Class 10 Maths · Chapter 1
Short answer:
Class 10 Maths Chapter 1 Real Numbers me sirf 2 exercises hain — Exercise 1.1 (7 questions, Fundamental Theorem of Arithmetic par based) aur Exercise 1.2 (3 questions, jisme ek question ke 3 parts hain, Revisiting Irrational Numbers par based). Total milaake 10 questions cover hote hain. Chapter prime factorisation se HCF/LCM nikalna aur irrational numbers ko proof by contradiction se prove karna sikhata hai.
Real Numbers chapter do core ideas ke around ghoomta hai — Fundamental Theorem of Arithmetic aur irrational numbers ka proof. Current NCERT syllabus me Euclid's Division Lemma/Algorithm wala purana section hata diya gaya hai — ab HCF/LCM sirf prime factorisation method se nikalte hain, aur irrationality proofs proof-by-contradiction se hote hain.
Chapter 1 Summary — 5 Minute Revision
1. Fundamental Theorem of Arithmetic
Ye theorem kehta hai ki har composite number ko primes ke product ki tarah likha ja sakta hai, aur ye factorisation unique hota hai — factors ka order chahe jo bhi ho.
Example: 140 ko hum 2 × 2 × 5 × 7 likhein ya 7 × 5 × 2 × 2, primes wahi rahenge — sirf order change hoga.
140 = 22 × 5 × 7
2. Prime Factorisation Method se HCF aur LCM
- HCF (Highest Common Factor): common primes ki smallest (lowest) power lo.
- LCM (Least Common Multiple): saare primes (common + non-common) ki greatest (highest) power lo.
336 = 24 × 3 × 7 54 = 2 × 33 → HCF = 2 × 3 = 6 LCM = 24 × 33 × 7 = 3024
3. HCF × LCM Relation — sirf DO numbers ke liye
Do positive integers p aur q ke liye:
HCF(p, q) × LCM(p, q) = p × q
Warning: ye relation teen ya zyada numbers ke liye directly apply nahi hoti — sirf do numbers ke case me valid hai.
4. Revisiting Irrational Numbers
Ek number irrational hota hai agar use p/q form me nahi likha ja sakta (jahan p, q integers hain aur q ≠ 0). √2, √3, √5 jaise numbers irrational hote hain — inhe proof by contradiction se prove karte hain.
5. Proof by Contradiction ka structure
- Assume: number rational hai — likho p/q form me, jahan p, q coprime integers hain aur q ≠ 0.
- Manipulate: equation ko square/simplify karke dikhao ki p ek particular prime se divisible hai.
- Substitute back: isse dikhao ki q bhi usi prime se divisible hai.
- Contradiction: agar p aur q dono ek hi prime se divisible hain, to woh coprime nahi ho sakte — jo humari assumption ke against hai.
- Conclude: isliye humari assumption galat thi — number irrational hai.
6. Useful facts jo proofs me kaam aate hain
- Rational + Irrational = Irrational
- Rational × Irrational (non-zero rational) = Irrational
- Agar p prime hai aur p, a2 ko divide karta hai, to p, a ko bhi divide karega (ye Fundamental Theorem of Arithmetic se follow hota hai)
Poore Class 10 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q10)
Q1. Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Har number ko baar-baar chhote primes se divide karke prime factorisation nikalte hain.
(i) 140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7 = 22 × 5 × 7
(ii) 156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13 = 22 × 3 × 13
(iii) 3825 = 3 × 1275 = 3 × 3 × 425 = 32 × 5 × 85 = 32 × 5 × 5 × 17 = 32 × 52 × 17
(iv) 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 437 = 17 × 19 × 23
Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers: (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Pehle dono numbers ka prime factorisation karo, phir HCF = common primes ki lowest power, LCM = saare primes ki highest power.
