7.2 Equivalence Relationsbtravers.weebly.com/uploads/6/7/2/9/6729909/... · Properties of Relations Deﬁnition A relation R : A !A is said to be reﬂexive if xRx for all x 2A. Example

§ 7.2 Equivalence Relations

Properties of Relations

DefinitionA relation R : A→ A is said to be reflexive if xRx for all x ∈ A.

ExampleWhich of the following relations are reflexive, where each is definedon 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)}




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)}




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)}




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)}




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)}




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)}




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)}


DefinitionA relation R : A→ A is said to be symmetric if xRy whenever yRx.

ExampleWhich of the following relations are symmetric, where each is definedon 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)}




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)}




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)}




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)}




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)}




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)}




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)}


DefinitionA relation R : A→ A is said to be antisymmetric if xRy whenevery 6 Rx.

ExampleWhich of the following relations are antisymmetric, where each isdefined 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)}




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)}




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)}




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)}




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)}




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)}




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)}


DefinitionA relation R : A→ A is said to be transitive if whenever xRy and yRz,xRz.

ExampleWhich of the following relations are transitive, where each is definedon 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.




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.










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.







Example1 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.



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.






This is transitive.




This is transitive.

3 R5 = {(a, a), (a, b)}This is transitive.







This is transitive.





This is transitive.







This is transitive.






This is transitive.






This is transitive.

Back To Common Examples

Some Common ExamplesLet Ri : Z→ Z. Determine if each of the following is reflexive,symmetric, antisymmetric, or transitive. Note: a relation could bemore 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}



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}



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}



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}



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

DefinitionA relation is called an equivalence relation if it is symmetric, reflexiveand 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}





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}





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}





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

ExampleA relation R is defined on Z by xRy if x + 3y is even. Prove R is anequivalence 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.



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.








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 issymmetric.


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)



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)



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)




b + 3a = b + 3(2k − 3b)

= b + 6k − 9b

= 6k − 8b

= 2(3k − 4b)




b + 3a = b + 3(2k − 3b)

= b + 6k − 9b

= 6k − 8b

= 2(3k − 4b)




b + 3a = b + 3(2k − 3b)

= b + 6k − 9b

= 6k − 8b

= 2(3k − 4b)




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 issymmetric.



b + 3a = b + 3(2k − 3b)

= b + 6k − 9b

= 6k − 8b

= 2(3k − 4b)



Proof.We now show R is transitive.

Assume aRb and bRc. Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 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 equivalencerelation.


Proof.We now show R is transitive. Assume aRb and bRc.

Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider

a + 3c = 2n− 3b + 2m− b

= 2n + 2m− 4b

= 2(n + m− 2b)




Proof.We now show R is transitive. Assume aRb and bRc. Hence, a + 3band b + 3c are even.

So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider

a + 3c = 2n− 3b + 2m− b

= 2n + 2m− 4b

= 2(n + m− 2b)




Proof.We now show R is transitive. Assume aRb and bRc. Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b.

Consider

a + 3c = 2n− 3b + 2m− b

= 2n + 2m− 4b

= 2(n + m− 2b)




Proof.We now show R is transitive. Assume aRb and bRc. Hence, a + 3band b + 3c are even. So, there exists integers n and m such thata + 3b = 2n and b + 3c = 2m. That is, a = 2n− 3b and3c = 2m− b. Consider

a + 3c = 2n− 3b + 2m− b

= 2n + 2m− 4b

= 2(n + m− 2b)





a + 3c = 2n− 3b + 2m− b

= 2n + 2m− 4b

= 2(n + m− 2b)





a + 3c = 2n− 3b + 2m− b

= 2n + 2m− 4b

= 2(n + m− 2b)





a + 3c = 2n− 3b + 2m− b

= 2n + 2m− 4b

= 2(n + m− 2b)





a + 3c = 2n− 3b + 2m− b

= 2n + 2m− 4b

= 2(n + m− 2b)



Another Example

Example

Let H = {4k | k ∈ Z}.A relation R is defined on Z by aRb ifa− 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 somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.

Another Example

Example


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 somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.

Another Example

Example


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 somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.

Another Example

Example


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 somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.

Another Example

Example


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 somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.

Another Example

Example


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 somek ∈ Z.

Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.

Another Example

Example


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 somek ∈ Z. Then, b− a = 4(−k), and since −k ∈ Z, it follows thatb− a ∈ H. Therefore, bRa.

Another Example

Proof.Transitive:

Assume aRb and bRc for a, b, c ∈ Z. Then a− b ∈ H andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand 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.

Another Example

Proof.Transitive: Assume aRb and bRc for a, b, c ∈ Z.

Then a− b ∈ H andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand 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 andb− c ∈ H.

Then, there exists integers k and l such that a− b = 4kand 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 andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand 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 andb− c ∈ H. Then, there exists integers k and l such that a− b = 4kand b− c = 4l. Therefore,

a− c = (a− b) + (b− c)

= 4k + 4l

= 4(k + l)


Another Example


a− c = (a− b) + (b− c)

= 4k + 4l

= 4(k + l)


Another Example


a− c = (a− b) + (b− c)

= 4k + 4l

= 4(k + l)


Another Example


a− c = (a− b) + (b− c)

= 4k + 4l

= 4(k + l)


And Another Example

Example

Determine if R is an equivalence relation if (x, y) ∈ R if 3 dividesx + 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 forsome 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

Example


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 forsome integer k. That is, x = 3k − 2y. Consider

y + 2x = y + 2(3k − 2y)

= 6k − 3y

= 3(2k − y)


And Another Example

Example


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 forsome integer k. That is, x = 3k − 2y. Consider

y + 2x = y + 2(3k − 2y)

= 6k − 3y

= 3(2k − y)


And Another Example

Example


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 forsome integer k.

