CBSE Class 9 Mathematics · Ganita Manjari (NEP 2026)

The World of Numbers The World of Numbers (संख्याओं का संसार)

From tally marks on the Ishango bone to Brahmagupta's revolution of Zero, rational density, decimal expansions, the discovery of irrationals ($\sqrt{2}, \pi$), and the continuous Real Number line. Ishango bone के counting marks से लेकर Brahmagupta के Zero के revolution, rational numbers की density, decimal expansions, irrationals ($\sqrt{2}, \pi$) और continuous Real Number line का complete अध्ययन।

1. The Evolution of Numbers & Real-Life Analogies 1. Numbers का Evolution और Real-Life Analogies

🏛️ The Merchant of Lothal & Tally Counting (Natural Numbers $\mathbb{N}$)

Thousands of years ago, a Harappan merchant in Lothal traded bags of spices for copper ingots. To keep count without abstract symbols, humans notched bones (like the 20,000-year-old Ishango bone) and used 12 finger joints on one hand. Counting items gives us Natural Numbers ($\mathbb{N} = \{1, 2, 3, \dots\}$) [NCERT p.41-45]. हजारों साल पहले, Lothal के port पर merchant spices के bags के बदले copper ingots exchange करते थे। Counting के लिए इंसान bones पर निशान लगाते थे (जैसे 20,000 साल पुरानी Ishango bone) और एक हाथ के 12 finger joints का इस्तेमाल करते थे। वस्तुओं को count करने से हमें Natural Numbers ($\mathbb{N} = \{1, 2, 3, \dots\}$) मिलते हैं [NCERT p.41-45]।

⚖️ Brahmagupta's Revolution: Fortunes vs. Debts (Zero & Integers $\mathbb{Z}$)

If you have 5 bags of grain and give away 5, what do you have left? Ancient Greeks saw a "void". In 628 CE, the Indian mathematician Brahmagupta transformed this void into the operational number Śhūnya (Zero). He established the laws of positive assets (dhana / fortunes) and negative liabilities (ṛiṇa / debts): "A debt subtracted from nothing is a fortune" ($0 - (-d) = +d$) and "The product of two debts is a fortune" ($(-a) \times (-b) = +ab$) [NCERT p.45-46]. अगर आपके पास 5 bags अनाज है और आपने पांचों दे दिए, तो क्या बचा? पहले इसे सिर्फ खालीपन माना जाता था। 628 CE में महान भारतीय गणितज्ञ Brahmagupta ने इस void को एक operational number Zero बनाया। उन्होंने profit/assets (dhana) और debt (ṛiṇa) के rules बनाए: "Debt को cancel करने का मतलब fortune का आना है" ($0 - (-d) = +d$) और "दो debts का product एक fortune होता है" ($(-a) \times (-b) = +ab$) [NCERT p.45-46]।

🍰 Sharing the Cake: Fractions & Density (Rational Numbers $\mathbb{Q}$)

When a farmer divides a field among 3 children, or a recipe requires half a cup of ghee, integers are not enough. We define Rational Numbers as numbers in the form $\frac{p}{q}$ where $p, q \in \mathbb{Z}$, $q \neq 0$, and $\gcd(p, q) = 1$. Rational numbers are dense: between any two rational numbers, there are infinitely many more rational numbers [NCERT p.47-52]. जब एक किसान अपने 3 बच्चों में खेत बांटता है या किसी recipe में आधा कप घी चाहिए होता है, तब integers काफी नहीं होते। हम Rational Numbers को $\frac{p}{q}$ के रूप में define करते हैं जहाँ $p, q \in \mathbb{Z}$, $q \neq 0$ और $p, q$ co-prime हैं। Rational numbers dense होते हैं: किन्हीं भी दो rational numbers के बीच अनगिनत rational numbers मौजूद होते हैं [NCERT p.47-52]।

The Real Number System Hierarchy

Natural ($\mathbb{N}$) ⊂ Whole ($\mathbb{W}$) ⊂ Integers ($\mathbb{Z}$) ⊂ Rationals ($\mathbb{Q}$) ⊂ Real Numbers ($\mathbb{R}$)

$\text{Real Numbers } (\mathbb{R}) = \text{Rational Numbers } (\mathbb{Q}) \cup \text{Irrational Numbers } (\mathbb{I}), \quad \mathbb{Q} \cap \mathbb{I} = \emptyset$

2. Visual Concept Lab & Interactive Simulation 2. Visual Concept Lab और Interactive Simulation

🔍 Real Number Classifier & Decimal Expansion Analyzer

NCERT Concepts pp. 47–61

Select Preset:

Selected: 7/20

Classification: Rational Number ($\mathbb{Q}$) & Real Number ($\mathbb{R}$)

Decimal Expansion: Terminating ($0.35$)

Mathematical Reason: Denominator $20 = 2^2 \times 5^1$. Since the prime factors are strictly in the form $2^m \times 5^n$, it terminates after $\max(2,1) = 2$ decimal places.

Natural ($\mathbb{N}$)

Whole ($\mathbb{W}$)

Integer ($\mathbb{Z}$)

Rational ($\mathbb{Q}$)

Yes

Irrational ($\mathbb{I}$)

Real ($\mathbb{R}$)

Yes

🌀 The Spiral of Theodorus & Geometric Root Mapper

NCERT Fig. 3.11 & 3.14 (pp. 55, 66)

Select Root:

Current Hypotenuse

OB = √2 ≈ 1.414

Pythagoras Step

1² + 1² = (√2)² = 2

Number Line Point

P ≈ 1.414 units

3. Master Formula & Key Properties Sheet 3. Master Formula और Key Properties Sheet

1. Rational Number Standard Form

$$\frac{p}{q}, \quad q \neq 0, \quad \gcd(p, q) = 1$$

Both $p, q \in \mathbb{Z}$. Can be terminating or non-terminating recurring [NCERT p.47].

2. Brahmagupta's Signed Laws

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

Removing a debt is equivalent to adding a fortune [NCERT p.46].

3. Density & Arithmetic Mean

$$a < \frac{a+b}{2} < b$$

The average of two rationals always lies strictly between them [NCERT p.52].

4. Terminating Decimal Condition

$$q = 2^m \cdot 5^n \quad (m, n \ge 0)$$

Terminates after exactly $\max(m, n)$ decimal places [NCERT p.58].

5. Pure Repeating Conversion ($p/q$)

$$0.\overline{d_1 d_2 \dots d_k} = \frac{d_1 d_2 \dots d_k}{99\dots9}$$

Example: $0.\overline{45} = \frac{45}{99} = \frac{5}{11}$ [NCERT p.59].

6. Spiral of Theodorus Recurrence

$$h_n = \sqrt{h_{n-1}^2 + 1^2} = \sqrt{n+1}$$

Iterative hypotenuses: $\sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \dots$ [NCERT p.56, 66].

7. Mādhava-Leibniz Series for $\pi$

$$\pi = 4 \left( 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \right)$$

Kerala School of Mathematics (14th Century) [NCERT p.56].

8. Interval Density Theorem

$$k_2 - k_1 > n + 1$$

Condition to fit $n$ distinct rationals with common denominator $m$ [NCERT p.65].

4. Solved Think & Reflect Conceptual Proofs 4. हल किए गए Think & Reflect Conceptual Proofs

Think & Reflect 1 Why does a negative times a negative equal a positive? Explain in terms of debt. [NCERT p.45] Think & Reflect 1 Negative times negative positive क्यों होता है? Debt (कर्ज) के real-life example से समझाएं। [NCERT p.45]

1. Concept A negative number represents debt (owing money), while a negative multiplier represents removing or cancelling that debt.

2. Mathematical Meaning If someone removes ($-$) four debts of ₹3 each ($-3$), they take away ₹12 of debt: $$(-4) \times (-3) = +12$$ Removing liability makes you financially richer.

🎯 Final Conclusion: Debt cancellation is wealth addition: $(-a) \times (-b) = +ab$.

