§ 7.2 Equivalence Relations
Properties of Relations
Definition
A relation R : A → A is said to be reflexive if xRx for all x ∈ A.
Properties of Relations
Definition
A relation R : A → A is said to be reflexive if xRx for all x ∈ A.
Example
Which of the following relations are reflexive, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
Properties of Relations
Definition
A relation R : A → A is said to be reflexive if xRx for all x ∈ A.
Example
Which of the following relations are reflexive, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
Properties of Relations
Definition
A relation R : A → A is said to be reflexive if xRx for all x ∈ A.
Example
Which of the following relations are reflexive, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
Properties of Relations
Definition
A relation R : A → A is said to be reflexive if xRx for all x ∈ A.
Example
Which of the following relations are reflexive, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
Properties of Relations
Definition
A relation R : A → A is said to be reflexive if xRx for all x ∈ A.
Example
Which of the following relations are reflexive, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
5
R5 = {(a, a), (a, b)}
Properties of Relations
Definition
A relation R : A → A is said to be reflexive if xRx for all x ∈ A.
Example
Which of the following relations are reflexive, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
5
R5 = {(a, a), (a, b)}
6
R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
Definition
A relation R : A → A is said to be symmetric if xRy whenever yRx.
Properties of Relations
Definition
A relation R : A → A is said to be symmetric if xRy whenever yRx.
Example
Which of the following relations are symmetric, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
Properties of Relations
Definition
A relation R : A → A is said to be symmetric if xRy whenever yRx.
Example
Which of the following relations are symmetric, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
Properties of Relations
Definition
A relation R : A → A is said to be symmetric if xRy whenever yRx.
Example
Which of the following relations are symmetric, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
Properties of Relations
Definition
A relation R : A → A is said to be symmetric if xRy whenever yRx.
Example
Which of the following relations are symmetric, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
Properties of Relations
Definition
A relation R : A → A is said to be symmetric if xRy whenever yRx.
Example
Which of the following relations are symmetric, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
5
R5 = {(a, a), (a, b), (b, a)}
Properties of Relations
Definition
A relation R : A → A is said to be symmetric if xRy whenever yRx.
Example
Which of the following relations are symmetric, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
5
R5 = {(a, a), (a, b), (b, a)}
6
R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
Definition
A relation R : A → A is said to be antisymmetric if xRy whenever
y 6 Rx.
Properties of Relations
Definition
A relation R : A → A is said to be antisymmetric if xRy whenever
y 6 Rx.
Example
Which of the following relations are antisymmetric, where each is
defined on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a), (a, c)}
Properties of Relations
Definition
A relation R : A → A is said to be antisymmetric if xRy whenever
y 6 Rx.
Example
Which of the following relations are antisymmetric, where each is
defined on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a), (a, c)}
2
R2 = {(a, b), (b, b), (b, c), (a, c)}
Properties of Relations
Definition
A relation R : A → A is said to be antisymmetric if xRy whenever
y 6 Rx.
Example
Which of the following relations are antisymmetric, where each is
defined on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a), (a, c)}
2
R2 = {(a, b), (b, b), (b, c), (a, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
Properties of Relations
Definition
A relation R : A → A is said to be antisymmetric if xRy whenever
y 6 Rx.
Example
Which of the following relations are antisymmetric, where each is
defined on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a), (a, c)}
2
R2 = {(a, b), (b, b), (b, c), (a, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
Properties of Relations
Definition
A relation R : A → A is said to be antisymmetric if xRy whenever
y 6 Rx.
Example
Which of the following relations are antisymmetric, where each is
defined on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a), (a, c)}
2
R2 = {(a, b), (b, b), (b, c), (a, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
5
R5 = {(a, a), (a, b), (b, a)}
Properties of Relations
Definition
A relation R : A → A is said to be antisymmetric if xRy whenever
y 6 Rx.
