Class 9 Maths · Chapter 3
Short answer: Class 9 Maths Chapter 3 "The World of Numbers" Ganita Manjari Part 1 ka chapter hai jisme rational, irrational aur real numbers, unki decimal expansions, number line par representation aur laws of exponents cover hote hain — ye Class 9 Maths chapter 3 the world of numbers NCERT solutions me poore step-by-step worked examples ke saath diya gaya hai, jo purane "Number Systems" chapter jaisa hi core concept hai lekin historical/decimal-expansion angle ke saath.
Chapter 3 "The World of Numbers" poore Class 9 Maths ki neev hai — jitna bhi aage algebra, coordinate geometry (Chapter 1 "Orienting Yourself") ya sequences (Chapter 8) padhoge, sab jagah rational-irrational numbers aur decimal expansions ka concept kaam aayega. Ganita Manjari Part 1 ke naye syllabus me is chapter ka naam purane "Number Systems" se badal kar "The World of Numbers" kiya gaya hai, aur focus thoda history aur decimal expansion ki taraf shift hua hai. Is spec me tumhe har concept — rational/irrational classification, terminating vs non-terminating decimals, number line par irrational numbers represent karna (successive magnification), aur real numbers ke laws of exponents — CBSE marking-scheme style step-by-step solved milega.
Chapter 3 Summary — 5 Minute Revision
Chapter 3 "The World of Numbers" me humne seekha ki rational numbers wo hain jinhe p/q form me likha ja sakta hai (q≠0), jabki irrational numbers ka decimal expansion non-terminating aur non-repeating hota hai — dono milkar real numbers banate hain. Denominator ke prime factorisation se pata chal sakta hai ki decimal expansion terminating hogi (agar denominator sirf 2 aur 5 ke powers se bana ho) ya non-terminating repeating. Number line par irrational numbers (jaise √2, √3, √5) ko successive magnification aur Pythagoras theorem based spiral method se accurately represent kiya ja sakta hai. Real numbers par operations karte waqt identities jaise (a+√b)(a−√b)=a²−b denominator ko rationalise karne me kaam aati hain. Aakhir me laws of exponents for real numbers — aᵐ×aⁿ=aᵐ⁺ⁿ, (aᵐ)ⁿ=aᵐⁿ, aᵖ/ᑫ = q-th root of aᵖ — is chapter ko complete karte hain.
In-Text Questions — Solutions
Kya zero (0) ek rational number hai? Justify karo.
Haan, 0 ek rational number hai. Rational number ki definition ke according, ise p/q form me likha jaana chahiye jahan p aur q integers hon aur q ≠ 0. Hum 0 ko 0/1, 0/2, 0/(−5) jaise kai forms me likh sakte hain — sab valid hain kyunki p=0 (koi bhi integer ho sakta hai) aur q non-zero integer hai. Isliye 0 rational hai.
3/5 aur 4/5 ke beech teen rational numbers dhoondo.
Step 1: 3/5 = 0.6 aur 4/5 = 0.8 hai.
Step 2: In dono ke beech easily 0.65, 0.7, aur 0.75 le sakte hain.
In p/q form: 13/20, 7/10, 3/4
Ye teeno numbers 3/5 aur 4/5 ke beech hain.
√9 rational hai ya irrational?
Step 1: √9 = 3 (kyunki 3×3=9).
Step 2: 3 ko 3/1 form me likha ja sakta hai, jo p/q form hai.
Conclusion: Isliye √9 rational hai — perfect square ka square root hamesha rational hota hai.
1/7 ka decimal expansion likho aur batao ye terminating hai ya non-terminating repeating.
Step 1: Division karne par: 1/7 = 0.142857142857… = 0.142857
Step 2: Yahan '142857' block baar-baar repeat ho raha hai, isliye ye non-terminating repeating decimal expansion hai — aur chunki ye repeat karta hai, 1/7 ek rational number hi hai.
Successive magnification (spiral) method se √5 ko number line par kaise locate karoge?
Step 1: O se A tak 2 units ka segment lo (OA=2), A par number line ke perpendicular AB=1 unit khींcho.
