1. A. Tick the correct option.
(a) The additive inverse of \( \frac{-3}{4} \) is- (i) \( \frac{-3}{4} \) (ii) \( \frac{-4}{3} \) (iii) \( \frac{4}{3} \) (iv) \( \frac{3}{4} \)
Answer \( \color{red}\frac{3}{4} \)
\[ \begin{align*} & = \frac{-3}{4} + \frac{3}{4} \\ \\ & = \frac{-3 + 3}{4} \\ \\ & = \frac{0}{4} \\ \\ & = 0 \\ \\ \end{align*} \]
(b) If x, y, and z are rational numbers,then the property \( (x + y) +z = x + (y + z)\) (i) commutative property (ii) associative property (iii) distributive property (iv) closure property
Answer \(\color{red} Associative \,\, property \)
(c) \( \frac{7}{12} \div \left(\frac{-7}{12} \right) \) is (i) 1 (ii) 7 (iii) -1 (iv) -7
Answer \(\color{red} -1 \)
\[ \begin{align*} & = \frac{7}{12} \div \left(\frac{-7}{12} \right) \\ \\ & = \frac{\cancelto{1}{7}}{\cancelto{1}{12}} \times \frac{-\cancelto{1}{12}}{\cancelto{1}{7}} \\ \\ & = -1 \end{align*} \]
(d) Identity element for subtraction of rational numbers is- (i) 1 (ii) 0 (iii) -1 (iv) does not exist
Answer \(\color{red} Does \, not \, exist \)
(e) The multiplicative inverse of \( 6\frac{1}{3} \) is - (i) \( \frac{-19}{3} \) (ii) \( \frac{-3}{19} \) (iii)\( \frac{3}{19} \) (iv) \( \frac{19}{3} \)
Answer \(\color{red}\frac{3}{19} \)
\[ \begin{align*} & =\frac{\cancelto{1}{19}}{\cancelto{1}{3}} \times \frac{\cancelto{1}{3}}{\cancelto{1}{19}} \\ \\ & = 1 \end{align*} \]
B. Answer the following questions.
(a) Write all rational numbers whose absolute vaule is \( \frac{5}{9} \)
\[ \begin{align*} \left| \frac{5}{9} \right| &= \frac{5}{9} \\ \\ \left| \frac{-5}{9} \right| &= \frac{5}{9} \\ \\ \end{align*} \]
Answer \(\color{red} \frac{5}{9} , \frac{-5}{9} \)
(b) Find the reciprocal of \( \frac{4}{5} \times \left( \frac{3}{-8} \right) \)
\[ \begin{align*} & = \frac{\cancelto{1}{4}}{5} \times \frac{-3}{\cancelto{2}{8}} \\ \\ &= \frac{-3}{10} \\\\ & \text{Reciprocal of } \frac{-3}{10} \text{ is } \frac{-10}{3} \end{align*} \]
Answer \(\color{red} \frac{-10}{3} \)
(c) What should be added to \( \frac{-5}{11} \) to get \( \frac{26}{33} \)?
\[ \begin{align*} x + \frac{-5}{11} &= \frac{26}{33} \\ \\ x - \frac{5}{11} &= \frac{26}{33} \\ \\ x &= \frac{26}{33} + \frac{5}{11} \\ \\ x &= \frac{26 + 15}{33} \\ \\ x &= \frac{41}{33} \\ \\ \end{align*} \]
Answer \(\color{red} \frac{41}{33} \)
(d) Subtract \( 6\frac{2}{3} \) from the sum of \( \frac{-3}{7} \) and 2.