Example
Which of the following relations are antisymmetric, where each is
defined on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a), (a, c)}
2
R2 = {(a, b), (b, b), (b, c), (a, c)}
3
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
4
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
5
R5 = {(a, a), (a, b), (b, a)}
6
R6 = {(a, a), (a, b), (b, b)}
Properties of Relations
Definition
A relation R : A → A is said to be transitive if whenever xRy and yRz,
xRz.
Properties of Relations
Definition
A relation R : A → A is said to be transitive if whenever xRy and yRz,
xRz.
Example
Which of the following relations are transitive, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
Properties of Relations
Definition
A relation R : A → A is said to be transitive if whenever xRy and yRz,
xRz.
Example
Which of the following relations are transitive, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
Not transitive because (a, b), (b, a) ∈ R1 but (a, a) 6∈ R1 .
Properties of Relations
Definition
A relation R : A → A is said to be transitive if whenever xRy and yRz,
xRz.
Example
Which of the following relations are transitive, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
Not transitive because (a, b), (b, a) ∈ R1 but (a, a) 6∈ R1 .
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
Properties of Relations
Definition
A relation R : A → A is said to be transitive if whenever xRy and yRz,
xRz.
Example
Which of the following relations are transitive, where each is defined
on the set S = {a, b, c}?
1
R1 = {(a, b), (b, a), (c, a)}
Not transitive because (a, b), (b, a) ∈ R1 but (a, a) 6∈ R1 .
2
R2 = {(a, b), (b, b), (b, c), (c, b), (c, c)}
Not reflexive because (a, b), (b, c) ∈ R2 but (a, c) 6∈ R2 .
Properties of Relations
Example
1
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
Properties of Relations
Example
1
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.
Properties of Relations
Example
1
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.
2
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
Properties of Relations
Example
1
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.
2
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.
Properties of Relations
Example
1
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.
2
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.
3
R5 = {(a, a), (a, b)}
Properties of Relations
Example
1
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.
2
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.
3
R5 = {(a, a), (a, b)}
This is transitive.
Properties of Relations
Example
1
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.
2
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.
3
R5 = {(a, a), (a, b)}
This is transitive.
4
R6 = {(a, b), (a, c)}
Properties of Relations
Example
1
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.
2
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.
3
R5 = {(a, a), (a, b)}
This is transitive.
4
R6 = {(a, b), (a, c)}
This is transitive.
Properties of Relations
Example
1
R3 = {(a, a), (a, c), (b, b), (c, a), (c, c)}
This is transitive.
2
R4 = {(a, a), (a, b), (b, b), (b, c), (a, c)}
This is transitive.
3
R5 = {(a, a), (a, b)}
This is transitive.
4
R6 = {(a, b), (a, c)}
This is transitive.
Back To Common Examples
Some Common Examples
Let Ri : Z → Z. Determine if each of the following is reflexive,
symmetric, antisymmetric, or transitive. Note: a relation could be
more than one of these ...
R1 = {(a, b) | a ≥ b}
Back To Common Examples
Some Common Examples
Let Ri : Z → Z. Determine if each of the following is reflexive,
symmetric, antisymmetric, or transitive. Note: a relation could be
more than one of these ...
R1 = {(a, b) | a ≥ b}
R2 = {(a, b) | a < b}
Back To Common Examples
Some Common Examples
Let Ri : Z → Z. Determine if each of the following is reflexive,
symmetric, antisymmetric, or transitive. Note: a relation could be
more than one of these ...
R1 = {(a, b) | a ≥ b}
R2 = {(a, b) | a < b}
R3 = {(a, b) | (a = b) ∨ (a 6= b)}
Back To Common Examples
Some Common Examples
Let Ri : Z → Z. Determine if each of the following is reflexive,
symmetric, antisymmetric, or transitive. Note: a relation could be
more than one of these ...
R1 = {(a, b) | a ≥ b}
R2 = {(a, b) | a < b}
R3 = {(a, b) | (a = b) ∨ (a 6= b)}
R4 = {(a, b) | a = b + 1}
Back To Common Examples
Some Common Examples
Let Ri : Z → Z. Determine if each of the following is reflexive,
symmetric, antisymmetric, or transitive. Note: a relation could be
more than one of these ...
