NCERT Solutions for Class 8 Maths Chapter 1: Rational Numbers — Complete Guide with All Exercises

Understanding where rational numbers sit in the number system helps you see the bigger picture:

Every natural number ($1, 2, 3, \ldots$) is a whole number. Every whole number ($0, 1, 2, \ldots$) is an integer. Every integer ($\ldots, -2, -1, 0, 1, 2, \ldots$) is a rational number. But not every rational number is an integer — for example, $\dfrac{3}{5}$ is rational but not an integer.

In Class 9, you will learn about irrational numbers (like $\sqrt{2}$ and $\pi$) that cannot be expressed as $\dfrac{p}{q}$. Together, rational and irrational numbers form the real numbers. But for now, our universe is $\mathbb{Q}$.

You might wonder why mathematicians care about closure. The reason is practical: if a set is closed under an operation, you can perform that operation freely without ever "leaving" the set. When you add, subtract, or multiply rational numbers, you are guaranteed to get a rational number back. This makes rational numbers a reliable workspace for algebra.

However, since division can lead to undefined results (division by zero), you must always check that the divisor is non-zero. This is a theme that will recur throughout your mathematical career — from solving equations to working with functions.

Problem: Verify that $\dfrac{-3}{7} + \dfrac{5}{9}$ is a rational number.

Solution:

Since $8$ and $63$ are integers and $63 \neq 0$, the result $\dfrac{8}{63}$ is a rational number. This verifies the closure property of addition for these two rational numbers.

Problem: Show that rational numbers are not closed under division.

Solution:

Consider $\dfrac{5}{3} \div \dfrac{0}{2}$. Here we are dividing $\dfrac{5}{3}$ by $0$ (since $\dfrac{0}{2} = 0$). Division by zero is undefined, so the result is not a rational number.

This single counterexample is sufficient to prove that rational numbers are not closed under division.

Problem: Verify the commutative property of addition for $\dfrac{-5}{8}$ and $\dfrac{3}{11}$.

Solution:

LHS $= \dfrac{-5}{8} + \dfrac{3}{11} = \dfrac{-5 \times 11 + 3 \times 8}{8 \times 11} = \dfrac{-55 + 24}{88} = \dfrac{-31}{88}$

RHS $= \dfrac{3}{11} + \dfrac{-5}{8} = \dfrac{3 \times 8 + (-5) \times 11}{11 \times 8} = \dfrac{24 - 55}{88} = \dfrac{-31}{88}$

Since LHS $=$ RHS $= \dfrac{-31}{88}$, the commutative property of addition is verified.

Problem: Show that subtraction is not commutative for rational numbers using $\dfrac{2}{5}$ and $\dfrac{4}{3}$.

Solution:

$\dfrac{2}{5} - \dfrac{4}{3} = \dfrac{6 - 20}{15} = \dfrac{-14}{15}$

$\dfrac{4}{3} - \dfrac{2}{5} = \dfrac{20 - 6}{15} = \dfrac{14}{15}$

Since $\dfrac{-14}{15} \neq \dfrac{14}{15}$, subtraction is not commutative for rational numbers.

Problem: Verify the associative property of addition for $\dfrac{1}{2}$, $\dfrac{-1}{3}$, and $\dfrac{1}{4}$.

Solution:

LHS $= \left(\dfrac{1}{2} + \dfrac{-1}{3}\right) + \dfrac{1}{4}$

$= \dfrac{3 - 2}{6} + \dfrac{1}{4} = \dfrac{1}{6} + \dfrac{1}{4} = \dfrac{2 + 3}{12} = \dfrac{5}{12}$

RHS $= \dfrac{1}{2} + \left(\dfrac{-1}{3} + \dfrac{1}{4}\right)$

$= \dfrac{1}{2} + \dfrac{-4 + 3}{12} = \dfrac{1}{2} + \dfrac{-1}{12} = \dfrac{6 - 1}{12} = \dfrac{5}{12}$

Since LHS $=$ RHS $= \dfrac{5}{12}$, the associative property of addition is verified.

