Rational Numbers

Where Rational Numbers Fit

We’ve already met natural numbers (1, 2, 3…), whole numbers (0, 1, 2…), and integers (…−2, −1, 0, 1, 2…). Rational numbers are the next step — any number that can be written as $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$ .

So $\frac{2}{3}$ is rational. So is $-\frac{5}{7}$ , $\frac{0}{4}$ , and even $5$ (which is $\frac{5}{1}$ ). Every integer is a rational number, but not every rational number is an integer.

This chapter teaches us how to add, subtract, multiply, and divide rational numbers, where they sit on the number line, and the special properties they obey. The skills here power everything from solving equations to working with proportions in higher classes.

Key Terms & Definitions

Rational number: A number of the form $\frac{p}{q}$ where $p, q$ are integers and $q \neq 0$ .

Standard form: A rational number is in standard form when $q > 0$ and HCF of $|p|$ and $q$ is 1. Example: $\frac{-3}{5}$ is in standard form; $\frac{6}{-10}$ is not (rewrite as $\frac{-3}{5}$ ).

Equivalent rational numbers: $\frac{p}{q} = \frac{kp}{kq}$ for any non-zero integer $k$ . So $\frac{2}{3} = \frac{4}{6} = \frac{6}{9}$ .

Additive inverse: For $\frac{p}{q}$ , the additive inverse is $-\frac{p}{q}$ (sum is 0).

Multiplicative inverse (reciprocal): For $\frac{p}{q}$ (with $p \neq 0$ ), the reciprocal is $\frac{q}{p}$ (product is 1).

Properties of Rational Numbers

Rational numbers are closed under addition, subtraction, and multiplication: if $a, b$ are rational, so is $a + b$ , $a - b$ , $a \times b$ .

For division: closed except by zero. $\frac{a}{b}$ is rational only when $b \neq 0$ .

Addition: $a + b = b + a$ ✓
Multiplication: $a \times b = b \times a$ ✓
Subtraction: $a - b \neq b - a$ in general ✗
Division: $a \div b \neq b \div a$ in general ✗

Addition: $(a + b) + c = a + (b + c)$ ✓
Multiplication: $(a \times b) \times c = a \times (b \times c)$ ✓
Subtraction and division: NOT associative.

$a \times (b + c) = a \times b + a \times c$

This is the rule that lets you “open brackets” in algebra.

Additive identity: 0 (because $a + 0 = a$ for all $a$ ).
Multiplicative identity: 1 (because $a \times 1 = a$ for all $a$ ).

Operations on Rational Numbers

Addition

To add $\frac{a}{b} + \frac{c}{d}$ :

Find LCM of $b$ and $d$ (or use $bd$ as a common denominator).
Convert both fractions to that denominator.
Add numerators.

Example: $\frac{2}{3} + \frac{1}{4}$ . LCM of 3, 4 = 12.

$\frac{2}{3} = \frac{8}{12}, \quad \frac{1}{4} = \frac{3}{12}$

$\frac{8}{12} + \frac{3}{12} = \frac{11}{12}$

Subtraction

Same as addition but subtract numerators. $\frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$ .

Multiplication

$\frac{a}{b} \times \frac{c}{d} = \frac{ac}{bd}$

Just multiply numerators and denominators directly. Simplify at the end.

Division

$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = \frac{ad}{bc}$

“Multiply by the reciprocal of the divisor.” Works because dividing by a number is the same as multiplying by its inverse.

Rational Numbers on the Number Line

Every rational number has a unique position on the number line. To plot $\frac{3}{4}$ :

Mark 0 and 1.
Divide the segment from 0 to 1 into 4 equal parts.
Count 3 parts from 0. That’s $\frac{3}{4}$ .

Negative rationals go to the left of zero. Mixed numbers like $1\frac{1}{2}$ are between 1 and 2.

Between any two rational numbers, there are infinitely many rational numbers. To find one between $\frac{1}{2}$ and $\frac{3}{4}$ : average them. $\frac{1}{2}(\frac{1}{2} + \frac{3}{4}) = \frac{1}{2} \times \frac{5}{4} = \frac{5}{8}$ . Halfway between.

Solved Examples

Easy

Add: $\frac{-3}{7} + \frac{2}{7}$ .

Same denominator, so just add numerators: $\frac{-3 + 2}{7} = \frac{-1}{7}$ .

Easy

Find the reciprocal of $\frac{-5}{8}$ .

Reciprocal flips the fraction: $\frac{-8}{5}$ (or equivalently $\frac{8}{-5}$ , but standard form keeps denominator positive: $\frac{-8}{5}$ ).

Medium

Simplify: $\frac{2}{3} \times \frac{-9}{14} \times \frac{7}{6}$ .

Multiply numerators: $2 \times (-9) \times 7 = -126$ . Denominators: $3 \times 14 \times 6 = 252$ .

$\frac{-126}{252} = \frac{-1}{2}$

(Divide top and bottom by their HCF, 126.)

Medium

Verify the distributive property for $a = \frac{2}{3}$ , $b = \frac{1}{2}$ , $c = \frac{-1}{4}$ .

LHS: $a \times (b + c) = \frac{2}{3} \times (\frac{1}{2} - \frac{1}{4}) = \frac{2}{3} \times \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$ .

