1. Add the following:
(i) \( \frac{2}{7} + \frac{5}{7} \)
Solution
\[ \begin{align*} &= \frac{2 + 5}{7}\\ \\ &= \frac{7}{7}\\ \\ &= 1 \end{align*} \]
Answer 1
(ii) \( \frac{-5}{9} + \frac{7}{9} \)
Solution
\[ \begin{align*} &= \frac{-5 + 7}{9}\\ \\ &= \frac{-5+7}{9}\\ \\ &= \frac{2}{9}\\ \\ \end{align*} \]
Answer \( \color{red} \frac{2}{9} \)
(iii) \( 2\frac{3}{11} + \frac{9}{-11} \)
Solution
\[ \begin{align*} &= \frac{25}{11} + \frac{(-9)}{11}\\ \\ &= \frac{25 + (-9)}{11}\\ \\ &= \frac{25 - 9}{11}\\ \\ &= \frac{16}{11}\\ \\ \end{align*} \]
Answer \( \color{red} \frac{16}{11} \)
(iv) \( \frac{-5}{4} + \frac{5}{4} \)
Solution
\[ \begin{align*} &= \frac{-5 + 5}{4}\\ \\ &= \frac{0}{4}\\ \\ &= 0 \end{align*} \]
Answer 0
(v) \( -2\frac{5}{6} + \frac{13}{-6} \)
Solution
\[ \begin{align*} &= \frac{-17}{6} + \frac{-13}{6}\\ \\ &= \frac{-17 + (-13)}{6}\\ \\ &= \frac{-30}{6}\\ \\ &= -5 \end{align*} \]
Answer \( \color{red} -5 \)
(vi) \( \frac{-21}{3} + \frac{18}{3} \)
Solution
\[ \begin{align*} & \boxed{\text{Method - } 1}\\ \\ &= \frac{-21 + 18}{3}\\ \\ &= \frac{-3}{3}\\ \\ &= -1 \\ \\ & \boxed{\text{Method - } 2}\\ \\ &= \frac{-\cancelto{7}{21}}{\cancelto{1}3} + \frac{\cancelto{6}{18}}{\cancelto{1}3}\\ \\ &= -7 + 6\\ \\ &= -1 \end{align*} \]
Answer \( \color{red} -1 \)
2. Add the following:
(i) \( \frac{4}{9} + \frac{7}{4} \)
Solution
\[ \begin{align*} &= \frac{4}{9} + \frac{7}{4} \\ \\ \text{LCM} & = 36 \\ \\ &= \frac{ (4 \times 4) + (7 \times 9)}{36}\\ \\ &= \frac{ 16 + 63}{36}\\ \\ &= \frac{ 79 }{36}\\ \\ \end{align*} \]
Answer \( \color{red} \frac{ 79 }{36} \)
(ii) \( \frac{-7}{11} + \frac{1}{4} \)
Solution
\[ \begin{align*} &= \frac{-7}{11} + \frac{1}{4} \\ \\ \text{LCM} & = 44 \\ \\ &= \frac{ (-7 \times 4) + (1 \times 11)}{44}\\ \\ &= \frac{ -28 + 11}{44}\\ \\ &= \frac{ -17 }{44}\\ \\ \end{align*} \]
Answer \( \color{red} \frac{ -17 }{44} \)
(iii) \( \frac{5}{8} + \frac{-3}{5} \)
Solution
\[ \begin{align*} &= \frac{5}{8} + \frac{-3}{5} \\ \\ \text{LCM} & = 40 \\ \\ &= \frac{ (5 \times 5) + (-3 \times 8)}{40}\\ \\ &= \frac{ 25 + (-24)}{40}\\ \\ &= \frac{ 25 - 24}{40}\\ \\ &= \frac{ 1 }{40}\\ \\ \end{align*} \]
Answer \( \color{red} \frac{ 1 }{40} \)
(iv) \( \frac{10}{63} + \frac{6}{7} \)
Solution
\[ \begin{align*} &= \frac{10}{63} + \frac{6}{7} \\ \\ \text{LCM} & = 63 \\ \\ &= \frac{(10 \times 1) + (6 \times 9)}{63}\\ \\ &= \frac{10 + 54}{63}\\ \\ &= \frac{64}{63}\\ \\ \end{align*} \]
Answer \( \color{red} \frac{64}{63} \)
(v) \( \frac{-7}{64} + \frac{3}{-16} \)
Solution
\[ \begin{align*} &= \frac{-7}{64} + \frac{-3}{16} \\ \\ \text{LCM} & = 64 \\ \\ &= \frac{(-7 \times 1) + (-3 \times 4)}{64}\\ \\ &= \frac{(-7) + (-12)}{64}\\ \\ &= \frac{-7 - 12}{64}\\ \\ &= \frac{-19}{64}\\ \\ \end{align*} \]
Answer \( \color{red} \frac{-19}{64} \)
(vi) \( \frac{5}{12} + \frac{-9}{20} \)
Solution
\[ \begin{align*} &= \frac{5}{12} + \frac{-9}{20} \\ \\ \text{LCM} & = 60 \\ \\ &= \frac{(5 \times 5) + (-9 \times 3)}{60}\\ \\ &= \frac{25 - 27}{60}\\ \\ &= \frac{-2}{60}\\ \\ &= \frac{-1}{30} \\ \\ \end{align*} \]
Answer \( \color{red} \frac{-1}{30} \)
