CBSE Class 9 Maths — Number Systems

Chapter Overview & Weightage

Number Systems is Chapter 1 of CBSE Class 9 Maths. It establishes the foundational understanding of real numbers — rational, irrational, and the real number line — that all subsequent chapters in the course build upon. Students who understand this chapter well find algebra, polynomials, and coordinate geometry much more accessible.

Exam Year	Marks Allocated	Question Types
2024	8–10 marks	1 MCQ + 2 short + 1 long
2023	8 marks	1 short + 1 SA (3M) + 1 LA (4M)
2022	6–8 marks	2 short + 1 rationalise
2021	6 marks	1 short + 1 long
2020	8 marks	2 short + 1 long

Number Systems typically contributes 6–10 marks in the Class 9 annual exam. The most frequently tested topics are: identifying rational/irrational numbers, representing irrational numbers on the number line, simplifying expressions with surds, and rationalizing denominators. These are scoring topics if you practice them systematically.

Key Concepts You Must Know

Types of numbers (hierarchy from smallest to largest set):

Natural numbers (N): 1, 2, 3, …
Whole numbers (W): 0, 1, 2, 3, …
Integers (Z): …, -2, -1, 0, 1, 2, …
Rational numbers (Q): numbers expressible as $p/q$ where $p, q$ are integers and $q \neq 0$ . Their decimal expansions are either terminating (1/4 = 0.25) or non-terminating repeating ( $1/3 = 0.333...$ )
Irrational numbers: numbers that cannot be expressed as $p/q$ . Decimal expansions are non-terminating and non-repeating. Examples: $\sqrt{2}, \sqrt{3}, \pi, e$
Real numbers (R): all rational + all irrational numbers = everything on the number line

Every real number corresponds to a unique point on the number line, and vice versa.

Key properties of irrational numbers:

Sum/difference of a rational and irrational number is irrational
Product of a non-zero rational and irrational number is irrational
Sum/product of two irrationals may be rational or irrational (e.g., $\sqrt{2} + (-\sqrt{2}) = 0$ , which is rational)

Laws of exponents (for real numbers $a > 0$ , $b > 0$ and rational exponents):

$a^m \cdot a^n = a^{m+n}$
$(a^m)^n = a^{mn}$
$a^m \cdot b^m = (ab)^m$
$a^0 = 1$
$a^{-n} = 1/a^n$
$a^{1/n} = \sqrt[n]{a}$

Important Formulas

To rationalize $\frac{1}{\sqrt{a}}$ : multiply by $\frac{\sqrt{a}}{\sqrt{a}}$ → result $= \frac{\sqrt{a}}{a}$

To rationalize $\frac{1}{a + \sqrt{b}}$ : multiply by $\frac{a - \sqrt{b}}{a - \sqrt{b}}$ → uses difference of squares: $(a+\sqrt{b})(a-\sqrt{b}) = a^2 - b$

Example: $\frac{1}{3 + \sqrt{2}} = \frac{3 - \sqrt{2}}{(3)^2 - (\sqrt{2})^2} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}$

$(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$

$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$

$(\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{ab} + b$

$a^{p/q} = (a^{1/q})^p = (\sqrt[q]{a})^p$

Solved Previous Year Questions

PYQ 1: (CBSE 2023, 3 marks)

Q: Represent $\sqrt{5}$ on the number line.

Solution: Draw a number line. Mark point O (0) and A (2). Construct OA = 2 units. At A, draw a perpendicular AB = 1 unit. Join OB.

By Pythagoras theorem: $OB = \sqrt{OA^2 + AB^2} = \sqrt{4 + 1} = \sqrt{5}$

With O as centre and OB as radius, draw an arc to cut the number line at P. OP = $\sqrt{5}$ .

This represents $\sqrt{5}$ on the number line.

PYQ 2: (CBSE 2024, 2 marks)

Q: Simplify: $\frac{3}{\sqrt{3} + 1} + \frac{1}{\sqrt{3} - 1}$

Solution:

Rationalize each term:

$\frac{3}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{3(\sqrt{3}-1)}{3-1} = \frac{3(\sqrt{3}-1)}{2}$

$\frac{1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} = \frac{\sqrt{3}+1}{3-1} = \frac{\sqrt{3}+1}{2}$

Adding: $\frac{3(\sqrt{3}-1) + (\sqrt{3}+1)}{2} = \frac{3\sqrt{3} - 3 + \sqrt{3} + 1}{2} = \frac{4\sqrt{3} - 2}{2} = 2\sqrt{3} - 1$

PYQ 3: (CBSE 2022, 4 marks)

Q: If $a = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ and $b = \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ , find $a^2 + b^2$ .

Solution:

Note that $b = 1/a$ , so $ab = 1$ .

Rationalizing $a$ : $a = \frac{(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} = \frac{3 - 2\sqrt{6} + 2}{3 - 2} = 5 - 2\sqrt{6}$

Similarly $b = 5 + 2\sqrt{6}$

$a + b = 10$ , $ab = (5-2\sqrt{6})(5+2\sqrt{6}) = 25 - 24 = 1$

$a^2 + b^2 = (a+b)^2 - 2ab = 100 - 2 = \mathbf{98}$

Difficulty Distribution

Difficulty	% of Chapter	Question Type
Easy (40%)	Identify rational/irrational, terminating/non-terminating decimals	MCQ, fill in blanks
Medium (40%)	Rationalise denominators, simplify surds, laws of exponents	2–3 mark short answers
Hard (20%)	Multi-step surd problems, represent irrational numbers on number line	4–5 mark long answers

Expert Strategy

For rationalisation problems, the conjugate of $(a + b)$ is $(a - b)$ . To rationalise $\frac{p}{a + \sqrt{b}}$ , multiply numerator and denominator by $(a - \sqrt{b})$ . The key step is $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$ — this eliminates the square root from the denominator. Practice this with at least 10 different denominators until it feels automatic.

For “represent on number line” questions: the method uses the geometric fact that the hypotenuse of a right triangle can represent an irrational. To represent $\sqrt{n}$ : form a right triangle with legs that give $n$ as the sum of perfect squares. Most common: $\sqrt{5} = \sqrt{4+1}$ (legs 2 and 1), $\sqrt{13} = \sqrt{4+9}$ (legs 2 and 3), $\sqrt{10} = \sqrt{9+1}$ (legs 3 and 1). Draw it neatly with compass and ruler.

For laws of exponents problems, remember: $a^{m/n} = \sqrt[n]{a^m}$ . Simplify by finding a common base or common exponent. For $2^3 \times 2^4 = 2^7$ (same base, add exponents). For $(2^3)^4 = 2^{12}$ (power of power, multiply exponents).

Common Traps

Trap 1: Assuming $\sqrt{2} + \sqrt{3} = \sqrt{5}$ . This is wrong — you cannot add surds with different radicands like regular addition. $\sqrt{2} + \sqrt{3}$ stays as $\sqrt{2} + \sqrt{3}$ ; it does not simplify. Only surds with the same radicand can be added: $3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$ .

Trap 2: Writing that all square roots are irrational. $\sqrt{4} = 2$ (rational), $\sqrt{9} = 3$ (rational). Only square roots of non-perfect-square positive integers are irrational. Similarly, $\sqrt[3]{8} = 2$ is rational.

Trap 3: Applying $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ when $a$ or $b$ is negative. This rule only works for non-negative numbers. $\sqrt{(-4)(-9)} \neq \sqrt{-4} \times \sqrt{-9}$ in real numbers.