Question

Write the additive inverse of each of the following rational numbers:
\[\dfrac{4}{9};\dfrac{-13}{7};\dfrac{5}{-11};\dfrac{-11}{-14}\]
A. \[\dfrac{-4}{9};\dfrac{13}{7};\dfrac{5}{11};\dfrac{-13}{14}\]
B. \[\dfrac{-4}{9};\dfrac{1}{7};\dfrac{5}{11};\dfrac{-11}{14}\]
C. \[\dfrac{-4}{9};\dfrac{13}{7};\dfrac{5}{11};\dfrac{-11}{14}\]
D. \[\dfrac{-5}{9};\dfrac{13}{7};\dfrac{5}{11};\dfrac{-11}{14}\]

Answer 1

Hint: We will be using the concepts of the number system to solve the problem. We will be using the concept of additive inverse to solve the problem. We know that additive inverse is a number that when added to the number yield zero.

Complete step-by-step answer:
Now, we have to find the additive inverse of,
\[\dfrac{4}{9};\dfrac{-13}{7};\dfrac{5}{-11};\dfrac{-11}{-14}\]
Now, we know that additive inverse of a number is a number which when added to the number itself results in zero.
So, we let the additive inverse of $\dfrac{4}{9}$as ${{x}_{1}}$. Therefore,
$\begin{align}
  & \dfrac{4}{9}+{{x}_{1}}=0 \\
 & {{x}_{1}}=\dfrac{-4}{9} \\
\end{align}$
Now, we let the additive inverse of $\dfrac{-13}{7}$as ${{x}_{2}}$. So, we have,
$\begin{align}
  & {{x}_{2}}-\dfrac{13}{7}=0 \\
 & {{x}_{2}}=\dfrac{13}{7} \\
\end{align}$
Now, we let the additive inverse of $\dfrac{-5}{11}$as ${{x}_{3}}$. So, we have,
$\begin{align}
  & {{x}_{3}}-\dfrac{5}{11}=0 \\
 & {{x}_{3}}=\dfrac{5}{11} \\
\end{align}$
Now, we let the additive inverse of $\dfrac{-11}{-14}$as ${{x}_{4}}$. So, we have,
$\begin{align}
  & {{x}_{4}}+\dfrac{11}{14}=0 \\
 & {{x}_{4}}=\dfrac{-11}{14} \\
\end{align}$
So, the additive inverse of \[\dfrac{4}{9};\dfrac{-13}{7};\dfrac{5}{-11};\dfrac{-11}{-14}\] are \[\dfrac{-4}{9};\dfrac{13}{7};\dfrac{5}{11};\dfrac{-11}{14}\]respectively.
Hence, the correct option is (C).

Note: To solve these types of questions it is important to note that the additive inverse of a number is a number which when added to the number result is zero. We should know the concept of rational and irrational numbers as well.