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Ex 2.1, 10 - The Cartesian product A x A has 9 elements - Number of elements

  1. Chapter 2 Class 11 Relations and Functions
  2. Serial order wise
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Ex 2.1, 10 (Method 1) The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A. Since A × A has 9 elements So, A would have 3 elements Now, A × A contains ( – 1,0) and (0,1) ( – 1,0) is in set A × A i.e. ( – 1,0) ∈ A × A So, –1 & 0 is in set A , i.e. – 1 , 0 ∈ A Similarly, (0, 1) is in set A × A i.e. (0, 1) ∈ A × A So, 0 & 1 is in set A , i.e. 0, 1 ∈ A From (1) & (2) – 1,0,1 is in set A , i.e. – 1,0,1 ∈ A Hence, the three elements of A are – 1,0 and 1 Hence, A = {– 1, 0, 1}. Now A × A = {– 1, 0, 1} × {– 1, 0, 1} A × A = {(– 1, – 1), (– 1, 0), ( – 1, 1) (0, – 1), (0, 0), (0,1), (1, – 1), (1, 0),(1, 1)} The remaining elements of set A × A are (– 1, – 1), (– 1, 1), (0, – 1), (0, 0),(1, – 1), (1, 0), and (1, 1) Ex 2.1, 10 (Method 2) The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A. Since A × A has 9 elements So, A would have 3 elements Let A = {a, b, c} Now, A × A = {a, b, c} × {a, b, c} = { (a, a) , (a, b) , (a, c), (b, a) , (b, b) , (b, c), (c, a) , (c, b) , (c, c) } Now, ( – 1,0) is in set A × A (0, 1) is in set A × A Thus, a = –1 , b = 0, c = 1 So, A = { a, b, c} = {– 1, 0, 1}. Now A × A = {– 1, 0, 1} × {– 1, 0, 1} A × A = {(– 1, – 1), (– 1, 1), ( – 1, 0) (0, – 1), (0, 0), (0,1), (1, – 1), (1, 0),(1, 1)} The remaining elements of set A × A are (– 1, – 1), (– 1, 1), (0, – 1), (0, 0),(1, – 1), (1, 0), and (1, 1)

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