CBSE Class 8 Mathematics Rational Numbers Assignment Set 02

1. Answer the following questions:

(i) Is there a rational number which is its own additive inverse? If yes ,write that rational number.

(ii) what is the standard form of ²²/_-55 ?

(iii) Write an equivalent rational number of ^-2/₇ with denominator 98.

(iv) Write multiplicative inverse of ^-11/₅ .

(v) Is the commutative law of division true for rational numbers ?

2. Represent the following rational numbers on number line

(a) -7/2

(b) 35/7

3. Write 5 rational numbers between 1/3 and 2/3 .

4. Taking some values of a & b, show that la+bl ≤ lal + lbl

5. Arrange in ascending order 1/2, 4/5, -2/3, -1/2, -5/7

6. If a = 3/4, b= -1/2 and c = 1/2 verify :

(I) a+b =b+a (ii) a+c = c+a (iii)(a+b)+c = a+(b+c)

7. What should be subtracted from the product of 3/7 & 2/5 to get -4/35 ?

8. Simplify 11/3 x 7/33 x -5/7 x 9/19

9. One coin weighs 5(3/4)g .Find the weight of 12 such coins.

10. Which of the following statement are True & which are false:
(i) (a - b) ÷ c ≠ a/c + (-b/c)
(ii) Zero is not a rational number .
(iii) –a ÷ a = -1.

ORDER OF RATIONAL NUMBERS

Question. Arrange the following fractions in ascending order. \( \frac{3}{8}, \frac{4}{12}, \frac{-7}{16}, \frac{-2}{3} \).
Answer: LCM of denominators 8, 12, 16, and 3 = \( 2 \times 2 \times 2 \times 2 \times 3 = 48 \).
Then \( \frac{3}{8} = \frac{3 \times 6}{8 \times 6} = \frac{18}{48} \);
\( \frac{-7}{16} = \frac{-7 \times 3}{16 \times 3} = \frac{-21}{48} \);
\( \frac{4}{12} = \frac{4 \times 4}{12 \times 4} = \frac{16}{48} \);
\( -\frac{2}{3} = \frac{-2 \times 16}{3 \times 16} = \frac{-32}{48} \).
The equivalent rational numbers are \( \frac{18}{48}, \frac{16}{48}, \frac{-21}{48} \) and \( \frac{-32}{48} \).
Therefore, the smallest rational number is \( \frac{-32}{48} \), then comes \( \frac{-21}{48} \), then comes \( \frac{16}{48} \), and the greatest rational number is \( \frac{18}{48} \). Hence, their ascending order is \( \frac{-2}{3}, \frac{-7}{16}, \frac{4}{12}, \frac{3}{8} \).

ADDITION OF RATIONAL NUMBERS

• When denominators are equal:

Question. Add \( \frac{5}{6} \) and \( \frac{7}{6} \).
Answer: \( \frac{5}{6} + \frac{7}{6} = \frac{5 + 7}{6} = \frac{12}{6} \)

Question. Add \( \frac{7}{5} \) and \( \frac{-13}{5} \).
Answer: \( \frac{7}{5} + \left( \frac{-13}{5} \right) = \frac{7 - 13}{5} = \frac{-6}{5} \)

• When one denominator is a multiple of the other denominator:

Question. Solve \( \frac{4}{3} \) and \( \frac{5}{6} \).
Answer: We know that \( \frac{4}{3} = \frac{4 \times 2}{3 \times 2} = \frac{8}{6} \)
(\( \frac{8}{6} \) is equivalent rational number of \( \frac{4}{3} \))
So, \( \frac{4}{3} + \frac{5}{6} = \frac{8}{6} + \frac{5}{6} = \frac{13}{6} \)

Question. Solve \( \frac{-3}{7} + \left( \frac{-5}{21} \right) \).
Answer: We know that \( \frac{-3}{7} = \frac{-3 \times 3}{7 \times 3} = \frac{-9}{21} \)
So, \( \frac{-3}{7} + \left( \frac{-5}{21} \right) = \frac{-9}{21} - \frac{5}{21} = \frac{-9 - 5}{21} = \frac{-14}{21} \)

• When denominator are co-prime:

Question. Find the sum of \( \frac{4}{5} \) and \( \frac{-6}{7} \).
Answer: \( \frac{4}{5} + \left( \frac{-6}{7} \right) = \frac{4 \times 7}{5 \times 7} - \frac{6 \times 5}{7 \times 5} \)
(Multiplying and dividing each fraction by the denominator of the other fraction)
\( = \frac{28}{35} - \frac{30}{35} = \frac{28 - 30}{35} = \frac{-2}{35} \)

• When denominator have a common factor:

Question. Solve \( \frac{5}{12} + \frac{7}{8} \).
Answer: Since 12 and 8 have common factors, we will proceed by finding the LCM of 12 and 8. LCM of 12 and 8 is \( 2 \times 2 \times 2 \times 3 = 24 \).
Now we will find equivalent fractions of the given numbers having 24 in the denominator.
Hence, \( \frac{5}{12} = \frac{5 \times 2}{12 \times 2} = \frac{10}{24} \) and \( \frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24} \).
So, \( \frac{5}{12} + \frac{7}{8} = \frac{10}{24} + \frac{21}{24} = \frac{10 + 21}{24} = \frac{31}{24} \)

PROPERTIES OF ADDITION OF RATIONAL NUMBERS

• Closure property: When two rational numbers are added, the result is always a rational number, i.e., if \( \frac{a}{b} \) and \( \frac{c}{d} \) is always a rational number.
  For example, \( \frac{2}{5} + \frac{3}{6} = \frac{12 + 15}{30} = \frac{27}{30} \), which is also a rational number.
• Commutative property: When two rational numbers are added, the order of addition does not matter, i.e., if \( \frac{a}{b} \) and \( \frac{c}{d} \) are two rational numbers, then \( \frac{a}{b} + \frac{c}{d} = \frac{c}{d} + \frac{a}{b} \).
  For example, \( \frac{3}{4} + \frac{4}{5} = \frac{15 + 16}{20} = \frac{31}{20} \) and \( \frac{4}{5} + \frac{3}{4} = \frac{16 + 15}{20} = \frac{31}{20} \). Both results are equal.
• Associative property: If \( \frac{a}{b}, \frac{c}{d} \) and \( \frac{e}{f} \) three rational numbers, then \( \left( \frac{a}{b} + \frac{c}{d} \right) + \frac{e}{f} = \frac{a}{b} + \left( \frac{c}{d} + \frac{e}{f} \right) \).
  Consider the fractions \( \frac{2}{5}, \frac{1}{4} \) and \( \frac{2}{3} \).
  \( \left( \frac{2}{5} + \frac{1}{4} \right) + \frac{2}{3} = \frac{8 + 5}{20} + \frac{2}{3} = \frac{13}{20} + \frac{2}{3} = \frac{39 + 40}{60} = \frac{79}{60} \)
  \( \frac{2}{5} + \left( \frac{1}{4} + \frac{2}{3} \right) = \frac{2}{5} + \frac{3 + 8}{12} = \frac{2}{5} + \frac{11}{12} = \frac{24 + 55}{60} = \frac{79}{60} \)
• Additive identity: If \( \frac{a}{b} \) is a rational number, then there exists a rational number zero such that \( \frac{a}{b} + 0 = \frac{a}{b} \). Zero is called the identity element of addition. Addition of zero does not change the value of the rational number.
• Additive inverse: If \( \frac{a}{b} \) is a rational number, then there exists a rational number \( \left( -\frac{a}{b} \right) \), called the additive inverse, such that \( \frac{a}{b} + \left( -\frac{a}{b} \right) = 0 \). The additive inverse is also referred to as ‘negative’ of the given number.

Question. \( \frac{3}{4} + \left( -\frac{3}{4} \right) = 0 \).
Answer: \( \therefore \left( -\frac{3}{4} \right) \) is the additive inverse of \( \frac{3}{4} \).

Question. \( \frac{-5}{6} + \frac{5}{6} = 0 \).
Answer: \( \therefore \frac{5}{6} \) is the additive inverse of \( \left( -\frac{5}{6} \right) \).

SUBTRACTION OF RATIONAL NUMBERS

When we have to subtract a rational number, say \( \frac{5}{9} \) from \( \frac{8}{9} \), we add the additive inverse of \( \frac{5}{9} \), i.e., \( \frac{-5}{9} \) to \( \frac{8}{9} \). Thus, \( \frac{8}{9} - \frac{5}{9} = \frac{8}{9} + \left( \frac{-5}{9} \right) = \frac{8 - 5}{9} = \frac{3}{9} = \frac{1}{3} \)

Question. Subtract \( \frac{-3}{7} \) from \( \frac{4}{11} \).
Answer: Here, \( \frac{4}{11} - \left( \frac{-3}{7} \right) = \frac{4}{11} + \left( \frac{+3}{7} \right) = \frac{4 \times 7}{11 \times 7} + \frac{3 \times 11}{7 \times 11} = \frac{28}{77} + \frac{33}{77} = \frac{61}{77} \)

MULTIPLICATION OF RATIONAL NUMBERS

Multiplication is the process of successive addition. Like \( 6 \times 8 = 8 + 8 + 8 + 8 + 8 + 8 = 48 \).
Similarly, \( 6 \times \frac{1}{3} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{6}{3} = 2 \)
Alternatively, \( 6 \times \frac{1}{3} = \frac{6}{1} \times \frac{1}{3} = \frac{6 \times 1}{1 \times 3} = \frac{6}{3} = 2 \).
So, when we multiply two rational numbers, we multiply the numerator with the numerator and the denominator with the denominator.
Thus, \( -5 \times (-7) = \frac{-5}{1} \times \left( \frac{-7}{1} \right) = \frac{(-5)(-7)}{1 \times 1} = 35 \)
and \( \frac{-2}{11} \times \frac{3}{5} = \frac{-2 \times 3}{11 \times 5} = \frac{-6}{55} \)

