Real Numbers

3 exercises15 questions solved

Exercise 1.1Euclid's Division Algorithm

Apply Euclid's division algorithm step-by-step to find the HCF of each pair: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Solution

(i) HCF of 135 and 225: Step 1: 225 = 1 × 135 + 90 Step 2: 135 = 1 × 90 + 45 Step 3: 90 = 2 × 45 + 0 Since remainder = 0, HCF(135, 225) = 45 (ii) HCF of 196 and 38220: Step 1: 38220 = 195 × 196 + 0 Since remainder = 0, HCF(196, 38220) = 196 (iii) HCF of 867 and 255: Step 1: 867 = 3 × 255 + 102 Step 2: 255 = 2 × 102 + 51 Step 3: 102 = 2 × 51 + 0 Since remainder = 0, HCF(867, 255) = 51

Prove that every positive odd integer can be written in one of the forms 6q + 1, 6q + 3, or 6q + 5, where q is an integer.

Solution

Let a be any positive integer and b = 6. By Euclid's division algorithm: a = 6q + r, where 0 ≤ r < 6. So r can be 0, 1, 2, 3, 4, or 5. This gives: a = 6q, 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. • 6q = 2(3q) → even • 6q+1 → odd ✓ • 6q+2 = 2(3q+1) → even • 6q+3 = 3(2q+1) → odd ✓ • 6q+4 = 2(3q+2) → even • 6q+5 = 6q+5 → odd ✓ Therefore, any positive odd integer is of the form 6q+1, 6q+3, or 6q+5.

A parade has two groups: 616 soldiers and a 32-member band. Both groups must march in an equal number of columns. Using HCF, find the greatest number of columns possible.

Solution

The maximum number of columns = HCF(616, 32) Using Euclid's division algorithm: Step 1: 616 = 19 × 32 + 8 Step 2: 32 = 4 × 8 + 0 Since the remainder is 0, HCF(616, 32) = 8 The maximum number of columns in which they can march is 8.

Using Euclid's division lemma, prove that the square of any positive integer is of the form 3m or 3m + 1 for some integer m. (Hint: write x = 3q + r and consider all possible remainders when dividing by 3.)

Solution

Let x be any positive integer. By Euclid's division algorithm: x = 3q + r, where 0 ≤ r < 3, so r = 0, 1, or 2. Case 1: r = 0 → x = 3q x² = 9q² = 3(3q²) = 3m, where m = 3q² Case 2: r = 1 → x = 3q + 1 x² = (3q+1)² = 9q² + 6q + 1 = 3(3q² + 2q) + 1 = 3m + 1, where m = 3q² + 2q Case 3: r = 2 → x = 3q + 2 x² = (3q+2)² = 9q² + 12q + 4 = 9q² + 12q + 3 + 1 = 3(3q² + 4q + 1) + 1 = 3m + 1, where m = 3q² + 4q + 1 In all cases, x² is either of the form 3m or 3m + 1.

Using Euclid's division lemma, prove that the cube of any positive integer must take one of the forms: 9m, 9m + 1, or 9m + 8.

Solution

Let x be any positive integer. By Euclid's division algorithm: x = 3q + r, where r = 0, 1, or 2. Case 1: r = 0 → x = 3q x³ = 27q³ = 9(3q³) = 9m, where m = 3q³ Case 2: r = 1 → x = 3q + 1 x³ = (3q+1)³ = 27q³ + 27q² + 9q + 1 = 9(3q³ + 3q² + q) + 1 = 9m + 1, where m = 3q³ + 3q² + q Case 3: r = 2 → x = 3q + 2 x³ = (3q+2)³ = 27q³ + 54q² + 36q + 8 = 9(3q³ + 6q² + 4q) + 8 = 9m + 8, where m = 3q³ + 6q² + 4q Therefore, the cube of any positive integer is of the form 9m, 9m+1, or 9m+8.

Exercise 1.2Fundamental Theorem of Arithmetic

Write each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution

(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 = 3 × 3 × 5 × 85 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17 (iv) 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13 (v) 7429 = 17 × 437 = 17 × 19 × 23

For each pair of integers, calculate the HCF and LCM, then verify that HCF × LCM equals the product of the two numbers: (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

Solution

(i) 26 and 91: 26 = 2 × 13; 91 = 7 × 13 HCF = 13; LCM = 2 × 7 × 13 = 182 Verification: 182 × 13 = 2366 = 26 × 91 ✓ (ii) 510 and 92: 510 = 2 × 3 × 5 × 17; 92 = 2² × 23 HCF = 2; LCM = 2² × 3 × 5 × 17 × 23 = 23460 Verification: 23460 × 2 = 46920 = 510 × 92 ✓ (iii) 336 and 54: 336 = 2⁴ × 3 × 7; 54 = 2 × 3³ HCF = 2 × 3 = 6; LCM = 2⁴ × 3³ × 7 = 3024 Verification: 3024 × 6 = 18144 = 336 × 54 ✓