Problem: Show that subtraction is not associative for rational numbers using $\dfrac{1}{2}$, $\dfrac{1}{3}$, and $\dfrac{1}{4}$.

Solution:

LHS $= \left(\dfrac{1}{2} - \dfrac{1}{3}\right) - \dfrac{1}{4} = \dfrac{1}{6} - \dfrac{1}{4} = \dfrac{2 - 3}{12} = \dfrac{-1}{12}$

RHS $= \dfrac{1}{2} - \left(\dfrac{1}{3} - \dfrac{1}{4}\right) = \dfrac{1}{2} - \dfrac{1}{12} = \dfrac{6 - 1}{12} = \dfrac{5}{12}$

Since $\dfrac{-1}{12} \neq \dfrac{5}{12}$, subtraction is not associative for rational numbers.

Problem: Using the distributive property, simplify $\dfrac{3}{5} \times \left(\dfrac{2}{3} + \dfrac{1}{7}\right)$.

Solution:

Apply distributivity: $a \times (b + c) = a \times b + a \times c$

Problem: Simplify $\dfrac{-2}{9} \times \dfrac{3}{5} + \dfrac{-2}{9} \times \dfrac{7}{5}$.

Solution:

Factor out $\dfrac{-2}{9}$ using the distributive property:

Problem: Verify that $\dfrac{4}{5} \times \left(\dfrac{3}{8} - \dfrac{1}{4}\right) = \dfrac{4}{5} \times \dfrac{3}{8} - \dfrac{4}{5} \times \dfrac{1}{4}$.

Solution:

LHS $= \dfrac{4}{5} \times \left(\dfrac{3}{8} - \dfrac{2}{8}\right) = \dfrac{4}{5} \times \dfrac{1}{8} = \dfrac{4}{40} = \dfrac{1}{10}$

RHS $= \dfrac{4}{5} \times \dfrac{3}{8} - \dfrac{4}{5} \times \dfrac{1}{4} = \dfrac{12}{40} - \dfrac{4}{20} = \dfrac{3}{10} - \dfrac{1}{5} = \dfrac{3}{10} - \dfrac{2}{10} = \dfrac{1}{10}$

Since LHS $=$ RHS $= \dfrac{1}{10}$, the distributive property over subtraction is verified.

Problem: (a) What is the additive identity for rational numbers? (b) What is the multiplicative identity? (c) Is there a rational number that is both?

Solution:

(a) The additive identity is $0$, because $a + 0 = a$ for all rational numbers $a$.

(b) The multiplicative identity is $1$, because $a \times 1 = a$ for all rational numbers $a$.

(c) No. $0$ and $1$ are different numbers. $0$ works for addition and $1$ works for multiplication. No single number serves as identity for both operations.

Problem: Find the additive inverse of each: (a) $\dfrac{-7}{9}$ (b) $\dfrac{11}{-15}$ (c) $\dfrac{0}{3}$.

Solution:

(a) Additive inverse of $\dfrac{-7}{9}$ is $\dfrac{7}{9}$.

Verification: $\dfrac{-7}{9} + \dfrac{7}{9} = \dfrac{-7 + 7}{9} = \dfrac{0}{9} = 0$ ✓

(b) First, write in standard form: $\dfrac{11}{-15} = \dfrac{-11}{15}$. Additive inverse is $\dfrac{11}{15}$.

Verification: $\dfrac{-11}{15} + \dfrac{11}{15} = 0$ ✓

(c) $\dfrac{0}{3} = 0$. Additive inverse of $0$ is $0$.

Verification: $0 + 0 = 0$ ✓

Problem: Find the multiplicative inverse of: (a) $\dfrac{-4}{7}$ (b) $\dfrac{-3}{-5}$ (c) $7$.

Solution:

(a) Multiplicative inverse of $\dfrac{-4}{7}$ is $\dfrac{7}{-4} = \dfrac{-7}{4}$.

