Since multiplication is commutative.
We have \(=\left( \frac { -3 }{ 8 } \times \frac { 4 }{ 5 } \right) +\left( \frac { -3 }{ 8 } \times \frac { 1 }{ 4 } \right) +\frac { 19 }{ 20 } \)
\(\\ =\left\{ \frac { -3 }{ 8 } \times \left( \frac { 4 }{ 5 } +\frac { 1 }{ 4 } \right) \right\} +\frac { 19 }{ 20 } \)
\(=\left\{ \frac { -3 }{ 8 } \times \left( \frac { (4\times 4)+(1\times 5) }{ 20 } \right) \right\} +\frac { 19 }{ 20 } \)
\(\\ =\left\{ \frac { -3 }{ 8 } \times \left( \frac { 16+5 }{ 20 } \right) \right\} +\frac { 19 }{ 20 } =\left\{ \frac { -3 }{ 8 } \times \frac { 21 }{ 20 } \right\} +\frac { 19 }{ 20 } \)
\(=\frac { -63 }{ 160 } +\frac { 19 }{ 20 } =\frac { (-63\times 1)+(19\times 8) }{ 160 } \)
\(=\frac { -63+152 }{ 160 } =\frac { 89 }{ 160 } \)
\(\therefore \frac { -3 }{ 8 } \times \frac { 4 }{ 5 } -\frac { 3 }{ 8 } \times \frac { 1 }{ 4 } +\frac { 19 }{ 20 } =\frac { 89 }{ 160 } \)

\(\left\{ \frac { 1 }{ 2 } \times \frac { -3 }{ 4 } \right\} -\left\{ \frac { 3 }{ 8 } \times \frac { -1 }{ 4 } \right\} +\left\{ \frac { -3 }{ 5 } \times \frac { -1 }{ 4 } \right\} \)
\(=\left\{ \frac { 1 }{ 2 } \times \frac { -3 }{ 4 } \right\} +\left\{ \frac { 3 }{ 8 } \times \frac { 1 }{ 4 } \right\} +\left\{ \frac { 3 }{ 8 } \times \frac { 1 }{ 4 } \right\} \)
\(=\left\{ \frac { 1 }{ 2 } \times \frac { -3 }{ 4 } \right\} +\left\{ \frac { 1 }{ 4 } \times \frac { 3 }{ 8 } \right\} +\left\{ \frac { 1 }{ 4 } \times \frac { 3 }{ 5 } \right\} \)
\(=\left\{ \frac { 1 }{ 2 } \times \frac { -3 }{ 4 } \right\} +\left[ \frac { 1 }{ 4 } \times \left\{ \frac { 3 }{ 8 } +\frac { 3 }{ 5 } \right\} \right] \)
\(=\left\{ \frac { 1 }{ 2 } \times \frac { -3 }{ 4 } \right\} +\left[ \frac { 1 }{ 4 } \times \left\{ \frac { (3\times 5)+(3\times 8) }{ 40 } \right\} \right] \)
\(=\left\{ \frac { 1 }{ 2 } \times \frac { -3 }{ 4 } \right\} +\left[ \frac { 1 }{ 4 } \times \left\{ \frac { 15+24 }{ 40 } \right\} \right] \)
\(=\left\{ \frac { 1 }{ 2 } \times \frac { -3 }{ 4 } \right\} +\left\{ \frac { 1 }{ 4 } \times \frac { 39 }{ 40 } \right\} =\frac { -3 }{ 8 } +\frac { 39 }{ 160 } \)
\(=\frac { (-3\times 20)+(39\times 1) }{ 160 } =\frac { -60+39 }{ 160 } =\frac { -21 }{ 160 } \)
\(\left\{ \frac { 1 }{ 2 } \times \frac { -3 }{ 4 } \right\} -\left\{ \frac { 3 }{ 8 } \times \frac { 1 }{ -4 } \right\} +\left\{ \frac { -3 }{ 5 } \times \frac { -1 }{ 4 } \right\} =\frac { -21 }{ 160 } \)
[∴ Multiplication is commutative for rational numbers]
[∴ Distributive property of multiplication over addition]

Given the rational number a = \(\frac{-1}{2};b=\frac{2}{3}\) c = \(\frac{-5}{6}\)
\(a\times (b+c)=\frac { -1 }{ 2 } \times \left( \frac { 2 }{ 3 } +\left( \frac { -5 }{ 6 } \right) \right) \)
\(=\frac { -1 }{ 2 } \times \left( \frac { (2\times 2)+(-5\times 1) }{ 6 } \right) \)
\(=\frac { -1 }{ 2 } \times \left( \frac { 4+(-5) }{ 6 } \right) =\frac { -1 }{ 2 } \times \left( \frac { -1 }{ 6 } \right) \)
\(a\times (b+c)=\frac { 1 }{ 12 } \quad \quad \quad ...(1)\)
\((a\times b)+(a\times c)=\left( \frac { -1 }{ 2 } \times \frac { 2 }{ 3 } \right) +\left( \frac { -1 }{ 2 } \times \left( \frac { -5 }{ 6 } \right) \right) \)
\(=\frac { -2 }{ 6 } +\frac { 5 }{ 12 } =\frac { (-2\times 2)+5\times 1 }{ 12 } =\frac { -4+5 }{ 12 } \)
\((a\times b)+(a\times c)=\frac { 1 }{ 12 } \quad \quad \quad \quad \quad \quad ...(2)\)
From (1) and (2) we have a x (b + c) = (a x b) + (a x c) is true.
Hence multiplication is distributive over addition for rational numbers Q.