1. Concept (अवधारणा) Negative number debt (कर्ज) को दर्शाता है, और negative multiplier उस debt को remove (हटाने) को दर्शाता है।

2. Step-by-Step Calculation अगर कोई व्यक्ति आपके ₹3 वाले 4 debts ($-3$) को cancel ($-$) कर देता है: $$(-4) \times (-3) = +12$$ Debt कम होने से आपकी net wealth ₹12 बढ़ जाती है।

🎯 Final Conclusion: Debt को cancel करना धन आने के बराबर है: $(-a) \times (-b) = +ab$ प्राप्त होता है।

Think & Reflect 2 Can you explain why we strictly require $q \neq 0$ in the definition of a rational number $\frac{p}{q}$? [NCERT p.47] Think & Reflect 2 Rational number $\frac{p}{q}$ की definition में $q \neq 0$ होना अनिवार्य क्यों है? [NCERT p.47]

1. Meaning of Division $\frac{p}{q} = x$ means finding a unique number $x$ such that $x \cdot q = p$.

2. Non-Zero Numerator ($p \neq 0, q = 0$) We would need $x \cdot 0 = p$. Since any number times 0 is 0, no such $x$ exists (undefined).

3. Zero Numerator ($p = 0, q = 0$) We would need $x \cdot 0 = 0$. Every real number satisfies this, so $x$ is indeterminate.

🎯 Final Conclusion: Division by zero is mathematically undefined. Therefore, $q \neq 0$ is mandatory.

1. Concept (अवधारणा) $\frac{p}{q} = x$ का अर्थ है एक ऐसा unique number $x$ ढूंढना ताकि $x \cdot q = p$ हो।

2. Case 1 ($p \neq 0, q = 0$) हमें $x \cdot 0 = p$ चाहिए। किसी भी number को 0 से multiply करने पर 0 मिलता है, इसलिए ऐसा कोई $x$ संभव नहीं है (undefined)।

3. Case 2 ($p = 0, q = 0$) $x \cdot 0 = 0$ हर real number के लिए true है, इसलिए value unique नहीं होती (indeterminate)।

🎯 Final Conclusion: 0 से division mathematically undefined है, इसलिए $q \neq 0$ होना आवश्यक है।

Think & Reflect 3 Prove by contradiction that $\sqrt{3}$ is an irrational number. [NCERT p.55] Think & Reflect 3 Proof by contradiction से prove करें कि $\sqrt{3}$ एक irrational number है। [NCERT p.55]

1. Assumption Assume $\sqrt{3}$ is rational: $\sqrt{3} = \frac{p}{q}$, where $p, q \in \mathbb{Z}, q \neq 0$, and $\gcd(p, q) = 1$ (co-prime).

2. Square Both Sides $$3 = \frac{p^2}{q^2} \implies p^2 = 3q^2$$ This means $p^2$ is divisible by 3, so $p$ is divisible by 3. Let $p = 3k$.

3. Substitute $p = 3k$ $$(3k)^2 = 3q^2 \implies 9k^2 = 3q^2 \implies q^2 = 3k^2$$ This implies $q^2$ (and hence $q$) is also divisible by 3.

4. Contradiction Both $p$ and $q$ share 3 as a common factor, contradicting co-prime assumption.

🎯 Final Conclusion: The initial assumption is false. Therefore, $\sqrt{3}$ is strictly irrational.

1. Assumption (मानना) मान लेते हैं कि $\sqrt{3}$ rational है: $\sqrt{3} = \frac{p}{q}$, जहाँ $p, q$ co-prime integers हैं ($q \neq 0$)।

2. Square करना $$3 = \frac{p^2}{q^2} \implies p^2 = 3q^2$$ इसका मतलब $p^2$ 3 से divisible है, इसलिए $p$ भी 3 का multiple होगा। $p = 3k$ put करते हैं।

3. Value Put करना $$(3k)^2 = 3q^2 \implies 9k^2 = 3q^2 \implies q^2 = 3k^2$$ इससे $q^2$ और $q$ भी 3 से divisible साबित होते हैं।

4. Contradiction (विरोधाभास) $p$ और $q$ दोनों में common factor 3 आ गया, जो कि co-prime होने की शर्त का उल्लंघन है।

🎯 Final Conclusion: अतः हमारी assumption गलत थी, $\sqrt{3}$ एक irrational number है।

Think & Reflect 4 Using algebra, explain why $0.\overline{9} = 0.99999\dots$ is exactly equal to 1. [NCERT p.62] Think & Reflect 4 Algebra का use करके prove करें कि $0.\overline{9} = 0.99999\dots$ ठीक 1 के equal है। [NCERT p.62]

1. Algebraic Setup Let $x = 0.99999\dots$ (Equation 1)

2. Multiply by 10 Since 1 digit repeats, multiply by 10: $$10x = 9.99999\dots \quad (\text{Equation 2})$$

3. Subtract Equations $$10x - x = 9.99999\dots - 0.99999\dots \implies 9x = 9$$

4. Solve for $x$ $x = \frac{9}{9} = 1$.

🎯 Final Conclusion: There is no real number between $0.\overline{9}$ and $1$. Hence, $0.\overline{9} = 1$ identically.

1. Variable मानना मान लेते हैं कि $x = 0.99999\dots$ (समीकरण 1)।

2. 10 से Multiply चूंकि 1 digit repeat हो रहा है, दोनों तरफ 10 से multiply करेंगे: $$10x = 9.99999\dots \quad (\text{समीकरण 2})$$

3. Subtract करना समीकरण 2 में से समीकरण 1 घटाने पर: $$10x - x = 9.99999\dots - 0.99999\dots \implies 9x = 9$$

4. $x$ Calculate करना $x = \frac{9}{9} = 1$ प्राप्त होता है।

🎯 Final Conclusion: $0.\overline{9}$ और $1$ के बीच कोई अंतर नहीं है, अतः $0.\overline{9} = 1$ सिद्ध होता है।

NCERT 5. Exercise Set 3.1 — Historical Counting & Natural Numbers 5. Exercise Set 3.1 — Historical Counting और Natural Numbers

Ex 3.1 · Q1 A merchant in the port city of Lothal receives 15 copper ingots for every 2 bags of spices. If he brings 12 bags of spices to the market, how many copper ingots will he leave with? [NCERT p.45] Ex 3.1 · Q1 Lothal के port पर एक merchant को हर 2 bags spices के बदले 15 copper ingots मिलते हैं। यदि वह 12 bags spices लाता है, तो उसे कितने copper ingots मिलेंगे? [NCERT p.45]

1. Given Ratio Exchange rate $= \frac{15\text{ ingots}}{2\text{ bags}}$.

2. Calculation Total spice bags $= 12$. $$\text{Total ingots} = 12 \times \frac{15}{2} = 6 \times 15 = 90\text{ ingots}$$

🎯 Final Answer: The merchant will leave with 90 copper ingots.

1. Given Ratio Exchange rate $= \frac{15\text{ ingots}}{2\text{ bags}}$ दिया गया है।

2. Calculation Total spices के bags $= 12$ हैं। $$\text{Total ingots} = 12 \times \frac{15}{2} = 6 \times 15 = 90\text{ ingots}$$

🎯 Final Answer: Merchant को कुल 90 copper ingots प्राप्त होंगे।

Ex 3.1 · Q2 Look at the sequence on one column of the Ishango bone: 11, 13, 17, 19. What do these numbers have in common? List the next three numbers that fit this pattern. [NCERT p.45] Ex 3.1 · Q2 Ishango bone के एक column की sequence देखें: 11, 13, 17, 19। इनमें क्या common pattern है? अगले 3 numbers identify करें। [NCERT p.45]

1. Pattern Identification The numbers $11, 13, 17, 19$ are consecutive Prime Numbers between 10 and 20.

2. Next Primes • Prime after 19 is 23.
• Prime after 23 is 29.
• Prime after 29 is 31.

🎯 Final Answer: Common property: Consecutive Prime Numbers. Next three: 23, 29, 31.

1. Pattern पहचान $11, 13, 17, 19$ सभी 10 और 20 के बीच आने वाले consecutive Prime Numbers (अभाज्य संख्याएं) हैं।

2. Next 3 Primes • 19 के बाद अगला prime 23 है।
• 23 के बाद अगला prime 29 है।
• 29 के बाद अगला prime 31 है।

🎯 Final Answer: Common property: Consecutive Prime Numbers। अगले 3 numbers 23, 29, 31 हैं।

Ex 3.1 · Q3 Natural Numbers are closed under addition. Are they closed under subtraction? Justify with examples. [NCERT p.45] Ex 3.1 · Q3 Natural Numbers addition में closed हैं। क्या वे subtraction में भी closed हैं? Examples देकर justify करें। [NCERT p.45]

1. Definition Closure under subtraction requires that for any $a, b \in \mathbb{N}$, the difference $a - b \in \mathbb{N}$.

