VIII.1. Let A = {1,2, 3, 4} and let R, S, T and U be the following relations: R= {(1,3), (3, 2), (2, 1), (4, 4)}, S = {(2,1), (3, 3), (4, 2)}, T = {(4,1), (4, 2), (3, 1), (3, 2), (1, 2)}, U = {(x, y) | x > y}. (a) For each of R, S, T and U determine whether they are functional, reflexive, symmetric, anti-symmetric or transitive. Explain your answer in each case, showing why your answer is correct. (b) What is the transitive closure of R? (c) Explain why R, the transitive closure of R, is an equivalence relation. De- scribe the equivalence classes E, into which the relation partitions the set A. VIII 2 Prove or give a counterexample to the following statement: for any relation R both

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Chapter2: Second-order Linear Odes
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VIII.1. Let A = {1,2, 3, 4} and let R, S, T and U be the following relations:
R= {(1,3), (3, 2), (2, 1), (4, 4)},
S = {(2,1), (3, 3), (4, 2)},
T = {(4,1), (4, 2), (3, 1), (3, 2), (1, 2)},
U = {(x, y) | x > y}.
(a) For each of R, S, T and U determine whether they are functional, reflexive,
symmetric, anti-symmetric or transitive.
Explain your answer in each case, showing why your answer is correct.
(b) What is the transitive closure of R?
(c) Explain why R", the transitive closure of R, is an equivalence relation. De-
scribe the equivalence classes E, into which the relation partitions the set A.
VIII.2. Prove or give a counterexample to the following statement: for any relation R both
R and RoR always have the same transitive closure.
VIII.3. Is there a mistake in the following proof that any transitive and symmetric relation
Ris reflexive? If so, what is it?
Let a Rb. By symmetry, bRa. By transitivity, if aRb and bRa, then aRa. This
proves reflexivity.
VIII.4. Determine for the following relations on the set of people if the relation is an equiv-
alence relation, a partial order, both an equivalence relation and a partial order, or
neither an equivalence relation nor a partial order.
(a) 'has the same parents (both) as'
(b) 'has at least one parent same as'
(c) 'is a brother of'
(d) 'is at least as clever as'.
VIII.5. Let Rand S be relations on a set A. Use proof by contradiction to show that if R
and S are partial orders then RnS is also a partial order.
Transcribed Image Text:VIII.1. Let A = {1,2, 3, 4} and let R, S, T and U be the following relations: R= {(1,3), (3, 2), (2, 1), (4, 4)}, S = {(2,1), (3, 3), (4, 2)}, T = {(4,1), (4, 2), (3, 1), (3, 2), (1, 2)}, U = {(x, y) | x > y}. (a) For each of R, S, T and U determine whether they are functional, reflexive, symmetric, anti-symmetric or transitive. Explain your answer in each case, showing why your answer is correct. (b) What is the transitive closure of R? (c) Explain why R", the transitive closure of R, is an equivalence relation. De- scribe the equivalence classes E, into which the relation partitions the set A. VIII.2. Prove or give a counterexample to the following statement: for any relation R both R and RoR always have the same transitive closure. VIII.3. Is there a mistake in the following proof that any transitive and symmetric relation Ris reflexive? If so, what is it? Let a Rb. By symmetry, bRa. By transitivity, if aRb and bRa, then aRa. This proves reflexivity. VIII.4. Determine for the following relations on the set of people if the relation is an equiv- alence relation, a partial order, both an equivalence relation and a partial order, or neither an equivalence relation nor a partial order. (a) 'has the same parents (both) as' (b) 'has at least one parent same as' (c) 'is a brother of' (d) 'is at least as clever as'. VIII.5. Let Rand S be relations on a set A. Use proof by contradiction to show that if R and S are partial orders then RnS is also a partial order.
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