Real Numbers — Karnataka (SSLC) Class 10 Mathematics Solutions (Free)

The Fundamental Theorem of Arithmetic states that every composite number can be expressed (factorised) as a product of prime numbers, and this factorisation is unique, apart from the order in which the prime factors occur. For example: - 60 = 2² × 3 × 5 - 84 = 2² × 3 × 7 This means that if we write a composite number as a product of primes in two different ways, the prime factors (with their powers) must be exactly the same.

Proof by contradiction: Assume √2 is rational. Then √2 = p/q, where p and q are coprime integers (HCF(p,q) = 1) and q ≠ 0. Squaring both sides: 2 = p²/q² Therefore: 2q² = p² ... (1) From equation (1), p² is even, which means p must be even. Let p = 2m for some integer m. Substituting in equation (1): 2q² = (2m)² 2q² = 4m² q² = 2m² This shows q² is even, so q must be even. But if both p and q are even, then HCF(p,q) ≥ 2, which contradicts our assumption that p and q are coprime. Therefore, our assumption is wrong, and √2 is irrational.

Using Euclid's division algorithm: Step 1: Divide 404 by 96 404 = 96 × 4 + 20 Step 2: Divide 96 by 20 96 = 20 × 4 + 16 Step 3: Divide 20 by 16 20 = 16 × 1 + 4 Step 4: Divide 16 by 4 16 = 4 × 4 + 0 Since the remainder is 0, the HCF is 4. Therefore, HCF(96, 404) = 4

Step 1: Find prime factorisation of both numbers 12 = 2² × 3 18 = 2 × 3² Step 2: Find HCF (take lowest powers of common prime factors) Common prime factors: 2 and 3 HCF = 2¹ × 3¹ = 6 Step 3: Find LCM (take highest powers of all prime factors) LCM = 2² × 3² = 4 × 9 = 36 Verification: HCF × LCM = 6 × 36 = 216 and 12 × 18 = 216 ✓ Therefore, HCF(12, 18) = 6 and LCM(12, 18) = 36

Let x = 0.353535... Since two digits repeat, multiply both sides by 100: 100x = 35.353535... Subtract the original equation: 100x - x = 35.353535... - 0.353535... 99x = 35 x = 35/99 Check if this can be simplified: HCF(35, 99) = HCF(5 × 7, 9 × 11) = 1 Therefore, 0.3̄5̄ = 35/99

Using prime factorisation: 84 = 2² × 3 × 7 120 = 2³ × 3 × 5 138 = 2 × 3 × 23 HCF = Product of lowest powers of common prime factors Common prime factors: 2 and 3 HCF = 2¹ × 3¹ = 6 Therefore, HCF(84, 120, 138) = 6

Let the number be N. Given: N = 407k + 4 for some integer k Also given: N is divisible by 37 Step 1: Express 407 in terms of 37 407 = 11 × 37 Step 2: Substitute in the equation N = 11 × 37 × k + 4 N = 37(11k) + 4 Since N is divisible by 37: 37(11k) + 4 ≡ 0 (mod 37) 4 ≡ 0 (mod 37) This is not possible. Let me recalculate: N = 407k + 4 Since N is divisible by 37: 407k + 4 ≡ 0 (mod 37) (11 × 37)k + 4 ≡ 0 (mod 37) 4 ≡ 0 (mod 37) — contradiction Let's check: if 407 = 11 × 37, and N = 407k + 4: N mod 11 = (407k + 4) mod 11 = (0 + 4) mod 11 = 4 Therefore, remainder when N is divided by 11 is 4

Given: LCM(a,b) = 180 HCF(a,b) = 12 One number (say a) = 36 Find the other number (say b) = ? Using the relation: HCF(a,b) × LCM(a,b) = a × b 12 × 180 = 36 × b 2160 = 36 × b b = 2160/36 b = 60 Verification: 60 = 2² × 3 × 5 36 = 2² × 3² HCF = 2² × 3 = 12 ✓ LCM = 2² × 3² × 5 = 180 ✓ Therefore, the other number is 60

Proof by contradiction: Assume 5 + 3√2 is rational. Then 5 + 3√2 = p/q, where p and q are coprime integers and q ≠ 0. Rearranging: 3√2 = p/q - 5 3√2 = (p - 5q)/q √2 = (p - 5q)/(3q) Since p and q are integers, (p - 5q)/(3q) is a rational number. This implies √2 is rational, which contradicts the fact that √2 is irrational. Therefore, our assumption is wrong, and 5 + 3√2 is irrational.