2. Counter-Examples • Take $a = 3, b = 5 \implies 3 - 5 = -2 \notin \mathbb{N}$ (Negative integer).
• Take $a = 4, b = 4 \implies 4 - 4 = 0 \notin \mathbb{N}$ (Zero).

🎯 Final Answer: NO, natural numbers are not closed under subtraction because the result can be negative or zero.

1. Closure Property Subtraction में closed होने के लिए जरूरी है कि किन्हीं भी दो natural numbers $a, b$ के लिए $a - b$ भी natural number ही आए।

2. Counter-Examples • $a = 3, b = 5$ लेने पर: $3 - 5 = -2 \notin \mathbb{N}$ (negative number आता है)।
• $a = 4, b = 4$ लेने पर: $4 - 4 = 0 \notin \mathbb{N}$ (Zero natural number नहीं है)।

🎯 Final Answer: NO, natural numbers subtraction में closed नहीं हैं क्योंकि $a - b$ negative या zero हो सकता है।

Ex 3.1 · Q4 Ancient Indians used finger joints to count (3 joints per finger, counted using the thumb). How many joints can you count on one hand? How does this relate to base-12 systems? [NCERT p.45] Ex 3.1 · Q4 प्राचीन काल में finger joints से counting की जाती थी (प्रति finger 3 joints, thumb से count करते हुए)। 1 हाथ पर कितने joints count होते हैं? यह base-12 system से कैसे संबंधित है? [NCERT p.45]

1. Count per Hand 4 fingers with 3 joints each $= 4 \times 3 = 12\text{ joints}$.

2. Base-12 & Base-60 Connection Counting 12 units on one hand created the dozen (base-12). Tracking completed dozens on the 5 fingers of the other hand gave $12 \times 5 = 60$ (the sexagesimal base-60 system).

🎯 Final Answer: Exactly 12 joints. It is the basis for the duodecimal (base-12) and sexagesimal (base-60) systems.

1. Joints की संख्या 4 fingers में हर finger के 3 joints होते हैं: $4 \times 3 = 12\text{ joints}$।

2. Base-12 और Base-60 संबंध एक हाथ से 12 count करने से 1 dozen (base-12) बना। दूसरे हाथ की 5 उंगलियों से dozen गिनने पर $12 \times 5 = 60$ (base-60 system) का जन्म हुआ।

🎯 Final Answer: कुल 12 joints। यह ancient base-12 और base-60 (घड़ी के 60 मिनट/सेकंड) का anatomical आधार है।

NCERT 6. Exercise Set 3.2 — Signed Numbers & Brahmagupta's Laws 6. Exercise Set 3.2 — Signed Numbers और ब्रह्मगुप्त के नियम

Ex 3.2 · Q1 The temperature in Ladakh is recorded as $4^\circ\text{C}$ at noon. By midnight, it drops by $15^\circ\text{C}$. What is the midnight temperature? [NCERT p.46] Ex 3.2 · Q1 Ladakh में दोपहर का temperature $4^\circ\text{C}$ था। आधी रात तक इसमें $15^\circ\text{C}$ की गिरावट आती है। Midnight का temperature क्या होगा? [NCERT p.46]

1. Given Values $T_{\text{noon}} = +4^\circ\text{C}$, Drop $\Delta T = -15^\circ\text{C}$.

2. Calculation $$T_{\text{midnight}} = 4 - 15 = -11^\circ\text{C}$$

🎯 Final Answer: The midnight temperature is $-11^\circ\text{C}$.

1. Given Data $T_{\text{noon}} = +4^\circ\text{C}$, temperature drop $\Delta T = -15^\circ\text{C}$ दिया है।

2. Calculation $$T_{\text{midnight}} = 4 - 15 = -11^\circ\text{C}$$

🎯 Final Answer: Midnight का temperature $-11^\circ\text{C}$ होगा।

Ex 3.2 · Q2 A spice trader takes a loan (debt) of ₹850. The next day, he makes a profit (fortune) of ₹1,200. The following week, he incurs a loss of ₹450. Write this sequence as an integer equation and find his final financial standing. [NCERT p.46] Ex 3.2 · Q2 एक spice trader ₹850 का debt लेता है। अगले दिन ₹1,200 का profit कमाता है और फिर ₹450 का loss होता है। Integer equation लिखकर final financial standing calculate करें। [NCERT p.46]

1. Sign Assignment Debt $= -850$, Profit $= +1200$, Loss $= -450$.

2. Integer Equation $$\text{Balance} = (-850) + (+1200) + (-450) = 1200 - 1300 = -100$$

🎯 Final Answer: Equation: $(-850) + 1200 - 450 = -100$ (Net debt of ₹100).

1. Sign Mapping Debt $= -850$, Profit $= +1200$, Loss $= -450$ assign करते हैं।

2. Integer Equation $$\text{Net Balance} = (-850) + (+1200) + (-450) = 1200 - 1300 = -100$$

🎯 Final Answer: Equation: $(-850) + 1200 - 450 = -100$ (व्यापारी पर ₹100 का net debt रहेगा)।

Ex 3.2 · Q3 Calculate using Brahmagupta's laws: (i) $(-12) \times 5$, (ii) $(-8) \times (-7)$, (iii) $0 - (-14)$, (iv) $(-20) \div 4$. [NCERT p.46] Ex 3.2 · Q3 Brahmagupta's rules से calculate करें: (i) $(-12) \times 5$, (ii) $(-8) \times (-7)$, (iii) $0 - (-14)$, (iv) $(-20) \div 4$। [NCERT p.46]

1. Evaluations • (i) $(-12) \times 5 = \mathbf{-60}$ (Negative $\times$ Positive = Negative)
• (ii) $(-8) \times (-7) = \mathbf{+56}$ (Negative $\times$ Negative = Positive)
• (iii) $0 - (-14) = 0 + 14 = \mathbf{+14}$ (Subtracting debt is adding fortune)
• (iv) $(-20) \div 4 = \mathbf{-5}$ (Negative $\div$ Positive = Negative)

🎯 Final Answers: (i) $-60$, (ii) $+56$, (iii) $+14$, (iv) $-5$.

1. Calculations • (i) $(-12) \times 5 = \mathbf{-60}$ (Negative $\times$ Positive $=$ Negative)
• (ii) $(-8) \times (-7) = \mathbf{+56}$ (Negative $\times$ Negative $=$ Positive)
• (iii) $0 - (-14) = 0 + 14 = \mathbf{+14}$ (Debt घटाना fortune जोड़ने के बराबर है)
• (iv) $(-20) \div 4 = \mathbf{-5}$ (Negative $\div$ Positive $=$ Negative)

🎯 Final Answers: (i) $-60$, (ii) $+56$, (iii) $+14$, (iv) $-5$ प्राप्त होते हैं।

Ex 3.2 · Q4 Explain, using a real-world example of debt, why subtracting a negative number is the same as adding a positive number (e.g. $10 - (-5) = 15$). [NCERT p.46] Ex 3.2 · Q4 Debt के real-world example से समझाएं कि negative number घटाना positive number जोड़ने के समान क्यों है ($10 - (-5) = 15$)। [NCERT p.46]

1. Practical Scenario You have ₹10 cash, but you owe a debt of ₹5 (net balance $= 10 - 5 = ₹5$).

2. Debt Forgiveness If a friend pays off and removes ($-$) your ₹5 debt ($-5$), you now keep the entire ₹10 without obligation, increasing your purchasing power by ₹5.

3. Mathematical Equivalent Removing a ₹5 debt has the exact same economic effect as receiving ₹5 extra cash: $$10 - (-5) = 10 + 5 = 15$$

🎯 Final Conclusion: Subtracting a negative removes a liability, which increases wealth identically to adding a positive number.