3. Verify \(\color{red} x + y = y + x \) for following values of \( x \text{ and } y \).
(i) \( x=\frac{5}{7} , y = \frac{-3}{2} \)
Answer
\[ \begin{array}{c|c} \begin{aligned} &\text{LHS} \\ & x + y \\ \\ & = \frac{5}{7} + \frac{-3}{2}\\ \\ \text{LCM} & = 14 \\ \\ & = \frac{(5 \times 2) + (-3 \times 7) }{14} \\ \\ & = \frac{10 -21}{14} \\ \\ & = \boxed{\frac{-11}{14}} \end{aligned} & \begin{aligned} &\text{RHS} \\ & y + x \\ \\ & = \frac{-3}{2} + \frac{5}{7} \\ \\ \text{LCM} & = 14 \\ \\ & = \frac{(-3 \times 7) + (5 \times 2) }{14} \\ \\ & = \frac{-21 + 10 }{14} \\ \\ & = \boxed{\frac{-11}{14}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x + y = y + x \end{align*} \]
(ii) \( x=5, y=\frac{3}{2} \)
Answer
\[ \begin{array}{c|c} \begin{aligned} &\text{LHS} \\ & x + y \\ \\ & = \frac{5}{1} + \frac{3}{2}\\ \\ \text{LCM} & = 2 \\ \\ & = \frac{(5 \times 2) + (3 \times 1) }{2} \\ \\ & = \frac{10 + 3}{2} \\ \\ & = \boxed{\frac{13}{2}} \end{aligned} & \begin{aligned} &\text{RHS} \\ & y + x \\ \\ & = \frac{3}{2} + \frac{5}{1} \\ \\ \text{LCM} & = 2 \\ \\ & = \frac{(3 \times 1) + (5 \times 2) }{2} \\ \\ & = \frac{3 + 10}{2} \\ \\ & = \boxed{\frac{13}{2}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x + y = y + x \end{align*} \]
(iii) \( x=\frac{-5}{14}, y=\frac{-1}{21} \)
Answer
\[ \begin{array}{c|c} \begin{aligned} &\text{LHS} \\ & x + y \\ \\ & = \frac{-5}{14} + \frac{-1}{21}\\ \\ \text{LCM} & = 42 \\ \\ & = \frac{(-5 \times 3) + (-1 \times 2) }{42} \\ \\ & = \frac{-15 - 2}{42} \\ \\ & = \frac{-17}{42} \\ \\ & = \boxed{\frac{-17}{42}} \end{aligned} & \begin{aligned} &\text{RHS} \\ & y + x \\ \\ & = \frac{-1}{21} + \frac{-5}{14} \\ \\ \text{LCM} & = 42 \\ \\ & = \frac{(-1 \times 2) + (-5 \times 3)}{42} \\ \\ & = \frac{-2 - 15}{42} \\ \\ & = \frac{-17}{42} \\ \\ & = \boxed{\frac{-17}{42}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x + y = y + x \end{align*} \]
(iv) \( x=8, y=\frac{9}{2} \)
Answer
\[ \begin{array}{c|c} \begin{aligned} &\text{LHS} \\ & x + y \\ \\ & = \frac{-8}{1} + \frac{9}{2}\\ \\ \text{LCM} & = 2 \\ \\ & = \frac{(-8 \times 2) + 9 }{2} \\ \\ & = \frac{-16 + 9}{2} \\ \\ & = \boxed{\frac{-7}{2}} \end{aligned} & \begin{aligned} &\text{RHS} \\ & y + x \\ \\ & = \frac{9}{2} + \frac{-8}{1} \\ \\ \text{LCM} & = 2 \\ \\ & = \frac{9 + (-8 \times 2)}{2} \\ \\ & = \frac{9 - 16}{2} \\ \\ & = \boxed{\frac{-7}{2}} \end{aligned} \end{array} \] \[ \begin{align*} \text{Hence verified} \\ \color{red} x + y = y + x \end{align*} \]
