Problems
Sets and relations
1
Let A, B, C be any three sets. And let D = (A \ (B ∪ C )), E = ((A \ C ) ∩ B), and
F = (A ∩ C ). Suppose each of D, E and F is nonempty. Prove that {D, E , F } is a
partition of A. That means to prove that (a) none of the sets in the collection is empty,
(b) the union of the sets in the collection is equal to A, and (c) the sets in the collection
are mutually disjointed, i.e., they are disjointed to each other.
2
Let S = {1, 2, 3, . . . , 18, 19}. Let R be the relation on S defined by “xy is a square,”
(a) Prove R is an equivalence relation.
(b) Find the equivalence class [1].
(c) List all equivalence classes with more than one element.
3
Let S = {1, 2, 3, . . . , 14, 15}. Let R be the equivalence relation on S defined by
x ≡ y (mod 5), that is, x − y is divisible by 5. Find the partition of S induced by R.
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October 27, 2016
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Problems
Counting
4
Find the number m of ways that 7 people can arrange themselves:
(a) In a row of chairs;
5
(b) around a circular table.
Find the number n of distinct permutations that can be formed from all the letters of
each word:
(a) THOSE;
(b) UNUSUAL;
(c) SOCIOLOGICAL.
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October 27, 2016
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Solutions: Problem 1
Let A, B, C be any three sets. And let D = (A \ (B ∪ C )), E = ((A \ C ) ∩ B), and F = (A ∩ C ).
Suppose each of D, E and F is nonempty. Prove that {D, E , F } is a partition of A. That means
to prove that (a) none of the sets in the collection is empty, (b) the union of the sets in the
collection is equal to A, and (c) the sets in the collection are mutually disjointed, i.e., they are
disjointed to each other.
Let’s first draw a Venn diagram, this is not a proof, but it will help us to understand the
problem.
B
A
C
Here the blue region corresponds to E = (A \ C ) ∩ B; the green region to D = A \ (B ∪ C ), and
the red region to F = A ∩ C .
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October 27, 2016
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Solutions: Problem 1
Let A, B, C be any three sets. And let D = (A \ (B ∪ C )), E = ((A \ C ) ∩ B), and F = (A ∩ C ).
Suppose each of D, E and F is nonempty. Prove that {D, E , F } is a partition of A. That means
to prove that (a) none of the sets in the collection is empty, (b) the union of the sets in the
collection is equal to A, and (c) the sets in the collection are mutually disjointed, i.e., they are
disjointed to each other.
(a) D, E nad F are given to be nonempty. Therefore, we only have to prove that (b) and (c)
are satisfied.
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October 27, 2016
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Solutions: Problem 1
Let A, B, C be any three sets. And let D = (A \ (B ∪ C )), E = ((A \ C ) ∩ B), and F = (A ∩ C ).
Suppose each of D, E and F is nonempty. Prove that {D, E , F } is a partition of A. That means
to prove that (a) none of the sets in the collection is empty, (b) the union of the sets in the
collection is equal to A, and (c) the sets in the collection are mutually disjointed, i.e., they are
disjointed to each other.
(b) To prove that (A \ (B ∪ C )) ∪ ((A \ C ) ∩ B) ∪ (A ∩ C ) = A:
Let a ∈ (A \ (B ∪ C )) ∪ ((A \ C ) ∩ B) ∪ (A ∩ C ). There are 3 cases:
1.
a ∈ A \ (B ∪ C ) → (a ∈ A) ∧ (a ∈
/ (B ∪ C ))
→a∈A
2.
a ∈ (A \ C ) ∩ B → (a ∈ (A \ C )) ∧ (a ∈ B)
→ a ∈ (A \ C )
→ (a ∈ A) ∧ (a ∈
/ C)
→a∈A
3.
a ∈ A ∩ C → (a ∈ A) ∧ (a ∈ C )
→a∈A
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Therefore, (A \ (B ∪ C )) ∪ ((A \ C ) ∩ B) ∪ (A ∩ C ) ⊆ A
October 27, 2016
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Solutions: Problem 1
Let A, B, C be any three sets. And let D = (A \ (B ∪ C )), E = ((A \ C ) ∩ B), and F = (A ∩ C ).