R1 = {(a, b) | a ≥ b}
R2 = {(a, b) | a < b}
R3 = {(a, b) | (a = b) ∨ (a 6= b)}
R4 = {(a, b) | a = b + 1}
R5 = {(a, b) | a + b ≤ 3}
Equivalence Relations
Definition
A relation is called an equivalence relation if it is symmetric, reflexive
and transitive.
Equivalence Relations
Definition
A relation is called an equivalence relation if it is symmetric, reflexive
and transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}
Equivalence Relations
Definition
A relation is called an equivalence relation if it is symmetric, reflexive
and transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}
R2 = {(a, b) | a < b}
Equivalence Relations
Definition
A relation is called an equivalence relation if it is symmetric, reflexive
and transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}
R2 = {(a, b) | a < b}
R3 = {(a, b) | (a = b) ∨ (a 6= b)}
Equivalence Relations
Definition
A relation is called an equivalence relation if it is symmetric, reflexive
and transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}
R2 = {(a, b) | a < b}
R3 = {(a, b) | (a = b) ∨ (a 6= b)}
R4 = {(a, b) | a = b + 1}
Equivalence Relations
Definition
A relation is called an equivalence relation if it is symmetric, reflexive
and transitive.
From our last example, which are equivalence relations?
Common Equivalence Relations
R1 = {(a, b) | a ≥ b}
R2 = {(a, b) | a < b}
R3 = {(a, b) | (a = b) ∨ (a 6= b)}
R4 = {(a, b) | a = b + 1}
R5 = {(a, b) | a + b ≤ 3}
An Equivalence Relation Proof
Example
A relation R is defined on Z by xRy if x + 3y is even. Prove R is an
equivalence relation.
An Equivalence Relation Proof
Example
A relation R is defined on Z by xRy if x + 3y is even. Prove R is an
equivalence relation.
Proof.
First, we show R is reflexive.
An Equivalence Relation Proof
Example
A relation R is defined on Z by xRy if x + 3y is even. Prove R is an
equivalence relation.
Proof.
First, we show R is reflexive. Let a ∈ Z. Then a + 3a = 4a = 2(2a)
is even since 2a ∈ Z. Therefore, R is reflexive.
An Equivalence Relation Proof
Example
A relation R is defined on Z by xRy if x + 3y is even. Prove R is an
equivalence relation.
Proof.
First, we show R is reflexive. Let a ∈ Z. Then a + 3a = 4a = 2(2a)
is even since 2a ∈ Z. Therefore, R is reflexive.
An Equivalence Relation Proof
Proof.
We now show R is symmetric.
An Equivalence Relation Proof
Proof.
We now show R is symmetric. Assume aRb. Then a + 3b is even.
An Equivalence Relation Proof
Proof.
We now show R is symmetric. Assume aRb. Then a + 3b is even. So,
there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.
An Equivalence Relation Proof
Proof.
We now show R is symmetric. Assume aRb. Then a + 3b is even. So,
there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.
Consider
b + 3a = b + 3(2k − 3b)
An Equivalence Relation Proof
Proof.
We now show R is symmetric. Assume aRb. Then a + 3b is even. So,
there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.
Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
An Equivalence Relation Proof
Proof.
We now show R is symmetric. Assume aRb. Then a + 3b is even. So,
there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.
Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
An Equivalence Relation Proof
Proof.
We now show R is symmetric. Assume aRb. Then a + 3b is even. So,
there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.
Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
An Equivalence Relation Proof
Proof.
We now show R is symmetric. Assume aRb. Then a + 3b is even. So,
there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.
Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even.
An Equivalence Relation Proof
Proof.
We now show R is symmetric. Assume aRb. Then a + 3b is even. So,
there exists k ∈ Z such that a + 3b = 2k, or that a = 2k − 3b.
