We have,
L.H.S.= \(\frac{1}{2} \div (\frac{-1}{3} \div \frac{2}{5})=\frac{1}{2} \div (\frac{-1}{3}\times \frac{5}{2})\) (reciprocal of \(\frac{2}{5}\) is \(\frac{5}{2}\))
\(=\frac{1}{2} \div (-\frac{5}{6})= \frac{1}{2}\times \frac{-6}{5}=\frac{-3}{5}\)
R.H.S.=\([\frac{1}{2}\div (\frac{-1}{3})]\div \frac{2}{5}\)
=\((\frac{1}{2}\times \frac{-3}{1})\div \frac{2}{5}=\frac{-3}{2} \div \frac{2}{5}\) (reciprocal of \(\frac{-1}{3}\) of \(\frac{-3}{1}\))
=\(\frac{-3}{2} \times \frac{5}{2}= \frac{-15}{4}\)
So,No; L.H.S.≠ R.H.S. 

Example 1.
\(\frac{-6}{5} \times (\frac{2}{3}\times \frac{1}{8})=\frac{-6}{5} \times \frac{1}{12}=\frac{-1}{10}\)
\((\frac{-6}{5} \times \frac{2}{3})\times \frac{1}{8}=\frac{-4}{5} \times \frac{1}{8}=\frac{-1}{10}\)
So, Yes; \(\frac{-6}{5}\times (\frac{2}{3}\times \frac{1}{8})=(\frac{-6}{5}\times \frac{2}{3})\times \frac{1}{8}\)
Example 2.
\(\frac{2}{7} \times (\frac{-5}{9}\times \frac{2}{3})=\frac{2}{7} \times (\frac{-10}{27})=\frac{-20}{189}\)
\((\frac{2}{7}\times \frac{-5}{9})\times \frac{2}{3}=\frac{-10}{63}\times \frac{2}{3}=\frac{-20}{189}\)
So, Yes; \(\frac{2}{7}\times (\frac{-5}{9}\times \frac{2}{3})=(\frac{2}{7} \times \frac{-5}{9}) \times \frac{2}{3}\)

Example 1. We have,
\(\frac{-3}{4}+[\frac{2}{3}+(\frac{-6}{7})]=\frac{-3}{4}+(\frac{-4}{21})=\frac{-79}{84}\)
\([\frac{-3}{4}+\frac{2}{3}]+(\frac{-6}{7})=\frac{-1}{12}+(\frac{-6}{7})=\frac{-79}{84}\)
So, Yes; \(\frac{-3}{4}+[\frac{2}{3}+(\frac{-6}{7})]=[\frac{-3}{4}+\frac{2}{3}]+(\frac{-6}{7})\).
Example 2. 
\(\frac{-1}{4}+[\frac{2}{9}+(\frac{-5}{11})] = \frac{-1}{4}+(\frac{-23}{99})=\frac{-191}{396}\)
\([\frac{-1}{4}+\frac{2}{9}]+(\frac{-5}{11})=\frac{-1}{36}+(\frac{-5}{11})=\frac{-191}{396}\)
So, Yes; \(\frac{-1}{4}+[\frac{2}{9}+(\frac{-5}{11})] = [\frac{-1}{4}+\frac{2}{9}]+(\frac{-5}{11})\)

(i) \(\frac{-3}{4} \times \frac{7}{9}=-\frac{21}{36}\) (a rational number)
(ii) \(\frac{5}{8} \times \frac{3}{7}=\frac{15}{56}\)(a rational number)
(iii) \(\frac{-3}{5} \times \frac{-7}{11}=\frac{21}{55}\) (a rational number).

(i) \(\frac{-3}{8}-\frac{4}{5}=\frac{-15-32}{40}=\frac{-47}{40}\) (a rational number)
(ii) \(\frac{8}{9}-\frac{3}{7}=\frac{-14-27}{63}=-\frac{13}{63}\) (a rational number)
(iii) \(\frac{2}{5}-(\frac{7}{9})=\frac{2}{5}+\frac{7}{9}=\frac{18+35}{45}=\frac{53}{45}\) (a rational number).