(i) 26 = 2 × 13 91 = 7 × 13 → HCF = 13, LCM = 2 × 7 × 13 = 182
Verify: HCF × LCM = 13 × 182 = 2366 aur 26 × 91 = 2366 ✓ match
(ii) 510 = 2 × 3 × 5 × 17 92 = 22 × 23 → HCF = 2, LCM = 22 × 3 × 5 × 17 × 23 = 23460
Verify: HCF × LCM = 2 × 23460 = 46920 aur 510 × 92 = 46920 ✓ match
(iii) 336 = 24 × 3 × 7 54 = 2 × 33 → HCF = 2 × 3 = 6, LCM = 24 × 33 × 7 = 3024
Verify: HCF × LCM = 6 × 3024 = 18144 aur 336 × 54 = 18144 ✓ match
Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method: (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Teen numbers ke case me bhi wahi rule: HCF = common primes ki lowest power, LCM = saare primes ki highest power. (Yaad rakho — yahan HCF × LCM ≠ product, kyunki woh relation sirf do numbers ke liye valid hai.)
(i) 12 = 22 × 3 15 = 3 × 5 21 = 3 × 7 → HCF = 3, LCM = 22 × 3 × 5 × 7 = 420
(ii) 17, 23, 29 teeno prime hain, koi common factor nahi → HCF = 1, LCM = 17 × 23 × 29 = 11339
(iii) 8 = 23 9 = 32 25 = 52 koi common prime nahi → HCF = 1, LCM = 23 × 32 × 52 = 1800
Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Do numbers ke liye HCF × LCM = product of the numbers wala relation use karo.
HCF × LCM = 306 × 657
LCM = (306 × 657) ÷ 9 = 34 × 657 = 22338
Q5. Check whether 6n can end with the digit 0 for any natural number n.
Koi bhi number 0 pe tabhi end hota hai jab woh 10 se divisible ho, aur 10 = 2 × 5. Iska matlab number ke prime factorisation me dono 2 aur 5 hone chahiye.
6n = (2 × 3)n = 2n × 3n
Is factorisation me sirf primes 2 aur 3 hain — 5 kabhi nahi aata (Fundamental Theorem of Arithmetic ke according prime factorisation unique hoti hai, to koi extra 5 factor ban hi nahi sakta). Isliye 6n kabhi bhi digit 0 pe end nahi ho sakta, kisi bhi natural number n ke liye.
Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Common factor nikalo aur dikhao ki number do factors ka product hai, dono 1 se bade.
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × 78 = 13 × 78
Ye 13 × 78 hai, jahan dono factors 1 se bade hain — isliye ye composite number hai.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009
Ye 5 × 1009 hai, jahan dono factors 1 se bade hain — isliye ye bhi composite number hai.
Q7. There is a circular path around a sports field. Priya takes 18 minutes to drive one round of the field, while Ravish takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Dono starting point pe tabhi wapas milenge jab dono ke complete rounds simultaneously ho — matlab time, dono ke individual times ka LCM hona chahiye.
18 = 2 × 32 12 = 22 × 3 → LCM = 22 × 32 = 36
Priya aur Ravish 36 minutes baad starting point pe dobara milenge.
Q8. Prove that √5 is irrational.
Proof by contradiction:
Step 1 — Assume: maano √5 rational hai. To ise p/q form me likh sakte hain jahan p aur q coprime integers hain (koi common factor nahi) aur q ≠ 0.
√5 = p/q ⇒ p = √5 q
Step 2 — Square karo:
p2 = 5q2 ...(i)
Isse pata chalta hai ki p2, 5 se divisible hai. Chunki 5 ek prime number hai, isliye p bhi 5 se divisible hoga (agar prime p, a2 ko divide kare, to p, a ko bhi divide karta hai).
Step 3 — Substitute: maano p = 5m, kisi integer m ke liye. Ise equation (i) me substitute karo.
(5m)2 = 5q2 ⇒ 25m2 = 5q2 ⇒ q2 = 5m2
Isse pata chalta hai ki q2 bhi 5 se divisible hai, isliye q bhi 5 se divisible hai.
Step 4 — Contradiction: ab p aur q dono 5 se divisible hain — matlab 5 unka ek common factor hai. Lekin humne shuru me assume kiya tha ki p aur q coprime hain (koi common factor nahi). Ye contradiction hai.
Step 5 — Conclude: isliye humari assumption galat thi. √5 ko p/q form me nahi likha ja sakta — √5 irrational hai. (Hence proved.)
Q9. Prove that 3 + 2√5 is irrational.
Proof by contradiction:
Step 1 — Assume: maano 3 + 2√5 rational hai. To ise a/b form me likh sakte hain jahan a, b integers hain aur b ≠ 0.