Consider
b + 3a = b + 3(2k − 3b)
= b + 6k − 9b
= 6k − 8b
= 2(3k − 4b)
Since 3k − 4b ∈ Z, b + 3a is even. Therefore, bRa and so R is
symmetric.
An Equivalence Relation Proof
Proof.
We now show R is transitive.
An Equivalence Relation Proof
Proof.
We now show R is transitive. Assume aRb and bRc.
An Equivalence Relation Proof
Proof.
We now show R is transitive. Assume aRb and bRc. Hence, a + 3b
and b + 3c are even.
An Equivalence Relation Proof
Proof.
We now show R is transitive. Assume aRb and bRc. Hence, a + 3b
and b + 3c are even. So, there exists integers n and m such that
a + 3b = 2n and b + 3c = 2m. That is, a = 2n − 3b and
3c = 2m − b.
An Equivalence Relation Proof
Proof.
We now show R is transitive. Assume aRb and bRc. Hence, a + 3b
and b + 3c are even. So, there exists integers n and m such that
a + 3b = 2n and b + 3c = 2m. That is, a = 2n − 3b and
3c = 2m − b. Consider
a + 3c = 2n − 3b + 2m − b
An Equivalence Relation Proof
Proof.
We now show R is transitive. Assume aRb and bRc. Hence, a + 3b
and b + 3c are even. So, there exists integers n and m such that
a + 3b = 2n and b + 3c = 2m. That is, a = 2n − 3b and
3c = 2m − b. Consider
a + 3c = 2n − 3b + 2m − b
= 2n + 2m − 4b
An Equivalence Relation Proof
Proof.
We now show R is transitive. Assume aRb and bRc. Hence, a + 3b
and b + 3c are even. So, there exists integers n and m such that
a + 3b = 2n and b + 3c = 2m. That is, a = 2n − 3b and
3c = 2m − b. Consider
a + 3c = 2n − 3b + 2m − b
= 2n + 2m − 4b
= 2(n + m − 2b)
An Equivalence Relation Proof
Proof.
We now show R is transitive. Assume aRb and bRc. Hence, a + 3b
and b + 3c are even. So, there exists integers n and m such that
a + 3b = 2n and b + 3c = 2m. That is, a = 2n − 3b and
3c = 2m − b. Consider
a + 3c = 2n − 3b + 2m − b
= 2n + 2m − 4b
= 2(n + m − 2b)
Since n + m − 2b ∈ Z, it follows that a + 3c is even, and so aRc.
Therefore, R is transitive.
An Equivalence Relation Proof
Proof.
We now show R is transitive. Assume aRb and bRc. Hence, a + 3b
and b + 3c are even. So, there exists integers n and m such that
a + 3b = 2n and b + 3c = 2m. That is, a = 2n − 3b and
3c = 2m − b. Consider
a + 3c = 2n − 3b + 2m − b
= 2n + 2m − 4b
= 2(n + m − 2b)
Since n + m − 2b ∈ Z, it follows that a + 3c is even, and so aRc.
Therefore, R is transitive.
Since R is reflexive, symmetric and transitive, R is an equivalence
relation.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb if
a − b ∈ H. Show that R is an equivalence relation.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb if
a − b ∈ H. Show that R is an equivalence relation.
Proof.
Reflexive:
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb if
a − b ∈ H. Show that R is an equivalence relation.
Proof.
Reflexive: Let a ∈ Z. Since a − a = 0 = 4 · 0, it follows that aRa.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb if
a − b ∈ H. Show that R is an equivalence relation.
Proof.
Reflexive: Let a ∈ Z. Since a − a = 0 = 4 · 0, it follows that aRa.
Symmetric:
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb if
a − b ∈ H. Show that R is an equivalence relation.
Proof.
Reflexive: Let a ∈ Z. Since a − a = 0 = 4 · 0, it follows that aRa.
Symmetric: Assume aRb, where a, b ∈ Z.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb if
a − b ∈ H. Show that R is an equivalence relation.
Proof.
