6f) For the set X = {m,n, p, q,r, s}, let R be a relation on P(X) (the power set of X), given by A R B iff A and B have the same number of elements. How many elements are in X? How many elements are in P(X)/R, or P(X) modulo R?
6f) For the set X = {m,n, p, q,r, s}, let R be a relation on P(X) (the power set of X), given by A R B iff A and B have the same number of elements. How many elements are in X? How many elements are in P(X)/R, or P(X) modulo R?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![6f) For the set X = {m,n, p, q, r, s}, let R be a relation on P(X) (the power set of X), given by A R B iff A and
B have the same number of elements. How many elements are in X? How many elements are in P(X)/R, or
P(X) modulo R?
Attempt:
If æ E A and y E B, then A E P(X) means A C X and BE X means BC X. If B= X, then we have A R X. Hence we
must find all subsets of X (using the definition A e X) such that they contain the same number of elements as X. Therefore,
we must find all subsets of X that contain 6 elements. Since only X contains 6 elements, one set is in
X = {X € P(X) : A R X}. Therefore, X has one element.
X can also be stated as
(). Moreover, since you are choosing 6 out of 6 elements for X, the total number number of elements
in P(X)/R = {A: A € P(X)} = {A: AC X} is therefore –1+ () (due to Ø having zero elements). Therefore we
have –1+ (() + (9) + (9) + () + (9) + (9) + () = -1+ (1+6+ 15+ 20 + 15 + 6 +1) = 64 – 1 = 63. Hence
P(X)/Rhas 63 elements.
Question: Am I correct?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fce901b6a-d7bd-4588-bd3a-47db4e805652%2F4e7c32fd-bc89-4ba0-948c-be9c6cb8730b%2Fj9ecso_processed.png&w=3840&q=75)
Transcribed Image Text:6f) For the set X = {m,n, p, q, r, s}, let R be a relation on P(X) (the power set of X), given by A R B iff A and
B have the same number of elements. How many elements are in X? How many elements are in P(X)/R, or
P(X) modulo R?
Attempt:
If æ E A and y E B, then A E P(X) means A C X and BE X means BC X. If B= X, then we have A R X. Hence we
must find all subsets of X (using the definition A e X) such that they contain the same number of elements as X. Therefore,
we must find all subsets of X that contain 6 elements. Since only X contains 6 elements, one set is in
X = {X € P(X) : A R X}. Therefore, X has one element.
X can also be stated as
(). Moreover, since you are choosing 6 out of 6 elements for X, the total number number of elements
in P(X)/R = {A: A € P(X)} = {A: AC X} is therefore –1+ () (due to Ø having zero elements). Therefore we
have –1+ (() + (9) + (9) + () + (9) + (9) + () = -1+ (1+6+ 15+ 20 + 15 + 6 +1) = 64 – 1 = 63. Hence
P(X)/Rhas 63 elements.
Question: Am I correct?
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