That is, x = 3k − 2y. Consider

y + 2x = y + 2(3k − 2y)

= 6k − 3y

= 3(2k − y)


And Another Example

Example


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 forsome integer k. That is, x = 3k − 2y.

Consider

y + 2x = y + 2(3k − 2y)

= 6k − 3y

= 3(2k − y)


And Another Example

Example



y + 2x = y + 2(3k − 2y)

= 6k − 3y

= 3(2k − y)


And Another Example

Example



y + 2x = y + 2(3k − 2y)

= 6k − 3y

= 3(2k − y)


And Another Example

Example



y + 2x = y + 2(3k − 2y)

= 6k − 3y

= 3(2k − y)


And Another Example

Example



y + 2x = y + 2(3k − 2y)

= 6k − 3y

= 3(2k − 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 − 2yand 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.

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 − 2yand 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 − 2yand 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 − 2yand 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 − 2yand 2z = 3l− y. Consider

x + 2z = (3k − 2y) + (3l− y)

= 3k + 3l− 3y

= 3(k + l− y)


And Another Example


x + 2z = (3k − 2y) + (3l− y)

= 3k + 3l− 3y

= 3(k + l− y)


And Another Example


x + 2z = (3k − 2y) + (3l− y)

= 3k + 3l− 3y

= 3(k + l− y)


And Another Example


x + 2z = (3k − 2y) + (3l− y)

= 3k + 3l− 3y

= 3(k + l− y)


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), thenx ≡ z (mod m).


Theorem

Let m ∈ Z+.





Theorem

Let m ∈ Z+.




The Proof

Proof.Proof of (a): x ∈ Z, Then m | 0,

Since x− x = 0, we have thatm | (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 (a): x ∈ Z, Then m | 0,Since x− x = 0, we have thatm | (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 thatm | (x− x). That is, x ≡ x (mod m).


The Proof


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 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 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



The Proof

Proof.Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) andy ≡ 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).

The Proof

Proof.Proof of (c): Let x, y, z ∈ Z. Assume that x ≡ y (mod m) andy ≡ 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) andy ≡ 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


Now,

x− z = (x− y) + (y− z)

= mk + ml

= m(k + l)


The Proof


Now,

x− z = (x− y) + (y− z)

= mk + ml

= m(k + l)


The Proof


Now,

x− z = (x− y) + (y− z)

= mk + ml

= m(k + l)


The Proof


Now,

x− z = (x− y) + (y− z)

= mk + ml

= m(k + l)


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

Example


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

Example


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.

Thus,2x + y ≡ 0 (mod 3) and so 3 | (2x + y), Therefore, 2x + y = 3r forsome 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.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 forsome 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 forsome 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 forsome r ∈ Z. Hence, y = 3r − 2x.So,

2y + x = 2(3r − 2x) + x

= 6r − 3x

= 3(2r + x)


Example


2y + x = 2(3r − 2x) + x

= 6r − 3x

= 3(2r + x)


Example


2y + x = 2(3r − 2x) + x

= 6r − 3x

= 3(2r + x)


Example


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


2y + x = 2(3r − 2x) + x

= 6r − 3x

= 3(2r + x)


Example

Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, 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 = 3rand 2y + z = 3s for some integers r and s. Adding these twoequations, we obtain

2x + 3y + z = 3r + 3s

so2x + z = 3r + 3s− 3y = 3(r + s− y)

Since r + s− y ∈ Z, 3 | (2x + z); so, 2x + z ≡ 0 (mod 3). Hence, xRzand so R is transitive.

Example

Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, 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 = 3rand 2y + z = 3s for some integers r and s. Adding these twoequations, we obtain

2x + 3y + z = 3r + 3s

so2x + z = 3r + 3s− 3y = 3(r + s− y)


Example

Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, 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 = 3rand 2y + z = 3s for some integers r and s. Adding these twoequations, we obtain

2x + 3y + z = 3r + 3s

so2x + z = 3r + 3s− 3y = 3(r + s− y)


Example

Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, 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 = 3rand 2y + z = 3s for some integers r and s.

Adding these twoequations, we obtain

2x + 3y + z = 3r + 3s

so2x + z = 3r + 3s− 3y = 3(r + s− y)


Example

Proof.Finally, we show R is transitive. Assume that xRy and yRz, wherex, 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 = 3rand 2y + z = 3s for some integers r and s. Adding these twoequations, we obtain

2x + 3y + z = 3r + 3s

so2x + z = 3r + 3s− 3y = 3(r + s− y)


Example


2x + 3y + z = 3r + 3s

so2x + z = 3r + 3s− 3y = 3(r + s− y)


Example


2x + 3y + z = 3r + 3s

so2x + z = 3r + 3s− 3y = 3(r + s− y)


Documents

7.2 Equivalence Relationsbtravers.weebly.com/uploads/6/7/2/9/6729909/... · Properties of Relations Deﬁnition A relation R : A !A is said to be reﬂexive if xRx for all x 2A. Example