Reflexive: Let a ∈ Z. Since a − a = 0 = 4 · 0, it follows that aRa.
Symmetric: Assume aRb, where a, b ∈ Z. Then, a − b = 4k for some
k ∈ Z.
Another Example
Example
Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb if
a − b ∈ H. Show that R is an equivalence relation.
Proof.
Reflexive: Let a ∈ Z. Since a − a = 0 = 4 · 0, it follows that aRa.
Symmetric: Assume aRb, where a, b ∈ Z. Then, a − b = 4k for some
k ∈ Z. Then, b − a = 4(−k), and since −k ∈ Z, it follows that
b − a ∈ H. Therefore, bRa.
Another Example
Proof.
Transitive:
Another Example
Proof.
Transitive: Assume aRb and bRc for a, b, c ∈ Z.
Another Example
Proof.
Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a − b ∈ H and
b − c ∈ H.
Another Example
Proof.
Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a − b ∈ H and
b − c ∈ H. Then, there exists integers k and l such that a − b = 4k
and b − c = 4l.
Another Example
Proof.
Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a − b ∈ H and
b − c ∈ H. Then, there exists integers k and l such that a − b = 4k
and b − c = 4l. Therefore,
a − c = (a − b) + (b − c)
Another Example
Proof.
Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a − b ∈ H and
b − c ∈ H. Then, there exists integers k and l such that a − b = 4k
and b − c = 4l. Therefore,
a − c = (a − b) + (b − c)
= 4k + 4l
Another Example
Proof.
Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a − b ∈ H and
b − c ∈ H. Then, there exists integers k and l such that a − b = 4k
and b − c = 4l. Therefore,
a − c = (a − b) + (b − c)
= 4k + 4l
= 4(k + l)
Another Example
Proof.
Transitive: Assume aRb and bRc for a, b, c ∈ Z. Then a − b ∈ H and
b − c ∈ H. Then, there exists integers k and l such that a − b = 4k
and b − c = 4l. Therefore,
a − c = (a − b) + (b − c)
= 4k + 4l
= 4(k + l)
Since k + l ∈ Z, it follows that a − c ∈ H and so aRc.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 divides
x + 2y.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 divides
x + 2y.
Proof.
Reflexive:
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 divides
x + 2y.
Proof.
Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 divides
x + 2y.
Proof.
Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.
Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k for
some integer k.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 divides
x + 2y.
Proof.
Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.
Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k for
some integer k. That is, x = 3k − 2y.
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 divides
x + 2y.
Proof.
Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.
Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k for
some integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 divides
x + 2y.
Proof.
Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.
Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k for
some integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 divides
x + 2y.
Proof.
Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.
Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k for
some integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
And Another Example
Example
Determine if R is an equivalence relation if (x, y) ∈ R if 3 divides
x + 2y.
Proof.
Reflexive: Let x ∈ Z. Then, x + 2x = 3x and 3 | 3x. So, xRx.
Symmetric: Suppose x, y ∈ Z such that xRy. Then, x + 2y = 3k for
some integer k. That is, x = 3k − 2y. Consider
y + 2x = y + 2(3k − 2y)
= 6k − 3y
= 3(2k − y)
Since 2k − y ∈ Z, 3 | (y + 2x) and so yRx.
And Another Example
Proof.
Transitive:
And Another Example
Proof.
Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz.
And Another Example
Proof.
Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,
for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l.
And Another Example
Proof.
Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,
for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2y
and 2z = 3l − y.
And Another Example
Proof.
Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,
for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2y
and 2z = 3l − y. Consider
x + 2z = (3k − 2y) + (3l − y)
And Another Example
Proof.
Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,
for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2y
and 2z = 3l − y. Consider
x + 2z = (3k − 2y) + (3l − y)
= 3k + 3l − 3y
And Another Example
Proof.
Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,
for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2y
and 2z = 3l − y. Consider
x + 2z = (3k − 2y) + (3l − y)
= 3k + 3l − 3y
= 3(k + l − y)
And Another Example
Proof.