\[ \begin{align*} & \color{Magenta} Method - 1 \\ \\ & \color{green} \text{Sum of } \frac{-3}{7} \text{ and } 2 \\ \\ & = \frac{-3}{7} + \frac{2}{1} \\ \\ & = \frac{-3 + 14}{7} \\ \\ & = \frac{11}{7} \\ \\ & \color{green} \text{Subtract } 6\frac{2}{3} \text{ from } \frac{11}{7} \\ \\ & = \frac{11}{7} - 6\frac{2}{3} \\ \\ & = \frac{11}{7} - \frac{20}{3} \\ \\ & = \frac{33 - 140}{21} \\ \\ & = \boxed{\frac{-107}{21}} \\ \\& \color{Magenta} Method - 2 \\ \\ & = \left(\frac{-3}{7} + \frac{2}{1} \right) - 6\frac{2}{3} \\ \\ & = \frac{-3}{7} + \frac{2}{1} - \frac{20}{3} \\ \\ & = \frac{-9 + 42 -140}{21} \\ \\ & = \frac{33 -140}{21} \\ \\ & = \boxed{\frac{-107}{21}} \\ \\ \end{align*} \]
Answer \(\color{red} \frac{-107}{21} \)
(e) Find the value of \( \color{green}1 + \frac{1}{1 + \frac{1}{6}} \)
Answer \(\color{red} \frac{13}{7} \)
2. State whether the following statements are true. If not, then give an example in support of your answer.
(i) If \( |x| = 0\), then x has no reciprocal
Answer True
\[ \begin{array}{c|c}\end{array} \]
(ii) If \( x < y \) then \( | x | < | y | \)
Answer False
\[ \begin{align*} \text{Let} \\ x = -2 \quad &, \quad y = 1 \\ \end{align*} \] \[ \begin{array}{c|c} -2 < 1 & 2 > 1 \\ \\ x < y \quad & \quad | x | > | y | \\ \\ \end{array} \]
(iii) If \( x < y \) then \( x^{-1} < y^{-1} \)
Answer False
\[ \begin{align*} \text{Let} \\ x = 2 \quad &, \quad y = 3 \\ \\ x^{-1} = \frac{1}{2} \quad &, \quad y^{-1} = \frac{1}{3} \\ \end{align*} \] \[ \begin{array}{c|c} \begin{aligned} & 2 < 3 \\ \\ & x < y \\ \\ \end{aligned} & \begin{aligned} & \frac{1}{2} > \frac{1}{3} \\ \\ & x^{-1} > y^{-1} \\ \\ \end{aligned} \end{array} \]
(iv) The negative of a negative rational number is a positive rational number.
Answer True
\[ \begin{align*} &= - \left(\frac{-1}{2} \right)\\ \\ &= \frac{1}{2} \\ \\ \end{align*} \]
(v) Product of two rational numbers can never be an integer.
Answer False
\[ \begin{align*} \text{Let} \\ x = \frac{4}{5} \quad &, \quad y = \frac{10}{4} \\ \\ & = x \times y \\ \\ & = \frac{\cancelto{1}{4}}{\cancelto{1}{5}} \times \frac{\cancelto{2}{10}}{\cancelto{1}{4}} \\ \\ &= \frac{2}{1} \\ \\ & = 2 \,\text{ (Integer)} \end{align*} \]
(vi) Product of two integers is never a fraction.
Answer True
\[ \begin{align*} \text{Let} \\ x = -2 \quad &, \quad y = 3 \\ \\ & = x \times y \\ \\ & = -2 \times 3 \\ \\ & = -6 \,\text{ (Integer)} \end{align*} \]
(vii) If x and y are two rational numbers such that \( x > y \), then \( x - y \) is always a positive rational number.
