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Exercise - 1.3

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

Solution :-
 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.

Solution :-
 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

Solution :-
(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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