| --- |
| source: ncert_exercises |
| topic: numbers |
| class_level: class_10 |
| chapter: ch1 |
| exercise: ex1.2 |
| difficulty: intermediate |
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| Class 10 | Ch1: Real Numbers | Exercise 1.2 |
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| Proving Irrationality: Assume rational p/q (lowest terms), derive contradiction. |
| Key: If p² is divisible by prime a, then p is divisible by a. |
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| Q1. Prove that √5 is irrational. |
| Answer: Assume √5 = p/q (lowest terms, HCF(p,q)=1). Squaring: 5 = p²/q² → p² = 5q². So 5|p² → 5|p (since 5 is prime). Let p=5m. Then 25m² = 5q² → q² = 5m² → 5|q² → 5|q. So 5|p AND 5|q. This contradicts HCF(p,q)=1. ∴ √5 is irrational. ✅ |
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| Q2. Prove that 3+2√5 is irrational. |
| Answer: Assume 3+2√5 is rational = p/q. Then 2√5 = p/q-3 = (p-3q)/q. √5 = (p-3q)/2q. Since p,q are integers → RHS is rational → √5 is rational. But √5 is irrational (contradiction). ∴ 3+2√5 is irrational. ✅ |
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| Q3. Prove the following are irrational: |
| (i) 1/√2 (ii) 7√5 (iii) 6+√2 |
| Answer: |
| (i) Assume 1/√2 = p/q. Then √2 = q/p (rational). But √2 is irrational. Contradiction. ✅ |
| (ii) Assume 7√5 = p/q. Then √5 = p/7q (rational). But √5 is irrational. Contradiction. ✅ |
| (iii) Assume 6+√2 = p/q. Then √2 = p/q-6 = (p-6q)/q (rational). But √2 is irrational. Contradiction. ✅ |