(i) \(\frac{3}{5}+\frac{7}{13}=\frac{39+35}{65}=\frac{74}{65}\) (a rational number)
(ii) \(\frac{3}{7}+\frac{(-4)}{9}=\frac{27+{(-28})}{63}=\frac{-1}{63}\) (a rational number)
(iii) \(\frac{-5}{9}+\frac{(-7)}{6}=\frac{-10+{(-21})}{18}=\frac{-31}{18}\) (a rational number).

We have
A rational number between -3 and -4
= \(\frac { (-3)+(-4) }{ 2 } =\frac { -7 }{ 2 } \)
A rational number between (-3) and \(\frac { -7 }{ 2 } \)
\(\left[ \left( -3 \right) +\left( \frac { -7 }{ 2 } \right) \right] \div 2=\left[ \frac { -6+(-7) }{ 2 } \right] \times \frac { 1 }{ 2 } \)
= \(\frac { -13 }{ 2 } \times \frac { 1 }{ 2 } =\frac { -13 }{ 4 } \)
A rational number between \(\left( \frac { -7 }{ 2 } \right) \) and (-4)
= \(\left[ \frac { -7 }{ 2 } +(-4) \right] \div 2=\left[ \frac { -7+(-8) }{ 2 } \right] \div 2\)
= \(\frac { -15 }{ 2 } \times \frac { 1 }{ 2 } =\frac { -15 }{ 4 } \)
Thus, the three rational numbers
\(\left( \frac { -7 }{ 2 } \right) ,\left( \frac { -13 }{ 4 } \right) \) and \(\left( \frac { -15 }{ 4 } \right) \) are between (-3) and (-4).

Using the closure property over addition, subtraction, multiplication and division for rational numbers, integers, whole-numbers and natural numbers, we have:

\(9\over7\)

Firstly, convert \({2\over3}and {4\over5}\) into rational numbers with the same denominators such that difference between the numerators is more than 5.
We have, \({2\over3}={2\times20\over3\times20}={40\over60}\)
[multiplying the numerator and denominator by 20] and \({4\over5}={4\times12\over5\times12}={48\over60}\)
[multiplying the numerator and denominator by 12]
Therefore, five rational numbers between \({40\over60}=({2\over3})\) and \({48\over60}=({4\over5})are{41\over60},{42\over60},{43\over60},{44\over60},{45\over60}\)

We have, \({-2\over5}={-2\times4 \over 5\times4}={-8\over20}\)
[multiplying the numerator and denominator both by 4] and \({1\over2}={1\times10 \over 2\times10}={10\over20}\)
Here, difference between 10 and -8 = 10 - (-8) = 18
(i.e. more than 10)
Hence, rational numbers between \({-8\over20}and {10\over20}\) are \({-7\over20},{-6\over20},{-5\over20},{-4\over20},{-3\over20}.{-2\over20},{-1\over20},0,{1\over20},{2\over20}\)

Here, the difference between any two consecutive points is
\({3\over5}-{2\over5}={7\over5}-{6\over5}={11\over5}-{10\over5}={12\over5}-{11\over5}i.e.,{1\over5}\)
Thus, equal space between any two adjacent points is \(1\over5\)
Now, we can easily find that
The rational number for the point A is\(1\over5\)
The rational number for the point B is \(4\over5\)
The rational number for the point C is\(5\over5\) or 1.
The rational number for the point D is \(8\over5\)
The rational number for the point E is \(9\over5\)
So, we can draw the number line accordingly.

\(1\over4\)

We have, \( \{{7\over5}\times ({-3\over12}) \}+ \{{7\over16}\times {5\over12} \}={7\over5}\times[{-3\over12}+{5\over12}]\)
[by distributivity, taking \({7\over5}\) as common factor]
\(={7\over5}\times[{-3+5 \over 12}]= {7\over5}\times {2\over 12}={7\over30}\)

All properties of operations on rational numbers also hold in case of integers except the following property:
a \(\div\) b is a rational number if b \(\neq\) 0 but a \(\div\)b is not necessarily an integer in case ab \(\in\) J,
All properties of operations on rational numbers also hold in case of whole numbers except the following properties:
(i) If a and b are rational numbers, then (a - b) may or may not be a whole number,
(ii) If a and b are rational numbers, then a\(\div\)b (where b \(\neq\) 0) is not necessarily a whole number,