Step 2: Pythagoras theorem se, OB = √(OA² + AB²) = √(2²+1²) = √(4+1) = √5.
Step 3: O ko center, OB(=√5) ko radius leke ek arc khींcho jo number line ko point P par kaate.
Conclusion: OP = √5, yahi required point hai.
Simplify karo: 2√3 + 3√3 − √3
Step 1: Sabhi terms me common factor √3 hai, isliye coefficients ko combine karo:
2√3 + 3√3 − √3 = (2+3−1)√3 = 4√3
1/√7 ka denominator rationalise karo.
Step 1: Numerator aur denominator ko √7 se multiply karo:
1/√7 × √7/√7 = √7/7
Isliye rationalised form = √7/7.
Laws of exponents use karke simplify karo: 2⁵ × 2⁻³
Step 1: Law: aᵐ × aⁿ = aᵐ⁺ⁿ
2⁵ × 2⁻³ = 2^(5+(−3)) = 2² = 4
Exercise Questions — Solutions (Q1–Q18)
1. Classify the following numbers as rational or irrational: 22/7, √2, 0.101001000100001…, −5, π.
Step 1: 22/7 — ye p/q form me directly likha hua hai (p=22, q=7, dono integers, q≠0) → rational.
Step 2: √2 — koi bhi p/q form me nahi likha ja sakta (proof by contradiction se pata chalta hai) → irrational.
Step 3: 0.101001000100001… — decimal expansion non-terminating hai AUR koi repeating block nahi hai (0 ki count har baar badh rahi hai) → irrational.
Step 4: −5 = −5/1, integer form me hai → rational.
Step 5: π ≈ 3.14159265… — non-terminating, non-repeating → irrational (π ko 22/7 sirf approximate value ke roop me use karte hain, exact nahi).
2. Kya har rational number ek real number hota hai? Kya har real number ek rational number hota hai? Example ke saath justify karo.
Step 1: Real numbers = rational numbers ∪ irrational numbers. Isliye har rational number real numbers ke set ka part hota hai. Example: 3/4 ek rational number hai aur real number bhi hai. → Haan.
Step 2: Lekin har real number rational nahi hota, kyunki irrational numbers (jaise √2, π) bhi real numbers hote hain par unhe p/q form me nahi likha ja sakta. Counter-example: √2 real hai par rational nahi. → Nahi.
3. Prove karo ki √3 ek irrational number hai.
Step 1 (Contradiction assume karo): Maan lo √3 rational hai, to √3 = p/q likh sakte hain jahan p, q co-prime integers hain (koi common factor nahi) aur q ≠ 0.
√3 = p/q ⟹ p² = 3q² ...(i)
Step 2: Equation (i) se pata chalta hai 3, p² ko divide karta hai. Isliye 3, p ko bhi divide karega (kyunki 3 ek prime number hai). Maan lo p = 3c kisi integer c ke liye.
(3c)² = 3q² ⟹ 9c² = 3q² ⟹ q² = 3c²
Step 3: Ab 3, q² ko divide karta hai, isliye 3, q ko bhi divide karta hai.
Step 4: Ye contradiction hai — 3, p aur q dono ko divide kar raha hai, jabki humne p, q ko co-prime maana tha.
Conclusion: Hamari assumption galat thi, isliye √3 rational nahi ho sakta. Hence √3 irrational hai. (Hence proved.)
4. Batao ki following rational hain ya irrational: (i) 3 + √5 (ii) √2 × √8 (iii) (2 − √2)(2 + √2)
(i) 3 + √5 — rational + irrational = irrational.
(ii) √2 × √8 = √(2×8) = √16 = 4 → rational
(iii) (2−√2)(2+√2) = 2² − (√2)² = 4 − 2 = 2 → rational
5. Bina actual division kiye batao ki 13/3125 ka decimal expansion terminating hai ya non-terminating repeating.
Step 1: Denominator ko prime factorise karo: 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵ = 2⁰ × 5⁵.
