Comprehensive Chapter Summary
1. Introduction
In Class IX, you began your exploration of the world of real numbers and encountered irrational numbers. We continue our discussion on real numbers in this chapter. We begin with two very important properties of positive integers, namely the Euclid’s division algorithm and the Fundamental Theorem of Arithmetic. The chapter explores divisibility, prime factorization, irrationality proofs, and decimal expansions of rationals.
Real numbers include both rational and irrational numbers. Rational numbers are those expressible as p/q where p and q are integers, q ≠ 0. Irrational numbers cannot be expressed this way, like √2, √3. The real number line represents all reals, with rationals dense but countable, irrationals uncountable. Historical context: Discovery of irrationals by Hippasus, Gauss on FTA.
Importance: Real numbers form the basis for calculus, analysis. Applications in physics (measurements), engineering (precise calculations).
2. Euclid’s Division Algorithm
Euclid’s division algorithm deals with divisibility of integers. Any positive integer a can be divided by another positive integer b leaving a remainder r smaller than b: a = bq + r, 0 ≤ r < b. This is used to find HCF using repeated division until remainder is 0.
Applications include computing HCF, which is the last non-zero remainder in the process. Also used in cryptography (Euclidean algorithm for gcd), number theory proofs.
Extended: For negative integers, take absolute values. Relation to modular arithmetic: r = a mod b.
3. The Fundamental Theorem of Arithmetic
Prime Factorization
Every composite number can be expressed as a product of primes uniquely (up to order). For example, 32760 = 2^3 × 3^2 × 5 × 7 × 13. This uniqueness is key for many proofs.
Composite numbers are natural numbers greater than 1 that are not prime, i.e., have factors other than 1 and itself. Primes are infinite, proof by Euclid: Assume finite, product+1 not divisible by any, new prime.
Method: Factor tree, trial division up to sqrt(n).
Applications
Proving irrationality: Used to show numbers like √2 are irrational by contradiction. Also, determines if decimal expansion of 1/p is terminating (if p has only 2 and 5 as prime factors) or repeating.
Infinitely many primes: If finite, product +1 would be new prime, contradiction. Used in RSA encryption, number of divisors formula d(n) = (e1+1)(e2+1)...
Radical simplification, rationalizing denominators.
HCF and LCM
HCF: Product of smallest powers of common primes. LCM: Product of highest powers. Relation: HCF(a,b) × LCM(a,b) = a × b. Extended to three numbers with formulas provided.
For example, HCF(6,20)=2, LCM=60, product=120=6*20. Applications: Simplifying fractions, solving Diophantine equations, scheduling problems.
Note for >2 numbers: Product ≠ HCF * LCM, use pairwise methods or formulas.
4. Irrational Numbers
Proof of Irrationality
Proof by contradiction for √p where p prime: Assume rational a/b coprime, square to a^2 = p b^2, p divides a^2 hence a, substitute, p divides b, contradiction.
Examples: √2, √3, √5 irrational. Properties: Rational + irrational = irrational, etc. Transcendental irrationals like π, e beyond scope but mentioned.
Historical: Pythagoreans, crisis in mathematics.
Properties
Sum/difference/product/quotient of rational and irrational is irrational (non-zero rational). Examples: 5 - √3, 3√2 irrational.
Locating irrationals on number line from Class IX recalled. Density: Between any two reals, infinite rationals and irrationals.
Irrational + irrational can be rational (√2 + (-√2)=0), product too (√2 * √2=2).
5. Decimal Expansions
For rational p/q in lowest terms, terminating if q = 2^m * 5^n, else non-terminating repeating. Prime factorization of q reveals this.
Examples: 1/2=0.5 terminating, 1/3=0.333... repeating. Converting repeating to fraction: Let x=0.333..., 10x=3.333..., 9x=3, x=1/3.
Non-repeating infinite decimals for irrationals like π=3.14159..., e=2.71828.... Applications in computing precision, financial calculations.
Solved Examples from Chapter
Example 1: Check if 4^n ends with 0 for some natural n.
Solution: \( 4^n = (2^2)^n = 2^{2n} \). To end with 0, divisible by 10=2*5, but no 5 in factors. By FTA uniqueness, impossible. Examples: \( 4^1=4 \), \( 4^2=16 \), \( 4^3=64 \), none end 0.
Example 2: HCF and LCM of 6 and 20 by prime factorization.
Solution: \( 6=2 \times 3 \), \( 20=2^2 \times 5 \). HCF=\( 2^1=2 \), LCM=\( 2^2 \times 3 \times 5=60 \). Verify \( 2 \times 60=120=6 \times 20 \). Application: Simplify 6/20 = 3/10.