Verification: $\dfrac{-4}{7} \times \dfrac{-7}{4} = \dfrac{(-4)(-7)}{(7)(4)} = \dfrac{28}{28} = 1$ ✓

(b) First simplify: $\dfrac{-3}{-5} = \dfrac{3}{5}$. Multiplicative inverse is $\dfrac{5}{3}$.

Verification: $\dfrac{3}{5} \times \dfrac{5}{3} = 1$ ✓

(c) $7 = \dfrac{7}{1}$. Multiplicative inverse is $\dfrac{1}{7}$.

Verification: $7 \times \dfrac{1}{7} = 1$ ✓

Problem: For $\dfrac{-5}{6}$, find both the additive inverse and the multiplicative inverse. Are they the same?

Solution:

Additive inverse of $\dfrac{-5}{6}$ is $\dfrac{5}{6}$ (so their sum is $0$).

Multiplicative inverse of $\dfrac{-5}{6}$ is $\dfrac{6}{-5} = \dfrac{-6}{5}$ (so their product is $1$).

They are not the same: $\dfrac{5}{6} \neq \dfrac{-6}{5}$. In general, the additive inverse and multiplicative inverse of a number are different (except for $-1$, where both are $-1$ itself, and $1$, where the multiplicative inverse is $1$ but the additive inverse is $-1$).

One of the most important ideas in this chapter is that between any two rational numbers, there are infinitely many rational numbers. This is called the density property of rational numbers.

Method 1 — Common Denominator: Convert both fractions to equivalent fractions with the same (larger) denominator, then pick numerators between the two.

Method 2 — Mean Method: The number $\dfrac{a + b}{2}$ always lies between $a$ and $b$. You can repeat this to find as many rational numbers as you want.

Example: Find five rational numbers between $\dfrac{1}{3}$ and $\dfrac{1}{2}$.

Convert to common denominator $48$: $\dfrac{1}{3} = \dfrac{16}{48}$ and $\dfrac{1}{2} = \dfrac{24}{48}$.

Five rational numbers between them: $\dfrac{17}{48}, \dfrac{18}{48}, \dfrac{19}{48}, \dfrac{20}{48}, \dfrac{21}{48}$.

These can be simplified: $\dfrac{18}{48} = \dfrac{3}{8}$, $\dfrac{20}{48} = \dfrac{5}{12}$, etc.

Problem: Represent $\dfrac{-3}{4}$ and $\dfrac{5}{4}$ on the same number line.

Solution:

Divide each unit interval into $4$ equal parts.

$\dfrac{-3}{4}$: lies between $-1$ and $0$, at the $3$rd mark from $0$ towards the left.

$\dfrac{5}{4} = 1\dfrac{1}{4}$: lies between $1$ and $2$, at the $1$st mark from $1$ towards the right.

On the number line:

Problem: Find ten rational numbers between $\dfrac{-3}{5}$ and $\dfrac{2}{5}$.

Solution:

Since both fractions already have the same denominator ($5$), we need rational numbers with numerators between $-3$ and $2$.

But there are only $4$ integers between $-3$ and $2$ (namely $-2, -1, 0, 1$). So we need a larger denominator.

Multiply numerator and denominator by $3$: $\dfrac{-3}{5} = \dfrac{-9}{15}$ and $\dfrac{2}{5} = \dfrac{6}{15}$.

Now integers between $-9$ and $6$ give us plenty of choices:

These are ten rational numbers between $\dfrac{-3}{5}$ and $\dfrac{2}{5}$.

Problem: Verify the commutative property of multiplication for $\dfrac{-5}{3}$ and $\dfrac{14}{9}$.

Solution:

LHS $= \dfrac{-5}{3} \times \dfrac{14}{9} = \dfrac{-5 \times 14}{3 \times 9} = \dfrac{-70}{27}$

RHS $= \dfrac{14}{9} \times \dfrac{-5}{3} = \dfrac{14 \times (-5)}{9 \times 3} = \dfrac{-70}{27}$

Since LHS $=$ RHS $= \dfrac{-70}{27}$, the commutative property of multiplication is verified.

