Given x = \(\frac{1}{3},y=\frac{-7}{9}\)
\(x+y=\frac { 1 }{ 3 } +\frac { -7 }{ 9 } =\frac { (1\times 3)+(-7\times 1) }{ 9 } =\frac { 3+(-7) }{ 9 } \)
\(=\frac { -4 }{ 9 } \)   ...(1)
\(y+x=\frac { -7 }{ 9 } +\frac { 1 }{ 3 } =\frac { -7+(1\times 3) }{ 9 } \)
\(=\frac { -7+3 }{ 9 } =\frac { -4 }{ 9 } \) ...(2)
From (1) and (2)
x + y = y + x, i.e commutative property is true.

If a, b and c be three rational numbers then (a + b) + c = a + (b + c).

If a, b and c be three rational numbers, then a x (b + c) = (a x b) + (a x c) is the distributive property of multiplication over addition.

\(\frac { 2 }{ 3 } \div \frac { -7 }{ 12 } =\frac { 2 }{ 3 } \times \frac { 12 }{ -7 } =\frac { 2\times 12 }{ 3\times -7 } =\frac { 24 }{ -21 } =-1\)

\(1\div \frac { 1 }{ 2 } =\frac { 1 }{ 1 } \times \frac { 2 }{ 1 } =\frac { 1\times 2 }{ 1\times 1 } =2\)

\(\frac { 3 }{ 5 } +\frac { 13 }{ 5 } =\frac { 3+13 }{ 5 } =\frac { 16 }{ 5 } =3\frac { 1 }{ 5 } \)

\(LHS=\frac { 3 }{ 5 } \div \frac { 7 }{ 8 } =\frac { 3 }{ 5 } \times \frac { 8 }{ 7 } =\frac { 24 }{ 35 } \)
\(RHS=\frac { 7 }{ 8 } \div \frac { 3 }{ 5 } =\frac { 7 }{ 8 } \times \frac { 5 }{ 3 } =\frac { 35 }{ 24 } \)
∴ LHS ≠ RHS
\(\therefore \ \frac { 3 }{ 5 } \div \frac { 7 }{ 8 } \neq \frac { 7 }{ 8 } \div \frac { 3 }{ 5 } \)
∴ Commutative property not hold good for division of rational numbers.

\(LHS=\frac { 3 }{ 5 } \div \frac { 7 }{ 8 } =\frac { (3\times 8)-(7\times 5) }{ 40 } =\frac { 24-35 }{ 40 } =\frac { -11 }{ 40 } \)
\(RHS=\frac { 7 }{ 8 } -\frac { 3 }{ 5 } =\frac { (7\times 5)-(3\times 8) }{ 40 } =\frac { 35-24 }{ 40 } =\frac { 11 }{ 40 } \)
LHS ≠ RHS
\(\therefore \frac { 3 }{ 5 } \div \frac { 7 }{ 8 } \neq \frac { 7 }{ 8 } -\frac { 3 }{ 5 } \)
∴ Subtraction of rational numbers is not commutative.

\(\\ \\ \frac { 27 }{ 18 } =1\frac { 9 }{ 18 } =1\frac { 1 }{ 2 } \)

\(\frac { -56 }{ -72 } =\frac { -56\div (-8) }{ -72\div (-8) } =\frac { 7 }{ 9 } \)

Let the total population = 1
Population of men = Total population - Women - Children
\(=1-\frac { 7 }{ 20 } -\frac { 1 }{ 4 } \)
\(=\frac { 20 }{ 20 } -\frac { 7 }{ 20 } -\frac { 5 }{ 20 } \)
\(=\frac { 20-7-5 }{ 20 } =\frac { 8 }{ 20 } \)
\(=\frac { 2 }{ 5 } \)
\(\therefore\) Population of men \(=\frac { 2 }{ 5 } \)

The rational number for the point marked on the number line is -\(\frac{2}{5}\)

∴ \(\frac{9}{4}\) lies between 2 and 3

\(\frac { 7 }{ -6 } =\frac { 7\times 2 }{ -6\times 2 } =\frac { 14 }{ -12 } \)
\(\frac { 7 }{ -6 } =\frac { 7\times 3 }{ -6\times 3 } =\frac { 21 }{ -18 } \)
\(\frac { 7 }{ -6 } =\frac { 7\times 4 }{ -6\times 4 } =\frac { 28 }{ -24 } \)
\(\frac { 7 }{ -6 } =\frac { 7\times 5 }{ -6\times 5 } =\frac { 35 }{ -30 } \)
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
\(\frac { -20 }{ 10 } ,\frac { -19 }{ 10 } ,\frac { -18 }{ 10 } ,\frac { -7 }{ 10 } ,\frac { -6 }{ 10 } ,\frac { -5 }{ 10 } ,\frac { 0 }{ 10 } (=0)\)