Q. 1. i ). A={2, -2, 3} and B = {1, -4} Find A X B, AX A and B X A
To find the Cartesian products A X B, AX A, and B×A, we need to understand that these operations involve forming all possible ordered pairs where the first element comes from the first set and the second element comes from the second set.
Given sets:
A={2,−2, 3}
B={1,−4}
Cartesian Product A×B: {2,−2, 3} X {1,−4}
A×B={(2,1),(2,−4), (−2,1), (−2,−4), (3,1),(3,−4)}
So, A×B={(2,1), (2,−4), (−2,1), (−2,−4), (3,1), (3,−4)}.
Cartesian Product A×A: {2,−2, 3} X {2,−2, 3}
A×A={(2,2),(2,−2),(2,3), (−2,2), (−2,−2), (−2,3), ( 3,2),(3,−2),(3,3)}
So, A×A= {(2,2),(2,−2),(2,3),(−2,2),(−2,−2),(−2,3),(3,2),(3,−2),(3,3)}.
Cartesian Product B×A: {1,−4} X {2,−2, 3}
B×A={(1,2),(1,−2),(1,3), (−4,2),(−4,−2),(−4,3)}
So, B×A={(1,2),(1,−2),(1,3),(−4,2),(−4,−2),(−4,3)}.
These are the Cartesian products of the sets A and B, A and A, and B and A, respectively. Each product consists of all possible ordered pairs formed by taking one element from each set in the pair.
Q 1.(ii) . A =B = {p, q} Find A X B, AX A and B X A
Given the sets A = B = {p, q} ), we can determine the Cartesian products A X B, A X A, and B X A. The Cartesian product A X B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.
Cartesian Product A X B:
Since A = {p, q}) and (B = {p, q}), we find (A X B) as follows:
A X B = {(a, b) mid a ∈ A, b ∈ B}
A X B = {(p, p), (p, q), (q, p), (q, q)}
Cartesian Product A X A:
Similarly, since A = {p, q}), we find (A X A) as follows:
A X A = {(a, a') | a ∈ A, a' ∈ A}
This means we need to consider all possible pairs of elements from A:
A X A = {(p, p), (p, q), (q, p), (q, q)}
Cartesian Product B X A:
Since B = {p, q}) and A = {p, q}), we find B X A as follows:
B X A = {(b, a) | b ∈ B, a ∈ A}
This means we need to consider all possible pairs of elements from B and A:
B X A = {(p, p), (p, q), (q, p), (q, q)}
Conclusion:
All three Cartesian products, A X B, A X A, and B X A, are equal and given by:
{ (p, p), (p, q), (q, p), (q, q)}
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Q 1.(iii) . A = {m, n}; B = ∅ Find A X B, AX A and B X A
