Chapter 1: Real Numbers

Mathematics · CBSE Class 10 · NCERT Solutions

📖 Chapter 1 — Real Numbers (CBSE Class 10 Maths): Detailed notes — Fundamental Theorem of Arithmetic, HCF & LCM, irrational numbers and their proofs.

Chapter 1 — Real Numbers

🔢 1. Fundamental Theorem of Arithmetic (FTA)

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order of the prime factors. e.g. 140 = 2² × 5 × 7.

🔗 2. HCF & LCM by Prime Factorisation

HCF = product of the smallest power of each common prime factor.
LCM = product of the greatest power of each prime factor involved.

For any two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b

⚠️ This relation holds only for two numbers (not for three or more).

🚫 3. Irrational Numbers

A number that cannot be written in the form p/q (q ≠ 0, p, q integers) is irrational, e.g. √2, √3, √5, π. Their decimal expansion is non-terminating and non-recurring.

Key results: If a prime p divides a², then p divides a. This is used to prove numbers like √2, √5 are irrational.

✍️ 4. Method to Prove Irrationality (by contradiction)

To prove √5 is irrational: assume √5 = p/q (rational, p, q coprime). Square → 5q² = p² → 5 divides p² → 5 divides p → write p = 5m → 5q² = 25m² → q² = 5m² → 5 divides q. So 5 divides both p and q — contradicts "coprime". Hence √5 is irrational.

🧠 Remember: FTA → unique prime factorisation. HCF = common primes (least power); LCM = all primes (highest power). HCF×LCM = a×b (two numbers only). To end in 0, a number needs factor 2 AND 5. Prove irrationality by contradiction.

⭐ Exam Tips: ① Prime factorisation (write as 2ᵃ×3ᵇ...). ② HCF & LCM + verify HCF×LCM=product. ③ Why 6ⁿ can't end in 0 (no factor 5). ④ Prove a number composite (take common factor). ⑤ Prove √2/√5/(a+b√p) irrational.

📘 EXERCISE 1.1

1Express each number as a product of its prime factors: 140, 156, 3825, 5005, 7429.

Divide each number by the smallest prime repeatedly (factor tree). 140 = 2×70 = 2×2×35 = 2×2×5×7 = 2² × 5 × 7. 156 = 2×78 = 2×2×39 = 2×2×3×13 = 2² × 3 × 13. 3825 = 3×1275 = 3×3×425 = 9×425 = 9×5×85 = 9×5×5×17 = 3² × 5² × 17. 5005 = 5×1001 = 5×7×143 = 5×7×11×13 = 5 × 7 × 11 × 13. 7429 = 17×437 = 17×19×23 = 17 × 19 × 23.

2Find the LCM and HCF of 26 and 91 and verify LCM × HCF = product of the numbers.

Prime factorise: 26 = 2 × 13 ; 91 = 7 × 13. HCF = product of common prime (lowest power) = 13. LCM = product of all primes (highest power) = 2 × 7 × 13 = 182. Verify: HCF × LCM = 13 × 182 = 2366 ; and 26 × 91 = 2366. ✓ ∴ HCF = 13, LCM = 182 (relation verified). [510,92 → HCF 2, LCM 23460 ; 336,54 → HCF 6, LCM 3024 — same method]

3Find the HCF and LCM of 12, 15 and 21 by prime factorisation.

12 = 2² × 3 ; 15 = 3 × 5 ; 21 = 3 × 7. HCF = common prime with lowest power = only 3 is common → 3. LCM = highest power of every prime = 2² × 3 × 5 × 7 = 420. ∴ HCF = 3, LCM = 420. [17,23,29 → HCF 1, LCM 11339 ; 8,9,25 → HCF 1, LCM 1800]

4Given HCF(306, 657) = 9, find LCM(306, 657).

Use LCM × HCF = product of the two numbers. LCM = (306 × 657) / HCF = 201042 / 9. ∴ LCM = 22338

5Check whether 6ⁿ can end with the digit 0 for any natural number n.

A number ends in 0 only if it is divisible by 10 = 2 × 5, i.e. it must contain both primes 2 and 5. 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ — its only prime factors are 2 and 3. There is no factor 5, so 6ⁿ is never divisible by 10. ∴ 6ⁿ can never end with the digit 0.

6Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite.

Take out the common factor. 7×11×13 + 13 = 13(7×11 + 1) = 13 × 78. It is a product of 13 and 78 (both > 1) → it has factors other than 1 and itself → composite. Similarly 7×6×5×4×3×2×1 + 5 = 5(7×6×4×3×2×1 + 1) = 5 × 1009 → composite. ∴ Both numbers are composite.