1. Real-Life Example आपके पास ₹10 हैं और आप पर ₹5 का debt है (actual worth $= ₹5$)।

2. Debt Cancel होना यदि कोई आपका ₹5 का debt cancel ($-$) कर दे, तो आपकी net worth ₹5 बढ़ जाती है।

3. Mathematical Proof किसी debt को हटाना सीधे cash मिलने के बराबर है: $$10 - (-5) = 10 + 5 = 15$$

🎯 Final Conclusion: Negative number घटाने से liability कम होती है, जो positive number जोड़ने के समान है।

NCERT 7. Exercise Set 3.3 — Rational Arithmetic & Properties 7. Exercise Set 3.3 — Rational Arithmetic और Properties

Ex 3.3 · Q1 Prove that the following rational numbers are equal: (i) $\frac{2}{3}$ and $\frac{4}{6}$, (ii) $\frac{5}{4}$ and $\frac{10}{8}$, (iii) $-\frac{3}{5}$ and $-\frac{6}{10}$, (iv) $\frac{9}{3}$ and $3$. [NCERT p.49] Ex 3.3 · Q1 Prove करें कि दिए गए rational numbers equal हैं: (i) $\frac{2}{3}$ और $\frac{4}{6}$, (ii) $\frac{5}{4}$ और $\frac{10}{8}$, (iii) $-\frac{3}{5}$ और $-\frac{6}{10}$, (iv) $\frac{9}{3}$ और $3$। [NCERT p.49]

1. Method Divide numerator and denominator by their greatest common divisor ($\gcd$).

2. Proofs • (i) $\frac{4}{6} = \frac{4 \div 2}{6 \div 2} = \frac{2}{3} \implies \frac{2}{3} = \frac{4}{6}$.
• (ii) $\frac{10}{8} = \frac{10 \div 2}{8 \div 2} = \frac{5}{4} \implies \frac{5}{4} = \frac{10}{8}$.
• (iii) $-\frac{6}{10} = -\frac{6 \div 2}{10 \div 2} = -\frac{3}{5} \implies -\frac{3}{5} = -\frac{6}{10}$.
• (iv) $\frac{9}{3} = \frac{9 \div 3}{3 \div 3} = 3 \implies \frac{9}{3} = 3$.

🎯 Final Conclusion: All pairs represent identical equivalent rational numbers.

1. Method (तरीका) Numerator और denominator को उनके $\gcd$ से divide करके simplest form में लाते हैं।

2. Step-by-Step Proof • (i) $\frac{4}{6} = \frac{4 \div 2}{6 \div 2} = \frac{2}{3} \implies \frac{2}{3} = \frac{4}{6}$।
• (ii) $\frac{10}{8} = \frac{10 \div 2}{8 \div 2} = \frac{5}{4} \implies \frac{5}{4} = \frac{10}{8}$।
• (iii) $-\frac{6}{10} = -\frac{6 \div 2}{10 \div 2} = -\frac{3}{5} \implies -\frac{3}{5} = -\frac{6}{10}$।
• (iv) $\frac{9}{3} = \frac{9 \div 3}{3 \div 3} = 3 \implies \frac{9}{3} = 3$।

🎯 Final Conclusion: सभी pairs आपस में identical equivalent rational numbers हैं।

Ex 3.3 · Q2-Q5 Evaluate: (i) Sum: $\frac{2}{5} + \frac{3}{10}$, (ii) Difference: $\frac{5}{6} - \frac{1}{4}$, (iii) Product: $\frac{2}{3} \times \frac{3}{10}$, (iv) Quotient: $\frac{2}{3} \div \frac{3}{10}$. [NCERT p.49-50] Ex 3.3 · Q2-Q5 Evaluate करें: (i) Sum: $\frac{2}{5} + \frac{3}{10}$, (ii) Difference: $\frac{5}{6} - \frac{1}{4}$, (iii) Product: $\frac{2}{3} \times \frac{3}{10}$, (iv) Quotient: $\frac{2}{3} \div \frac{3}{10}$। [NCERT p.49-50]

1. Sum $\text{LCM}(5, 10) = 10 \implies \frac{4}{10} + \frac{3}{10} = \mathbf{\frac{7}{10}}$.

2. Difference $\text{LCM}(6, 4) = 12 \implies \frac{10}{12} - \frac{3}{12} = \mathbf{\frac{7}{12}}$.

3. Product $\frac{2 \times 3}{3 \times 10} = \frac{6}{30} = \mathbf{\frac{1}{5}}$.

4. Quotient $\frac{2}{3} \times \frac{10}{3} = \mathbf{\frac{20}{9}}$.

🎯 Final Answers: (i) $\frac{7}{10}$, (ii) $\frac{7}{12}$, (iii) $\frac{1}{5}$, (iv) $\frac{20}{9}$.

1. Sum Calculation $\text{LCM}(5, 10) = 10 \implies \frac{4}{10} + \frac{3}{10} = \mathbf{\frac{7}{10}}$।

2. Difference Calculation $\text{LCM}(6, 4) = 12 \implies \frac{10}{12} - \frac{3}{12} = \mathbf{\frac{7}{12}}$।

3. Product Calculation $\frac{2 \times 3}{3 \times 10} = \frac{6}{30} = \mathbf{\frac{1}{5}}$।

4. Quotient Calculation Reciprocal से multiply करने पर: $\frac{2}{3} \times \frac{10}{3} = \mathbf{\frac{20}{9}}$।

🎯 Final Answers: (i) $\frac{7}{10}$, (ii) $\frac{7}{12}$, (iii) $\frac{1}{5}$, (iv) $\frac{20}{9}$ प्राप्त होते हैं।

Ex 3.3 · Q6-Q8 (a) Verify distributive law: $\left(\frac{1}{2} + \frac{3}{4}\right) \times \frac{8}{3} = \frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}$. (b) Solve for rational $x$: $x\left(\frac{5}{6} + \frac{3}{5}\right) = \frac{5}{6}x + \frac{1}{2}$. [NCERT p.50] Ex 3.3 · Q6-Q8 (a) Distributive law verify करें: $\left(\frac{1}{2} + \frac{3}{4}\right) \times \frac{8}{3} = \frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}$। (b) $x$ solve करें: $x\left(\frac{5}{6} + \frac{3}{5}\right) = \frac{5}{6}x + \frac{1}{2}$। [NCERT p.50]

1. Distributive Verification $$\text{LHS} = \left(\frac{2+3}{4}\right) \times \frac{8}{3} = \frac{5}{4} \times \frac{8}{3} = \frac{10}{3}$$ $$\text{RHS} = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \implies \text{LHS} = \text{RHS}$$

2. Solve Linear Rational Equation $$\frac{5}{6}x + \frac{3}{5}x = \frac{5}{6}x + \frac{1}{2}$$ Subtract $\frac{5}{6}x$ from both sides: $$\frac{3}{5}x = \frac{1}{2} \implies x = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$

🎯 Final Answer: Distributive law verified ($\text{LHS} = \text{RHS} = \frac{10}{3}$). Value of $x = \mathbf{\frac{5}{6}}$.

1. Distributive Law Verify करना $$\text{LHS} = \left(\frac{2+3}{4}\right) \times \frac{8}{3} = \frac{5}{4} \times \frac{8}{3} = \frac{10}{3}$$ $$\text{RHS} = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \implies \text{LHS} = \text{RHS} \quad (\text{Verified})$$

2. Equation Solve करना $$\frac{5}{6}x + \frac{3}{5}x = \frac{5}{6}x + \frac{1}{2}$$ दोनों sides से $\frac{5}{6}x$ subtract करने पर: $$\frac{3}{5}x = \frac{1}{2} \implies x = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$$

🎯 Final Answer: Distributive property verify हो गई ($\text{LHS} = \text{RHS} = \frac{10}{3}$)। $x$ की value $\frac{5}{6}$ प्राप्त होती है।

NCERT 8. Exercise Set 3.4 — Density & Number Line Representation 8. Exercise Set 3.4 — Density और Number Line Representation

Ex 3.4 · Q1-Q2 (i) Represent $\frac{2}{3}$, $-\frac{5}{4}$, and $1\frac{1}{2}$ on a single number line. (ii) Find three distinct rational numbers that lie strictly between $-\frac{1}{2}$ and $\frac{1}{4}$. [NCERT p.52] Ex 3.4 · Q1-Q2 (i) $\frac{2}{3}$, $-\frac{5}{4}$ और $1\frac{1}{2}$ को एक ही number line पर दर्शाएं। (ii) $-\frac{1}{2}$ और $\frac{1}{4}$ के बीच 3 distinct rational numbers identify करें। [NCERT p.52]

1. Number Line Locations • $-\frac{5}{4} = -1.25$ lies between $-1$ and $-2$.
• $\frac{2}{3} \approx 0.67$ lies between $0$ and $1$.
• $1\frac{1}{2} = 1.5$ lies exactly halfway between $1$ and $2$.

