---
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. ✅