Let us add \(\frac{6}{5}\ and \frac{(-2)}{5}.We\ find\frac{6}{5}+(\frac{-2}{5})\) on the number line, we have

The distance between two consecutive points is \(\frac{1}{5}\). So adding \(\frac{-2}{5}\ to\frac{1}{5}\) Will mean moving to the left making 2 jumps from \(\frac{6}{5}.\) We reach at \(\frac{4}{5}.\)  So \(\frac{6}{5}+(\frac{-2}{5})=\frac{4}{5}\)
Let us now try this another way \(\frac{6}{5}+(\frac{-2}{5})=\frac{6+(-2)}{5}=\frac{4}{5}\)
We get the same answer.

Since, a and b are two different numbers.
Let a = 15 and b = 10.
\(\therefore\) \(\frac{a-b}{a+b}\)=\(\frac{15-10}{15+10}\)=\(\frac{5}{25}=\frac{1}{5}\)
and \(\frac{a+b}{a-b}\)=\(\frac{15+10}{15-10}=\frac{25}{5}\)=5
So, (a + b) always greater than (a - b) when denominator is less number is greater,
then \(\left(\frac{a+b}{a-b}\right)>\left(\frac{a-b}{a+b}\right)\)

\(\frac { -5 }{ 7 } ,\frac { 2 }{ -3 } ,\frac { -4 }{ 9 } \)

8

\(\frac { 1 }{ 2 } \)

\(\frac { 7 }{ 6 } \)

\(\frac { 9 }{ 11 } \) is a positive rational number 
\(\because \) Both 9 and 11 have positive signs 

Given, \(\frac{-3}{5} \) as a rational number
with numerator  -15,
\(\frac { -3\times 5 }{ 5\times 5 } =\frac { -15 }{ 25 } \)

We have \(\frac { -9 }{ 10 } +\frac { 22 }{ 15 } \) 
\(\because \) LCM of denominators 10 and 15 = 30.
\(\therefore \) \(\frac { -9 }{ 10 } =\frac { -9\times 3 }{ 10\times 3 } =\frac { -27 }{ 30 } \) and \(\frac { 22 }{ 15 } =\frac { 22\times 2 }{ 15\times } =\frac { 44 }{ 30 } \)
\(\therefore \)\(\frac { -9 }{ 10 } +\frac { 22 }{ 15 } =\frac { -27 }{ 30 } +\frac { 44 }{ 30 } =\frac { -27+44 }{ 30 } =\frac { 17 }{ 30 } \) 

\(\\ \frac { -8 }{ 5 } >\frac { -7 }{ 4 } \)

The required four more rational numbers are
\(\frac { 5 }{ -30 } ,\frac { 6 }{ -36 } ,\frac { 7 }{ -42 }\) and \(\frac { 8 }{ -48 }\)

Given rational number is \(\frac { 1 }{ 2 } \).
For greatest negative integer, we divide by \(\frac { -1 }{ 2 } \) by \(\frac { 1 }{ 2 } \)
\(\frac { -1 }{ 2 } \div \left( \frac { 1 }{ 2 } \right) =\frac { 1 }{ 2 } \times \frac { 2 }{ -1 } =\frac { 2 }{ -2 } =-1\)
Hence, \(\frac { -1 }{ 2 } \) should be divided by \(\frac { 1 }{ 2 } \) to obtain the greatest integer.

Given p=mand q=n\(\times \)t
\(\therefore \) \(\frac { p }{ q } =\frac { m\times t }{ n\times t } \Rightarrow \frac { p }{ q } =\frac { m }{ n } \)

Given rational numbers are  \(\frac { -3 }{ 7 } ,\frac { 7 }{ 10 } ,\frac { -5 }{ 6 }.\)
\(\because \) LCM of 5,10 and 6 is 30.
\(\frac { -3\times 6 }{ 5\times 6 }, \frac { 7\times 3 }{ 10\times 3 } ,\frac { -5\times 5 }{ 6\times 5 } =\frac { -18 }{ 30 } ,\frac { 21 }{ 30 } ,\frac { -25 }{ 30 } \)
Denominators are same, so descending order of numerators are 21>-18>-25.
So,\(\frac { 7 }{ 10 } >\frac { -3 }{ 5 } >\frac { -5 }{ 6 } \) .

(a) On the number line, it will mean four jumps of \(\frac{6}{5}\) to the left from zero. We reach at \(\frac{-24}{5}.\)

So, \(\frac{-6}{5}\times4=\frac{-24}{5}\)
(b) \(\frac{-6}{5}\times4=\frac{-6\times4}{5}=\frac{-24}{5}\)
We observe that we arrive at the same rational number.

(a)\(\frac{9}{2}(\frac{-7}{4})\)=\(\frac{9}{2}\times\frac{-7}{4}=-(\frac{9}{2}\times\frac{7}{4})\)
=\(\frac{-63}{8}\)
(b)\(\frac{3}{10}\times(-9)\) =\(\frac{3}{10}\times\frac{-9}{1}=-(\frac{3}{10}\times\frac{9}{1})\)
=\(-\frac{27}{10}\)
(c) \(\frac{-6}{5}\times\frac{9}{11}\)=-\((\frac{6}{5}\times\frac{9}{11})=-(\frac{54}{55})\)
=\(\frac{-54}{55}\)
(d) \(\frac{3}{7}\times(\frac{-2}{5})\)=-\((\frac{3}{7}\times\frac{2}{5})=\frac{-6}{35}\)
(e) \(\frac{3}{11}\times\frac{2}{5}\)=\(\frac{3\times2}{11\times5}=\times\frac{6}{55}\)
(f) \(\frac{3}{-5}\times\frac{-5}{3}\)=\(\frac{3\times(-5)}{-5\times3}=1\)