PROPERTIES OF MULTIPLICATION OF RATIONAL NUMBER

• Closure property: The rational number are closed under multiplication. It means that the product of two rational numbers is always a rational number, i.e., if \( \frac{a}{b} \) and \( \frac{c}{d} \) are two rational numbers, \( \frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd} \) is always a rational number.
  For example, \( \frac{-3}{7} \times \frac{5}{8} = \frac{-15}{56} \) which is rational number.
• Commutative property: If \( \frac{a}{b} \) and \( \frac{c}{d} \) are two rational numbers, then \( \frac{a}{b} \times \frac{c}{d} = \frac{c}{d} \times \frac{a}{b} \), i.e., \( \frac{ac}{bd} = \frac{ca}{db} \).

Question. Solve \( \frac{4}{5} \times \left( \frac{-3}{7} \right) \) and \( \left( \frac{-3}{7} \right) \times \frac{4}{5} \).
Answer: \( \frac{4 \times (-3)}{5 \times 7} = \frac{-12}{35} \) and \( \frac{(-3) \times 4}{7 \times 5} = \frac{-12}{35} \).
So, \( \frac{4}{5} \times \left( \frac{-3}{7} \right) = \left( \frac{-3}{7} \right) \times \frac{4}{5} \)

• Associative property: If \( \frac{a}{b}, \frac{c}{d} \) and \( \frac{e}{f} \) are three rational numbers, then \( \left( \frac{a}{b} \times \frac{c}{d} \right) \times \frac{e}{f} = \frac{a}{b} \times \left( \frac{c}{d} \times \frac{e}{f} \right) \) i.e., \( \frac{ac}{bd} \times \frac{e}{f} = \frac{a}{b} \times \frac{ce}{df} \) or \( \frac{ace}{bdf} = \frac{ace}{bdf} \). Thus, rational numbers can be multiplied in any order.

Question. Solve \( \left( \frac{-3}{7} \times \frac{4}{5} \right) \times \left( \frac{-5}{8} \right) \) and \( \frac{-3}{7} \times \left( \frac{4}{5} \times \frac{-5}{8} \right) \).
Answer: \( \frac{(-3) \times 4}{7 \times 5} \times \left( \frac{-5}{8} \right) = \frac{-12}{35} \times \left( \frac{-5}{8} \right) = \frac{60}{280} = \frac{3}{14} \)
\( \frac{-3}{7} \times \frac{4 \times (-5)}{5 \times 8} = \frac{-3}{7} \times \left( \frac{-20}{40} \right) = \frac{60}{280} = \frac{3}{14} \)

• Multiplicative identity: When any rational number, say \( \frac{a}{b} \), is multiplied by the rational number 1, the product is always \( \frac{a}{b} \). \( \frac{a}{b} \times 1 = \frac{a \times 1}{b} = \frac{a}{b} \) or \( 1 \times \frac{a}{b} = \frac{1 \times a}{b} = \frac{a}{b} \).

Question. Solve \( \frac{21}{35} \times 1 \).
Answer: \( \frac{21}{35} \times \frac{1}{1} = \frac{21 \times 1}{35 \times 1} = \frac{21}{35} \)

Question. Solve \( \frac{-3}{7} \times 1 \).
Answer: \( \frac{-3}{7} \times \frac{1}{1} = \frac{(-3) \times 1}{7 \times 1} = \frac{-3}{7} \). ‘One’ is called the multiplicative identity or identity element of multiplication for rational numbers.

• Multiplicative inverse, or reciprocal: For every non-zero rational number \( \frac{a}{b} \), there exists a rational number \( \frac{b}{a} \) such that \( \frac{a}{b} \times \frac{b}{a} = 1 \). This is so, because \( \frac{a}{b} \times \frac{b}{a} = \frac{a \times b}{b \times a} = \frac{ab}{ba} = 1 \).

Question. Solve \( \frac{2}{3} \times \frac{3}{2} \).
Answer: \( \frac{2 \times 3}{3 \times 2} = \frac{6}{6} = 1 \). So \( \frac{3}{2} \) is the multiplicative inverse of \( \frac{2}{3} \) and \( \frac{2}{3} \) is the multiplicative inverse of \( \frac{3}{2} \).

Question. Solve \( \left( -\frac{4}{7} \right) \times \left( -\frac{7}{4} \right) \).
Answer: \( \frac{(-4)(-7)}{7 \times 4} = \frac{28}{28} = 1 \). So \( -\frac{7}{4} \) is the multiplicative inverse of \( -\frac{4}{7} \) and vice versa.