Problem: Verify the associative property of multiplication for $\dfrac{2}{3}$, $\dfrac{-4}{5}$, and $\dfrac{1}{2}$.

Solution:

LHS $= \left(\dfrac{2}{3} \times \dfrac{-4}{5}\right) \times \dfrac{1}{2} = \dfrac{-8}{15} \times \dfrac{1}{2} = \dfrac{-8}{30} = \dfrac{-4}{15}$

RHS $= \dfrac{2}{3} \times \left(\dfrac{-4}{5} \times \dfrac{1}{2}\right) = \dfrac{2}{3} \times \dfrac{-4}{10} = \dfrac{2}{3} \times \dfrac{-2}{5} = \dfrac{-4}{15}$

Since LHS $=$ RHS $= \dfrac{-4}{15}$, the associative property of multiplication is verified.

Problem: Find the value of $\dfrac{-4}{5} \times \dfrac{3}{7} + \dfrac{-4}{5} \times \dfrac{4}{7}$.

Solution:

Using the distributive property: $a \times b + a \times c = a \times (b + c)$

Problem: Find a rational number $x$ such that $x + \dfrac{3}{7} = \dfrac{-2}{7}$.

Solution:

Transpose $\dfrac{3}{7}$ to the RHS:

Verification: $\dfrac{-5}{7} + \dfrac{3}{7} = \dfrac{-5 + 3}{7} = \dfrac{-2}{7}$ ✓

Problem: Find the multiplicative inverse of $-1\dfrac{3}{5}$.

Solution:

First convert to improper fraction: $-1\dfrac{3}{5} = \dfrac{-8}{5}$.

The multiplicative inverse of $\dfrac{-8}{5}$ is $\dfrac{5}{-8} = \dfrac{-5}{8}$.

Verification: $\dfrac{-8}{5} \times \dfrac{-5}{8} = \dfrac{40}{40} = 1$ ✓

Problem: Represent $\dfrac{-5}{8}$ and $\dfrac{3}{8}$ on the number line.

Solution:

Divide the segment between each pair of consecutive integers into $8$ equal parts.

$\dfrac{3}{8}$: lies between $0$ and $1$ — mark the $3$rd division point to the right of $0$.

$\dfrac{-5}{8}$: lies between $-1$ and $0$ — mark the $5$th division point to the left of $0$.

On the number line: $\quad -1 \quad \dfrac{-5}{8} \quad \ldots \quad 0 \quad \ldots \quad \dfrac{3}{8} \quad \ldots \quad 1$

Problem: Find five rational numbers between $\dfrac{1}{3}$ and $\dfrac{1}{2}$.

Solution:

We need a common denominator large enough to give at least $5$ numerators between the two.

$\dfrac{1}{3} = \dfrac{8}{24}$ and $\dfrac{1}{2} = \dfrac{12}{24}$.

Numerators between $8$ and $12$: $9, 10, 11$. That gives only $3$ — not enough.

Multiply by $2$: $\dfrac{1}{3} = \dfrac{16}{48}$ and $\dfrac{1}{2} = \dfrac{24}{48}$.

Five rational numbers: $\dfrac{17}{48}, \dfrac{18}{48}, \dfrac{19}{48}, \dfrac{20}{48}, \dfrac{21}{48}$.

Simplifying where possible: $\dfrac{18}{48} = \dfrac{3}{8}$, $\dfrac{20}{48} = \dfrac{5}{12}$.

Problem: Find three rational numbers between $\dfrac{1}{4}$ and $\dfrac{1}{2}$ using the mean method.

Solution:

Step 1: Find the mean of $\dfrac{1}{4}$ and $\dfrac{1}{2}$:

So $\dfrac{1}{4} < \dfrac{3}{8} < \dfrac{1}{2}$.