4. Verify \(\color{red} x + (y + z) = (x+y) + z \) for following values of \( x ,y \text{ and } z \).
(i) \( x=\frac{3}{4}, y=\frac{5}{6}, z=\frac{-7}{8} \)
Answer
\[ \begin{align*} \text{LHS} &= x + (y + z) \\ \\ & = \frac{3}{4} + \left(\frac{5}{6} + \frac{-7}{8}\right)\\ \\ &\text{LCM of 6 and 8} = 24 \\ \\ & = \frac{3}{4} + \left(\frac{5 \times 4}{24} + \frac{-7 \times 3}{24}\right)\\ \\ & = \frac{3}{4} + \left(\frac{20}{24} + \frac{-21}{24}\right)\\ \\ & = \frac{3}{4} + \left(\frac{20 - 21}{24}\right)\\ \\ & = \frac{3}{4} + \frac{-1}{24}\\ \\ &\text{LCM} = 24 \\ \\ & = \frac{(3 \times 6) + (-1 \times 1)}{24}\\ \\ & = \frac{18 - 1}{24}\\ \\ & = \boxed{\frac{17}{24}} \\ \\ \text{RHS} &= (x + y) + z \\ \\ & = \left(\frac{3}{4} + \frac{5}{6}\right) + \frac{-7}{8} \\ \\ & \text{LCM of 4 and 6} = 12 \\ \\ & = \left(\frac{3 \times 3}{12} + \frac{5 \times 2}{12}\right) + \frac{-7}{8}\\ \\ & = \left(\frac{9}{12} + \frac{10}{12}\right) + \frac{-7}{8}\\ \\ & = \left(\frac{9 + 10}{12}\right) + \frac{-7}{8}\\ \\ & = \frac{19}{12} + \frac{-7}{8}\\ \\ & \text{LCM} = 24 \\ \\ & = \frac{(19 \times 2) + (-7 \times 3)}{24}\\ \\ & = \frac{38 - 21}{24}\\ \\ & = \boxed{\frac{17}{24}} \\ \\ &\text{Hence verified} \\ &\color{red} x + (y + z) = (x+y) + z \end{align*} \]
(ii) \( x=\frac{2}{3}, y=\frac{-5}{6}, z=\frac{-7}{9} \)
Answer
\[ \begin{align*} \text{LHS} &= x + (y + z) \\ \\ & = \frac{2}{3} + \left(\frac{-5}{6} + \frac{-7}{9}\right)\\ \\ &\text{LCM of 6 and 9} = 18 \\ \\ & = \frac{2}{3} + \left(\frac{-5 \times 3}{18} + \frac{-7 \times 2}{18}\right)\\ \\ & = \frac{2}{3} + \left(\frac{-15}{18} + \frac{-14}{18}\right)\\ \\ & = \frac{2}{3} + \left(\frac{-15 - 14}{18}\right)\\ \\ & = \frac{2}{3} + \frac{-29}{18}\\ \\ &\text{LCM} = 18 \\ \\ & = \frac{(2 \times 6) + (-29 \times 1)}{18}\\ \\ & = \frac{12 - 29}{18}\\ \\ & = \boxed{\frac{-17}{18}} \\ \\ \text{RHS} &= (x + y) + z \\ \\ & = \left(\frac{2}{3} + \frac{-5}{6}\right) + \frac{-7}{9} \\ \\ & \text{LCM of 3 and 6} = 6 \\ \\ & = \left(\frac{2 \times 2}{6} + \frac{-5 \times 1}{6}\right) + \frac{-7}{9}\\ \\ & = \left(\frac{4}{6} + \frac{-5}{6}\right) + \frac{-7}{9}\\ \\ & = \left(\frac{4 - 5}{6}\right) + \frac{-7}{9}\\ \\ & = \frac{-1}{6} + \frac{-7}{9}\\ \\ & \text{LCM} = 18 \\ \\ & = \frac{(-1 \times 3) + (-7 \times 2)}{18}\\ \\ & = \frac{-3 - 14}{18}\\ \\ & = \boxed{\frac{-17}{18}} \\ \\ &\text{Hence verified} \\ &\color{red} x + (y + z) = (x+y) + z \end{align*} \]
(iii) \( x=\frac{3}{5}, y=\frac{-6}{9}, z=\frac{2}{10} \)
Answer
\[ \begin{align*} \text{LHS} &= x + (y + z) \\ \\ & = \frac{3}{5} + \left(\frac{-\cancelto{2}{6}}{\cancelto{3}{9}} + \frac{\cancelto{1}{2}}{\cancelto{5}{10}}\right)\\ \\ & = \frac{3}{5} + \left(\frac{-2}{3} + \frac{1}{5}\right)\\ \\ &\text{LCM of 3 and 5} = 15 \\ \\ & = \frac{3}{5} + \left(\frac{-2 \times 5}{15} + \frac{1 \times 3}{15}\right)\\ \\ & = \frac{3}{5} + \left(\frac{-10}{15} + \frac{3}{15}\right)\\ \\ & = \frac{3}{5} + \left(\frac{-10 + 3}{15}\right)\\ \\ & = \frac{3}{5} + \frac{-7}{15}\\ \\ &\text{LCM} = 15 \\ \\ & = \frac{(3 \times 3) + (-7 \times 1)}{15}\\ \\ & = \frac{9 - 7}{15}\\ \\ & = \boxed{\frac{2}{15}} \\ \\ \text{RHS} &= (x + y) + z \\ \\ & = \left(\frac{3}{5} + \frac{-\cancelto{2}{6}}{\cancelto{3}{9}}\right) + \frac{\cancelto{1}{2}}{\cancelto{5}{10}} \\ \\ & = \left(\frac{3}{5} + \frac{-2}{3}\right) + \frac{1}{5} \\ \\ & \text{LCM of 5 and 3} = 15 \\ \\ & = \left(\frac{3 \times 3}{15} + \frac{-2 \times 5}{15}\right) + \frac{1}{5}\\ \\ & = \left(\frac{9}{15} + \frac{-10}{15}\right) + \frac{1}{5}\\ \\ & = \left(\frac{9 -10}{15}\right) + \frac{1}{5}\\ \\ & = \frac{-1}{15} + \frac{1}{5}\\ \\ &\text{LCM} = 15 \\ \\ & = \frac{(-1 \times 1) + (1 \times 3)}{15}\\ \\ & = \frac{-1 + 3}{15}\\ \\ & = \boxed{\frac{2}{15}} \\ \\ &\text{Hence verified} \\ &\color{red} x + (y + z) = (x+y) + z \end{align*} \]
(iv) \( x=\frac{-3}{5}, y=\frac{-7}{10}, z=\frac{-8}{15} \)
Answer
\[ \begin{align*} \text{LHS} &= x + (y + z) \\ \\ & = \frac{-3}{5} + \left(\frac{-7}{10} + \frac{-8}{15}\right)\\ \\ &\text{LCM of 10 and 15} = 30 \\ \\ & = \frac{-3}{5} + \left(\frac{-7 \times 3}{30} + \frac{-8 \times 2}{30}\right)\\ \\ & = \frac{-3}{5} + \left(\frac{-21}{30} + \frac{-16}{30}\right)\\ \\ & = \frac{-3}{5} + \left(\frac{-21 - 16}{30}\right)\\ \\ & = \frac{-3}{5} + \frac{-37}{30}\\ \\ &\text{LCM} = 30 \\ \\ & = \frac{(-3 \times 6) + (-37 \times 1)}{30}\\ \\ & = \frac{-18 - 37}{30}\\ \\ & = \frac{-55}{30}\\ \\ & = \boxed{\frac{-11}{6}} \\ \\ \text{RHS} &= (x + y) + z \\ \\ & = \left(\frac{-3}{5} + \frac{-7}{10}\right) + \frac{-8}{15} \\ \\ & \text{LCM of 5 and 10} = 10 \\ \\ & = \left(\frac{-3 \times 2}{10} + \frac{-7 \times 1}{10}\right) + \frac{-8}{15}\\ \\ & = \left(\frac{-6}{10} + \frac{-7}{10}\right) + \frac{-8}{15}\\ \\ & = \left(\frac{-6 - 7}{10}\right) + \frac{-8}{15}\\ \\ & = \frac{-13}{10} + \frac{-8}{15}\\ \\ & \text{LCM} = 30 \\ \\ & = \frac{(-13 \times 3) + (-8 \times 2)}{30}\\ \\ & = \frac{-39 - 16}{30}\\ \\ & = \frac{-55}{30}\\ \\ & = \boxed{\frac{-11}{6}} \\ \\ &\text{Hence verified} \\ &\color{red} x + (y + z) = (x+y) + z \end{align*} \]