2. Three Rationals between $-\frac{1}{2}$ and $\frac{1}{4}$ Convert to common denominator 16: $$-\frac{1}{2} = -\frac{8}{16}, \quad \frac{1}{4} = \frac{4}{16}$$ Three intermediate rationals are: $$-\frac{4}{16} = -\frac{1}{4}, \quad \frac{0}{16} = 0, \quad \frac{2}{16} = \frac{1}{8}$$

🎯 Final Answer: Three rational numbers: $-\frac{1}{4}, 0, \frac{1}{8}$.

1. Number Line पर Position • $-\frac{5}{4} = -1.25$, $-1$ और $-2$ के बीच स्थित है।
• $\frac{2}{3} \approx 0.67$, $0$ और $1$ के बीच स्थित है।
• $1\frac{1}{2} = 1.5$, $1$ और $2$ का exact midpoint है।

2. $-\frac{1}{2}$ और $\frac{1}{4}$ के बीच 3 Rationals Denominator 16 में convert करते हैं: $$-\frac{1}{2} = -\frac{8}{16}, \quad \frac{1}{4} = \frac{4}{16}$$ बीच के 3 rational numbers: $$-\frac{4}{16} = -\frac{1}{4}, \quad \frac{0}{16} = 0, \quad \frac{2}{16} = \frac{1}{8}$$

🎯 Final Answer: तीन rational numbers $-\frac{1}{4}, 0, \frac{1}{8}$ प्राप्त होते हैं।

Ex 3.4 · Q4 A tailor has $15\frac{3}{4}\text{ metres}$ of fine silk. If making one kurta requires $2\frac{1}{4}\text{ metres}$, exactly how many kurtas can he make? [NCERT p.53] Ex 3.4 · Q4 एक tailor के पास $15\frac{3}{4}\text{ metres}$ silk है। यदि 1 kurta बनाने में $2\frac{1}{4}\text{ metres}$ कपड़ा लगता है, तो वह कितने kurtas बना सकता है? [NCERT p.53]

1. Convert Mixed Fractions $$\text{Total Silk} = \frac{15 \times 4 + 3}{4} = \frac{63}{4}\text{ m}$$ $$\text{Silk per Kurta} = \frac{2 \times 4 + 1}{4} = \frac{9}{4}\text{ m}$$

2. Division $$\text{Kurtas} = \frac{63}{4} \div \frac{9}{4} = \frac{63}{4} \times \frac{4}{9} = \frac{63}{9} = 7$$

🎯 Final Answer: The tailor can make exactly 7 kurtas.

1. Improper Fraction में बदलना $$\text{Total Silk} = \frac{15 \times 4 + 3}{4} = \frac{63}{4}\text{ m}$$ $$\text{Per Kurta Silk} = \frac{2 \times 4 + 1}{4} = \frac{9}{4}\text{ m}$$

2. Division Calculation $$\text{Total Kurtas} = \frac{63}{4} \div \frac{9}{4} = \frac{63}{4} \times \frac{4}{9} = 7$$

🎯 Final Answer: Tailor कुल 7 kurtas बना सकता है।

NCERT 9. Exercise Set 3.5 — Decimal Expansions & p/q Conversions 9. Exercise Set 3.5 — Decimal Expansions और p/q Conversions

Ex 3.5 · Q1 Without long division, determine whether $\frac{7}{20}$, $\frac{4}{15}$, and $\frac{13}{250}$ are terminating or repeating decimals. Verify by long division. [NCERT p.61] Ex 3.5 · Q1 बिना long division के predict करें कि $\frac{7}{20}$, $\frac{4}{15}$ और $\frac{13}{250}$ terminating हैं या repeating। Long division से verify करें। [NCERT p.61]

1. Factorise Denominators • $\frac{7}{20}$: Denominator $20 = 2^2 \times 5^1 \implies$ Terminating ($0.35$).
• $\frac{4}{15}$: Denominator $15 = 3 \times 5 \implies$ Repeating ($0.2\overline{6}$).
• $\frac{13}{250}$: Denominator $250 = 2^1 \times 5^3 \implies$ Terminating ($0.052$).

🎯 Final Answers: (i) $\frac{7}{20}$: Terminating ($0.35$), (ii) $\frac{4}{15}$: Repeating ($0.2\overline{6}$), (iii) $\frac{13}{250}$: Terminating ($0.052$).

1. Prime Factors Check • $\frac{7}{20}$: Denominator $20 = 2^2 \times 5^1$ (केवल 2 और 5 हैं) $\implies$ Terminating ($0.35$)।
• $\frac{4}{15}$: Denominator $15 = 3 \times 5$ (3 मौजूद है) $\implies$ Non-terminating Repeating ($0.2\overline{6}$)।
• $\frac{13}{250}$: Denominator $250 = 2^1 \times 5^3$ (केवल 2 और 5 हैं) $\implies$ Terminating ($0.052$)।

🎯 Final Answers: (i) Terminating ($0.35$), (ii) Repeating ($0.2\overline{6}$), (iii) Terminating ($0.052$)।

Ex 3.5 · Q3 Classify as rational or irrational: (i) $\sqrt{81}$, (ii) $\sqrt{12}$, (iii) $0.3333\dots$, (iv) $0.1234512345\dots$, (v) $1.010010001\dots$, (vi) $23.560185612239874790120$. [NCERT p.62] Ex 3.5 · Q3 Rational या irrational classify करें: (i) $\sqrt{81}$, (ii) $\sqrt{12}$, (iii) $0.3333\dots$, (iv) $0.1234512345\dots$, (v) $1.010010001\dots$, (vi) $23.560185612239874790120$। [NCERT p.62]

1. Classification • (i) $\sqrt{81} = 9 = \frac{9}{1} \implies$ Rational ($\mathbb{Q}$).
• (ii) $\sqrt{12} = 2\sqrt{3} \implies$ Irrational ($\mathbb{I}$).
• (iii) $0.\overline{3} = \frac{1}{3} \implies$ Rational ($\mathbb{Q}$).
• (iv) $0.\overline{12345} = \frac{12345}{99999} \implies$ Rational ($\mathbb{Q}$).
• (v) $1.010010001\dots$ (Non-terminating, non-repeating) $\implies$ Irrational ($\mathbb{I}$).
• (vi) Finite terminating decimal $\implies$ Rational ($\mathbb{Q}$).

🎯 Final Summary: Rational: (i), (iii), (iv), (vi). Irrational: (ii), (v).

1. Classification Check • (i) $\sqrt{81} = 9 \implies$ Rational ($\mathbb{Q}$)।
• (ii) $\sqrt{12} = 2\sqrt{3} \implies$ Irrational ($\mathbb{I}$)।
• (iii) $0.\overline{3} = \frac{1}{3} \implies$ Rational ($\mathbb{Q}$)।
• (iv) $0.\overline{12345} = \frac{12345}{99999} \implies$ Rational ($\mathbb{Q}$)।
• (v) $1.010010001\dots$ (non-repeating pattern) $\implies$ Irrational ($\mathbb{I}$)।
• (vi) Finite terminating decimal $\implies$ Rational ($\mathbb{Q}$)।

🎯 Final Summary: Rational numbers: (i), (iii), (iv), (vi); Irrational numbers: (ii), (v) प्राप्त होते हैं।

NCERT 10. NCERT Solutions — End-of-Chapter Exercises (Q1 to Q16) 10. NCERT Solutions — End-of-Chapter Exercises (Q1 से Q16)

End Ex · Q1 Convert by long division into terminating or repeating decimals: (i) $\frac{3}{50}$, (ii) $\frac{2}{9}$. [NCERT p.64] End Ex · Q1 Long division से decimal form identify करें: (i) $\frac{3}{50}$, (ii) $\frac{2}{9}$। [NCERT p.64]

1. Calculations • (i) $3 \div 50 = \mathbf{0.06}$ (Terminating decimal).
• (ii) $2 \div 9 = 0.222\dots = \mathbf{0.\overline{2}}$ (Non-terminating repeating decimal with period 1).

🎯 Final Answer: (i) $0.06$ (Terminating), (ii) $0.\overline{2}$ (Non-terminating Repeating).

