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Exercise - 1.3

Real Numbers

**Question-1 :-** Prove that √5 is irrational.

Let √5 is a rational number. Therefore, we can find two integers a, b (b ≠ 0) such that √5 = a/b Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime. a = b√5 squaring both sides a² = 5b² Therefore, a² is divisible by 5 and it can be said that a is divisible by 5. Let a = 5k, where k is an integer (5k)² = 5b² b² = 5k² This means that b² is divisible by 5 and hence, b is divisible by 5. This implies that a and b have 5 as a common factor. And this is a contradiction to the fact that a and b are co-prime. Hence, √5 cannot be expressed as p/q or it can be said that √5 is irrational.

**Question-2 :-** Prove that 3+2√5 is irrational.

Let 3+2√5 is rational. Therefore, we can find two integers a, b (b ≠ 0) such that 3+2√5 = a/b 2√5 = (a/b) - 3 √5 = 1/2(a/b - 3) Since a and b are integers, 1/2(a/b - 3) will also be rational and therefore,√5 is rational. This contradicts the fact that √5 is irrational. Hence, our assumption that 3+2√5 is rational is false. Therefore, 3+2√5 is irrational.

**Question-3 :-** Prove that the following are irrationals:

(i) 1/√2, (ii) 7√5, (iii) 6+√2

(i) 1/√2 Let 1/√2 is rational. Therefore, we can find two integers a, b (b ≠ 0) such that 1/√2 = a/b √2 = b/a b/a is rational as a and b are integers. Therefore, √2 is rational which contradicts to the fact that √2 is irrational. Hence, our assumption is false and 1/√2 is irrational.

(ii) 7√5 Let 7√5 is rational. Therefore, we can find two integers a, b (b ≠ 0) such that 7√5 = a/b √5 = a/7b a/7b is rational as a and b are integers. Therefore, √5 should be rational. This contradicts the fact that √5 is irrational. Therefore, our assumption that 7√5 is rational is false. Hence, 7√5 is irrational.

(iii) 6+√2 Let 6+√2 is rational. Therefore, we can find two integers a, b (b ≠ 0) such that 6+√2 = a/b √2 = (a/b)-6 Since a and b are integers, (a/b)-6 is also rational and hence, √2 should be rational. This contradicts the fact that √2 is irrational. Therefore, our assumption is false and hence, 6+√2 is irrational.

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