• Distributive property: If \( \frac{a}{b}, \frac{c}{d} \) and \( \frac{e}{f} \) are three rational numbers, then \( \frac{a}{b} \times \left( \frac{c}{d} + \frac{e}{f} \right) = \frac{a}{b} \times \frac{c}{d} + \frac{a}{b} \times \frac{e}{f} \).

Question. Solve \( 3(4 + 5) \).
Answer: \( 3 \times 4 + 3 \times 5 = 12 + 15 = 27 \); \( 3 \times 9 = 27 \). 27 = 27

Question. Solve \( \frac{4}{7} \left( \frac{2}{3} + \frac{3}{4} \right) \).
Answer: \( \frac{4}{7} \times \frac{2}{3} + \frac{4}{7} \times \frac{3}{4} = \frac{8}{21} + \frac{12}{28} = \frac{32 + 36}{84} = \frac{68}{84} \).
Alternatively: \( \frac{4}{7} \left( \frac{8 + 9}{12} \right) = \frac{4}{7} \times \frac{17}{12} = \frac{68}{84} \). 68/84 = 68/84.

Question. Solve \( \frac{-3}{5} \left( \frac{3}{4} + \frac{-8}{9} \right) \).
Answer: \( \left( \frac{-3}{5} \times \frac{3}{4} \right) + \left( \frac{-3}{5} \times \frac{-8}{9} \right) = \frac{-9}{20} + \frac{24}{45} = \frac{-81 + 96}{180} = \frac{15}{180} \).
Alternatively: \( \frac{-3}{5} \left( \frac{27 - 32}{36} \right) = \frac{-3}{5} \times \frac{-5}{36} = \frac{15}{180} \). 15/180 = 15/180.

MULTIPLICATION OF A RATIONAL NUMBER BY ZERO

When any rational number \( \frac{a}{b} \) is multiplied by 0, the product is always zero. \( \frac{a}{b} \times 0 = \frac{a \times 0}{b} = \frac{0}{b} = 0 \).

Question. Solve \( \frac{7}{8} \times 0 \).
Answer: \( \frac{7 \times 0}{8} = \frac{0}{8} = 0 \)

Question. Solve \( \frac{-3}{4} \times 0 \).
Answer: \( \frac{-3 \times 0}{4} = \frac{0}{4} = 0 \)

DIVISION OF RATIONAL NUMBERS

Division is the inverse process of multiplication. If \( \frac{a}{b} \) and \( \frac{c}{d} \) are two rational numbers, then \( \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} \).

Question. Solve \( \frac{2}{7} \div \frac{5}{9} \).
Answer: \( \frac{2}{7} \times \frac{9}{5} = \frac{18}{35} \)

Question. Solve \( \frac{3}{8} \div \frac{-4}{9} \).
Answer: \( \frac{3}{8} \times \left( -\frac{9}{4} \right) = \frac{-27}{32} \)

PROPERTIES OF DIVISION OF RATIONAL NUMBERS

• CLOSURE PROPERTY: When a rational number is divided by another rational number, the quotient is always a rational number. Thus, if \( \frac{a}{b} \) and \( \frac{c}{d} \) are two rational numbers, then \( \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc} \), which is again a rational number since b, c, d are non-zero integers.

Question. Solve \( \frac{3}{4} \div \left( -\frac{1}{3} \right) \).
Answer: \( \frac{3}{4} \times \left( -\frac{3}{1} \right) = \frac{-9}{4} \)

• Division is not commutative: If \( \frac{a}{b} \) and \( \frac{c}{d} \) are two rational numbers in which b, c and d ≠ 0, then \( \frac{a}{b} \div \frac{c}{d} \neq \frac{c}{d} \div \frac{a}{b} \) because \( \frac{a}{b} \div \frac{c}{d} = \frac{ad}{bc} \) and \( \frac{c}{d} \div \frac{a}{b} = \frac{cb}{da} \).

Question. Is \( \frac{4}{7} \div \frac{1}{3} \) equal to \( \frac{1}{3} \div \frac{4}{7} \)?
Answer: No. \( \frac{4}{7} \div \frac{1}{3} = \frac{4}{7} \times \frac{3}{1} = \frac{12}{7} \), whereas \( \frac{1}{3} \div \frac{4}{7} = \frac{1}{3} \times \frac{7}{4} = \frac{7}{12} \). So \( \frac{4}{7} \div \frac{1}{3} \neq \frac{1}{3} \div \frac{4}{7} \).

So Addition, Subtraction and Multiplication are closed for rationales. Addition, multiplication are commutative and associative for rationals.

Question. Solve \( \frac{5}{7} + \frac{8}{13} \).
Answer: \( \frac{65 + 56}{91} = \frac{121}{91} \). Since \( \frac{5}{7} \), \( \frac{8}{13} \) are rational no. and \( \frac{121}{91} \) is also rational, it is closed.

