What types of problems appear in the Class 9 Maths Chapter 3 exercises?

Chapter 3 exercises ask students to classify numbers into N, W, Z, Q and irrationals; decide whether decimal expansions terminate or repeat; convert recurring decimals like 0.overline{6} into p/q form; place sqrt(2), sqrt(3), sqrt(5) on a number line; and rationalise denominators such as 1/(sqrt(2.

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The World of Numbers — Exercises

🎓 Class 9 Mathematics CBSE Theory Ch 3 — The World of Numbers ⏱ ~40 min

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3.9 Properties of Real Numbers

Real numbers obey the same arithmetic rules you have used since primary school — but now those rules apply across rationals AND irrationals. The key properties are:

Closure	For any \(a, b \in \mathbb{R}\): \(a+b, a-b, a\cdot b\) and (when \(b \neq 0\)) \(a \div b\) are real numbers.
Commutativity	\(a + b = b + a;\ \ a \cdot b = b \cdot a\).
Associativity	\((a+b)+c = a+(b+c)\); \((ab)c = a(bc)\).
Distributivity	\(a(b+c) = ab + ac\).
Identity	\(a + 0 = a\); \(a \cdot 1 = a\).
Inverse	For every \(a\): \(a + (-a) = 0\). For every \(a \neq 0\): \(a \cdot \frac{1}{a} = 1\).

3.9.1 Operations Mixing Rationals and Irrationals

What happens when you combine a rational and an irrational using +, −, ×, or ÷?

Key Rules

(rational) + (irrational) = irrational. E.g. \(2 + \sqrt{3}\) is irrational.
(rational ≠ 0) × (irrational) = irrational. E.g. \(5\sqrt{2}\) is irrational.
(irrational) ± (irrational) MAY be rational or irrational. E.g. \(\sqrt{2} + (-\sqrt{2}) = 0\) (rational).
(irrational) × (irrational) MAY be rational. E.g. \(\sqrt{2} \cdot \sqrt{2} = 2\).

Activity: Which Mixtures Stay Irrational?

L4 Analyse

Materials: Pen, paper, calculator (optional).

Predict: If you add two irrational numbers, must the result be irrational? Test five cases.

Compute \(\sqrt{2} + \sqrt{3}\) (≈ 3.146...). Is it rational or irrational?
Compute \((3 + \sqrt{5}) + (3 - \sqrt{5})\). Surprise — equal to 6, rational!
Compute \(\sqrt{2} \cdot \sqrt{8}\). Equal to \(\sqrt{16} = 4\), rational!
Compute \(\pi + 1\). Irrational.
Compute \(\pi - \pi\). Zero, rational.

Insight: The irrational numbers are NOT closed under addition, subtraction, or multiplication. They are a "leaky" set. Only \(\mathbb{R}\) (the union with rationals) is closed under all four operations.

Exercise 3.1 — The Number System

Q1. Is zero a rational number? Justify.

Yes. Zero can be written as \(\frac{0}{1}, \frac{0}{2}, \frac{0}{(-7)}\), etc. — i.e. \(\frac{p}{q}\) with \(p = 0\) (an integer) and \(q\) any non-zero integer. Hence \(0 \in \mathbb{Q}\).

Q2. Find six rational numbers between 3 and 4.

Multiply both by 7 (just need any common denominator that gives "room"): \(3 = \frac{21}{7},\ 4 = \frac{28}{7}\). Six rationals in between: \(\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}\).

Q3. Find five rational numbers between \(\frac{3}{5}\) and \(\frac{4}{5}\).

Multiply numerator & denominator by 6: \(\frac{3}{5} = \frac{18}{30},\ \frac{4}{5} = \frac{24}{30}\). Five between: \(\frac{19}{30}, \frac{20}{30}, \frac{21}{30}, \frac{22}{30}, \frac{23}{30}\) (i.e. \(\frac{19}{30}, \frac{2}{3}, \frac{7}{10}, \frac{11}{15}, \frac{23}{30}\)).

Q4. State whether each is true or false. Give a reason for your answer.
(i) Every natural number is a whole number. (ii) Every integer is a whole number. (iii) Every rational number is a whole number.

(i) True — \(\mathbb{N} \subset \mathbb{W}\). (ii) False — negatives like \(-3\) are integers but not whole numbers. (iii) False — \(\frac{1}{2}\) is rational but not a whole number.

Q5. State whether each is true or false. Justify.
(i) Every irrational number is a real number. (ii) Every point on the number line is a rational number. (iii) Every real number is irrational.