Step 2: Rule: agar denominator sirf 2ᵐ × 5ⁿ form me likha ja sakta ho (koi aur prime factor na ho), to decimal expansion terminating hoti hai.
Conclusion: Yahan denominator = 2⁰ × 5⁵ hai, sirf 2 aur 5 ke powers hain, isliye 13/3125 ka decimal expansion terminating hoga.
6. 0.666… (i.e. 0.6̄) ko p/q form me likho.
Step 1: Maan lo x = 0.666… ...(i)
Step 2: Kyunki sirf ek digit (6) repeat ho raha hai, dono taraf 10 se multiply karo:
10x = 6.666… ...(ii)
Step 3: (ii) − (i) karo:
10x − x = 6.666… − 0.666… ⟹ 9x = 6 ⟹ x = 6/9 = 2/3
Isliye 0.6̄ = 2/3.
7. 0.235235235… (repeating block '235') ko p/q form me likho.
Step 1: Maan lo x = 0.235235235… ...(i). Block '235' ki length 3 digits hai, isliye dono taraf 1000 se multiply karo:
1000x = 235.235235… ...(ii)
Step 2: (ii) − (i):
1000x − x = 235 ⟹ 999x = 235 ⟹ x = 235/999
Isliye required p/q form = 235/999.
8. Successive magnification / Pythagoras spiral method se √6 ko number line par represent karo.
Step 1: Pehle O se A(2,0) tak 2 units ka line segment lo, A par perpendicular AB=1 unit khींcho. OB = √(2²+1²) = √5 milta hai (spiral ka pehla step).
Step 2: Compass se O ko center, radius OB(=√5) leke number line par point P mark karo — OP = √5.
Step 3: Ab P par number line ke perpendicular 1 unit ka segment PQ khींcho. Pythagoras theorem se: OQ = √((√5)² + 1²) = √(5+1) = √6.
Step 4: O ko center, radius OQ(=√6) leke arc lagao jo number line ko point R par kaate — OR = √6, ye required point hai.
9. Successive magnification method se 3.765 ko number line par visualise karo.
Step 1: Sabse pehle 3 aur 4 ke beech ka segment lo, kyunki 3.765, 3 aur 4 ke beech hai.
Step 2: Us segment ko 10 barabar bhaago me baanto — 3.765, 3.7 aur 3.8 ke beech aayega, isliye 3.7–3.8 wale portion ko magnify (zoom) karo.
Step 3: 3.7–3.8 ko phir 10 bhaago me baanto — 3.765, 3.76 aur 3.77 ke beech aayega, is portion ko phir magnify karo.
Step 4: 3.76–3.77 ko phir 10 bhaago me baanto — 5th division point exactly 3.765 ko represent karega. Isi process ko 'successive magnification' kehte hain — har baar interval ko zoom karke exact point tak pahunchte hain.
10. Simplify: (√5 + √2)²
Step 1: Identity use karo: (a+b)² = a² + 2ab + b²
(√5+√2)² = (√5)² + 2(√5)(√2) + (√2)² = 5 + 2√10 + 2
= 7 + 2√10
11. Simplify: (5 + √7)(5 − √7)
Step 1: Identity use karo: (a+b)(a−b) = a² − b²
(5+√7)(5−√7) = 5² − (√7)² = 25 − 7 = 18
12. Rationalise the denominator of 1/(7 + 3√2)
Step 1: Numerator aur denominator ko conjugate (7 − 3√2) se multiply karo:
1/(7+3√2) × (7−3√2)/(7−3√2) = (7−3√2)/(7² − (3√2)²)
Step 2: Denominator simplify karo: 7² − (3√2)² = 49 − 9×2 = 49 − 18 = 31.
Result = (7 − 3√2)/31
13. Agar (√5 + √2)/(√5 − √2) = a + b√10 hai, to a aur b ki values nikalo.
Step 1: Denominator ko rationalise karne ke liye numerator aur denominator ko (√5+√2) se multiply karo:
(√5+√2)/(√5−√2) × (√5+√2)/(√5+√2) = (√5+√2)²/((√5)²−(√2)²)
Step 2: Numerator = (√5+√2)² = 5 + 2√10 + 2 = 7 + 2√10. Denominator = 5 − 2 = 3.
