​ Which of the two rational numbers is greater in the given pair? (i) -3/4 or -8/7 (ii) 7/-9 or -5/8 (iii) 9/-13 or 7/-12

Question

Which of the two rational numbers is greater in the given pair?

(i) \(\frac{-4}{3}\) or \(\frac{-8}{7}\)

(ii) \(\frac{7}{-9}\) or \(\frac{-5}{8}\)

(iii) \(\frac{-1}{3}\) or \(\frac{4}{-5}\)

(iv) \(\frac{9}{-13}\) or \(\frac{7}{-12}\)

(v) \(\frac{4}{-5}\) or \(\frac{-7}{10}\) 

(vi) \(\frac{-12}{5}\) or \(-3\)

Answer 1

(i) \(\frac{-4}{3}\) and \(\frac{-8}{7}\) have different denominators. 

Therefore, we take LCM of 3 and 7 that is 21. 

Now,

\(\frac{-4}{3}\) = \(\frac{-4\times7}{3\times7}\) = \(\frac{-28}{21}\)

And,

\(\frac{-8}{7} = \frac{-8\times3}{7\times3} = \frac{-24}{21}\)

Since, -24 > -28 

Therefore, \(\frac{-24}{21}>\frac{-28}{21}\)

Hence, \(\frac{-8}{7}>\frac{-4}{3}\)

(ii) \(\frac{7}{-9} = \frac{7\times1}{-9\times-1}= \frac{-7}{9}\)

\(\frac{-7}{9}\) and \(\frac{-5}{8}\) have different denominators.

Therefore, we take LCM of 9 and 8 that is 72.

Now,

\(\frac{-7}{9}\) = \(\frac{-7\times8}{9\times8}=\frac{-56}{72}\)

And,

\(\frac{-5}{8}\) = \(\frac{-5\times9}{8\times9} = \frac{-45}{72}\)

Since, -45 > -56 

Therefore, \(\frac{-45}{72}>\frac{-56}{72}\)

Hence, \(\frac{-5}{8}>\frac{-7}{9}\)

(iii) \(\frac{4}{-5} = \frac{4\times-1}{-5\times-1} = \frac{-4}{5}\)

\(\frac{-1}{3} \) and \(\frac{-4}{5}\)have different denominators.

Therefore, we take LCM of 3 and 5 that is 15. 

Now,

\(\frac{-1}{3} = \frac{-1\times5}{3\times5} = \frac{-5}{15}\)

And,

\(\frac{-4}{5}\) = \(\frac{-4\times3}{5\times3} = \frac{-12}{15}\)

Since, -5 > -12 

Therefore, \(\frac{-5}{15}>\frac{-12}{15}\)

Hence, \(\frac{-1}{3}>\frac{-4}{5}\)

(iv) \(\frac{9}{-13} = \frac{9\times-1}{-13\times-1} = \frac{-9}{13}\)

And,

\(\frac{7}{-12} = \frac{9\times-1}{-13\times-1} =\frac{-7}{12}\)

\(\frac{-9}{13}\) and \(\frac{-7}{12}\) have different denominators.

Therefore, we take LCM of 13 and 12 that is 156. 

Now,

\(\frac{-9}{13}\) = \(\frac{-9\times12}{13\times12} = \frac{-108}{156}\)

And,

\(\frac{-7}{12}\) = \(\frac{-7\times13}{12\times13} = \frac{-91}{156}\)

Since, -91 > -108 

Therefore, \(\frac{-91}{156}>\frac{-108}{156}\)

Hence, \(\frac{-7}{12}>\frac{-9}{13}\)

(v) \(\frac{4}{-5} = \frac{4\times-1}{-5\times-1}= \frac{-4}{5}\)

\(\frac{-7}{10}\) and \(\frac{-4}{5}\) have different denominators.

Therefore, 

we take LCM of 10 and 5 that is 10. 

Now,

\(\frac{-4}{5}\) = \(\frac{-4\times2}{5\times2} = \frac{-8}{10}\)

Since, -7 > -8 

Therefore, \(\frac{-7}{10}\) > \(\frac{-8}{10}\)

Hence, \(\frac{-7}{10}\) > \(\frac{-4}{5}\)

(vi) We can write \(-3 = \frac{-3}{1}\)

\( \frac{-3}{1}\) and \(\frac{-12}{5}\) have different denominators. 

Therefore, we take LCM of 1 and 5 that is 5. 

Now,

\(\frac{-12}{5}\) = \(\frac{-12\times1}{5\times1} = \frac{-12}{5}\)

And,

\( \frac{-3}{1}\) = \(\frac{-3\times5}{1\times1} = \frac{-12}{5}\)

Since, -12 > -15 

Therefore, \(\frac{-12}{5}\) > \(\frac{-15}{5}\)

Hence, \(\frac{-12}{5}\) > \(-3\)