Transitive: Suppose for x, y, z ∈ Z, we have that xRy and yRz. Then,
for some k, l ∈ Z, x + 2y = 3k and y + 2z = 3l. That is, x = 3k − 2y
and 2z = 3l − y. Consider
x + 2z = (3k − 2y) + (3l − y)
= 3k + 3l − 3y
= 3(k + l − y)
Since k + l − y ∈ Z, 3 | (x + 2z) and so xRz.
Congruences and Equivalence Relations
Theorem
Let m ∈ Z+ .
(a) For every x ∈ Z, x ≡ x (mod m).
Congruences and Equivalence Relations
Theorem
Let m ∈ Z+ .
(a) For every x ∈ Z, x ≡ x (mod m).
(b) For all x, y ∈ Z, if x ≡ y (mod m) then y ≡ x (mod m).
Congruences and Equivalence Relations
Theorem
Let m ∈ Z+ .
(a) For every x ∈ Z, x ≡ x (mod m).
(b) For all x, y ∈ Z, if x ≡ y (mod m) then y ≡ x (mod m).
(c) For all x, y, z ∈ Z, if x ≡ y (mod m) and y ≡ z (mod m), then
x ≡ z (mod m).
The Proof
Proof.
Proof of (a): x ∈ Z, Then m | 0,
The Proof
Proof.
Proof of (a): x ∈ Z, Then m | 0,Since x − x = 0, we have that
m | (x − x).
The Proof
Proof.
Proof of (a): x ∈ Z, Then m | 0,Since x − x = 0, we have that
m | (x − x). That is, x ≡ x (mod m).
The Proof
Proof.
Proof of (a): x ∈ Z, Then m | 0,Since x − x = 0, we have that
m | (x − x). That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m).
The Proof
Proof.
Proof of (a): x ∈ Z, Then m | 0,Since x − x = 0, we have that
m | (x − x). That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m). By definition,
m | (x − y). Then, ∃k ∈ Z such that mk = x − y.
The Proof
Proof.
Proof of (a): x ∈ Z, Then m | 0,Since x − x = 0, we have that
m | (x − x). That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m). By definition,
m | (x − y). Then, ∃k ∈ Z such that mk = x − y. But, m(−k) = y − x,
so m | (y − x).
The Proof
Proof.
Proof of (a): x ∈ Z, Then m | 0,Since x − x = 0, we have that
m | (x − x). That is, x ≡ x (mod m).
Proof of (b): Let x, y ∈ Z. Assume that x ≡ y (mod m). By definition,
m | (x − y). Then, ∃k ∈ Z such that mk = x − y. But, m(−k) = y − x,
so m | (y − x). That is, y ≡ x (mod m).
The Proof
Proof.
Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) and
y ≡ z (mod m).
The Proof
Proof.
Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) and
y ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x − y.
The Proof
Proof.
Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) and
y ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x − y.
Since y ≡ z (mod m), ∃l ∈ Z such that ml = y − z.
The Proof
Proof.
Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) and
y ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x − y.
Since y ≡ z (mod m), ∃l ∈ Z such that ml = y − z.
Now,
x − z = (x − y) + (y − z)
The Proof
Proof.
Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) and
y ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x − y.
Since y ≡ z (mod m), ∃l ∈ Z such that ml = y − z.
Now,
x − z = (x − y) + (y − z)
= mk + ml
The Proof
Proof.
Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) and
y ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x − y.
Since y ≡ z (mod m), ∃l ∈ Z such that ml = y − z.
Now,
x − z = (x − y) + (y − z)
= mk + ml
= m(k + l)
The Proof
Proof.
Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) and
y ≡ z (mod m). Since x ≡ y (mod m), ∃k ∈ Z such that mk = x − y.
Since y ≡ z (mod m), ∃l ∈ Z such that ml = y − z.
Now,
x − z = (x − y) + (y − z)
= mk + ml
= m(k + l)
Since k + l ∈ Z, x − z = m(k + l) implies that m | (x − z). That is,
x ≡ z (mod m).
Example
Example
Let R be the relation defined on Z by aRb if 2a + b ≡ 0 (mod 3).
Prove R is an equivalence relation.
Example
Example
Let R be the relation defined on Z by aRb if 2a + b ≡ 0 (mod 3).
Prove R is an equivalence relation.
Proof.
Let x ∈ Z. Since 3 | (3x), it follows that 3x ≡ 0 (mod 3).
Example
Example
Let R be the relation defined on Z by aRb if 2a + b ≡ 0 (mod 3).
Prove R is an equivalence relation.
Proof.
Let x ∈ Z. Since 3 | (3x), it follows that 3x ≡ 0 (mod 3). So,
2x + x ≡ 0 (mod 3). Thus, R is reflexive.
Example
Proof.
Next, we show R is symmetric. Assume xRy, where x, y ∈ Z.
Example
Proof.
Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,
2x + y ≡ 0 (mod 3) and so 3 | (2x + y),
Example
Proof.
Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,
2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r for
some r ∈ Z.
Example
Proof.
Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,
2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r for
some r ∈ Z. Hence, y = 3r − 2x.
So,
2y + x = 2(3r − 2x) + x
Example
Proof.
Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,
2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r for
some r ∈ Z. Hence, y = 3r − 2x.
So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
Example
Proof.
Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,
2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r for
some r ∈ Z. Hence, y = 3r − 2x.
So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Example
Proof.
Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,
2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r for
some r ∈ Z. Hence, y = 3r − 2x.
So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Since 2r − x ∈ Z, 3 | (2y + x). So, 2y + x ≡ 0 (mod 3).
Example
Proof.
Next, we show R is symmetric. Assume xRy, where x, y ∈ Z. Thus,
2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r for
some r ∈ Z. Hence, y = 3r − 2x.
So,
2y + x = 2(3r − 2x) + x
= 6r − 3x
= 3(2r + x)
Since 2r − x ∈ Z, 3 | (2y + x). So, 2y + x ≡ 0 (mod 3). Therefore,
yRx and R is symmetric.
Example
Proof.
Finally, we show R is transitive. Assume that xRy and yRz, where
x, y, z ∈ Z.
Example
Proof.
Finally, we show R is transitive. Assume that xRy and yRz, where
x, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3).
Example
Proof.
Finally, we show R is transitive. Assume that xRy and yRz, where
x, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,
3 | (2x + y) and 3 | (2y + z).
Example
Proof.
Finally, we show R is transitive. Assume that xRy and yRz, where
x, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,
3 | (2x + y) and 3 | (2y + z). From this, it follows that 2x + y = 3r
and 2y + z = 3s for some integers r and s.
Example
Proof.
Finally, we show R is transitive. Assume that xRy and yRz, where
x, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,
3 | (2x + y) and 3 | (2y + z). From this, it follows that 2x + y = 3r
and 2y + z = 3s for some integers r and s. Adding these two
equations, we obtain
2x + 3y + z = 3r + 3s
Example
Proof.
Finally, we show R is transitive. Assume that xRy and yRz, where
x, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,
3 | (2x + y) and 3 | (2y + z). From this, it follows that 2x + y = 3r
and 2y + z = 3s for some integers r and s. Adding these two
equations, we obtain
2x + 3y + z = 3r + 3s
so
2x + z = 3r + 3s − 3y = 3(r + s − y)
Example
Proof.
Finally, we show R is transitive. Assume that xRy and yRz, where
x, y, z ∈ Z. Then 2x + y ≡ 0 (mod 3) and 2y + z ≡ 0 (mod 3). Thus,
3 | (2x + y) and 3 | (2y + z). From this, it follows that 2x + y = 3r
and 2y + z = 3s for some integers r and s. Adding these two
equations, we obtain
2x + 3y + z = 3r + 3s
so
2x + z = 3r + 3s − 3y = 3(r + s − y)
Since r + s − y ∈ Z, 3 | (2x + z); so, 2x + z ≡ 0 (mod 3). Hence, xRz
and so R is transitive.