Answer True
\[ \begin{align*} &\text{Let} \\ &x = \frac{5}{1} \, , \, y = \frac{3}{1} \\ \\ & = x - y \\ \\ & = \frac{5}{1} - \frac{3}{1} \\ \\ & = \frac{2}{1} \,\text{ (Positive rational number)} \end{align*} \]
3. For \( x = \frac{3}{4} \) and \( y = \frac{-9}{8} \) insert a rational number between
(i) \( (x + y)^{-1} \) and \( x^{-1} + y^{-1} \)
Solution
\[ \begin{array}{c|c} \begin{aligned} &\color{green} (x + y)^{-1} \\ \\ & = \left(\frac{3}{4} + \frac{-9}{8} \right)^{-1} \\ \\ & = \left(\frac{6 -9}{8} \right)^{-1} \\ \\ & = \left(\frac{-3}{8} \right)^{-1} \\ \\ & = \frac{8}{-3} \\ \\ & = \boxed{\color{red}\frac{-8}{3}} \\ \\ \end{aligned} & \begin{aligned} &\color{green} x^{-1} + y^{-1} \\ \\ & = \left(\frac{3}{4} \right)^{-1} + \left(\frac{-9}{8} \right)^{-1}\\ \\ & = \frac{4}{3} + \frac{8}{-9} \\ \\ & = \frac{4}{3} + \frac{-8}{9} \\ \\ & = \frac{12 -8}{9} \\ \\ &= \boxed{\color{red}\frac{4}{9}} \\ \\ \end{aligned} \end{array} \] \[ \begin{align*} & \color{green}\text{Rational number between } \frac{-8}{3} \text{ and } \frac{4}{9} \\ \\ & = \frac{1}{2} \times \left(\frac{-8}{3} + \frac{4}{9} \right) \\ \\ & = \frac{1}{2} \times \left(\frac{-24 + 4}{9} \right) \\ \\ & = \frac{1}{\cancelto{1}{2}} \times \frac{-\cancelto{10}{20}}{9} \\ \\ & = \boxed{\color{red}\frac{-10}{9}} \\ \\ \end{align*} \]
Answer\( \boxed{\color{red}\frac{-10}{9}} \)
(ii) \( (x - y)^{-1} \) and \( x^{-1} - y^{-1} \)
Solution
\[ \begin{array}{c|c} \begin{aligned} &\color{green} (x - y)^{-1} \\ \\ & = \left(\frac{3}{4} - \frac{-9}{8} \right)^{-1} \\ \\ & = \left(\frac{6 + 9}{8} \right)^{-1} \\ \\ & = \left(\frac{15}{8} \right)^{-1} \\ \\ & = \boxed{\color{red}\frac{8}{15}} \\ \\ \end{aligned} & \begin{aligned} &\color{green} x^{-1} - y^{-1} \\ \\ & = \left(\frac{3}{4} \right)^{-1} - \left(\frac{-9}{8} \right)^{-1}\\ \\ & = \frac{4}{3} - \frac{-8}{9} \\ \\ & = \frac{12 + 8}{9} \\ \\ & = \boxed{\color{red}\frac{20}{9}} \\ \\ \end{aligned} \end{array} \] \[ \begin{align*} & \color{green}\text{Rational number between } \frac{8}{15} \text{ and } \frac{20}{9} \\ \\ & = \frac{1}{2} \times \left(\frac{8}{15} + \frac{20}{9} \right) \\ \\ & = \frac{1}{2} \times \left(\frac{24 + 100}{45} \right) \\ \\ & = \frac{1}{\cancelto{1}{2}} \times \frac{\cancelto{62}{124}}{45} \\ \\ & = \boxed{\color{red}\frac{62}{45}} \\ \\ \end{align*} \]
Answer\( \boxed{\color{red}\frac{62}{45}} \)
4. Verify that \( \color{black}(x \div y)^{-1} = x^{-1} \div y^{-1} \) by taking \( \color{black}x = \frac{-5}{11} , y =\frac{7}{3} \)
Answer