1. Long Division हल • (i) $3 \div 50 = \mathbf{0.06}$ (Terminating decimal प्राप्त होता है)।
• (ii) $2 \div 9 = 0.222\dots = \mathbf{0.\overline{2}}$ (Non-terminating repeating decimal, period $= 1$)।

🎯 Final Answer: (i) $0.06$ (Terminating), (ii) $0.\overline{2}$ (Repeating)।

End Ex · Q2 Prove that $\sqrt{5}$ is an irrational number. [NCERT p.64] End Ex · Q2 Prove करें कि $\sqrt{5}$ एक irrational number है। [NCERT p.64]

1. Assumption Assume $\sqrt{5}$ is rational: $\sqrt{5} = \frac{p}{q}$ ($p, q \in \mathbb{Z}, q \neq 0, \gcd(p, q) = 1$).

2. Square Both Sides $5 = \frac{p^2}{q^2} \implies p^2 = 5q^2$. Thus, $p^2$ (and $p$) is divisible by 5. Let $p = 5k$.

3. Substitute $p = 5k$ $(5k)^2 = 5q^2 \implies 25k^2 = 5q^2 \implies q^2 = 5k^2$. Thus, $q$ is also divisible by 5.

4. Contradiction Both $p, q$ share 5 as common factor, contradicting co-prime assumption.

🎯 Final Conclusion: Hence, $\sqrt{5}$ is an irrational number (Proved).

1. Assumption (मानना) मान लेते हैं कि $\sqrt{5}$ rational है: $\sqrt{5} = \frac{p}{q}$ ($p, q$ co-prime integers हैं, $q \neq 0$)।

2. Square करना $5 = \frac{p^2}{q^2} \implies p^2 = 5q^2$। इसलिए $p^2$ और $p$ 5 से divisible होंगे। $p = 5k$ put करते हैं।

3. Value Put करना $(5k)^2 = 5q^2 \implies 25k^2 = 5q^2 \implies q^2 = 5k^2$। अतः $q$ भी 5 से divisible है।

4. Contradiction $p$ और $q$ दोनों का common factor 5 आ गया, जो assumption का उल्लंघन है।

🎯 Final Conclusion: अतः हमारी assumption गलत थी, $\sqrt{5}$ एक irrational number सिद्ध होता है।

End Ex · Q3 Convert into $\frac{p}{q}$ form: (i) $12.6$, (ii) $0.0120$, (iii) $3.052$, (iv) $1.235$, (v) $0.\overline{23}$, (vi) $2.0\overline{5}$, (vii) $2.125$, (viii) $3.12\overline{5}$, (ix) $2.\overline{1625}$. [NCERT p.64] End Ex · Q3 $\frac{p}{q}$ form में convert करें: (i) $12.6$, (ii) $0.0120$, (iii) $3.052$, (iv) $1.235$, (v) $0.\overline{23}$, (vi) $2.0\overline{5}$, (vii) $2.125$, (viii) $3.12\overline{5}$, (ix) $2.\overline{1625}$। [NCERT p.64]

1. Terminating Decimals • (i) $12.6 = \frac{126}{10} = \mathbf{\frac{63}{5}}$.
• (ii) $0.012 = \frac{12}{1000} = \mathbf{\frac{3}{250}}$.
• (iii) $3.052 = \frac{3052}{1000} = \mathbf{\frac{763}{250}}$.
• (iv) $1.235 = \frac{1235}{1000} = \mathbf{\frac{247}{200}}$.
• (vii) $2.125 = \frac{2125}{1000} = \mathbf{\frac{17}{8}}$.

2. Recurring Decimals • (v) $0.\overline{23} = \mathbf{\frac{23}{99}}$.
• (vi) $x = 2.0\overline{5} \implies 100x - 10x = 205.5\dots - 20.5\dots \implies 90x = 185 \implies x = \mathbf{\frac{37}{18}}$.
• (viii) $x = 3.12\overline{5} \implies 1000x - 100x = 3125.5\dots - 312.5\dots \implies 900x = 2813 \implies x = \mathbf{\frac{2813}{900}}$.
• (ix) $2.\overline{1625} = 2 + \frac{1625}{9999} = \mathbf{\frac{21623}{9999}}$.

🎯 Final Summary: (i) $\frac{63}{5}$, (ii) $\frac{3}{250}$, (iii) $\frac{763}{250}$, (iv) $\frac{247}{200}$, (v) $\frac{23}{99}$, (vi) $\frac{37}{18}$, (vii) $\frac{17}{8}$, (viii) $\frac{2813}{900}$, (ix) $\frac{21623}{9999}$.

1. Terminating Decimals का हल • (i) $12.6 = \frac{126}{10} = \mathbf{\frac{63}{5}}$।
• (ii) $0.012 = \frac{12}{1000} = \mathbf{\frac{3}{250}}$।
• (iii) $3.052 = \frac{3052}{1000} = \mathbf{\frac{763}{250}}$।
• (iv) $1.235 = \frac{1235}{1000} = \mathbf{\frac{247}{200}}$।
• (vii) $2.125 = \frac{2125}{1000} = \mathbf{\frac{17}{8}}$।

2. Recurring Decimals का हल • (v) $0.\overline{23} = \mathbf{\frac{23}{99}}$।
• (vi) $x = 2.0\overline{5} \implies 90x = 185 \implies x = \mathbf{\frac{37}{18}}$।
• (viii) $x = 3.12\overline{5} \implies 900x = 2813 \implies x = \mathbf{\frac{2813}{900}}$।
• (ix) $2.\overline{1625} = 2 + \frac{1625}{9999} = \mathbf{\frac{21623}{9999}}$।

End Ex · Q5-Q7 (a) Find 6 rational numbers between 3 and 4. (b) Find 5 rational numbers between $\frac{2}{5}$ and $\frac{3}{5}$. (c) Find 5 rational numbers between $\frac{1}{6}$ and $\frac{2}{5}$. [NCERT p.65] End Ex · Q5-Q7 (a) 3 और 4 के बीच 6 rational numbers calculate करें। (b) $\frac{2}{5}$ और $\frac{3}{5}$ के बीच 5 rational numbers निकालें। (c) $\frac{1}{6}$ और $\frac{2}{5}$ के बीच 5 rational numbers निकालें। [NCERT p.65]

1. 6 Rationals between 3 and 4 Multiply by $6+1 = 7$: $$3 = \frac{21}{7}, \quad 4 = \frac{28}{7} \implies \mathbf{\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}}$$

2. 5 Rationals between $\frac{2}{5}$ and $\frac{3}{5}$ Multiply by $5+1 = 6$: $$\frac{2}{5} = \frac{12}{30}, \quad \frac{3}{5} = \frac{18}{30} \implies \mathbf{\frac{13}{30}, \frac{14}{30}, \frac{15}{30}, \frac{16}{30}, \frac{17}{30}}$$

3. 5 Rationals between $\frac{1}{6}$ and $\frac{2}{5}$ $\text{LCM}(6, 5) = 30$, multiply by 6: $$\frac{1}{6} = \frac{30}{180}, \quad \frac{2}{5} = \frac{72}{180} \implies \mathbf{\frac{35}{180}, \frac{40}{180}, \frac{45}{180}, \frac{50}{180}, \frac{55}{180}}$$

🎯 Final Summary: Calculated cleanly using denominator scaling.

1. 3 और 4 के बीच 6 Rationals Multiplier $6+1 = 7$ लेते हैं: $$3 = \frac{21}{7}, \quad 4 = \frac{28}{7} \implies \mathbf{\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}}$$

2. $\frac{2}{5}$ और $\frac{3}{5}$ के बीच 5 Rationals Multiplier $5+1 = 6$ लेते हैं: $$\frac{2}{5} = \frac{12}{30}, \quad \frac{3}{5} = \frac{18}{30} \implies \mathbf{\frac{13}{30}, \frac{14}{30}, \frac{15}{30}, \frac{16}{30}, \frac{17}{30}}$$

3. $\frac{1}{6}$ और $\frac{2}{5}$ के बीच 5 Rationals Common denominator $180$ में convert करने पर: $$\frac{1}{6} = \frac{30}{180}, \quad \frac{2}{5} = \frac{72}{180} \implies \mathbf{\frac{35}{180}, \frac{40}{180}, \frac{45}{180}, \frac{50}{180}, \frac{55}{180}}$$

🎯 Final Summary: Denominator scaling method से intermediate rational numbers calculate होते हैं।

End Ex · Q8-Q10 (a) Solve $\frac{x}{3} + \frac{x}{5} = \frac{16}{15}$. (b) If $a+b=0$ ($a, b \neq 0$), determine whether $ab$ is positive or negative. (c) If a rational terminates at the 4th decimal place, show it is $\frac{p}{10^4}$ with $p$ not divisible by 10. [NCERT p.65] End Ex · Q8-Q10 (a) Solve करें: $\frac{x}{3} + \frac{x}{5} = \frac{16}{15}$। (b) यदि $a+b=0$ ($a, b \neq 0$), तो बताएं $ab$ positive है या negative। (c) यदि कोई rational 4th decimal पर terminate होता है, तो prove करें कि यह $\frac{p}{10^4}$ है। [NCERT p.65]

1. Q8 Equation $$\frac{5x + 3x}{15} = \frac{16}{15} \implies 8x = 16 \implies \mathbf{x = 2}$$

2. Q9 Sign of $ab$ $b = -a \implies ab = a(-a) = -a^2$. Since $a^2 > 0$ for all non-zero reals, $-a^2 < 0$. Hence $ab$ is strictly negative.

3. Q10 4th Decimal Place $x = 0.d_1 d_2 d_3 d_4 = \frac{d_1 d_2 d_3 d_4}{10^4} = \frac{p}{10^4}$. If $p$ were divisible by 10, it would reduce to 3 places. In lowest terms $\frac{p}{2^4 \cdot 5^4}$, at least one power of 4 ($2^4$ or $5^4$) survives in denominator.

🎯 Final Answers: Q8: $x = 2$; Q9: $ab < 0$ (Negative); Q10: $\frac{p}{10^4}$ with denominator having $2^4$ or $5^4$.

1. Q8 Equation का हल $$\frac{5x + 3x}{15} = \frac{16}{15} \implies 8x = 16 \implies \mathbf{x = 2}$$

2. Q9 $ab$ का Sign $b = -a \implies ab = -a^2$। किसी भी non-zero real number का square हमेशा positive ($a^2 > 0$) होता है, इसलिए $-a^2 < 0$ (Negative) होगा।

3. Q10 4th Decimal Theorem $x = \frac{p}{10^4}$। यदि $p$ 10 से divisible होता तो decimal 3 places पर सिमट जाता। Lowest form $\frac{p}{2^4 \cdot 5^4}$ में $2^4$ या $5^4$ का denominator में रहना आवश्यक है।

🎯 Final Answers: Q8: $x = 2$; Q9: $ab < 0$ (Negative); Q10: $\frac{p}{10^4}$ रूप सिद्ध हुआ।

End Ex · Q11-Q12 (a) Determine whether $\frac{18}{125}$ terminates and state the number of decimal places. (b) A rational in lowest terms has denominator $2^3 \times 5$. How many decimal places will it have? [NCERT p.65] End Ex · Q11-Q12 (a) बताएं $\frac{18}{125}$ terminating है या नहीं और कितने decimal places होंगे। (b) यदि denominator $2^3 \times 5$ है, तो कितने decimal places होंगे? [NCERT p.65]

1. Part (a) $\frac{18}{125}$ Denominator $125 = 5^3 = 2^0 \times 5^3 \implies$ Terminating. Places $= \max(0, 3) = \mathbf{3\text{ places}}$ ($\frac{18}{125} = 0.144$).

2. Part (b) Denominator $2^3 \times 5^1$ Highest power $= \max(3, 1) = 3$. Multiplying by $\frac{5^2}{5^2}$ gives denominator $10^3$, yielding exactly 3 decimal places.

🎯 Final Answer: (a) Terminating, 3 decimal places. (b) 3 decimal places.

1. Part (a) $\frac{18}{125}$ का हल Denominator $125 = 5^3 = 2^0 \times 5^3$ (factors 2 और 5 हैं) $\implies$ Terminating। Decimal places $= \max(0, 3) = \mathbf{3\text{ places}}$ ($0.144$)।

2. Part (b) Denominator $2^3 \times 5^1$ Highest power $= \max(3, 1) = 3$ है। $\frac{5^2}{5^2}$ से multiply करने पर denominator $10^3$ बन जाता है, इसलिए ठीक 3 decimal places होंगे।

🎯 Final Answer: (a) Terminating, 3 decimal places; (b) 3 decimal places प्राप्त होते हैं।

End Ex · Q14 Three rational numbers $x, y, z$ satisfy $x + y + z = 0$ and $xy + yz + zx = 0$. Show that $x, y, z$ must be simultaneously zero. [NCERT p.65] End Ex · Q14 तीन rational numbers $x, y, z$ equations $x + y + z = 0$ और $xy + yz + zx = 0$ को satisfy करते हैं। Prove करें कि $x = y = z = 0$ होगा। [NCERT p.65]

1. Algebraic Identity $$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$$

2. Substitute Given Values $$0^2 = x^2 + y^2 + z^2 + 2(0) \implies x^2 + y^2 + z^2 = 0$$

3. Sum of Squares For real numbers, $x^2 \ge 0, y^2 \ge 0, z^2 \ge 0$. A sum of non-negative squares is 0 if and only if each variable is 0.

🎯 Final Conclusion: $x = y = z = 0$ simultaneously (Proved).

1. Algebraic Identity $$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$$

2. Given Values Put करना $$0^2 = x^2 + y^2 + z^2 + 2(0) \implies x^2 + y^2 + z^2 = 0$$

3. Sum of Squares Property किसी भी real number का square कभी negative नहीं होता ($x^2 \ge 0$)। Squares का sum 0 तभी हो सकता है जब हर term 0 हो: $x = 0, y = 0, z = 0$।

🎯 Final Conclusion: $x = y = z = 0$ simultaneously सिद्ध होता है।

End Ex · Q15 Show that the rational number $\frac{a+b}{2}$ lies strictly between rational numbers $a$ and $b$ (where $a < b$). [NCERT p.66] End Ex · Q15 Prove करें कि rational number $\frac{a+b}{2}$ हमेशा $a$ और $b$ ($a < b$) के ठीक बीच स्थित होता है। [NCERT p.66]

1. Left Inequality Since $a < b$, add $a$: $2a < a + b \implies a < \frac{a+b}{2}$.

2. Right Inequality Since $a < b$, add $b$: $a + b < 2b \implies \frac{a+b}{2} < b$.

3. Combined Result $$a < \frac{a+b}{2} < b$$

🎯 Final Conclusion: The mean $\mathbf{\frac{a+b}{2}}$ always lies strictly between $a$ and $b$ (Proved).

1. Left Inequality Prove करना $a < b$ में $a$ जोड़ने पर: $a + a < a + b \implies 2a < a + b \implies a < \frac{a+b}{2}$।

2. Right Inequality Prove करना $a < b$ में $b$ जोड़ने पर: $a + b < b + b \implies a + b < 2b \implies \frac{a+b}{2} < b$।

3. Combining Inequalities $$a < \frac{a+b}{2} < b$$

🎯 Final Conclusion: Arithmetic mean $\mathbf{\frac{a+b}{2}}$ हमेशा $a$ और $b$ के बीच स्थित होता है।

End Ex · Q16 Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 (Square Root Spiral). [NCERT p.66] End Ex · Q16 Fig. 3.14 (Square Root Spiral) के सभी right triangles के hypotenuses की length calculate करें। [NCERT p.66]

1. Step-by-Step Hypotenuses • Triangle 1: $h_1 = \sqrt{1^2 + 1^2} = \mathbf{\sqrt{2}}$.
• Triangle 2: $h_2 = \sqrt{(\sqrt{2})^2 + 1^2} = \mathbf{\sqrt{3}}$.
• Triangle 3: $h_3 = \sqrt{(\sqrt{3})^2 + 1^2} = \mathbf{\sqrt{4} = 2}$.
• Triangle 4: $h_4 = \sqrt{2^2 + 1^2} = \mathbf{\sqrt{5}}$.
• Triangle $n$: $h_n = \mathbf{\sqrt{n+1}}$.

🎯 Final Answer: Hypotenuse sequence: $\sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}, \dots, \sqrt{n+1}$.