Question. Solve \( \frac{1}{2} - \frac{3}{8} \).
Answer: \( \frac{4 - 3}{8} = \frac{1}{8} \). Since \( \frac{1}{2}, \frac{3}{8} \) are rational & \( \frac{1}{8} \) is also rational, subtraction is closed.

Question. Find \( \frac{3}{7} + \left( -\frac{6}{11} \right) + \left( -\frac{8}{21} \right) + \left( \frac{5}{22} \right) \).
Answer: \( \frac{3}{7} + \left( -\frac{6}{11} \right) + \left( -\frac{8}{21} \right) + \left( \frac{5}{22} \right) = \frac{198}{462} + \left( -\frac{252}{462} \right) + \left( -\frac{176}{462} \right) + \left( \frac{105}{462} \right) \)
(Note that 462 is the LCM of 7, 11, 21 and 22)
\( = \frac{198 - 252 - 176 + 105}{462} = \frac{-125}{462} \).
We can also solve it as:
\( \left[ \frac{3}{7} + \left( -\frac{8}{21} \right) \right] + \left[ -\frac{6}{11} + \frac{5}{22} \right] \) (by using commutative and associativity)
\( = \left[ \frac{9 + (-8)}{21} \right] + \left[ \frac{-12 + 5}{22} \right] \)
\( = \frac{1}{21} + \left( -\frac{7}{22} \right) = \frac{22 - 147}{462} = \frac{-125}{462} \).

Question. Find \( \frac{-4}{5} \times \frac{3}{7} \times \frac{15}{16} \times \left( -\frac{14}{9} \right) \).
Answer: We have \( \left( \frac{-4 \times 3}{5 \times 7} \right) \times \left( \frac{15 \times (-14)}{16 \times 9} \right) = \frac{-12}{35} \times \left( -\frac{35}{24} \right) = \frac{-12 \times (-35)}{35 \times 24} = \frac{1}{2} \).
We can also do it as:
\( \left( \frac{-4}{5} \times \frac{15}{16} \right) \times \left[ \frac{3}{7} \times \left( -\frac{14}{9} \right) \right] \) (Using commutativity and associativity)
\( = \frac{-3}{4} \times \left( -\frac{2}{3} \right) = \frac{1}{2} \).

THE ROLE OF ZERO (0) AND ONE (1)

Zero is called the identity for the addition of rational numbers. It is the additive identity for integers and whole numbers as well. 1 is the multiplicative identity for rational numbers.
a + 0 = 0 + a = a
& a × 1 = 1 × a = a

REPRESENTATION OF RATIONAL NUMBERS ON THE NUMBER LINE

• Natural Numbers: 1 2 3 4 5 6 7 8 (Note: The line extends indefinitely only to the right side of 1).
• Whole Numbers: 0 1 2 3 4 5 6 7 8 (Note: The line extends indefinitely to the right, but from 0. There are no numbers to the left of 0).
• Integers: –3 –2 –1 0 1 2 3 4 (Note: The line extends indefinitely on both sides).
• Rational Numbers: \( -\frac{1}{2}, -1, 0, \frac{1}{2}, 1 \) etc. You can see numbers between –1, 0; 0, 1 etc.

The point to be labeled is twice as far from and to the right of 0 as the point labeled \( \frac{1}{3} \). So it is two times \( \frac{1}{3} \), i.e., \( \frac{2}{3} \). The next marking is 1. You can see that 1 is the same as \( \frac{3}{3} \). Then comes \( \frac{4}{3}, \frac{5}{3}, \frac{6}{3} \) (or 2), \( \frac{7}{3} \) and so on.

Similarly, to represent \( \frac{1}{8} \), the number line may be divided into eight equal parts. We use the number \( \frac{1}{8} \) to name the first point of this division. The second point of division will be labeled \( \frac{2}{8} \), the third point \( \frac{3}{8} \), and so on.

Any rational number can be represented on the number line in this way. In a rational number, the numeral below the bar, i.e., the denominator, tells the number of equal parts into which the first unit has been divided. The numeral above the bar i.e., the numerator, tells ‘how many’ of these parts are considered. So, a rational number such as \( \frac{4}{9} \) means four of nine equal parts on the right of 0 and for \( \frac{-7}{4} \), we make 7 marking of distance \( \frac{1}{4} \) each on the left of zero and starting from 0. The seventh marking is \( \frac{-7}{4} \).

Question. Represent \( \frac{13}{3} \) and \( -\frac{13}{3} \) on number line.
Answer: \( \frac{13}{3} = 4 + \frac{1}{3} \) and \( -\frac{13}{3} = -\left( 4 + \frac{1}{3} \right) \). Therefore, from O mark OA, AB, BC, CD and DE to the right of O. Such that OA = AB = BC = CD = DE = 1 unit. Since we have to consider 4 complete units and a part of the fifth unit, therefore divide the fifth unit DE into 3 equal parts. Take 1 part out of these 3 parts. Then point P is the representation of number \( \frac{13}{3} \) on the number line. Similarly, take 4 full unit lengths to the left of 0 and divide the fifth unit D'E' into 3 equal parts. Take 1 part out of these three equal parts. Thus, P' represents the rational number \( -\frac{13}{3} \).

