Let X be a nonempty set and P(X) the power set of X. Define the "not equal to" relation R on P(X) as follows. For every A, BE P(X), ARBA B. (a) Is R reflexive? Justify your answer. R is reflexive for every set A in P(X), A R A. By definition of R this means that for every set A in P(X), -Select--- ✓ This is --Select--- for every set in P(X) because --Select--- ✓ set --Select-- equal to itself. Thus, R -Select--- reflexive. (b) Is R symmetric? Justify your answer. R is symmetric for all sets A and B in P(X), if AR B then B R A. By definition of R, this means that for all sets A and B in P(X), if AB, then BA, which is ---Select--- V Thus, R -Select-- symmetric. (c) Is R transitive? Justify your answer. R is transitive for all sets A, B, and C in P(X), if AR B and B R C then AR C. By definition of R this means that for all sets A, B, and C in P(X), if A + B and B & C, then A # C. For X = {1, 2, 3}, one or more of the following examples can be used to show that R --Select-- transitive according to this definition. (Select all that apply.) A = {1, 2, 3}, B = {1, 2, 3}, C = {1, 2} A = {1, 2}, B = {1, 2, 3}, C = {1, 2, 3} A = {1, 2, 3}, B = {1, 2, 3}, C = {1, 2, 3} A = {1, 2, 3}, B = {1, 2}, C = {1, 2, 3} Then A ? B and B ? ✓ C, and A ? ✓ C. Thus, R -Select--- transitive.

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Chapter2: Second-order Linear Odes
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Let X be a nonempty set and P(X) the power set of X. Define the "not equal to" relation R on P(X) as follows.
For every A, BE P(X), ARBA B.
(a) Is R reflexive? Justify your answer.
R is reflexive for every set A in P(X), A R A. By definition of R this means that for every set A in P(X),
-Select--- ✓ This is --Select---
for every set in P(X) because --Select--- ✓ set --Select-- equal to itself. Thus, R -Select--- reflexive.
(b) Is R symmetric? Justify your answer.
R is symmetric for all sets A and B in P(X), if AR B then B R A. By definition of R, this means that for all sets A and B in P(X), if AB, then BA, which is ---Select--- V
Thus, R
-Select-- symmetric.
(c) Is R transitive? Justify your answer.
R is transitive for all sets A, B, and C in P(X), if AR B and B R C then AR C. By definition of R this means that for all sets A, B, and C in P(X), if A + B and B & C, then A # C. For X = {1, 2, 3}, one or more of the following examples can be used to show that R --Select-- transitive according
to this definition. (Select all that apply.)
A = {1, 2, 3}, B = {1, 2, 3}, C = {1, 2}
A = {1, 2}, B = {1, 2, 3}, C = {1, 2, 3}
A = {1, 2, 3}, B = {1, 2, 3}, C = {1, 2, 3}
A = {1, 2, 3}, B = {1, 2}, C = {1, 2, 3}
Then A ? B and B ? ✓ C, and A ? ✓ C. Thus, R -Select--- transitive.
Transcribed Image Text:Let X be a nonempty set and P(X) the power set of X. Define the "not equal to" relation R on P(X) as follows. For every A, BE P(X), ARBA B. (a) Is R reflexive? Justify your answer. R is reflexive for every set A in P(X), A R A. By definition of R this means that for every set A in P(X), -Select--- ✓ This is --Select--- for every set in P(X) because --Select--- ✓ set --Select-- equal to itself. Thus, R -Select--- reflexive. (b) Is R symmetric? Justify your answer. R is symmetric for all sets A and B in P(X), if AR B then B R A. By definition of R, this means that for all sets A and B in P(X), if AB, then BA, which is ---Select--- V Thus, R -Select-- symmetric. (c) Is R transitive? Justify your answer. R is transitive for all sets A, B, and C in P(X), if AR B and B R C then AR C. By definition of R this means that for all sets A, B, and C in P(X), if A + B and B & C, then A # C. For X = {1, 2, 3}, one or more of the following examples can be used to show that R --Select-- transitive according to this definition. (Select all that apply.) A = {1, 2, 3}, B = {1, 2, 3}, C = {1, 2} A = {1, 2}, B = {1, 2, 3}, C = {1, 2, 3} A = {1, 2, 3}, B = {1, 2, 3}, C = {1, 2, 3} A = {1, 2, 3}, B = {1, 2}, C = {1, 2, 3} Then A ? B and B ? ✓ C, and A ? ✓ C. Thus, R -Select--- transitive.
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