Example 3: HCF of 96 and 404, then LCM.
Solution: \( 96=2^5 \times 3 \), \( 404=2^2 \times 101 \). HCF=\( 2^2=4 \). LCM=(96*404)/4=9696. Euclidean: 404=4*96 +20, 96=4*20+16, 20=1*16+4, 16=4*4+0, HCF=4.
Example 4: HCF and LCM of 6,72,120.
Solution: \( 6=2 \times 3 \), \( 72=2^3 \times 3^2 \), \( 120=2^3 \times 3 \times 5 \). HCF=\( 2 \times 3=6 \), LCM=\( 2^3 \times 3^2 \times 5=360 \). Note 6*72*120 = 51840 ≠ 6*360=2160. Use pairwise LCM.
Example 5: Prove √3 irrational.
Solution: Assume \( \sqrt{3} = \frac{a}{b} \) coprime. \( a^2=3b^2 \), 3 divides a, a=3c, 9c^2=3b^2, b^2=3c^2, 3 divides b. Contradiction. Approximation: 1.7320508...
Example 6: Show 5 - √3 irrational.
Solution: Assume rational \( \frac{a}{b} \). Then \( \sqrt{3} = 5 - \frac{a}{b} \) rational, contradiction. Similarly for 5 + √3.
Example 7: Show 3√2 irrational.
Solution: Assume \( \frac{a}{b} \). Then \( \sqrt{2} = \frac{a}{3b} \) rational, contradiction. General: k√m irrational if √m irrational, k rational nonzero.
Example 8: Decimal of 1/7.
Solution: 7 has prime 7, so repeating: 0.142857 repeating. Period 6.
Example 9: Factorize 123456789.
Solution: \( 3^2 \times 3803 \times 3607 \), where 3803 and 3607 primes.
Example 10: HCF(306,657).
Solution: Euclidean: 657=2*306+45, 306=6*45+36, 45=1*36+9, 36=4*9+0, HCF=9.
Questions and Answers
Exercise 1.1
1. Express each number as a product of its prime factors:
(i) 140
Solution: \( 140 = 2^{2} \times 5 \times 7 \)
(ii) 156
Solution: \( 156 = 2^{2} \times 3 \times 13 \)
(iii) 3825
Solution: \( 3825 = 3^{2} \times 5^{2} \times 17 \)
(iv) 5005
Solution: \( 5005 = 5 \times 7 \times 11 \times 13 \)
(v) 7429
Solution: \( 7429 = 17 \times 19 \times 23 \)
2. 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
Solution: \( 26 = 2 \times 13 \), \( 91 = 7 \times 13 \), HCF = 13, LCM = \( 2 \times 7 \times 13 = 182 \), Verify: \( 13 \times 182 = 2366 = 26 \times 91 \)
(ii) 510 and 92
Solution: \( 510 = 2 \times 3 \times 5 \times 17 \), \( 92 = 2^{2} \times 23 \), HCF = 2, LCM = \( 2^{2} \times 3 \times 5 \times 17 \times 23 = 23460 \), Verify: \( 2 \times 23460 = 46920 = 510 \times 92 \)
(iii) 336 and 54
Solution: \( 336 = 2^{4} \times 3 \times 7 \), \( 54 = 2 \times 3^{3} \), HCF = \( 2 \times 3 = 6 \), LCM = \( 2^{4} \times 3^{3} \times 7 = 3024 \), Verify: \( 6 \times 3024 = 18144 = 336 \times 54 \)
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
Solution: \( 12 = 2^{2} \times 3 \), \( 15 = 3 \times 5 \), \( 21 = 3 \times 7 \), HCF = 3, LCM = \( 2^{2} \times 3 \times 5 \times 7 = 420 \)
(ii) 17, 23 and 29
Solution: All primes, HCF = 1, LCM = \( 17 \times 23 \times 29 = 11339 \)
(iii) 8, 9 and 25
Solution: \( 8 = 2^{3} \), \( 9 = 3^{2} \), \( 25 = 5^{2} \), HCF = 1, LCM = \( 2^{3} \times 3^{2} \times 5^{2} = 1800 \)
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution: LCM = \( \frac{306 \times 657}{9} = 22338 \)
5. Check whether 6n can end with the digit 0 for any natural number n.
Solution: \( 6^{n} = (2 \times 3)^{n} = 2^{n} \times 3^{n} \), to end with 0 need 2 and 5, no 5, impossible.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution: \( 7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13 \times 78 \) composite. \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times 1009 \) composite.