Question. Represent the rational number \( \frac{7}{4} \) on the number line.
Answer: In order to represent \( \frac{7}{4} \) on the number line, we first draw a number line and mark a point O on it which represent ‘0’. Now we have to find a point, say, N on the number line which represents the numerator 7 of the rational number \( \frac{7}{4} \). So, N is the point that represents the integer 7 on the number line and is on the right hand side of the point O. Divide the segment ON into four (Denominator of \( \frac{7}{4} \)) equal parts. Let A, B, C be the points of division. Then OA = AB = BC = CN. By construction, each segment OA, AB, BC and CN represents \( \frac{1}{4} \)th of segment ON. Therefore, the point A represents the rational number \( \frac{7}{4} \). Similarly, \( -\frac{7}{4} \) can be represented on the number line on the left hand side of ‘O’.

Question. Draw the number line and represent the following rational numbers on it: (i) \( \frac{3}{8} \) (ii) \( -\frac{5}{3} \).
Answer: (i) In order to represent \( \frac{3}{8} \) on number line, we first draw a number line and mark the point O on it representing ‘0’ (zero), we find the point P on the number line representing the positive integer 3. Now divide the segment OP into 8 equal parts. By construction, OA is \( \frac{1}{8} \) of OP. Therefore, A represents the rational number \( \frac{3}{8} \).
(ii) In order to represent \( -\frac{5}{3} \) on number line, we first draw a number line and mark a point O on it representing zero. We find the point Q on the number line representing the integer –5 on the left side of O. Now divide OQ into 3 equal parts. Then OA = AB = BQ. By construction, each segment OA, AB and BQ represents \( \frac{1}{3} \) of OQ. Therefore, the point A represents the rational number \( -\frac{5}{3} \).

Note: There are countless rational number between any two given rational numbers.

Question. Write any 3 rational numbers between –2 and 0.
Answer: –2 can be written as \( \frac{-20}{10} \) and 0 as \( \frac{0}{10} \). Thus we have \( \frac{-19}{10}, \frac{-18}{10}, \frac{-17}{10}, \frac{-16}{10}, \frac{-15}{10}, ....... \frac{-1}{10} \) between – 2 and 0.

Question. Find any ten rational numbers between \( \frac{-5}{6} \) and \( \frac{5}{8} \).
Answer: We first convert \( \frac{-5}{6} \) and \( \frac{5}{8} \) to rational numbers with the same denominators. \( \frac{-5 \times 4}{6 \times 4} = \frac{-20}{24} \) and \( \frac{5 \times 3}{8 \times 3} = \frac{15}{24} \). Thus we have, \( \frac{-19}{24}, \frac{-18}{24}, \frac{-17}{24}, ........ \frac{14}{24} \) as the rational numbers between \( \frac{-20}{24} \) and \( \frac{15}{24} \).

Question. Find a rational number between \( \frac{1}{4} \) and \( \frac{1}{2} \).
Answer: We find the mean of the given rational numbers. \( \left( \frac{1}{4} + \frac{1}{2} \right) \div 2 = \left( \frac{1 + 2}{4} \right) \div 2 = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8} \). \( \frac{3}{8} \) lies between \( \frac{1}{4} \) and \( \frac{1}{2} \). We find \( \frac{1}{4} < \frac{3}{8} < \frac{1}{2} \). If a and b are two rational numbers, then \( \frac{a + b}{2} \) is a rational number between a and b such that a < \( \frac{a + b}{2} \) < b.

Question. Find three rational numbers between \( \frac{1}{4} \) and \( \frac{1}{2} \).
Answer: The mean is \( \frac{3}{8} \) and \( \frac{1}{4} < \frac{3}{8} < \frac{1}{2} \). Now we find another rational number between \( \frac{1}{4} \) and \( \frac{3}{8} \). For this, we again find the mean of \( \frac{1}{4} \) and \( \frac{3}{8} \). That is, \( \left( \frac{1}{4} + \frac{3}{8} \right) \div 2 = \frac{5}{8} \times \frac{1}{2} = \frac{5}{16} \). So \( \frac{1}{4} < \frac{5}{16} < \frac{3}{8} < \frac{1}{2} \). Now find the mean of \( \frac{3}{8} \) and \( \frac{1}{2} \). We have, \( \left( \frac{3}{8} + \frac{1}{2} \right) \div 2 = \frac{7}{8} \times \frac{1}{2} = \frac{7}{16} \). Thus we get \( \frac{1}{4} < \frac{5}{16} < \frac{3}{8} < \frac{7}{16} < \frac{1}{2} \). Thus, \( \frac{5}{16}, \frac{3}{8}, \frac{7}{16} \) are the three rational numbers between \( \frac{1}{4} \) and \( \frac{1}{2} \).