Suppose each of D, E and F is nonempty. Prove that {D, E , F } is a partition of A. That means
to prove that (a) none of the sets in the collection is empty, (b) the union of the sets in the
collection is equal to A, and (c) the sets in the collection are mutually disjointed, i.e., they are
disjointed to each other.
(b) . . . To prove that (A \ (B ∪ C )) ∪ ((A \ C ) ∩ B) ∪ (A ∩ C ) = A:
Let a ∈ (A \ (B ∪ C )) ∪ ((A \ C ) ∩ B) ∪ (A ∩ C ). There are 3 cases:
Let a ∈ A. Consider the following two cases:
1. a ∈ C . If so, a ∈ (A ∩ C ).
2. a ∈
/ C . There are two sub-cases (a ∈ B and a ∈
/ B):
1
(a ∈ A) ∧ (a ∈
/ C ) ∧ (a ∈ B) → (a ∈ (A \ C )) ∧ (a ∈ B)
→ a ∈ ((A \ C ) ∩ B)
2
(a ∈ A) ∧ (a ∈
/ B) ∧ (a ∈
/ C ) → (a ∈ A) ∧ (a ∈
/ (B ∪ C ))
→ a ∈ (A \ (B ∪ C ))
From this, and the definition of union, we have
a ∈ (A \ (B ∪ C )) ∪ ((A \ C ) ∩ B) ∪ (A ∩ C )
Therefore, A ⊆ (A \ (B ∪ C )) ∪ ((A \ C ) ∩ B) ∪ (A ∩ C ). That prove (b).
October 27, 2016
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Solutions: Problem 1
Let A, B, C be any three sets. And let D = (A \ (B ∪ C )), E = ((A \ C ) ∩ B), and F = (A ∩ C ).
Suppose each of D, E and F is nonempty. Prove that {D, E , F } is a partition of A. That means
to prove that (a) none of the sets in the collection is empty, (b) the union of the sets in the
collection is equal to A, and (c) the sets in the collection are mutually disjointed, i.e., they are
disjointed to each other.
(c) To prove that D, E and F are mutually disjointed:
1.
a ∈ (A \ (B ∪ C )) → (a ∈ A) ∧ (a ∈
/ (B ∪ C ))
→ a ∈ A∧a ∈
/ B ∧a ∈
/C →a∈
/ ((A \ C ) ∩ B) ∧ a ∈
/ (A ∩ C )
2.
a ∈ ((A \ C ) ∩ B) → a ∈ (A \ C ) ∧ a ∈ B → a ∈ A ∧ a ∈
/ C ∧a ∈ B
→ a ∈ A ∧ a ∈ (B ∪ C ) ∧ a ∈
/ (A ∩ C )
→a∈
/ (A \ (B ∪ C )) ∧ a ∈
/ (A ∩ C )
3.
a ∈ (A ∩ C ) → a ∈ A ∧ a ∈ C → a ∈ A ∧ a ∈ (B ∪ C ) ∧ a ∈
/ (A \ C )
→a∈
/ (A \ (B ∪ C )) ∧ a ∈
/ ((A \ C ) ∩ B)
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That proves the claim.
October 27, 2016
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Solutions: Problem 2
Let S = {1, 2, 3, . . . , 18, 19}. Let R be the relation on S defined by “xy is a square,”
(a) Prove R is an equivalence relation.
(b) Find the equivalence class [1].
(c) List all equivalence classes with more than one element.
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October 27, 2016
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Solutions: Problem 2
Let S = {1, 2, 3, . . . , 18, 19}. Let R be the relation on S defined by “xy is a square,”
(a) Prove R is an equivalence relation.