3 + 2√5 = a/b
Step 2 — Isolate √5:
2√5 = a/b − 3 = (a − 3b)/b
√5 = (a − 3b) / 2b
Step 3 — Contradiction: a, b integers hain, isliye (a − 3b)/2b bhi ek rational number hai. Iska matlab √5 rational ho jaata hai. Lekin humein pehle hi pata hai (Q8 se proved) ki √5 irrational hai — ye contradiction hai.
Step 4 — Conclude: isliye humari assumption galat thi. 3 + 2√5 irrational hai. (Hence proved.)
Q10. Prove that the following are irrationals: (i) 1/√2 (ii) 7√5 (iii) 6 + √2
(i) 1/√2 irrational hai
Maano 1/√2 rational hai — likho a/b form me (a, b integers, coprime, b ≠ 0).
1/√2 = a/b ⇒ √2 = b/a
a, b integers hain to b/a rational hoga, matlab √2 rational ho jaata hai — jo ki galat hai (√2 ek known irrational number hai, isi tarah proof-by-contradiction se prove hota hai jaise Q8 me √5 ke liye kiya). Contradiction. Isliye 1/√2 irrational hai.
(ii) 7√5 irrational hai
Maano 7√5 rational hai — likho a/b form me (a, b integers, b ≠ 0).
7√5 = a/b ⇒ √5 = a / 7b
a, b integers hain to a/7b rational hoga, matlab √5 rational ho jaata hai — jo ki Q8 me disprove ho chuka hai. Contradiction. Isliye 7√5 irrational hai.
(iii) 6 + √2 irrational hai
Maano 6 + √2 rational hai — likho a/b form me (a, b integers, b ≠ 0).
6 + √2 = a/b ⇒ √2 = a/b − 6 = (a − 6b)/b
a, b integers hain to (a − 6b)/b rational hoga, matlab √2 rational ho jaata hai — jo ki galat hai. Contradiction. Isliye 6 + √2 irrational hai.
Important Equations — Ek Nazar Me
| Concept | Statement / Formula | Example |
|---|---|---|
| Fundamental Theorem of Arithmetic | Har composite number ko primes ke product ki tarah uniquely likha ja sakta hai (order chahe jo bhi ho) | 36 = 22 × 32 |
| HCF by prime factorisation | Common primes ki lowest power ka product | HCF(12, 18) = 2 × 3 = 6 |
| LCM by prime factorisation | Saare primes (common + non-common) ki highest power ka product | LCM(12, 18) = 22 × 32 = 36 |
| HCF × LCM relation | HCF(p, q) × LCM(p, q) = p × q — sirf do numbers ke liye valid | HCF(26,91)=13, LCM=182, 13×182=26×91=2366 |
| Divisibility by 10 | Koi number 0 pe end hota hai sirf tab jab woh 2 aur 5 dono se divisible ho | 6n = 2n×3n — 5 kabhi factor nahi banta, isliye kabhi 0 pe end nahi hota |
| p divides a2 ⇒ p divides a | Agar p prime hai aur p | a2, to p | a | Irrationality proofs ka core step |
| Rational + Irrational | Hamesha Irrational hota hai | 3 + 2√5 is irrational |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- HCF × LCM = product wala rule teen numbers pe laga dena. Ye relation SIRF do numbers ke liye valid hai. Teen ya zyada numbers ke case me (jaise Q3) is shortcut ka use mat karo — direct prime factorisation se hi HCF/LCM nikalo.
- Proof-by-contradiction ka structure incomplete chhodna. Sirf 'assume rational hai' likh kar seedha answer pe kood jaana — har step likhna zaroori hai: assumption, p/q coprime hone ki condition, manipulation, substitution, aur final contradiction statement.
- p aur q ko coprime bolna bhool jaana. Proof shuru karte waqt 'p aur q coprime hain (no common factor)' likhna miss kar dena — bina isके contradiction step complete nahi hota, marks katte hain.
- Prime factorisation me arithmetic slip. Bade numbers (jaise 3825, 7429) factorise karte waqt beech me divide karne me galti — hamesha last step pe multiply karke original number se verify karo.
