Let A = {1, {1}, 3, 4, {5, 6}} be a set, and let B and C be the following relations on A: B = {(1, 3), (4, 1), (3, {5, 6}), ({1}, 4), ({5, 6}, {1})} C = {(1, 1), (3, 4), (4, 3), ({5, 6}, {1})The composition relation C;B is {(1, 3), (3, 1), (3, {5, 6}), ({5, 6}, 4)}.a.Trueb.False
Question
Let A = {1, {1}, 3, 4, {5, 6}} be a set, and let B and C be the following relations on A: B = {(1, 3), (4, 1), (3, {5, 6}), ({1}, 4), ({5, 6}, {1})} C = {(1, 1), (3, 4), (4, 3), ({5, 6}, {1})The composition relation C;B is {(1, 3), (3, 1), (3, {5, 6}), ({5, 6}, 4)}.a.Trueb.False
Solution
False. The composition of two relations B and C, denoted as C;B, is defined as follows: for any a, c in A, (a, c) is in C;B if and only if there exists an element b in A such that (a, b) is in C and (b, c) is in B.
Let's calculate C;B step by step:
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Start with the first pair in C, (1, 1). There is no pair in B where 1 is the second element. So, there are no pairs from (1, 1) in C;B.
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The second pair in C is (3, 4). The pairs in B where 4 is the second element are (4, 1). So, we get (3, 1) in C;B.
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The third pair in C is (4, 3). The pairs in B where 3 is the second element are (1, 3). So, we get (4, 3) in C;B.
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The fourth pair in C is ({5, 6}, {1}). The pairs in B where {1} is the second element are ({5, 6}, {1}). So, we get ({5, 6}, {1}) in C;B.
So, the composition relation C;B is {(3, 1), (4, 3), ({5, 6}, {1})}, not {(1, 3), (3, 1), (3, {5, 6}), ({5, 6}, 4)}. Therefore, the statement is false.
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