(b) Find the equivalence class [1].
(c) List all equivalence classes with more than one element.
(a) The relation is
1. Reflexive, as x · x = x 2 for any x.
2. Symmetric, as ∀x∀y ∃t(xy = t 2 = yx).
3. Transitive, as if xy = t 2 and yz = u 2 , then there is an integer v , such
that xz = v 2 . Note, that x and y are related iff there exists integers a, b
and c, such that x = ab 2 and y = ac 2 . (Then xy = (abc)2 or t = abc.)
And y and z can be related iff there is an integer d, such that z = ad 2 .
(That is u = acd.) Then xz = (abd)2 (or v = abd), therefore x and z
are related.
Due to 1., 2., and 3. above, this is an equivalence relation.
(b) [1] = {1, 4, 9, 16}.
(c)
[1] = {1, 4, 9, 16}
[2] = {2, 8, 18}
[3] = {3, 12}
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October 27, 2016
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Solutions: Problem 2
Let S = {1, 2, 3, . . . , 18, 19}. Let R be the relation on S defined by “xy is a square,”
(a) Prove R is an equivalence relation.
(b) Find the equivalence class [1].
(c) List all equivalence classes with more than one element.
(a) The relation is
1. Reflexive, as x · x = x 2 for any x.
2. Symmetric, as ∀x∀y ∃t(xy = t 2 = yx).
3. Transitive, as if xy = t 2 and yz = u 2 , then there is an integer v , such
that xz = v 2 . Note, that x and y are related iff there exists integers a, b
and c, such that x = ab 2 and y = ac 2 . (Then xy = (abc)2 or t = abc.)
And y and z can be related iff there is an integer d, such that z = ad 2 .
(That is u = acd.) Then xz = (abd)2 (or v = abd), therefore x and z
are related.
Due to 1., 2., and 3. above, this is an equivalence relation.
(b) [1] = {1, 4, 9, 16}.
(c)
[1] = {1, 4, 9, 16}
[2] = {2, 8, 18}
[3] = {3, 12}
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October 27, 2016
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Solutions: Problem 2
Let S = {1, 2, 3, . . . , 18, 19}. Let R be the relation on S defined by “xy is a square,”
(a) Prove R is an equivalence relation.
(b) Find the equivalence class [1].
(c) List all equivalence classes with more than one element.
(a) The relation is
1. Reflexive, as x · x = x 2 for any x.
2. Symmetric, as ∀x∀y ∃t(xy = t 2 = yx).
3. Transitive, as if xy = t 2 and yz = u 2 , then there is an integer v , such
that xz = v 2 . Note, that x and y are related iff there exists integers a, b
and c, such that x = ab 2 and y = ac 2 . (Then xy = (abc)2 or t = abc.)
And y and z can be related iff there is an integer d, such that z = ad 2 .
(That is u = acd.) Then xz = (abd)2 (or v = abd), therefore x and z
are related.
Due to 1., 2., and 3. above, this is an equivalence relation.
(b) [1] = {1, 4, 9, 16}.
(c)
[1] = {1, 4, 9, 16}
[2] = {2, 8, 18}
[3] = {3, 12}
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October 27, 2016
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Solutions: Problem 3
Let S = {1, 2, 3, . . . , 14, 15}. Let R be the equivalence relation on S defined by
x ≡ y (mod 5), that is, x − y is divisible by 5. Find the partition of S induced by R.
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October 27, 2016
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Solutions: Problem 3
Let S = {1, 2, 3, . . . , 14, 15}. Let R be the equivalence relation on S defined by
x ≡ y (mod 5), that is, x − y is divisible by 5. Find the partition of S induced by R.
The equivalence class [n] contains the numbers which have the reminder n when divided by 5.
So, there are five classes:
[0] = {5, 10, 15}
[1] = {1, 6, 11}
[2] = {2, 7, 12}
[3] = {3, 8, 13}
[4] = {4, 9, 14}
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