Using prime factorisation, find the HCF and LCM of each set of numbers: (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

Solution

(i) 12, 15 and 21: 12 = 2² × 3; 15 = 3 × 5; 21 = 3 × 7 HCF = 3 (common factor) LCM = 2² × 3 × 5 × 7 = 420 (ii) 17, 23 and 29: All three are prime numbers. HCF = 1 (no common factor) LCM = 17 × 23 × 29 = 11339 (iii) 8, 9 and 25: 8 = 2³; 9 = 3²; 25 = 5² HCF = 1 (no common factor) LCM = 2³ × 3² × 5² = 1800

If HCF(306, 657) = 9, calculate LCM(306, 657) using the HCF–LCM relationship.

Solution

We know that for any two positive integers a and b: LCM × HCF = a × b LCM(306, 657) × HCF(306, 657) = 306 × 657 LCM(306, 657) × 9 = 306 × 657 LCM(306, 657) = (306 × 657) / 9 LCM(306, 657) = 201042 / 9 LCM(306, 657) = 22338

Can 6ⁿ ever end with the digit 0 for any natural number n? Justify your answer using prime factorisation.

Solution

For 6ⁿ to end with 0, it must be divisible by 10 = 2 × 5. So 6ⁿ must have 5 as a prime factor. Prime factorisation of 6 = 2 × 3 Therefore, 6ⁿ = 2ⁿ × 3ⁿ The only prime factors of 6ⁿ are 2 and 3. Since 5 is not a factor of 6ⁿ, 6ⁿ cannot be divisible by 5. By the Fundamental Theorem of Arithmetic, the prime factorisation of any number is unique. Therefore, 6ⁿ can never end with the digit 0 for any natural number n.

Show that 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are both composite numbers. Give a mathematical justification for each.

Solution

(i) 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78 = 13 × 2 × 39 = 13 × 2 × 3 × 13 This number has factors 2, 3, and 13 (other than 1 and itself), so it is a COMPOSITE number. ✓ (ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009 This number has factors 5 and 1009 (other than 1 and itself), so it is a COMPOSITE number. ✓

Two athletes jog on a circular track. One completes a lap in 18 minutes, the other in 12 minutes. Both start together from the same point going in the same direction. After how many minutes will they be at the starting point together again?

Solution

They will meet at the starting point after a time equal to the LCM of 18 and 12 minutes. Prime factorisation: 18 = 2 × 3² 12 = 2² × 3 LCM(18, 12) = 2² × 3² = 4 × 9 = 36 Sonia and Ravi will meet again at the starting point after 36 minutes. (Check: In 36 minutes, Sonia completes 36/18 = 2 rounds and Ravi completes 36/12 = 3 rounds — both back at start. ✓)

Exercise 1.3Irrational Numbers

Using proof by contradiction, establish that √5 is irrational.

Solution

Proof by contradiction: Assume that √5 is rational. Then it can be written as p/q, where p and q are integers with HCF(p, q) = 1 (i.e., p/q is in lowest terms). √5 = p/q Squaring both sides: 5 = p²/q² ⟹ p² = 5q² ... (1) This means 5 divides p², and since 5 is prime, 5 divides p. Let p = 5m for some integer m. Substituting in (1): (5m)² = 5q² ⟹ 25m² = 5q² ⟹ q² = 5m² This means 5 divides q², and since 5 is prime, 5 divides q. But we now have 5 dividing both p and q, which contradicts HCF(p, q) = 1. Our assumption was wrong. Therefore, √5 is irrational.

Prove that 3 + 2√5 cannot be expressed as a rational number.

Solution

Given that √5 is irrational (proved in Q1). Assume that 3 + 2√5 is rational. Then: 3 + 2√5 = p/q, where p, q are integers, q ≠ 0, HCF(p, q) = 1 2√5 = p/q − 3 2√5 = (p − 3q)/q √5 = (p − 3q)/(2q) Since p, q are integers, (p − 3q)/(2q) is rational. This means √5 is rational. But this contradicts the fact that √5 is irrational. Therefore, our assumption is wrong. 3 + 2√5 is irrational.

Establish that each of the following is irrational: (i) 1/√2 (ii) 7√5 (iii) 6 + √2

Solution

(i) Prove 1/√2 is irrational: Assume 1/√2 is rational = p/q (lowest terms). Then √2 = q/p, which is rational. Contradiction (√2 is irrational). Therefore, 1/√2 is irrational. (ii) Prove 7√5 is irrational: Assume 7√5 = p/q (rational, lowest terms). Then √5 = p/(7q), which is rational. Contradiction (√5 is irrational). Therefore, 7√5 is irrational. (iii) Prove 6 + √2 is irrational: Assume 6 + √2 = p/q (rational). Then √2 = p/q − 6 = (p − 6q)/q, which is rational. Contradiction (√2 is irrational). Therefore, 6 + √2 is irrational.

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