\(\frac { 3 }{ -4 } ,\frac { -1 }{ 2 } \)
LCM of 4 and 2 = 4
\(\frac { 3 }{ -4 } =\frac { -3 }{ 4 } \)
\(\frac { -1 }{ 2 } =\frac { -1\times 2 }{ 2\times 2 } =\frac { -2 }{ 4 } \)
\(\frac { 3 }{ -4 } <\frac { -2 }{ 4 } \)
\(-\frac { 3 }{ 4 } <\frac { -1 }{ 2 }\)

\(\frac{-21}{5} \div \frac{-7}{-10} =\frac{-21}{5} \times \frac{10}{7} \)
\(=\frac{-3}{1} \times \frac{2}{1}=-6 \)

\(\frac { 7 }{ -6 } =\frac { 7\times 2 }{ -6\times 2 } =\frac { 14 }{ -12 } \)
\(\frac { 7 }{ -6 } =\frac { 7\times 3 }{ -6\times 3 } =\frac { 21 }{ -18 } \)
\(\frac { 7 }{ -6 } =\frac { 7\times 4 }{ -6\times 4 } =\frac { 28 }{ -24 } \)
\(\frac { 7 }{ -6 } =\frac { 7\times 5 }{ -6\times 5 } =\frac { 35 }{ -30 } \)
Four equivalent rational numbers of \(\frac { -7 }{ 6 } \) are \(\frac { 14 }{ -12 } ,\frac { 21 }{ -18 } ,\frac { 28 }{ -24 } ,\frac { 35 }{ -30 } \)

-2 and 0 
i,e. \(\frac{-2}{1}\) and \(\frac{0}{1}\)
\(\frac { -2 }{ 1 } =\frac { -2\times 10 }{ 1\times 10 } =\frac { -20 }{ 10 } \)
\(\frac { 0 }{ 1 } =\frac { 0\times 10 }{ 1\times 10 } =\frac { 0 }{ 10 } \)
∴ Five rational numbers between \(\frac{-20}{10}\) ( = -2) and \(\frac{0}{10}\)( = 0)are
\(\frac { -20 }{ 10 } ,\frac { -19 }{ 10 } ,\frac { -18 }{ 10 } ,\frac { -7 }{ 10 } ,\frac { -6 }{ 10 } ,\frac { -5 }{ 10 } ,\frac { 0 }{ 10 } (=0)\)

Let a = \(\frac{-10}{11},b=\frac{5}{6}\) and c = \(\frac{-4}{3}\) be the given rational numbers
\((a+b)+c=\left( \frac { -10 }{ 11 } +\frac { 5 }{ 6 } \right) +\left( \frac { -4 }{ 3 } \right) =\left( \frac { (-10\times 6)+(5\times 11) }{ 66 } \right) +\left( \frac { -4 }{ 3 } \right) \)
\(=\frac { -66+55 }{ 66 } +\left( \frac { -4 }{ 3 } \right) \)
\(=\left( \frac { -5 }{ 66 } \right) +\left( \frac { -4 }{ 3 } \right) =\frac { -5+(-4\times 22) }{ 66 } \)
\(=\frac { -5+(-88) }{ 66 } =\frac { -93 }{ 66 } \)
\((a+b)+c=\frac { -31 }{ 22 } \)          .....(1)
Also \(a+(b+c)=\frac { -10 }{ 11 } +\left( \frac { 5 }{ 6 } +\left( \frac { -4 }{ 3 } \right) \right) =\frac { -10 }{ 11 } +\left( \frac { 5+(-4\times 2) }{ 6 } \right) \)
\(=\frac { -10 }{ 11 } +\left( \frac { 5+(-8) }{ 6 } \right) =\frac { -10 }{ 11 } +\left( \frac { -3 }{ 6 } \right) \)
\(=\frac { (-10\times 6)+(-3)\times -11 }{ 66 } =\frac { -60+(-33) }{ 66 } =\frac { -93 }{ 66 } \)
\(\\ a+(b+c)=\frac { -31 }{ 22 } \)  .....(2)
From (1)and (2), (a + b) + c = a + (b + c) is true for rational numbers.
Now (a x b) x c= 
(a x b) x c = \(\frac{100}{99}\)   .....(1)
\(a\times (b\times c)=\frac { -10 }{ 11 } \times \left( \frac { 5 }{ 6 } \times \left( \frac { -4 }{ 3 } \right) \right) =\frac { -10 }{ 11 } \times \left( \frac { -20 }{ 18 } \right) \)
\(=\frac { -10 }{ 11 } \times \left( \frac { -10 }{ 9 } \right) \)
\(a\times (b\times c)=\frac { 100 }{ 99 } \)  .....(2)
From (1) and (2) a x (b x c) = (a x b) x c is true for rational numbers.
Thus associative property is true for addition and multiplication of rational numbers.     

