}}{\cancelto{1}{16}} \times \frac{-\cancelto{1}{16}}{\cancelto{1}{3}} \\ \\ & = 1 \end{aligned} \]