= (7 + 2√10)/3 = 7/3 + (2/3)√10
Step 3: Compare karne par a + b√10 = 7/3 + (2/3)√10 se milta hai:
a = 7/3, b = 2/3
14. Simplify: (64)^(1/2)
Step 1: 64 = 8² hai, isliye
(64)^(1/2) = (8²)^(1/2) = 8^(2×1/2) = 8^1 = 8
15. Simplify: 2^(2/3) × 2^(1/3)
Step 1: Law of exponents use karo: aᵐ × aⁿ = aᵐ⁺ⁿ
2^(2/3) × 2^(1/3) = 2^(2/3 + 1/3) = 2^(1) = 2
16. Simplify: (1/3²)⁷
Step 1: Power of a power rule use karo: (aᵐ)ⁿ = aᵐⁿ
(1/3²)⁷ = (3⁻²)⁷ = 3⁻¹⁴ = 1/3¹⁴
17. Simplify: 7^(1/2) × 8^(1/2)
Step 1: Law use karo: aᵐ × bᵐ = (ab)ᵐ
7^(1/2) × 8^(1/2) = (7×8)^(1/2) = 56^(1/2) = √56 = √(4×14) = 2√14
18. x ki value nikalo agar: 2^x = 1/32
Step 1: 32 ko power of 2 me likho: 32 = 2⁵, isliye 1/32 = 2⁻⁵.
2^x = 2⁻⁵
Step 2: Base same hone par exponents compare karo:
x = −5
Important Equations — Ek Nazar Me
| Concept | Rule / Formula |
|---|---|
| Rational number | p/q form, p aur q integers, q ≠ 0 |
| Terminating decimal condition | Denominator (lowest form) = 2m × 5n ho to expansion terminating hai |
| Non-terminating repeating | Denominator me 2, 5 ke alawa koi aur prime factor ho to expansion non-terminating repeating (rational) hai |
| Irrational number | Decimal expansion non-terminating, non-repeating; p/q form me nahi likha ja sakta |
| √a irrational check | Agar a perfect square nahi hai to √a irrational hai |
| Identity 1 | (√a + √b)(√a − √b) = a − b |
| Identity 2 | (a + √b)(a − √b) = a² − b |
| Identity 3 | (√a + √b)² = a + 2√(ab) + b |
| Rationalisation of 1/√a | 1/√a × √a/√a = √a/a |
| Rationalisation of 1/(a+√b) | 1/(a+√b) × (a−√b)/(a−√b) = (a−√b)/(a²−b) |
| Law of exponents 1 | am × an = am+n |
| Law of exponents 2 | am ÷ an = am−n |
| Law of exponents 3 | (am)n = amn |
| Law of exponents 4 | am × bm = (ab)m |
| Zero & negative exponent | a0 = 1 (a≠0); a−n = 1/an |
| Rational exponent | ap/q = (q-th root of a)p, a > 0 |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Ye maan lena ki har square root irrational hota hai — lekin √4 = 2, √9 = 3 jaise perfect squares ke roots actually rational hote hain. Pehle check karo number perfect square hai ya nahi.
- Terminating aur non-terminating repeating decimal expansion me confuse ho jaana — dono hi rational numbers ki expansions hain, sirf non-terminating NON-repeating hi irrational hoti hai.
- Denominator rationalise karte waqt sirf denominator ko conjugate se multiply karna, numerator ko bhool jaana — dono ko multiply karna zaroori hai, warna value badal jaati hai.
- Ye galat assumption karna ki do irrational numbers ka sum ya product hamesha irrational hoga — jaise √2 × √2 = 2 (rational) ya (2+√3)+(2−√3)=4 (rational).
- Laws of exponents apply karte waqt negative ya fractional exponents me sign/power galat kar dena, jaise a⁻ⁿ ko 1/aⁿ ki jagah −aⁿ likh dena.