Step 2: Find the mean of $\dfrac{1}{4}$ and $\dfrac{3}{8}$:

Step 3: Find the mean of $\dfrac{3}{8}$ and $\dfrac{1}{2}$:

Three rational numbers between $\dfrac{1}{4}$ and $\dfrac{1}{2}$: $\dfrac{5}{16}, \dfrac{3}{8}, \dfrac{7}{16}$.

Problem: Find five rational numbers between $\dfrac{-3}{4}$ and $\dfrac{-1}{2}$.

Solution:

Convert to common denominator: $\dfrac{-3}{4} = \dfrac{-3}{4}$ and $\dfrac{-1}{2} = \dfrac{-2}{4}$.

Only one integer ($-2$ is not between $-3$ and $-2$... wait, we need integers strictly between $-3$ and $-2$, but there are none). We need a larger denominator.

Multiply by $6$: $\dfrac{-3}{4} = \dfrac{-18}{24}$ and $\dfrac{-1}{2} = \dfrac{-12}{24}$.

Five rational numbers: $\dfrac{-17}{24}, \dfrac{-16}{24}, \dfrac{-15}{24}, \dfrac{-14}{24}, \dfrac{-13}{24}$.

Simplifying: $\dfrac{-16}{24} = \dfrac{-2}{3}$, $\dfrac{-15}{24} = \dfrac{-5}{8}$.

Problem: Represent $-2\dfrac{1}{3}$ on the number line.

Solution:

Convert to improper fraction: $-2\dfrac{1}{3} = \dfrac{-7}{3}$.

This lies between $-3$ and $-2$ (since $\dfrac{-9}{3} = -3$ and $\dfrac{-6}{3} = -2$).

Divide the segment between $-3$ and $-2$ into $3$ equal parts.

Starting from $-3$, count $2$ parts to the right (or equivalently, from $-2$, count $1$ part to the left).

The point at the $2$nd division from $-3$ represents $\dfrac{-7}{3} = -2\dfrac{1}{3}$.

Problem: Simplify $\dfrac{2}{3} \times \dfrac{-3}{4} + \dfrac{1}{2} \times \dfrac{-3}{4}$.

Solution:

Notice that $\dfrac{-3}{4}$ is common in both terms. Using the distributive property (in reverse):

Problem: Find $x$ if $\dfrac{-3}{5} \times x = \dfrac{6}{25}$.

Solution:

Multiply both sides by the multiplicative inverse of $\dfrac{-3}{5}$, which is $\dfrac{-5}{3}$:

Verification: $\dfrac{-3}{5} \times \dfrac{-2}{5} = \dfrac{6}{25}$ ✓

Problem: Name the property illustrated by each statement:
(a) $\dfrac{-4}{7} + 0 = \dfrac{-4}{7}$
(b) $\dfrac{2}{3} \times \dfrac{3}{2} = 1$
(c) $\dfrac{-1}{5} + \dfrac{1}{5} = 0$
(d) $\dfrac{3}{7} \times (\dfrac{-2}{9} + \dfrac{5}{9}) = \dfrac{3}{7} \times \dfrac{-2}{9} + \dfrac{3}{7} \times \dfrac{5}{9}$

Solution:

(a) Additive identity property — adding $0$ to any rational number gives the same number.

(b) Multiplicative inverse property — $\dfrac{3}{2}$ is the multiplicative inverse (reciprocal) of $\dfrac{2}{3}$.

(c) Additive inverse property — $\dfrac{1}{5}$ is the additive inverse of $\dfrac{-1}{5}$.

(d) Distributive property of multiplication over addition.

Problem: Find six rational numbers between $3$ and $4$.

Solution:

Write $3 = \dfrac{21}{7}$ and $4 = \dfrac{28}{7}$.

Six rational numbers between them: $\dfrac{22}{7}, \dfrac{23}{7}, \dfrac{24}{7}, \dfrac{25}{7}, \dfrac{26}{7}, \dfrac{27}{7}$.

Note: $\dfrac{22}{7}$ is a famous approximation of $\pi$!

Problem: Express $\dfrac{-48}{60}$ in standard form.

Solution: