Math 311W
Russell Falls
HOMEWORK ASSIGNMENT #4
20-POINTERS
Using LATEX for Worksheets # 8, 11
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10.11) Let a and b be integers. Let A = {x ∈ Z : a|x} and B = {x ∈ Z : b|x}. Find and
prove a necessary and sufficient condition for A ⊆ B. In other words, given the notation
developed, find and prove a theorem of the form “A ⊆ B if and only if (condition involving
a and b.”
Let a be any integer greater than or equal to b, and let b be any integer except zero.
Prove A ⊆ B if and only if (a is divisible by b.)
We first show that if A ⊆ B, then b|a.
Let x ∈ A and let A ⊆ B be true. By the definition of subset, x ∈ A and x ∈ B.
Furthermore A ⊆ B implies that a|x and b|x.
By the definition of divisibility there exists n ∈ Z such that x = a ∗ n.
By the definition of divisibility there exists m ∈ Z such that x = b ∗ m.
Observe:
x=a∗n
x=b∗m
a∗n=b∗m
m
a=b∗
n
Hence, there exists p ∈ Z, namely p =
Therefore, b|a.
Thus, if A ⊆ B, then b|a.
m
,
n
such that b ∗ p = a.
We then show that if b|a, then A ⊆ B.
Let b|a be true. By the definition of divisibility, there exists k ∈ Z such that a = k ∗ b.
By the definition of A, A is the set of all integers x such that a|x.
By the definition of divisibility, there exists q ∈ Z such that x = a ∗ q.
Observe:
a=k∗b
x=a∗q
x = (k ∗ b) ∗ q
x = b ∗ (kq)
Hence, there exists p ∈ Z, namely p = kq, such that x = b ∗ p.
Therefore, b|x| is also true, which means that x ∈ B.
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So, A ⊆ B.
Thus, A ⊆ B if and only if b|a.
—————————————————————————12.5) Prove the statements from Theorems # 12.3.
1. A ∪ B = B ∪ A and A ∩ B = B ∩ A (Commutative Property)
2. A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ (B ∩ C) = (A ∩ B) ∩ C (Associative Property)
3. A ∪ ∅ = A and A ∩ ∅ = ∅
4. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) and A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) (Distributive
Property)
Use the definitions [in]:
ˆ A ∪ B = {x : (x ∈ A) ∨ (x ∈ B)} [all]
ˆ A ∩ B = {x : (x ∈ A) ∧ (x ∈ B)} [all]
ˆ x ∨ y = y ∨ x (Communicative) [1]
ˆ x ∧ y = y ∧ x (Communicative) [1]
ˆ x ∨ (y ∨ z) = (x ∨ y) ∨ z (Associative) [2]
ˆ x ∧ (y ∧ z) = (x ∧ y) ∧ z (Associative) [2]
ˆ x ∧ TRUE = x (Identity elements)
ˆ x ∨ TRUE = x (Identity elements)
ˆ x ∨ FALSE = x (Identity elements) [3]
ˆ x ∧ FALSE = FALSE (Identity elements) [3]
ˆ x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z) (Distributive property) [4]
ˆ x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) (Distributive property) [4]
1.
A ∪ B = {x : (x ∈ A) ∨ (x ∈ B)}
= {x : (x ∈ B) ∨ (x ∈ A)}
=B∪A
A ∩ B = {x : (x ∈ A) ∧ (x ∈ B)}
= {x : (x ∈ B) ∧ (x ∈ A)}
=B∩A
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2.
A ∪ (B ∪ C) = {x : (x ∈ A) ∨ (x ∈ B ∪ C)}
= {x : (x ∈ A) ∨ ((x ∈ B) ∨ (x ∈ C))}
= {x : ((x ∈ A) ∨ (x ∈ B)) ∨ (x ∈ C)}
= {x : (x ∈ A ∪ B) ∨ (x ∈ C)}
= (A ∪ B) ∪ C
A ∩ (B ∩ C) = {x : (x ∈ A) ∧ (x ∈ B ∩ C)}
= {x : (x ∈ A) ∧ ((x ∈ B) ∧ (x ∈ C))}
= {x : ((x ∈ A) ∧ (x ∈ B)) ∧ (x ∈ C)}
= {x : (x ∈ A ∩ B) ∧ (x ∈ C)}
= (A ∩ B) ∩ C
3.
A ∪ ∅ = {x : (x ∈ A) ∨ (x ∈ ∅)}
= {x : (x ∈ A)}
=A
A ∩ ∅ = {x : (x ∈ A) ∧ (x ∈ ∅)}
= {x : (x ∈ ∅)}
=∅
4.
A ∪ (B ∩ C) = {x : (x ∈ A) ∨ (x ∈ B ∩ C)}
= {x : (x ∈ A) ∨ ((x ∈ B) ∧ (x ∈ C)}
= {x : ((x ∈ A) ∨ (x ∈ B)) ∧ ((x ∈ A) ∨ (x ∈ C))}
= {x : (x ∈ A ∪ B) ∧ (x ∈ A ∪ C)}
= (A ∪ B) ∩ (A ∪ C)
A ∩ (B ∪ C) = {x : (x ∈ A) ∧ (x ∈ B ∪ C)}
= {x : (x ∈ A) ∧ ((x ∈ B) ∨ (x ∈ C)}
= {x : ((x ∈ A) ∧ (x ∈ B)) ∨ ((x ∈ A) ∧ (x ∈ C))}
= {x : (x ∈ A ∩ B) ∨ (x ∈ A ∩ C)}
= (A ∩ B) ∪ (A ∩ C)
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