- Number line par √n represent karte waqt successive magnification ya Pythagoras spiral method me perpendicular ki length ya radius measurement me chhoti si error kar dena, jisse point galat jagah mark ho jaata hai.
Board-Style Important Questions
- Kya 0 ek rational number hai? Ek line me justify karo.
- √16 rational hai ya irrational? Reason do.
- 3/8 ko decimal me convert karo aur batao ye terminating hai ya non-terminating repeating, bina actual division kiye reason bhi do.
- Simplify karo: (3+√3)(2+√2)
- Rationalise the denominator: 1/(√7 − √5), aur simplified answer do.
- Prove karo ki √2 ek irrational number hai (contradiction method se, poore steps ke saath).
Aksar Poochhe Jaane Wale Sawaal
Class 9 Maths Chapter 3 'The World of Numbers' NCERT solutions me kya-kya cover hota hai?
Ye chapter rational aur irrational numbers, real numbers ki decimal expansions (terminating vs non-terminating repeating), number line par irrational numbers ko successive magnification se represent karna, real numbers ke operations, aur laws of exponents for real numbers cover karta hai — historical/decimal-expansion angle is chapter ka core focus hai. Ganita Manjari Class 9 Maths NCERT solutions me har concept ke saath step-by-step solved examples milte hain.
Class 9 Maths new syllabus 2026-27 me chapter list kya hai, PDF kahan milegi?
Session 2026-27 se Class 9 Maths ki NCERT book poori tarah badal gayi hai — purani 'Mathematics' ki jagah naya NCF-2023-based textbook 'Ganita Manjari' aaya hai. Abhi sirf Part 1 (8 chapters) published aur taught ho raha hai — jaise Chapter 1 'Orienting Yourself: Use of Coordinates', Chapter 2 'Introduction to Linear Polynomials', Chapter 3 'The World of Numbers', Chapter 5 'I'm Up and Down, and Round and Round (Circles)', Chapter 6 'Measuring Space: Perimeter and Area', Chapter 7 'Probability', Chapter 8 'Sequences and Progressions'. Official chapter list PDF ncert.nic.in par available hai.
Ganita Manjari Part 2 kab aayega, aur usme kya topics honge?
Abhi tak (4 Aug 2026) Ganita Manjari Part 2 release nahi hua hai aur official release date bhi announce nahi hui. Kai secondary sources expect karte hain isme Linear Equations in Two Variables, Introduction to Euclid's Geometry, Lines and Angles, Triangles, Quadrilaterals, Surface Area & Volume, Statistics jaise topics honge — lekin ye sirf expected list hai, koi official confirmation abhi tak nahi mila hai.
'Number Systems' chapter ka naam badal kar 'The World of Numbers' kyun kiya gaya?
Naya NCF-2023 based Ganita Manjari textbook me is chapter ka naam 'The World of Numbers' rakha gaya hai aur scope me historical aur decimal-expansion angle zyada emphasize kiya gaya hai, jabki core concept (rational, irrational, real numbers) purane 'Number Systems' chapter jaisa hi hai.
√2, √3 jaise numbers irrational kyun hote hain — proof kaise yaad rakhein?
Inhe contradiction method se prove karte hain: pehle maan lo number rational hai (p/q form), fir algebra se dikhate hain ki p aur q dono ek common prime factor share karte hain — jo co-prime hone ki assumption se contradict karta hai. Isliye original assumption galat thi aur number irrational sabit hota hai. Ye chapter me full step-by-step solved hai.
Ganita Manjari ke saare chapters ke NCERT solutions PDF ek jagah kaise milenge?
Ganita Manjari Part 1 ke saare 8 chapters ke solutions (Ch 1 coordinates, Ch 2 linear polynomials, Ch 3 world of numbers, Ch 4 algebraic identities, Ch 5 circles, Ch 6 perimeter & area, Ch 7 probability, Ch 8 sequences & progressions) chapter-wise PDF download ke roop me available hote hain — important questions aur step-by-step solved examples ke saath, jo board-style marking scheme follow karte hain.
Class 9 Maths — Saare Chapters
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