(i) True — irrationals form a subset of reals. (ii) False — points like \(\sqrt{2}\) are irrational. (iii) False — every rational is real but not irrational.

Q6. Are square roots of all positive integers irrational? If not, give an example.

No. If the integer is a perfect square, its root is rational. Example: \(\sqrt{4} = 2,\ \sqrt{9} = 3,\ \sqrt{36} = 6\) — all rational.

Exercise 3.2 — Decimal Expansions

Q1. Write the following in decimal form and say what kind of decimal expansion each has:
(i) \(\frac{36}{100}\) (ii) \(\frac{1}{11}\) (iii) \(4\frac{1}{8}\) (iv) \(\frac{3}{13}\) (v) \(\frac{2}{11}\) (vi) \(\frac{329}{400}\)

(i) \(0.36\) — terminating. (ii) \(0.\overline{09}\) — non-terminating recurring. (iii) \(\frac{33}{8} = 4.125\) — terminating. (iv) \(0.\overline{230769}\) — period 6. (v) \(0.\overline{18}\). (vi) \(\frac{329}{400} = 0.8225\) — terminating (denom \(400 = 2^4 \cdot 5^2\)).

Q2. Express the following in the form \(\frac{p}{q}\):
(i) \(0.\overline{6}\) (ii) \(0.4\overline{7}\) (iii) \(0.\overline{001}\)

(i) Let \(x = 0.\overline{6}\). Then \(10x = 6.\overline{6}\); \(10x - x = 6\); \(x = \frac{6}{9} = \frac{2}{3}\).
(ii) Let \(x = 0.4\overline{7} = 0.4777\ldots\). \(10x = 4.\overline{7}, 100x = 47.\overline{7}\). Subtract: \(100x - 10x = 47.\overline{7} - 4.\overline{7} = 43\); \(90x = 43\); \(x = \frac{43}{90}\).
(iii) Let \(x = 0.\overline{001}\). \(1000x = 1.\overline{001}\); \(999x = 1\); \(x = \frac{1}{999}\).

Q3. Find three different irrational numbers between \(\frac{5}{7}\) and \(\frac{9}{11}\).

\(\frac{5}{7} = 0.\overline{714285},\ \frac{9}{11} = 0.\overline{81}\). Any non-terminating, non-recurring decimal between 0.7142… and 0.8181… qualifies. Three examples:
0.7505005000500005…
0.7670670067000067…
0.7898989898998999…
(Each digit pattern shows no repeating block.)

Q4. Classify the following as rational or irrational:
(i) \(\sqrt{23}\) (ii) \(\sqrt{225}\) (iii) \(0.3796\) (iv) \(7.478478\ldots\) (v) \(1.101001000100001\ldots\)

(i) Irrational (23 not a perfect square). (ii) \(\sqrt{225} = 15\) — Rational. (iii) Terminating — Rational. (iv) Recurring 478 — Rational = \(\frac{7471}{999}\). (v) Non-terminating, non-recurring (gap between 1's increases) — Irrational.

Exercise 3.3 — Operations on Real Numbers

Q1. Classify the following as rational or irrational and explain:
(i) \(2 - \sqrt{5}\) (ii) \((3 + \sqrt{23}) - \sqrt{23}\) (iii) \(\frac{2\sqrt{7}}{7\sqrt{7}}\) (iv) \(\frac{1}{\sqrt{2}}\) (v) \(2\pi\)

(i) Irrational (rational − irrational). (ii) \(= 3\), Rational. (iii) \(= \frac{2}{7}\), Rational. (iv) \(= \frac{\sqrt{2}}{2}\), Irrational. (v) Irrational (rational × irrational, with rational ≠ 0).

Q2. Simplify: (i) \((3 + \sqrt{3})(2 + \sqrt{2})\) (ii) \((3 + \sqrt{3})(3 - \sqrt{3})\) (iii) \((\sqrt{5} + \sqrt{2})^2\) (iv) \((\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})\)

(i) \(6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}\).
(ii) \(9 - 3 = 6\).
(iii) \(5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}\).
(iv) \(5 - 2 = 3\).