POWERS

• Exponential Notation and Rational Numbers: Exponential notation can be extended to rational numbers. For example: \( \left( \frac{4}{5} \right) \times \left( \frac{4}{5} \right) \times \left( \frac{4}{5} \right) \) can be written as \( \left( \frac{4}{5} \right)^3 \) which is read as \( \frac{4}{5} \) raised to the power 3.

Question. Evaluate (i) \( \left( \frac{3}{4} \right)^3 \) (ii) \( \left( \frac{-5}{6} \right)^2 \).
Answer: (i) \( \left( \frac{3}{4} \right)^3 = \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{3^3}{4^3} = \frac{27}{64} \).
(ii) \( \left( \frac{-5}{6} \right)^2 = \left( -\frac{5}{6} \right) \times \left( -\frac{5}{6} \right) = \frac{(-5)^2}{6^2} = \frac{25}{36} \).

Question. Solve \( \left( -\frac{2}{3} \right)^3 \).
Answer: \( \left( -\frac{2}{3} \right) \times \left( -\frac{2}{3} \right) \times \left( -\frac{2}{3} \right) = \frac{(-2)^3}{3^3} = \frac{-8}{27} \). In general, if \( \frac{x}{y} \) is a rational number and a is a positive integer, then \( \left( \frac{x}{y} \right)^a = \frac{x^a}{y^a} \).

Question. Evaluate \( \left( \frac{-4}{5} \right)^3 \).
Answer: \( \left( -\frac{4}{5} \right) \times \left( -\frac{4}{5} \right) \times \left( -\frac{4}{5} \right) = \frac{(-4)^3}{5^3} = \frac{-64}{125} \).

Question. Express \( \frac{27}{64} \) and \( \frac{-8}{27} \) as the powers of rational numbers.
Answer: \( \frac{27}{64} = \frac{3 \times 3 \times 3}{4 \times 4 \times 4} = \frac{3^3}{4^3} = \left( \frac{3}{4} \right)^3 \).
\( \frac{-8}{27} = \frac{(-2) \times (-2) \times (-2)}{3 \times 3 \times 3} = \frac{(-2)^3}{3^3} = \left( -\frac{2}{3} \right)^3 \).

• Reciprocals with Positive Integral Exponents: The reciprocal of 2 is \( \frac{1}{2} \), reciprocal of \( 2^3 \) is \( \frac{1}{2^3} \).
  Reciprocal of \( \left( \frac{2}{3} \right)^4 = \frac{1}{\left( \frac{2}{3} \right)^4} = \frac{3^4}{2^4} = \left( \frac{3}{2} \right)^4 \).
  Reciprocal of \( \left( \frac{-4}{5} \right)^4 = \left( \frac{-5}{4} \right)^4 \) and Reciprocal of \( \left( \frac{1}{3} \right)^5 = \left( \frac{3}{1} \right)^5 = 3^5 \).
• Reciprocals with Negative Integral Exponents: Reciprocal of \( 2 = \frac{1}{2} = 2^{-1} \). Therefore, the reciprocal of 2 is \( 2^{-1} \). The reciprocal of \( 3^2 = \frac{1}{3^2} = 3^{-2} \). Reciprocal of \( \left( \frac{4}{5} \right)^2 = \left( \frac{4}{5} \right)^{-2} \).

Reciprocal of \( \left( \frac{-2}{3} \right)^3 = \left( \frac{-2}{3} \right)^{-3} \), etc.

In general, if x is any rational number other than zero and a is any positive integer, then:
\[ x^{-a} = \frac{1}{x^a} \]

Question. Simplify \( \left( \frac{2}{3} \right)^{-3} \div \left( \frac{4}{3} \right)^{-2} \).
Answer: \( \left( \frac{2}{3} \right)^{-3} \div \left( \frac{4}{3} \right)^{-2} = \left( \frac{3}{2} \right)^3 \div \left( \frac{3}{4} \right)^2 = \frac{3 \times 3 \times 3}{2 \times 2 \times 2} \div \frac{3 \times 3}{4 \times 4} = \frac{27}{8} \div \frac{9}{16} = \frac{27}{8} \times \frac{16}{9} = 6 \)

Laws of Exponents :

Consider the following.
(i) \( 3^3 \times 3^4 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^7 = 3^{3+4} \)
(ii) \( \left( \frac{5}{2} \right)^2 \times \left( \frac{5}{2} \right)^3 = \frac{5}{2} \times \frac{5}{2} \times \frac{5}{2} \times \frac{5}{2} \times \frac{5}{2} = \left( \frac{5}{2} \right)^5 = \left( \frac{5}{2} \right)^{2+3} \)
\( \therefore x^a \times x^b = x^{a+b} \)