The average of a and b is \(\frac{1}{2}(a+b)\)
The average of \(\frac{14}{5}\) and \(\frac{16}{3}\) is C₁=\(\frac { 1 }{ 2 } \left( \frac { 14 }{ 5 } +\frac { 16 }{ 3 } \right) \)
\({ C }_{ 1 }=\frac { 1 }{ 2 } \left( \frac { 42+80 }{ 15 } \right) \)
\({ C }_{ 1 }=\frac { 122 }{ 30 } \)
\({ C }_{ 1 }=\frac { 61 }{ 15 } \)
\(\\ \frac { 14 }{ 5 } <\frac { 61 }{ 15 } <\frac { 16 }{ 3 } \)   ..(1)
The average of \(\frac { 14 }{ 5 }\) and \(\frac{61}{15}\)is C₂=\(\frac { 1 }{ 2 } \left( \frac { 14 }{ 5 } +\frac { 61 }{ 15 } \right) \)
\(\frac { 1 }{ 2 } \left( \frac { 14 }{ 5 } +\frac { 61 }{ 15 } \right) \)
\({ C }_{ 2 }=\frac { 1 }{ 2 } \times \left( \frac { 42+61 }{ 15 } \right) \)
\({ C }_{ 2 }=\frac { 1 }{ 2 } \times \frac { 103 }{ 15 } \)
\(\frac { 103 }{ 15 } \)
\(\therefore \frac { 14 }{ 5 } <\frac { 103 }{ 30 } <\frac { 61 }{ 15 } \)  ..(2)
The average of \(\frac{103}{30}\) and \(\frac{61}{15}\) is C₃=\(\frac { 1 }{ 2 } \left( \frac { 103 }{ 30 } +\frac { 61 }{ 15 } \right) \)
\({ C }_{ 3 }=\frac { 1 }{ 2 } \times \left( \frac { 103+122 }{ 30 } \right) \)
\({ C }_{ 3 }=\frac { 1 }{ 2 } \times \frac { 225 }{ 30 } \)
\({ C }_{ 3 }=\frac { 1 }{ 2 } \times \frac { 15 }{ 2 } \)
\(\\ { C }_{ 3 }=\frac { 15 }{ 4 } \)
\(\therefore \frac { 103 }{ 30 } <\frac { 15 }{ 4 } <\frac { 61 }{ 15 } \)...(3)
From (1), (2) and (3) we get,       
\(\frac { 14 }{ 5 } <\frac { 103 }{ 30 } <\frac { 15 }{ 4 } <\frac { 61 }{ 15 } <\frac { 16 }{ 3 } \)

Let a = \(\frac{-10}{11}\) and b = \(\frac{-8}{33}\) be the given rational numbers.
Now, \(\\ \\ a+b=\frac { -10 }{ 11 } +\left( \frac { -8 }{ 33 } \right) =\frac { (-10\times 3)+(-8\times 1) }{ 33 } =\frac { -30+(-8) }{ 33 } \)
\(a+b=\frac { -38 }{ 33 } \quad \quad \quad \quad .....(1)\)
\(b+a=\frac { -8 }{ 33 } +\left( \frac { -10 }{ 11 } \right) =\frac { (-8\times 1)+((-10)\times 3) }{ 33 } =\frac { -8+(-30) }{ 33 } \)
\(b+a=\frac { -38 }{ 33 } \quad \quad \quad .....(2)\)
From (1) and (2)
a + b = b + a and hence addition is commutative for rational numbers.
Further a x b \(=\frac { -10 }{ 11 } \times \left( \frac { -8 }{ 33 } \right) =\frac { 80 }{ 363 } \)
\(\\ a\times b=\frac { 80 }{ 363 } \quad \quad \quad \quad \quad \quad \quad \quad ...(3)\)
\(b\times a=\frac { -8 }{ 33 } \times \left( \frac { -10 }{ 11 } \right) =\frac { 80 }{ 363 } \)
\(b\times a\quad =\frac { 80 }{ 363 } \quad \quad \quad\quad \quad ....(4)\)
From (3) and (4) a x b = b x a
Hence multiplication is commutative for rational numbers.