7Sonia takes 18 min and Ravi 12 min per round of a circular path. After how many minutes do they meet again at the start?

They meet at the start after a time that is a multiple of both 18 and 12 → take the LCM. 18 = 2 × 3² ; 12 = 2² × 3 → LCM = 2² × 3² = 36. ∴ They meet again after 36 minutes.

📗 EXERCISE 1.2 (Irrational Numbers)

1Prove that √5 is irrational.

Assume the opposite: let √5 be rational, so √5 = p/q where p, q are coprime integers (q ≠ 0). Squaring: 5 = p²/q² → p² = 5q². So 5 divides p². Hence 5 divides p. Write p = 5m. Then (5m)² = 5q² → 25m² = 5q² → q² = 5m². So 5 divides q² → 5 divides q. Now 5 divides both p and q — but we assumed they are coprime. Contradiction. ∴ Our assumption is wrong; √5 is irrational.

2Prove that 3 + 2√5 is irrational.

Assume 3 + 2√5 is rational, say = a (a rational). Then 2√5 = a − 3 → √5 = (a − 3)/2. The right side is rational (difference/quotient of rationals), so √5 would be rational. But √5 is irrational (proved above). Contradiction. ∴ 3 + 2√5 is irrational.

3Prove that (i) 1/√2 (ii) 7√5 (iii) 6 + √2 are irrational.

(i) If 1/√2 were rational, then √2 = 1 ÷ (that rational) would be rational — but √2 is irrational. Contradiction → irrational. (ii) If 7√5 = rational r, then √5 = r/7 would be rational — contradiction → irrational. (iii) If 6 + √2 = rational r, then √2 = r − 6 would be rational — contradiction → irrational. ∴ All three are irrational.

Extra practice (4× the textbook) — MCQ, short answer & long answer.

🎯 Extra MCQ

1The HCF of the smallest prime and the smallest composite number is —

Smallest prime = 2, smallest composite = 4; HCF(2,4) = 2.

2If p and q are two coprime numbers, their LCM is —

p × q (their product).

3The number 2 × 3 × 5 × 7 + 7 is —

Composite [= 7(2×3×5+1) = 7×31].

4√12 can be simplified to —

2√3 (still irrational).

📝 Short Answer

5Find the HCF and LCM of 96 and 404, and verify the relation.

96=2⁵×3, 404=2²×101 → HCF=2²=4, LCM=2⁵×3×101=9696. Check: 4×9696=38784=96×404 ✓.

6Show that any positive odd integer is of the form 2q+1.

Any integer divided by 2 leaves remainder 0 or 1. Even → 2q; odd → 2q+1.

7Two tankers contain 850 L and 680 L of diesel. Find the maximum capacity of a container that can measure both exactly.

Required = HCF(850,680). 850=2×5²×17, 680=2³×5×17 → HCF=2×5×17=170 L.

8Can the LCM of two numbers be smaller than their HCF? Why?

No. The LCM is always a multiple of the HCF, so LCM ≥ HCF.

9Find the largest number that divides 245 and 1029 leaving remainder 5 in each case.

HCF(245−5, 1029−5) = HCF(240,1024) = 16.

📖 Long Answer

10Prove that √3 is irrational.

Assume √3 = p/q (coprime). 3q²=p² → 3|p² → 3|p → p=3m → 3q²=9m² → q²=3m² → 3|q. So 3 divides both p and q, contradicting coprime. Hence √3 is irrational.

11Prove that 5 − √3 is irrational, given √3 is irrational.

If 5−√3 = rational r, then √3 = 5−r = rational, contradicting that √3 is irrational. So 5−√3 is irrational.

12Three bells ring at intervals of 9, 12 and 15 minutes. If they ring together at 8 a.m., when will they next ring together?

LCM(9,12,15): 9=3², 12=2²×3, 15=3×5 → LCM=2²×3²×5=180 min = 3 hours. Next together at 11 a.m.

13Find the HCF and LCM of 6, 72 and 120 using prime factorisation.

6=2×3, 72=2³×3², 120=2³×3×5 → HCF=2×3=6, LCM=2³×3²×5=360.

14Explain why 17 + 11×13×17×19 is a composite number.

= 17(1 + 11×13×19) = 17 × 2718, which has factors other than 1 and itself → composite.

15The HCF of two numbers is 18 and their LCM is 720. If one number is 90, find the other.

Other = (HCF×LCM)/one = (18×720)/90 = 144.