- LCM nikalte waqt lowest power le lena (HCF ka rule LCM pe apply karna) ya vice versa. Yaad rakho: HCF = lowest power of common primes, LCM = highest power of ALL primes.
- 'Irrational + Irrational = Irrational' galat samajhna. Ye hamesha true nahi hota (jaise √2 + (−√2) = 0, jo rational hai) — sirf 'Rational + Irrational = Irrational' aur 'non-zero Rational × Irrational = Irrational' hi guaranteed results hain jo exam me use karne chahiye.
Board-Style Important Questions
- 1 mark: HCF (33, 81, 45) find karo prime factorisation method se.
- 2 marks: Prove that √3 is an irrational number.
- 2 marks: Do numbers ka product 1050 hai aur unka HCF 5 hai. Unka LCM find karo.
- 3 marks: Explain kyun 5 × 7 × 11 + 7 aur 5 × 7 × 11 × 3 + 11 composite numbers hain.
- 3 marks: Prove that 2 − 3√5 is an irrational number.
Quick Quiz — Score Check Karein
Q1. Fundamental Theorem of Arithmetic ke according, har composite number ko primes ke product ki tarah kaise likha ja sakta hai?
Q2. 336 = 2⁴ × 3 × 7 aur 54 = 2 × 3³ diya hai. In dono ka HCF kya hoga?
Q3. HCF(p, q) × LCM(p, q) = p × q wala relation kab valid hota hai?
Q4. HCF(306, 657) = 9 diya hai. LCM(306, 657) kya hoga?
Q5. 6ⁿ kabhi bhi digit 0 pe end kyun nahi hota, kisi bhi natural number n ke liye?
Q6. 7 × 11 × 13 + 13 composite number kyun hai?
Q7. Priya circular field ka ek round 18 minutes me aur Ravish 12 minutes me complete karte hain, same point se same direction me start karke. Woh dono dobara starting point pe kab milenge?
Q8. √5 ko irrational prove karne wale proof by contradiction me pehla step kya hota hai?
Q9. 3 + 2√5 ko irrational prove karne ke liye kaunsa pehle se proved fact use hota hai?
Q10. Chapter ke according 'Rational + Irrational' ka result kya hota hai, aur 'Irrational + Irrational' ke baare me kya common mistake batayi gayi hai?
Aksar Poochhe Jaane Wale Sawaal
Euclid's Division Lemma ab bhi Class 10 syllabus me hai kya?
Nahi. Current NCERT (rationalised) syllabus me Euclid's Division Lemma aur Euclid's Division Algorithm hata diya gaya hai. Ab HCF/LCM sirf Fundamental Theorem of Arithmetic (prime factorisation method) se nikalte hain.
HCF x LCM = product of numbers, ye rule kab use kar sakte hain?
Ye rule sirf DO positive integers ke liye valid hai. Teen ya zyada numbers ke liye ye directly apply nahi hota — prime factorisation se hi HCF aur LCM alag-alag nikalne padte hain.
Proof by contradiction me sabse important step kya hai?
Assumption ke saath 'p aur q coprime hain' likhna, aur end me clearly ye dikhana ki p aur q dono ek common prime factor share kar rahe hain — jo coprime hone ke against hai. Yehi contradiction proof ko complete karta hai.
6^n kabhi bhi 0 pe end kyun nahi hota, chahe n kitna bhi bada ho?
Kyunki 6 = 2 × 3, to 6^n = 2^n × 3^n. Number 0 pe tabhi end hota hai jab uske prime factors me 2 aur 5 dono hon. 6^n ke factorisation me 5 kabhi aata hi nahi, isliye n ki koi bhi value ho, ye 0 pe end nahi hoga.
Composite number prove karne ka sabse aasan tareeka kya hai?
Expression me se common factor nikal kar dikhao ki woh do integers ka product hai, jahan dono factors 1 se bade hon. Agar aisa ho jaaye to number definitely composite hai.
Is chapter me kitne exercises hain aur kya farak hai?
Sirf 2 exercises: Exercise 1.1 (Fundamental Theorem of Arithmetic — HCF/LCM related, 7 questions) aur Exercise 1.2 (Revisiting Irrational Numbers — proofs, 3 questions). Koi in-text questions nahi hain is chapter me.
Class 10 Maths — Saare Chapters
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