5. Simplify
(i) \( \frac{-3}{10} + \frac{12}{-10} + \frac{14}{10} \)
Answer
\[ \begin{align*} & = \frac{-3}{10}, \frac{-12}{10}, \frac{14}{10} \\ \\ & = \frac{-3 -12 + 14}{10} \\ \\ & = \frac{-15 + 14}{10} \\ \\ & = \boxed{\color{red}\frac{-1}{10}} \\ \\ \end{align*} \]
(ii) \( \frac{-5}{10} + \frac{6}{13} + 8 \)
Answer
\[ \begin{align*} &\color{green} \text{Method - } 1 \\ \\ & = \frac{-5}{10} + \frac{6}{13} + \frac{8}{1} \\ \\ & \text{LCM} = 130 \\ \\ & = \frac{(-5 \times 13) + (6 \times 10) + (8 \times 130)}{130} \\ \\ & = \frac{-65 + 60 +1040}{130} \\ \\ & = \frac{-65 + 1100}{130} \\ \\ & = \frac{1035}{130} \\ \\ & = \boxed{\color{red}\frac{207}{26}} \\ \\ &\color{green} \text{Method - } 2 \\ \\ & = \frac{-\cancelto{1}5}{\cancelto{2}{10}} + \frac{6}{13} + \frac{8}{1} \\ \\ & = \frac{-1}{2} + \frac{6}{13} + \frac{8}{1} \\ \\ & \text{LCM} = 26 \\ \\ & = \frac{(-1 \times 13) + (6 \times 2) + (8 \times 26)}{26} \\ \\ & = \frac{-13 + 12 + 208}{26} \\ \\ & = \frac{-13 + 220}{26} \\ \\ & = \frac{207}{26} \\ \\ & = \boxed{\color{red}\frac{207}{26}} \\ \\ \end{align*} \]
(iii) \( \frac{-5}{10} + \frac{9}{7} + \frac{3}{20} + \frac{-11}{14} \)
Answer
\[ \begin{align*} &\color{green} \text{Method - } 1 \text{: Grouping } \\ \\ & \left(\frac{-5}{10} + \frac{3}{20}\right)+ \left(\frac{9}{7} + \frac{-11}{14}\right) \\ \\ & = \left(\frac{-10 + 3}{20}\right)+ \left(\frac{18 - 11}{14}\right) \\ \\ & = \frac{-7}{20} + \frac{\cancelto{1}7}{\cancelto{2}{14}} \\ \\ & = \frac{-7}{20} + \frac{1}{2} \\ \\ & = \frac{-7 + 10}{20} \\ \\ & = \boxed{\color{red}\frac{3}{20}} \\ \\ &\color{green} \text{Method - } 2 \\ \\ &= \frac{-5}{10} + \frac{9}{7} + \frac{3}{20} + \frac{-11}{14} \\ \\ & \text{LCM} = 140 \\ \\ &= \frac{(-5 \times 14) + (9 \times 20) + (3 \times 7) + (-11 \times 10)}{140} \\ \\ &= \frac{-70 + 180 + 21 - 110}{140} \\ \\ &= \frac{\cancel{110} + 21 \cancel{-110}}{140} \\ \\ &= \frac{21}{140} \\ \\ &= \boxed{\color{red}\frac{3}{20}} \\ \\ \end{align*} \]
(iv) \( \frac{5}{36} + \frac{-7}{8} + \frac{6}{-72} + \frac{-3}{-12} \)
Answer
