Real Numbers — Telangana (SSC) Class 9 Mathematics Solutions (Free)

Step 1: √16 = 4 (integer) → Rational Step 2: √17 cannot be simplified (17 is prime) → Irrational Step 3: π ≈ 3.14159... (non-terminating, non-repeating) → Irrational Step 4: 22/7 (ratio of integers) → Rational (approximation of π) Step 5: 0.333... = 1/3 (repeating decimal) → Rational Step 6: √(4/9) = 2/3 (ratio of integers) → Rational Final Answer: Rational: √16, 22/7, 0.333..., √(4/9); Irrational: √17, π

Step 1: √72 = √(36 × 2) = 6√2 Step 2: √50 = √(25 × 2) = 5√2 Step 3: √8 = √(4 × 2) = 2√2 Step 4: 6√2 - 5√2 + 2√2 = (6 - 5 + 2)√2 = 3√2 Final Answer: 3√2

Step 1: Multiply numerator and denominator by (√5 - √3) Step 2: [1/(√5 + √3)] × [(√5 - √3)/(√5 - √3)] Step 3: = (√5 - √3)/[(√5 + √3)(√5 - √3)] Step 4: = (√5 - √3)/(5 - 3) [using (a+b)(a-b) = a² - b²] Step 5: = (√5 - √3)/2 Final Answer: (√5 - √3)/2

Step 1: Use law aᵐ × aⁿ = aᵐ⁺ⁿ Step 2: 2³ × 2⁴ = 2⁽³⁺⁴⁾ = 2⁷ Step 3: Use law aᵐ/aⁿ = aᵐ⁻ⁿ Step 4: 2⁷/2⁵ = 2⁽⁷⁻⁵⁾ = 2² = 4 Final Answer: 4

Step 1: Assume √2 is rational Step 2: Then √2 = p/q where p, q are integers with no common factors (lowest terms) Step 3: Squaring: 2 = p²/q² Step 4: 2q² = p² Step 5: p² is even, so p is even. Let p = 2k Step 6: 2q² = (2k)² = 4k² Step 7: q² = 2k² Step 8: q² is even, so q is even Step 9: Both p and q are even (have common factor 2), contradicting our assumption Step 10: Therefore √2 is irrational Final Answer: √2 is irrational (proof by contradiction)