Q3. Rationalise the denominator: (i) \(\frac{1}{\sqrt{7}}\) (ii) \(\frac{1}{\sqrt{7} - \sqrt{6}}\) (iii) \(\frac{1}{\sqrt{5} + \sqrt{2}}\) (iv) \(\frac{1}{\sqrt{7} - 2}\)

(i) Multiply top & bottom by \(\sqrt{7}\): \(\frac{\sqrt{7}}{7}\).
(ii) Multiply by conjugate \((\sqrt{7}+\sqrt{6})\): \(\frac{\sqrt{7}+\sqrt{6}}{7-6} = \sqrt{7}+\sqrt{6}\).
(iii) Multiply by \((\sqrt{5}-\sqrt{2})\): \(\frac{\sqrt{5}-\sqrt{2}}{5-2} = \frac{\sqrt{5}-\sqrt{2}}{3}\).
(iv) Multiply by \((\sqrt{7}+2)\): \(\frac{\sqrt{7}+2}{7-4} = \frac{\sqrt{7}+2}{3}\).

Exercise 3.4 — Real Numbers and Their Decimal Expansions

Q1. Without actually performing long division, state whether the following will have a terminating or non-terminating recurring decimal expansion:
(i) \(\frac{13}{3125}\) (ii) \(\frac{17}{8}\) (iii) \(\frac{64}{455}\) (iv) \(\frac{15}{1600}\) (v) \(\frac{29}{343}\)

Check the prime factorisation of each denominator:
(i) \(3125 = 5^5\) — only 5. Terminating.
(ii) \(8 = 2^3\) — only 2. Terminating.
(iii) \(455 = 5 \cdot 7 \cdot 13\) — primes 7, 13. Non-terminating recurring.
(iv) \(1600 = 2^6 \cdot 5^2\) — only 2 and 5. Terminating.
(v) \(343 = 7^3\) — prime 7. Non-terminating recurring.

Q2. Write down the decimal expansions (without long division) of the terminating fractions in Q1.

(i) \(\frac{13}{3125} = \frac{13 \cdot 2^5}{10^5} = \frac{416}{100000} = 0.00416\).
(ii) \(\frac{17}{8} = \frac{17 \cdot 125}{1000} = \frac{2125}{1000} = 2.125\).
(iv) \(\frac{15}{1600} = \frac{15}{2^6 \cdot 5^2} = \frac{15 \cdot 5^4}{2^6 \cdot 5^6} = \frac{15 \cdot 625}{10^6} \cdot 100 = \frac{9375}{1000000} = 0.009375\). (Or directly \(= 0.009375\).)

Chapter Summary

Key Takeaways — Chapter 3

Counting was born from need: tally marks on bones (Lebombo, ~35,000 BCE) — the first mathematics.
Natural numbers: \(\mathbb{N} = \{1, 2, 3, \ldots\}\) — the counting numbers.
Whole numbers: \(\mathbb{W} = \mathbb{N} \cup \{0\}\) — Brahmagupta's revolutionary inclusion of zero (628 CE).
Integers: \(\mathbb{Z} = \{\ldots, -2, -1, 0, 1, 2, \ldots\}\) — fortunes (positive) and debts (negative).
Rational numbers: \(\mathbb{Q} = \{p/q : p, q \in \mathbb{Z}, q \neq 0\}\) — closed under +, −, ×, ÷ (≠ 0).
Density: Between any two distinct rationals lie infinitely many other rationals.
Decimal expansion of rationals: Either terminating or non-terminating recurring; terminates iff denominator (in lowest form) has prime factors only 2 and/or 5.
Irrational numbers: Cannot be written as p/q. Their decimals are non-terminating, non-recurring. Examples: \(\sqrt{2}, \sqrt{3}, \pi, e\).
Real numbers: \(\mathbb{R} = \mathbb{Q} \cup \text{Irrationals}\) — every point on the number line is a real number, and vice-versa.
Geometric construction of \(\sqrt{n}\) uses Pythagoras' theorem and the spiral of Theodorus.
Rationalisation: Multiply numerator and denominator by the conjugate to clear surds from the denominator.

Competency-Based Questions

Scenario: A bookshop tracks weekly inventory using fractional cartons. They received \(\frac{17}{8}\) cartons of Maths textbooks, \(\frac{29}{40}\) carton of Science books, \(\frac{1}{6}\) carton of damaged returns, and \(\frac{22}{7}\) cartons of stationery. The accountant wants exact decimal records.

Q1. Without long division, classify each of the four fractions as terminating or non-terminating recurring.

