source: ncert_exercises
topic: numbers
class_level: class_9
chapter: ch1
exercise: ex1.5
difficulty: intermediate
Class 9 | Ch1: Number Systems | Exercise 1.5
Q1. Classify as rational or irrational: (i) 2-√5 (ii) (3+√23)-√23 (iii) 2√7/7√7 (iv) 1/√2 (v) 2π Answer: (i) Irrational—rational minus irrational=irrational. (ii) Rational—simplifies to 3. (iii) Rational—simplifies to 2/7. (iv) Irrational—=√2/2. (v) Irrational—2×irrational=irrational. ✅
Q2. Simplify: (i) (3+√3)(2+√2) (ii) (3+√3)(3-√3) (iii) (√5+√2)² (iv) (√5-√2)(√5+√2) Answer: (i) 6+3√2+2√3+√6. (ii) 9-3=6. (iii) 5+2√10+2=7+2√10. (iv) 5-2=3. ✅
Q3. Does π=c/d contradict π being irrational? Answer: No contradiction. Physical measurement gives approximate rational values. Exact π is irrational. We can only measure approximately — rational approximations like 22/7, not the exact value. ✅
Q4. Represent √9.3 on number line. (Same as Ex 1.4 Q1) Answer: Semicircle method — 9.3 units extended by 1, midpoint, semicircle, perpendicular gives √9.3. ✅
Q5. Rationalise denominators: (i) 1/√7 (ii) 1/(√7-√6) (iii) 1/(√5+√2) (iv) 1/(√7-2) Answer: (i) √7/7. (ii) (√7+√6)/(7-6)=√7+√6. (iii) (√5-√2)/(5-2)=(√5-√2)/3. (iv) (√7+2)/(7-4)=(√7+2)/3. ✅