\[ \begin{align*} \text{LHS} &= \color{green} (x \div y)^{-1}\\ \\ & = \left( \frac{-5}{11} \div \frac{7}{3} \right)^{-1} \\ \\ & = \left( \frac{-5}{11} \times \frac{3}{7} \right)^{-1} \\ \\ & = \left( \frac{-15}{77} \right)^{-1} \\ \\ & = \boxed{\color{red}\frac{-77}{15}} \\ \\ \text{RHS} &= \color{green} x^{-1} \div y^{-1} \\ \\ & = \left( \frac{-5}{11} \right)^{-1} \div \left( \frac{7}{3} \right)^{-1} \\ \\ & = \frac{-11}{5} \div \frac{3}{7} \\ \\ & = \frac{-11}{5} \times \frac{7}{3} \\ \\ & = \boxed{\color{red}\frac{-77}{15}} \\ \\ \end{align*} \] \[ \begin{aligned} \text{LHS} &= \text{RHS} \\ \color{red} (x \div y)^{-1} &= \color{red}x^{-1} \div y^{-1} \\ \text{Hence } & \text{ Verified} \end{aligned} \]
5. Verify that \( \color{black} | x + y| \le |x| + |y| \) by taking \( \color{black}x = \frac{2}{3} , y =\frac{-3}{5} \)
Answer
\[ \begin{array}{c|c} \begin{aligned} &\text{LHS} = \color{green} | x + y| \\ \\ & = \left| \frac{2}{3} + \frac{-3}{5} \right| \\ \\ & = \left| \frac{10 - 9}{15} \right| \\ \\ & = \left| \frac{1}{15} \right| \\ \\ & = \boxed{\color{red}\frac{1}{15}} \\ \\ \end{aligned} & \begin{aligned} &\text{RHS} = \color{green} |x| + |y| \\ \\ & = \left| \frac{2}{3} \right| + \left| \frac{-3}{5} \right| \\ \\ & = \frac{2}{3} + \frac{3}{5} \\ \\ & = \frac{10 + 9}{15} \\ \\ & = \boxed{\color{red}\frac{19}{15}} \\ \\ \end{aligned} \end{array} \] \[ \begin{aligned} \text{LHS} \quad & {<} \quad \text{RHS} \\ \color{red} | x + y| \quad & \le \quad \color{red} |x| + |y| \\ \text{Hence } & \text{ Verified} \end{aligned} \]
6. Find the reciprocals of:
(i) \( \frac{2}{-5} \times \frac{3}{-7} \)
Answer
\[ \begin{aligned} & = \frac{2}{-5} \times \frac{3}{-7}\\ \\ & = \frac{6}{35} \\ \\ &\text{Reciprocal of } \frac{6}{35} \text{ is } \boxed{\color{red} \frac{35}{6}} \end{aligned} \]
(ii) \( \left( \frac{-6+5}{9} \right) \div \frac{(-1)}{6} \)
Answer
\[ \begin{aligned} & = \left( \frac{-6+5}{9} \right) \div \frac{(-1)}{6} \\ \\ & = \frac{-1}{9} \div \frac{(-1)}{6} \\ \\ & = \frac{-1}{\cancelto{3}{9}} \times \frac{(-\cancelto{2}{6})}{1} \\ \\ & = \frac{-1}{3} \times \frac{-2}{1} \\ \\ & = \frac{2}{3} \\ \\ &\text{Reciprocal of } \frac{2}{3} \text{ is } \boxed{\color{red} \frac{3}{2}} \end{aligned} \]
7. Simplify
(i) \( \left| \frac{5}{7} - \frac{2}{3} \right| + \left| \frac{3}{14} - \frac{5}{7} \right| \)
Answer
\[ \begin{aligned} & = \left| \frac{5}{7} - \frac{2}{3} \right| + \left| \frac{3}{14} - \frac{5}{7} \right| \\ \\ & = \left| \frac{15-14}{21} \right| + \left| \frac{3 - 10}{14} \right| \\ \\ & = \left| \frac{1}{21} \right| + \left| \frac{-\cancelto{1}{7}}{\cancelto{2}{14}} \right| \\ \\ & = \frac{1}{21} + \frac{1}{2} \\ \\ & = \frac{2 + 21}{42} \\ \\ & = \boxed{\color{red} \frac{23}{42}} \end{aligned} \]
(ii) \( \left( \frac{5}{11} \right)^{-1} - \frac{13}{5} + \frac{3}{15} \)
Answer