RHS: $a \times b + a \times c = \frac{2}{3} \times \frac{1}{2} + \frac{2}{3} \times \frac{-1}{4} = \frac{1}{3} - \frac{1}{6} = \frac{2}{6} - \frac{1}{6} = \frac{1}{6}$ .

LHS = RHS. ✓

Hard

Find three rational numbers between $\frac{1}{4}$ and $\frac{1}{2}$ .

One method: convert to common denominator with extra zeroes. $\frac{1}{4} = \frac{4}{16}$ and $\frac{1}{2} = \frac{8}{16}$ . Numbers between: $\frac{5}{16}, \frac{6}{16}, \frac{7}{16}$ .

In standard form: $\frac{5}{16}, \frac{3}{8}, \frac{7}{16}$ .

Class 8 Exam Tips

CBSE Class 8 board exams test rational number operations through 1-mark MCQs (identify reciprocal, additive inverse), 2-mark short answers (verify a property), and 3-mark word problems. Memorise:

Standard form rules (positive denominator, HCF = 1)
Reciprocal vs additive inverse (often confused)
Closure properties (which operations are closed)

For “find rationals between two given rationals”, convert to a common denominator with a large enough multiplier. To find $n$ rationals between, use a denominator with at least $n+1$ “spaces” between numerators.

Common Mistakes to Avoid

Confusing reciprocal with additive inverse: Reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$ ; additive inverse is $-\frac{2}{3}$ . Different operations.
Adding fractions without common denominator: $\frac{1}{2} + \frac{1}{3} \neq \frac{2}{5}$ . Convert first.
Standard form errors: Negative sign goes with the numerator, not the denominator. $\frac{-3}{5}$ is standard form; $\frac{3}{-5}$ is not.
Reciprocal of zero: Doesn’t exist. Zero has an additive inverse (itself), but no multiplicative inverse.
Multiplying then forgetting to simplify: Always reduce to standard form. $\frac{6}{12}$ should become $\frac{1}{2}$ .

Practice Questions

Q1. Express $\frac{-12}{15}$ in standard form.

HCF of 12 and 15 is 3. So $\frac{-12 \div 3}{15 \div 3} = \frac{-4}{5}$ .

Q2. Find: $\frac{3}{5} + \frac{-7}{10}$ .

LCM of 5, 10 = 10. $\frac{3}{5} = \frac{6}{10}$ . So $\frac{6}{10} + \frac{-7}{10} = \frac{-1}{10}$ .

Q3. Verify: $\frac{2}{3} + \frac{4}{5} = \frac{4}{5} + \frac{2}{3}$ (commutative property).

Both equal $\frac{10 + 12}{15} = \frac{22}{15}$ . ✓

Q4. Find the multiplicative inverse of $\frac{-3}{8}$ .

$\frac{-8}{3}$ . (In standard form: keep negative on top.)

Q5. Find a rational number between $\frac{1}{3}$ and $\frac{1}{2}$ .

Average: $\frac{1}{2}(\frac{1}{3} + \frac{1}{2}) = \frac{1}{2} \times \frac{5}{6} = \frac{5}{12}$ .

Q6. Compute: $\frac{-2}{7} \times \frac{14}{-3}$ .

$\frac{(-2) \times 14}{7 \times (-3)} = \frac{-28}{-21} = \frac{4}{3}$ .

Q7. The sum of two rational numbers is $\frac{-7}{11}$ . If one number is $\frac{-3}{4}$ , find the other.

Other = $\frac{-7}{11} - \frac{-3}{4} = \frac{-7}{11} + \frac{3}{4} = \frac{-28 + 33}{44} = \frac{5}{44}$ .

Q8. By what number should $\frac{-5}{8}$ be multiplied to get $\frac{15}{16}$ ?

Required = $\frac{15}{16} \div \frac{-5}{8} = \frac{15}{16} \times \frac{8}{-5} = \frac{120}{-80} = \frac{-3}{2}$ .

FAQs

Is every fraction a rational number? Yes — every fraction $\frac{p}{q}$ with integer $p, q$ and $q \neq 0$ is rational. But not every rational number is written as a “fraction” in the everyday sense (e.g., 5 is rational but written as a whole number).

Is 0 a rational number? Yes. $0 = \frac{0}{1}$ . But $\frac{0}{0}$ is not defined; division by zero is forbidden.

Are decimals rational? Terminating decimals (0.5, 1.25) and repeating decimals ( $0.\overline{3} = 1/3$ ) are rational. Non-terminating, non-repeating decimals (like $\pi$ or $\sqrt{2}$ ) are irrational.

What’s the difference between rational and integer? Every integer is a rational number (e.g., $7 = 7/1$ ). Not every rational is an integer (e.g., $1/2$ isn’t).

Why is $\frac{a}{0}$ undefined? If $\frac{a}{0} = b$ for some number $b$ , then $a = 0 \times b = 0$ , contradicting $a \neq 0$ . For $a = 0$ , we’d have $\frac{0}{0}$ equalling any number, so it’s also undefined.

Can I have a rational number with 0 in the numerator? Yes! $\frac{0}{5} = 0$ . The denominator just can’t be 0.