\[ \begin{align*} &= \frac{5}{36} + \frac{-7}{8} + \frac{-6}{72} + \frac{3}{12} \\ \\ \end{align*} \] \[ \begin{array}{c|c} 2 & 36, 8, 72, 12 \\ \hline 2 & 18, 4, 36, 6 \\ \hline 2 & 9, 2, 18, 3 \\ \hline 3 & 9, 1, 9, 3 \\ \hline 3 & 3, 1, 3, 1 \\ \hline & 1, 1, 1, 1 \\ \end{array} \] \[ \begin{align*} & \text{LCM} = 72 \\ \\ &= \frac{5 \times 2}{72} + \frac{-7 \times 9}{72} + \frac{-6 \times 1}{72} + \frac{3 \times 6}{72} \\ \\ &= \frac{10 - 63 - 6 + 18}{72} \\ \\ &= \frac{28 - 69}{72} \\ \\ &= \frac{-41}{72} \\ \\ &= \boxed{\color{red} \frac{-41}{72}} \\ \\ \end{align*} \]
(6) For \( x = \frac{1}{5} \) and \( y = \frac{3}{7}\) , verify that \( \color{red} -(x + y) = (-x) + (-y) \)
Answer
\[ \begin{align*} \text{LHS} &= -(x + y) \\ \\ & = -\left(\frac{1}{5} + \frac{3}{7} \right) \\ \\ \text{LCM} & = 35 \\ \\ & = -\left(\frac{7 + 15}{35} \right) \\ \\ & = -\left(\frac{22}{35} \right) \\ \\ & = \boxed{\frac{-22}{35}} \\ \\ \text{RHS} &= (-x) + (-y) \\ \\ & = \left(-\frac{1}{5} \right) + \left(-\frac{3}{7} \right) \\ \\ & = \frac{-1}{5} + \frac{-3}{7} \\ \\ \text{LCM} & = 35 \\ \\ & = \frac{-7 + (-15)}{35} \\ \\ & = \frac{-7 -15}{35} \\ \\ & = \boxed{\frac{-22}{35}} \\ \\ \text{Hence} & \text{ verified} \\ \color{red} -(x + y) & \color{red} = (-x) + (-y) \end{align*} \]
7. Write True or False for the following statements
(i) \( \frac{-2}{-3} \) is the additive inverse of \( \frac{2}{3} \)
Answer False
\[ \begin{align*} \frac{-2}{3} \text{ is the the additive inverse of } \frac{2}{3} \\ \\ \end{align*} \]
(ii) \( \frac{2}{3} + \frac{4}{5} \) is a rational number.
Answer True
\[ \begin{align*} & = \frac{2}{3} + \frac{4}{5} \\ \\ & = \frac{10 + 12}{15} \\ \\ & = \frac{22}{15} \text{ is a rational number}\\ \\ \end{align*} \]
(iii) \( \frac{-5}{3} + \frac{5}{-3} \) is equal to zero.
Answer False
\[ \begin{align*} & = \frac{-5}{3} + \frac{-5}{3} \\ \\ & = \frac{-5 -5}{3} \\ \\ & \frac{-10}{3} \neq 0 \\ \\ \end{align*} \]
(iv) 1 is the identity element of addition.
Answer False
\( \color{red} 0 \) is the identity element of addition.
(v) \( \frac{-9}{7} + 0 = 0 \)
Answer False
\[ \begin{align*} \frac{-9}{7} + 0 & = \frac{-9}{7} \\ \\ \end{align*} \]
(vi) Additive inverse of \( \frac{-3}{5} \) is \( \frac{3}{5} \)
Answer True
\[ \begin{align*} & = \frac{-3}{5} + \frac{3}{5} \\ \\ & = \frac{-3 + 3}{5} \\ \\ & = 0 \end{align*} \]
(vii) Negative of a negative rational number is negative.
Answer False
\[ \begin{align*} & = -\left(\frac{-1}{5} \right) \\ \\ & = +\frac{1}{5} \\ \\ \end{align*} \]