(i) \( 2^5 \div 2^2 = \frac{2 \times 2 \times 2 \times 2 \times 2}{2 \times 2} = 2 \times 2 \times 2 = 2^3 = 2^{5-2} \)
(ii) \( \left( \frac{2}{3} \right)^6 \div \left( \frac{2}{3} \right)^2 = \frac{\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}}{\frac{2}{3} \times \frac{2}{3}} = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \left( \frac{2}{3} \right)^4 = \left( \frac{2}{3} \right)^{6-2} \)
\( \therefore x^a \div x^b = x^{a-b} \)

(i) \( (2^3)^2 = (2 \times 2 \times 2)^2 = (2 \times 2 \times 2) \times (2 \times 2 \times 2) = 2^6 = 2^{3 \times 2} \)
(ii) \( \left\{ \left( \frac{2}{3} \right)^3 \right\}^2 = \left( \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \right)^2 = \left( \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \right) \times \left( \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \right) = \left( \frac{2}{3} \right)^6 = \left( \frac{2}{3} \right)^{3 \times 2} \)
\( \therefore (x^a)^b = x^{ab} \)

(i) \( 2^4 \times 3^4 = (2 \times 2 \times 2 \times 2) \times (3 \times 3 \times 3 \times 3) = (2 \times 3) \times (2 \times 3) \times (2 \times 3) \times (2 \times 3) = (2 \times 3)^4 \)
(ii) \( \left( \frac{3}{5} \right)^4 \times \left( \frac{1}{2} \right)^4 = \left( \frac{3}{5} \times \frac{3}{5} \times \frac{3}{5} \times \frac{3}{5} \right) \times \left( \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \right) = \left( \frac{3}{5} \times \frac{1}{2} \right) \times \left( \frac{3}{5} \times \frac{1}{2} \right) \times \left( \frac{3}{5} \times \frac{1}{2} \right) \times \left( \frac{3}{5} \times \frac{1}{2} \right) = \left( \frac{3}{5} \times \frac{1}{2} \right)^4 \)
\( \therefore x^a \times y^a = (x \times y)^a \)

(i) \( 2^4 \div 3^4 = \frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 3 \times 3} = \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \left( \frac{2}{3} \right)^4 \)
(ii) \( \left( \frac{3}{5} \right)^4 \div \left( \frac{1}{2} \right)^4 = \frac{\frac{3}{5} \times \frac{3}{5} \times \frac{3}{5} \times \frac{3}{5}}{\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}} = \left( \frac{\frac{3}{5}}{\frac{1}{2}} \right) \times \left( \frac{\frac{3}{5}}{\frac{1}{2}} \right) \times \left( \frac{\frac{3}{5}}{\frac{1}{2}} \right) \times \left( \frac{\frac{3}{5}}{\frac{1}{2}} \right) = \left( \frac{3/5}{1/2} \right)^4 \)
\( \therefore x^a \div y^a = \left( \frac{x}{y} \right)^a \)

Question. Find x so that \( \left( \frac{2}{3} \right)^{-5} \times \left( \frac{2}{3} \right)^{-11} = \left( \frac{2}{3} \right)^{8x} \).
Answer: \( \left( \frac{2}{3} \right)^{-5} \times \left( \frac{2}{3} \right)^{-11} = \left( \frac{2}{3} \right)^{8x} \Rightarrow \left( \frac{2}{3} \right)^{(-5)+(-11)} = \left( \frac{2}{3} \right)^{8x} \Rightarrow \left( \frac{2}{3} \right)^{-5-11} = \left( \frac{2}{3} \right)^{8x} \Rightarrow \left( \frac{2}{3} \right)^{-16} = \left( \frac{2}{3} \right)^{8x} \Rightarrow 8x = -16 \therefore x = -2 \)

So, If x is any rational number different from zero and a, b are any integers, then,
Law I: \( x^a \times x^b = x^{a+b} \)
Law II: \( x^a \div x^b = x^{a-b} \)
Law III: \( (x^a)^b = x^{ab} \)
Law IV: \( x^a \times y^a = (x \times y)^a \) (where y is also a non-zero rational number)
Law V: \( x^a \div y^a = \left( \frac{x}{y} \right)^a \) (where y is also a non-zero rational number)

Question. Evaluate \( \left( \frac{2}{3} \right)^4 \div \left( \frac{2}{3} \right)^4 \).
Answer: \( \left( \frac{2}{3} \right)^4 \div \left( \frac{2}{3} \right)^4 = \left( \frac{2}{3} \right)^{4-4} = \left( \frac{2}{3} \right)^0 \).
but \( \left( \frac{2}{3} \right)^4 \div \left( \frac{2}{3} \right)^4 = \frac{\left( \frac{2}{3} \right)^4}{\left( \frac{2}{3} \right)^4} = 1 \).
\( \therefore \) the expression \( = \left( \frac{2}{3} \right)^0 = 1 \)