Question

Write the additive inverse of each of the following rational numbers:
$\frac{4}{9}$; $\frac{-13}{7}$; $\frac{5}{-11}$; $\frac{-11}{-14}$

• A. $\frac{-4}{9};\frac{13}{7};\frac{5}{11};\frac{-13}{14}$
• B. $\frac{-4}{9};\frac{1}{7};\frac{5}{11};\frac{-11}{14}$
• C. $\frac{-4}{9};\frac{13}{7};\frac{5}{11};\frac{-11}{14}$
• D. $\frac{-5}{9};\frac{13}{7};\frac{5}{11};\frac{-11}{14}$

Answer 1

The correct option is C $\frac{-4}{9};\frac{13}{7};\frac{5}{11};\frac{-11}{14}$

The additive inverse of $\frac{4}{9}$ is $-\frac{4}{9}=\frac{-4}{9}$
The additive inverse of $\frac{-13}{7}$ is $-\left(\begin{array}{c}\frac{-13}{7}\end{array}\right)=-\left(\begin{array}{c}-\frac{13}{7}\end{array}\right)=\frac{13}{7}$
We have, $\frac{5}{-11}=\frac{-5}{11}$
The additive inverse of $\frac{-5}{11}$ is $-\left(\begin{array}{c}\frac{-5}{11}\end{array}\right)$$=-\left(\begin{array}{c}-\frac{5}{11}\end{array}\right)=\frac{5}{11}$
We have, $\frac{-11}{-14}=\frac{11}{14}$
The additive inverse of $\frac{11}{14}$ is $-\left(\begin{array}{c}\frac{11}{14}\end{array}\right)$ $=\frac{-11}{14}$