L3 Apply

\(\frac{17}{8}\): denom \(2^3\) → terminating. \(\frac{29}{40}\): denom \(2^3 \cdot 5\) → terminating. \(\frac{1}{6} = \frac{1}{2 \cdot 3}\) → recurring (3 ≠ 2,5). \(\frac{22}{7}\) → recurring (7 ≠ 2,5).

Q2. The accountant prefers data that fits in 3 decimal places exactly. Analyse which entries can be recorded faithfully and which cannot.

L4 Analyse

\(\frac{17}{8} = 2.125\) (3 dp ✓). \(\frac{29}{40} = 0.725\) (3 dp ✓). \(\frac{1}{6} = 0.1\overline{6}\) — needs rounding. \(\frac{22}{7} = 3.\overline{142857}\) — needs rounding. Two entries cannot be stored exactly with 3 decimal places.

Q3. The accountant rounds \(\frac{22}{7}\) to 3.143. Evaluate the absolute and percentage error compared to the exact value.

L5 Evaluate

Exact \(\frac{22}{7} \approx 3.142857\). Error = \(3.143 - 3.142857 = 0.000143\). Percentage = \(\frac{0.000143}{3.142857} \times 100\% \approx 0.0046\%\) — quite acceptable for inventory but unacceptable for, say, satellite navigation.

Q4. Design an inventory unit (i.e. choose a denominator) that guarantees ALL fractional cartons in this bookshop will give terminating decimals. Justify your choice mathematically.

L6 Create

Solution: Choose any denominator of the form \(2^m \cdot 5^n\). For example, use 1/1000 of a carton (= \(2^3 \cdot 5^3\)) as the smallest unit. Then any fractional carton will be \(\frac{k}{1000}\) — a 3-decimal terminating value. Other valid bases: 1/100, 1/10, 1/8, 1/200, etc. — all powers of 2 × powers of 5.

Assertion–Reason Questions

Assertion (A): The number \(0.\overline{37}\) is a rational number.
Reason (R): A non-terminating but recurring decimal can always be expressed in the form \(\frac{p}{q}\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

(a) — \(0.\overline{37} = \frac{37}{99}\). R explains A.

Assertion (A): \(\sqrt{2} + \sqrt{3}\) is irrational.
Reason (R): Sum of two irrationals is always irrational.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

(c) — A is TRUE (\(\sqrt{2}+\sqrt{3}\) cannot be written as p/q; provable). R is FALSE — \(\sqrt{2}+(-\sqrt{2}) = 0\) is rational.

Assertion (A): \(\frac{15}{1600}\) has a terminating decimal expansion.
Reason (R): 1600 = \(2^6 \cdot 5^2\) — only primes 2 and 5.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

(a) — \(\frac{15}{1600} = 0.009375\) terminates. R provides exactly the criterion required.

Frequently Asked Questions

How do you convert a recurring decimal like 0.6666... into p/q form?

Let x = 0.6666.... Multiply both sides by 10 to get 10x = 6.6666.... Subtract: 10x - x = 6.6666... - 0.6666..., giving 9x = 6, so x = 6/9 = 2/3. The same multiply-and-subtract trick works for any repeating decimal.

Is 0 a rational number? Explain for Class 9.

Yes. 0 is rational because it can be written as 0/1 (or 0/q for any non-zero integer q). It satisfies the definition p/q with integer p = 0 and integer q != 0.

How do you find a rational number between 1/4 and 1/3?

One quick method is to take the average: (1/4 + 1/3)/2 = (3/12 + 4/12)/2 = (7/12)/2 = 7/24. Since 1/4 = 6/24 and 1/3 = 8/24, the number 7/24 lies between them. You can find infinitely many rationals between any two distinct rationals.

How do you locate sqrt(5) on a number line?

Draw segment OA = 2 on the number line, then a perpendicular AB = 1. By Pythagoras OB = sqrt(2^2 + 1^2) = sqrt(5). With centre O and radius OB draw an arc cutting the number line at P. Then OP = sqrt(5).

How do you rationalise 1/(sqrt(2) + 1)?

Multiply numerator and denominator by the conjugate sqrt(2) - 1: 1/(sqrt(2)+1) x (sqrt(2)-1)/(sqrt(2)-1) = (sqrt(2)-1)/((sqrt(2))^2 - 1^2) = (sqrt(2)-1)/(2 - 1) = sqrt(2) - 1.

Is the sum of a rational and irrational number always irrational?

Yes. If r is rational and i is irrational, then r + i must be irrational. If r + i were rational, then i = (r + i) - r would also be rational (difference of two rationals), contradicting that i is irrational.

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