Real Numbers

Answer

Step 1: Divide 140 by smallest prime (2) → 140 ÷ 2 = 70 Step 2: 70 ÷ 2 = 35 Step 3: 35 ÷ 5 = 7 Step 4: 7 is prime Therefore: 140 = 2² × 5 × 7 Answer: 140 = 2² × 5¹ × 7¹

Answer

Step 1: Prime factorization 96 = 2⁵ × 3¹ 404 = 2² × 101¹ Step 2: HCF = Product of lowest powers of common factors Common factor: 2 HCF = 2² = 4 Step 3: LCM = Product of highest powers of all factors L

Answer

Step 1: Prime factorization 26 = 2¹ × 13¹ 91 = 7¹ × 13¹ Step 2: Find HCF and LCM HCF = 13¹ = 13 (common factor) LCM = 2¹ × 7¹ × 13¹ = 182 Step 3: Verify the relationship HCF × LCM = 13 × 182 = 2366 Pr

Answer

Step 1: For a number to end in 0, it must be divisible by 10 Step 2: 10 = 2 × 5, so the number must have both 2 and 5 as factors Step 3: Find prime factorization of 4ⁿ 4ⁿ = (2²)ⁿ = 2²ⁿ Step 4: Analysi

Answer

Step 1: Prime factorization 12 = 2² × 3¹ 15 = 3¹ × 5¹ 21 = 3¹ × 7¹ Step 2: HCF = Product of lowest powers of common factors Common factor: 3¹ HCF = 3 Step 3: LCM = Product of highest powers of all fac

Answer

Step 1: Use the relationship HCF × LCM = Product of numbers Step 2: Substitute known values 9 × LCM = 306 × 657 Step 3: Calculate product 306 × 657 = 201,042 Step 4: Find LCM LCM = 201,042 ÷ 9 = 22,33

Answer

Step 1: Assume √2 is rational, so √2 = a/b where a, b are coprime integers Step 2: Square both sides: 2 = a²/b², so 2b² = a² Step 3: Therefore 2 divides a², so 2 divides a (by theorem) Step 4: Let a =

Answer

Step 1: Assume √3 is rational, so √3 = a/b where a, b are coprime Step 2: Square both sides: 3 = a²/b², so 3b² = a² Step 3: Therefore 3 divides a², so 3 divides a Step 4: Let a = 3c, then 3b² = 9c², s