1. Step-by-Step Hypotenuse Calculation • Triangle 1: $h_1 = \sqrt{1^2 + 1^2} = \mathbf{\sqrt{2}}$।
• Triangle 2: $h_2 = \sqrt{(\sqrt{2})^2 + 1^2} = \mathbf{\sqrt{3}}$।
• Triangle 3: $h_3 = \sqrt{(\sqrt{3})^2 + 1^2} = \mathbf{\sqrt{4} = 2}$।
• Triangle 4: $h_4 = \sqrt{2^2 + 1^2} = \mathbf{\sqrt{5}}$।
• General $n$-th Triangle: $h_n = \mathbf{\sqrt{n+1}}$।

🎯 Final Answer: Hypotenuses की sequence $\sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}, \dots, \sqrt{n+1}$ प्राप्त होती है।

CBSE Exam Prep 11. Past 5 Years CBSE Board & Exam Questions (2020–2025) 11. पिछले 5 वर्षों के CBSE Board & Exam Questions (2020–2025)

CBSE 2024 [1 Mark] The decimal expansion of the rational number $\frac{14587}{1250}$ will terminate after how many decimal places? (A) 1 (B) 2 (C) 3 (D) 4 Rational number $\frac{14587}{1250}$ का decimal expansion कितने decimal places के बाद terminate होगा? (A) 1 (B) 2 (C) 3 (D) 4

1. Factorise Denominator [½ Mark] $$1250 = 2^1 \times 5^4$$

2. Determine Decimal Places [½ Mark] Highest exponent $= \max(1, 4) = 4$. Terminates after 4 decimal places.

🎯 Correct Option: (D) 4 decimal places.

1. Denominator का Prime Factorisation [½ Mark] $$1250 = 2^1 \times 5^4$$

2. Decimal Places की संख्या [½ Mark] Highest power $= \max(1, 4) = 4$ है, इसलिए decimal expansion ठीक 4 places बाद terminate होगा।

🎯 Correct Option: (D) 4 decimal places।

CBSE 2023 [2 Marks] Express $0.2\overline{35}$ in the simplest rational form $\frac{p}{q}$. $0.2\overline{35}$ को simplest rational form $\frac{p}{q}$ में express करें।

1. Set up Algebraic Equations [1 Mark] Let $x = 0.2353535\dots$ $$10x = 2.353535\dots \quad (\text{Eq. 1})$$ $$1000x = 235.353535\dots \quad (\text{Eq. 2})$$

2. Subtract & Solve [1 Mark] $$1000x - 10x = 235.35\dots - 2.35\dots \implies 990x = 233 \implies x = \frac{233}{990}$$

🎯 Final Answer: $\frac{233}{990}$.

1. Equations Setup [1 Mark] मान लेते हैं कि $x = 0.2353535\dots$ $$10x = 2.353535\dots \quad (\text{समीकरण 1})$$ $$1000x = 235.353535\dots \quad (\text{समीकरण 2})$$

2. Subtract करके $x$ निकालना [1 Mark] $$1000x - 10x = 233 \implies 990x = 233 \implies x = \frac{233}{990}$$

🎯 Final Answer: Simplest rational form $\frac{233}{990}$ प्राप्त होती है।

CBSE 2022 [3 Marks] Prove that $3 + 2\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is irrational. Prove करें कि $3 + 2\sqrt{5}$ एक irrational number है, दिया गया है कि $\sqrt{5}$ irrational है।

1. Assumption [1 Mark] Assume $3 + 2\sqrt{5} = \frac{a}{b}$ is rational ($a, b \in \mathbb{Z}, b \neq 0, \gcd(a, b) = 1$).

2. Rearrange for $\sqrt{5}$ [1 Mark] $$2\sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b} \implies \sqrt{5} = \frac{a - 3b}{2b}$$

3. Contradiction [1 Mark] Since $a, b$ are integers, $\frac{a - 3b}{2b}$ is rational. This implies $\sqrt{5}$ is rational, contradicting that $\sqrt{5}$ is irrational.

🎯 Final Conclusion: $3 + 2\sqrt{5}$ is an irrational number (Proved).

1. Assumption (मानना) [1 Mark] मान लेते हैं कि $3 + 2\sqrt{5} = \frac{a}{b}$ एक rational number है ($a, b$ co-prime integers हैं, $b \neq 0$)।

2. $\sqrt{5}$ को Isolate करना [1 Mark] $$2\sqrt{5} = \frac{a - 3b}{b} \implies \sqrt{5} = \frac{a - 3b}{2b}$$

3. Contradiction [1 Mark] चूंकि $a, b$ integers हैं, $\frac{a - 3b}{2b}$ rational होगा। इससे $\sqrt{5}$ rational साबित होता है, जो कि दी गई condition का विरोधाभास है।

🎯 Final Conclusion: अतः $3 + 2\sqrt{5}$ एक irrational number सिद्ध होता है।

CBSE 2024 [4 Marks Case Study] Case Study: Himalayan Meteorological Station
A high-altitude weather station in Dras records base temperature at 6:00 AM as $-8^\circ\text{C}$. Solar radiation increases temperature by $14^\circ\text{C}$ by 2:00 PM. A blizzard drops it by $11^\circ\text{C}$ at 8:00 PM.
(i) Calculate the 8:00 PM temperature. [1 Mark]
(ii) Determine if atmospheric pressure $\frac{23}{40}\text{ bar}$ terminates. [1 Mark]
(iii) Convert thermal factor $0.\overline{54}$ into $\frac{p}{q}$ form. [2 Marks] Case Study: Himalayan Meteorological Station
Dras के weather station में 6:00 AM पर temperature $-8^\circ\text{C}$ था। 2:00 PM तक धूप से temperature $14^\circ\text{C}$ बढ़ा और 8:00 PM पर बर्फबारी से $11^\circ\text{C}$ गिर गया।
(i) 8:00 PM का temperature calculate करें। [1 Mark]
(ii) बताएं atmospheric pressure $\frac{23}{40}\text{ bar}$ terminating है या नहीं। [1 Mark]
(iii) Thermal decay factor $0.\overline{54}$ को $\frac{p}{q}$ रूप में बदलें। [2 Marks]

Part (i) [1 Mark] $$T = (-8) + (+14) - 11 = 6 - 11 = \mathbf{-5^\circ\text{C}}$$

Part (ii) [1 Mark] Denominator $40 = 2^3 \times 5^1$. Since prime factors are strictly 2 and 5, it is terminating ($0.575\text{ bar}$).

Part (iii) [2 Marks] Let $x = 0.5454\dots \implies 100x - x = 54 \implies 99x = 54 \implies x = \frac{54}{99} = \mathbf{\frac{6}{11}}$.

🎯 Final Answers: (i) $-5^\circ\text{C}$, (ii) Terminating ($0.575\text{ bar}$), (iii) $\frac{6}{11}$.

Part (i) [1 Mark] $$T = (-8) + (+14) - 11 = 6 - 11 = \mathbf{-5^\circ\text{C}}$$

Part (ii) [1 Mark] Denominator $40 = 2^3 \times 5^1$ (factors 2 और 5 हैं) $\implies$ Terminating ($0.575\text{ bar}$) होगा।

Part (iii) [2 Marks] $x = 0.5454\dots \implies 100x - x = 54 \implies 99x = 54 \implies x = \frac{54}{99} = \mathbf{\frac{6}{11}}$।

🎯 Final Answers: (i) $-5^\circ\text{C}$, (ii) Terminating ($0.575\text{ bar}$), (iii) $\frac{6}{11}$ प्राप्त होते हैं।

12. Chapter Sources & Textbook Authority 12. Chapter Sources और प्रामाणिक संदर्भ

Primary Authority: NCERT Class 9 Mathematics Textbook — Ganita Manjari (Part I, NEP 2026), Chapter 3: The World of Numbers, pp. 41–67.
Historical Treatises: Brāhmasphuṭasiddhānta (Brahmagupta, 628 CE) for Zero & signed arithmetic rules; Baudhāyana Śhulbasūtra (c. 800 BCE) for diagonal of square $\sqrt{2}$; Āryabhaṭīya (Āryabhaṭa, 499 CE) & Mādhava of Sangamagrama (14th Century Kerala School) for $\pi$ series.
CBSE Examination References: CBSE Class 9 Annual Examination Papers & Exemplar Problems (2020–2025).