7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi 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?
Solution: LCM(18,12) = 36 minutes.
Exercise 1.2
1. Prove that √5 is irrational.
Solution: Assume \( \sqrt{5} = \frac{a}{b} \) coprime. \( a^{2} = 5b^{2} \), 5 divides a, a=5c, \( 25c^{2} = 5b^{2} \), \( b^{2} = 5c^{2} \), 5 divides b, contradiction.
2. Prove that 3 + 2√5 is irrational.
Solution: Assume rational r, then \( 2\sqrt{5} = r - 3 \) rational, \( \sqrt{5} = \frac{r-3}{2} \) rational, contradiction.
3. Prove that the following are irrationals:
(i) \( \frac{1}{\sqrt{2}} \)
Solution: Assume \( \frac{a}{b} \), then \( \sqrt{2} = \frac{b}{a} \) rational, contradiction.
(ii) 7√5
Solution: Assume \( \frac{a}{b} \), then \( \sqrt{5} = \frac{a}{7b} \) rational, contradiction.
(iii) 6 + √2
Solution: Assume \( \frac{a}{b} \), then \( \sqrt{2} = \frac{a}{b} - 6 \) rational, contradiction.
Short Questions
Q1. Prime factors of 12.
Answer: \( 2^{2} \times 3 \).
Q2. HCF(8,12).
Answer: 4.
Q3. LCM(4,6).
Answer: 12.
Q4. Is √4 rational?
Answer: Yes, 2.
Q5. Terminating? 1/4.
Answer: Yes, 0.25.
Q6. Prime factors of 30.
Answer: \( 2 \times 3 \times 5 \).
Q7. HCF(15,25).
Answer: 5.
Q8. LCM(5,7).
Answer: 35.
Q9. Is π rational?
Answer: No.
Q10. Terminating? 1/6.
Answer: No, 0.1666...
Q11. Factors of 42.
Answer: \( 2 \times 3 \times 7 \).
Q12. HCF(18,24).
Answer: 6.
Q13. LCM(8,12).
Answer: 24.
Q14. Rational? 0.5.
Answer: Yes, \( \frac{1}{2} \).
Q15. Repeating? 1/9.
Answer: Yes, 0.111...
Q16. Factors of 100.
Answer: \( 2^{2} \times 5^{2} \).
Q17. HCF(20,30).
Answer: 10.
Q18. LCM(9,15).
Answer: 45.
Q19. Irrational? e.
Answer: Yes.
Q20. Terminating? 3/8.
Answer: Yes, 0.375.
Medium Questions
Q1. Prove √7 irrational.
Solution: Assume \( \sqrt{7} = \frac{a}{b} \) coprime. \( a^{2} = 7b^{2} \), 7 divides a, a=7c, 49c^{2}=7b^{2}, b^{2}=7c^{2}, 7 divides b. Contradiction.
Q2. HCF(48,60,72).
Solution: \( 48=2^{4} \times 3 \), \( 60=2^{2} \times 3 \times 5 \), \( 72=2^{3} \times 3^{2} \). HCF=\( 2^{2} \times 3=12 \).
Q3. LCM(10,15,25).
Solution: \( 10=2 \times 5 \), \( 15=3 \times 5 \), \( 25=5^{2} \). LCM=\( 2 \times 3 \times 5^{2}=150 \).
Q4. Show 2 + √5 irrational.
Solution: Assume rational, \( \sqrt{5} = \) rational -2 rational, contradiction.
Q5. Decimal type 1/11.
Solution: 11 prime not 2 or 5, repeating 0.0909...
Q6. Factorize 980.
Solution: \( 2^{2} \times 5 \times 7^{2} \).
Q7. HCF(105,175).
Solution: Euclidean: 175=1*105+70, 105=1*70+35, 70=2*35+0, HCF=35.
Q8. LCM(12,18,24).
Solution: \( 12=2^{2} \times 3 \), \( 18=2 \times 3^{2} \), \( 24=2^{3} \times 3 \). LCM=\( 2^{3} \times 3^{2}=72 \).
Q9. Prove 4√2 irrational.
Solution: Assume rational, \( \sqrt{2} = \) rational/4 rational, contradiction.
Q10. Terminating? 5/16.
Solution: 16=\( 2^{4} \), yes, 0.3125.
Q11. Factors of 315.
Solution: \( 3^{2} \times 5 \times 7 \).
Q12. HCF(84,96).
Answer: 12.
Q13. LCM(14,21).
Answer: 42.
Q14. Show √3 + √5 irrational.