\[ \begin{aligned} & = \left( \frac{5}{11} \right)^{-1} - \frac{13}{5} + \frac{3}{15} \\ \\ & = \frac{11}{5} - \frac{13}{5} + \frac{3}{15} \\ \\ & = \frac{33 -39+3}{15} \\ \\ & = \frac{-6 + 3}{15} \\ \\ & = \frac{-\cancelto{1}{3}}{\cancelto{5}{15}} \\ \\ & = \boxed{\color{red} \frac{-1}{5}} \end{aligned} \]
(iii) \( \frac{9}{5} \times \frac{-2}{27} + \frac{7}{30} \)
Answer
\[ \begin{aligned} & = \frac{\cancelto{1}{9}}{5} \times \frac{-2}{\cancelto{3}{27}} + \frac{7}{30} \\ \\ & = \frac{-2}{15} + \frac{7}{30} \\ \\ & = \frac{-4 + 7}{30} \\ \\ & = \frac{\cancelto{1}{3}}{\cancelto{10}{30}} \\ \\ & = \boxed{\color{red} \frac{1}{10}} \end{aligned} \]
(iv) \( \frac{-7}{15} \div \left( \frac{50}{3} \right)^{-1} \)
Answer
\[ \begin{aligned} & = \frac{-7}{15} \div \left( \frac{50}{3} \right)^{-1} \\ \\ & = \frac{-7}{15} \div \frac{3}{50} \\ \\ & = \frac{-7}{\cancelto{3}{15}} \times \frac{\cancelto{10}{50}}{3} \\ \\ & = \boxed{\color{red}\frac{-70}{9}} \\ \\ \end{aligned} \]
8. Divide
(i) The sum of \( \frac{5}{21} \) and \( \frac{4}{7} \) by thier difference.
Answer
\[ \begin{aligned} & = \left( \frac{5}{21} + \frac{4}{7} \right) \div \left( \frac{5}{21} - \frac{4}{7} \right) \\ \\ & = \left( \frac{5 + 12}{21} \right) \div \left( \frac{5 - 12}{21} \right) \\ \\ & = \frac{17}{21} \div \frac{-7}{21} \\ \\ & = \frac{17}{\cancelto{1}{21}} \times \frac{-\cancelto{1}{21}}{7} \\ \\ & = \boxed{\color{red} \frac{-17}{7}} \end{aligned} \]
(ii) The difference of \( \frac{12}{5} , \frac{-16}{20} \) by thier product.
Answer
\[ \begin{aligned} & = \left( \frac{12}{5} - \frac{-16}{20} \right) \div \left( \frac{12}{5} \times \frac{-16}{20} \right) \\ \\ & = \left( \frac{48 + 16}{20} \right) \div \left( \frac{12}{5} \times \frac{-\cancelto{4}{16}}{\cancelto{5}{20}} \right) \\ \\ & = \left( \frac{\cancelto{16}{64}}{\cancelto{5}{20}} \right) \div \left( \frac{-48}{25} \right) \\ \\ & = \frac{16}{5} \div \frac{-48}{25} \\ \\ & = \frac{\color{green}\cancelto{1}{16}}{\cancelto{1}{5}} \times \frac{-\cancelto{5}{25}}{\color{green}\cancelto{3}{48}} \\ \\ & = \boxed{\color{red} \frac{-5}{3}} \end{aligned} \]
9. Find reciprocal of \( \frac{-2}{3} \times \frac{5}{7} +\frac{2}{9} \div \frac{1}{3} \times \frac{6}{7} \)
Answer
\[ \begin{aligned} & = \frac{-2}{3} \times \frac{5}{7} + \left(\frac{2}{9} \div \frac{1}{3} \right) \times \frac{6}{7} \\ \\ & = \frac{-2}{3} \times \frac{5}{7} + \left(\frac{2}{\cancelto{3}{9}} \times \frac{\cancelto{1}{3}}{1} \right) \times \frac{6}{7} \\ \\ & = \left( \frac{-2}{3} \times \frac{5}{7} \right) + \left(\frac{2}{\cancelto{1}{3}} \times \frac{\cancelto{2}{6}}{7} \right) \\ \\ & = \frac{-10}{21} +\frac{4}{7} \\ \\ & = \frac{-10 +12}{21} \\ \\ & = \boxed{ \frac{2}{21}} \\ \\ &\text{Reciprocal of } \frac{2}{21} \text{ is } \boxed{\color{red} \frac{21}{2}}\end{aligned} \]