Question

Which of the following rational numbers are equal?
(i) $\frac{-9}{12}\mathrm{and}\frac{8}{-12}$
(ii) $\frac{-16}{20}\mathrm{and}\frac{20}{-25}$
(iii) $\frac{-7}{21}\mathrm{and}\frac{3}{-9}$
(iv) $\frac{-8}{14}\mathrm{and}\frac{13}{21}$

Answer 1

(i)
$$\begin{gathered} \mathrm{The}\mathrm{standard}\mathrm{form}\mathrm{of}\frac{-9}{12}\mathrm{is}\frac{-9/3}{12/3}=\frac{-3}{4} \\ \mathrm{The}\mathrm{standard}\mathrm{form}\mathrm{of}\frac{8}{-12}\mathrm{is}\frac{8/-4}{-12/-4}=\frac{-2}{3} \\ \mathrm{Since},\mathrm{the}\mathrm{standard}\mathrm{forms}\mathrm{of}\mathrm{two}\mathrm{rational}\mathrm{numbers}\mathrm{are}\mathrm{not}\mathrm{same}. \\ \mathrm{Hence},\mathrm{they}\mathrm{are}\mathrm{not}\mathrm{equal}. \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Since},\mathrm{LCM}\hspace{0.17em}\mathrm{of}20\mathrm{and}25\mathrm{is}100. \\ \mathrm{Therefore}\mathrm{making}\mathrm{the}\mathrm{denominators}\mathrm{equal}, \\ \frac{-16}{20}=\frac{-16\times 5}{20\times 5}=\frac{-80}{100}\mathrm{and}\frac{20}{-25}=\frac{-20\times 4}{25\times 4}=\frac{-80}{100}. \\ \mathrm{Therefore},\frac{-16}{20}=\frac{20}{-25}. \end{gathered}$$


(iii)
$$\begin{gathered} \mathrm{Since},\mathrm{LCM}\hspace{0.17em}\mathrm{of}21\mathrm{and}9\mathrm{is}63. \\ \mathrm{Therefore}\mathrm{making}\mathrm{the}\mathrm{denominators}\mathrm{equal}, \\ \frac{-7}{21}=\frac{-7\times 3}{21\times 3}=\frac{-21}{63}\mathrm{and}\frac{3}{-9}=\frac{-3\times 7}{9\times 7}=\frac{-21}{63}. \\ \mathrm{Therefore},\frac{-7}{21}=\frac{3}{-9}. \end{gathered}$$


(iv)
$$\begin{gathered} \mathrm{Since},\mathrm{LCM}\hspace{0.17em}\mathrm{of}14\mathrm{and}21\mathrm{is}42. \\ \mathrm{Therefore}\mathrm{making}\mathrm{the}\mathrm{denominators}\mathrm{equal}, \\ \frac{-8}{14}=\frac{-8\times 3}{14\times 3}=\frac{-24}{42}\mathrm{and}\frac{13}{21}=\frac{13\times 2}{21\times 2}=\frac{26}{42}. \\ \mathrm{Therefore},\frac{-8}{14}\mathrm{is}\mathrm{not}\mathrm{equal}\mathrm{to}\frac{13}{21}. \end{gathered}$$