Solution: Assume rational r. Square: 8 + 2√15 = r^2 rational, 2√15 rational, √15 rational, contradiction.
Q15. Decimal 2/5.
Solution: 0.4 terminating.
Q16. Factorize 1232.
Solution: \( 2^{4} \times 7 \times 11 \).
Q17. HCF(225,300).
Answer: 75.
Q18. LCM(20,25,30).
Answer: 300.
Q19. Prove √10 irrational.
Solution: 10=2*5, not square, assume a/b, a^2=10b^2, 2 and 5 divide a, then b, contradiction.
Q20. Repeating? 1/12.
Solution: 12=\( 2^{2} \times 3 \), repeating 0.08333...
Long Questions
Q1. Use Euclid algorithm to find HCF(1234,5678), show steps.
Solution: 5678=4*1234+742, 1234=1*742+492, 742=1*492+250, 492=1*250+242, 250=1*242+8, 242=30*8+2, 8=4*2+0, HCF=2.
Q2. Prove that if p prime divides ab, then p divides a or b.
Solution: Assume not a, gcd(p,a)=1, extended Euclidean px + ay=1, multiply b: p xb + a b y = b, p divides ab hence left, so b.
Q3. Find LCM(12,18,24,30) using prime factors, verify with formula.
Solution: \( 12=2^{2} \times 3 \), \( 18=2 \times 3^{2} \), \( 24=2^{3} \times 3 \), \( 30=2 \times 3 \times 5 \). LCM=\( 2^{3} \times 3^{2} \times 5=360 \). Pairwise verify.
Q4. Prove √6 irrational using FTA.
Solution: Assume a/b, a^2=6b^2, 2*3 b^2 = a^2, 2 and 3 divide a^2 hence a, a=6c, 36c^2=6b^2, 6c^2=b^2, 2 and 3 divide b, contradiction.
Q5. Determine decimal type for 7/40, compute.
Solution: 40=\( 2^{3} \times 5 \), terminating. 7/40=0.175.
Q6. Factorize 987654321, check if prime factors unique.
Solution: \( 3^{2} \times 37 \times 333667 \), unique by FTA.
Q7. Show that √2 + √3 irrational.
Solution: Assume r rational. Square: 5 + 2√6 = r^2 rational, 2√6 rational, √6 rational, contradiction as 6 not square.
Q8. Find HCF(1001,1000) using Euclidean, explain steps.
Solution: 1001=1*1000+1, 1000=1000*1+0, HCF=1. Coprime.
Q9. Prove there are infinitely many primes of form 4k+3.
Solution: Assume finite p1..pn all 4k+3. Let n=4(p1..pn)-1=4m+3. n odd >1, prime factors 4k+1 or 4k+3. If all 4k+1, product 4k+1 mod 4, but n=3 mod 4, contradiction. So new 4k+3 prime.
Q10. Convert 0.123123... to fraction.
Solution: x=0.123123..., 1000x=123.123..., 999x=123, x=\( \frac{123}{999} = \frac{41}{333} = \frac{13}{111} \).
Q11. Factorize 2025.
Solution: \( 45^{2} = (9 \times 5)^{2} = 3^{4} \times 5^{2} \).
Q12. HCF(385,462).
Solution: 462=1*385+77, 385=5*77+0, HCF=77.
Q13. LCM(16,24,36,48).
Solution: Prime max: \( 2^{4} \times 3^{2} = 144 \).
Q14. Prove √2 * √3 = √6 irrational.
Solution: Product of irrationals can be irrational. Assume rational, square 6 rational, but √6 irrational.
Q15. Decimal of 3/7.
Solution: 0.428571 repeating, period 6.
Q16. Factorize 5544.
Solution: \( 2^{3} \times 3^{2} \times 7 \times 11 \).
Q17. Show 1/√3 irrational.
Solution: Assume \( \frac{a}{b} \), \( \sqrt{3} = \frac{b}{a} \) rational, contradiction.
Q18. HCF(999,1001).
Solution: 1001=1*999+2, 999=499*2+1, 2=2*1+0, HCF=1.
Q19. LCM(35,49,63).
Solution: \( 35=5 \times 7 \), \( 49=7^{2} \), \( 63=3^{2} \times 7 \). LCM=\( 3^{2} \times 5 \times 7^{2}=2205 \).
Q20. Prove log10 2 irrational.
Solution: Assume p/q, \( 10^{p/q}=2 \), \( 10^{p} = 2^{q} \), left 2 or 5 factors only, right 2 only, contradiction unless p=q=0.
