Equivalence Relations

[Zelda22]

R₁ = {(a, b) | a ≡ b (mod 3)}
R₂ = {(a, b) | a ≡ b (mod 2)}.

Find
a) R₁ ∪ R₂.
Equivalence Relation on Z? —-Yes —-No

b) R₁ ∩ R₂.
Equivalence Relation on Z? —-Yes —-No



a. R₁ ∩ R₂.= {(a, b) | a ≡ b (mod 6)}— Yes

b. R₁ ∪ R₂.= ?

I’m no sure about the union, please help.

R₁ ∪ R₂.= {(a, b) | a ≡ b (mod 1)} - Yes ?

Thanks.

 

[topsquark]

R₁ = {(a, b) | a ≡ b (mod 3)}
R₂ = {(a, b) | a ≡ b (mod 2)}.

Find
a) R₁ ∪ R₂.
Equivalence Relation on Z? —-Yes —-No

b) R₁ ∩ R₂.
Equivalence Relation on Z? —-Yes —-No



a. R₁ ∩ R₂.= {(a, b) | a ≡ b (mod 6)}— Yes

b. R₁ ∪ R₂.= ?

I’m no sure about the union, please help.

R₁ ∪ R₂.= {(a, b) | a ≡ b (mod 1)} - Yes ?

Thanks.

[imath]R_1 \cup R_2 = \{ (a, b) | a \equiv \text{ (mod 3) or } a \equiv \text{ (mod 2)} \}[/imath]

Can we say that a ~ a? Yes. Because this is true for both [imath]R_1[/imath] and [imath]R_2[/imath].

Can we say that if a ~ b then b ~ a? Yes. (Why?)

Can we say that if a ~ b and b ~ c that a ~ c? Yes/No? (Why?)

-Dan

 

[Zelda22]

[imath]R_1 \cup R_2 = \{ (a, b) | a \equiv \text{ (mod 3) or } a \equiv \text{ (mod 2) \}[/imath]

Can we say that a ~ a? Yes. Because this is true for both [imath]R_1[/imath] and [imath]R_2[/imath].

Can we say that if a ~ b then b ~ a? Yes. (Why?)

Can we say that if a ~ b and b ~ c that a ~ c? Yes/No? (Why?)

-Dan

Yes, to be an equivalence relation it need to be , reflexive, symmetric and transitive.

But I don’t understand what is the union of two congruent relations. Can you please explain to me? Thank you

 

[Zelda22]

Yes, to be an equivalence relation it need to be , reflexive, symmetric and transitive.

But I don’t understand what is the union of two congruent relations. Can you please explain to me? Thank you

if a ~ b and b ~ c that a ~ c? Yes, is transitive
6|(a-b) and 6|(b-c) then 6|(a-c)
6|(22-10) and 6|(10-4) then 6|(22-4)

 

[topsquark]

Yes, to be an equivalence relation it need to be , reflexive, symmetric and transitive.

But I don’t understand what is the union of two congruent relations. Can you please explain to me? Thank you

It means that a ~ b if either [imath]a \equiv b \text{ (mod 3)}[/imath] or if [imath]a \equiv b \text{ (mod 2)}[/imath]. Or equivalently a ~ b if a - b = 3n or a - b = 2n. There is no way to write it as a single modular relation if that's what you are asking.

For your transitive argument you are using the set [imath]R_1 \cap R_2[/imath]. I had thought you were asking only about [imath]R_1 \cup R_2[/imath]?

-Dan

 

[blamocur]

But I don’t understand what is the union of two congruent relations. Can you please explain to me? Thank you

Relations are defined as sets of pairs, e.g., [imath]R = \{(x,y): x \bowtie y\}[/imath] satisfying the relations' axioms, where [imath]\bowtie[/imath] is another fancy symbol for a relation. The union of two relations is a union of their corresponding sets. Whether the resulting set satisfies the relations' axioms is something you have to figure out.

 

[pka]

R₁ = {(a, b) | a ≡ b (mod 3)}
R₂ = {(a, b) | a ≡ b (mod 2)}.
Find
a) R₁ ∪ R₂. Equivalence Relation on Z? —-Yes —-No

b) R₁ ∩ R₂. Equivalence Relation on Z? —-Yes —-No

@zelda, it would have been best had you shown some effort at understanding relations. Do you?
You are given [imath]\mathcal{R}_1=\{(a,b):a\equiv b \mod(3)\}[/imath] and [imath]\mathcal{R}_2=\{(p,q):p\equiv q \mod(2)\}[/imath].
Is it clear to you that [imath] (7,4)\in\mathcal{R}_1~\&~(4,8)\in\mathcal{R}_2~?[/imath]
If you are clear on that, is true that [imath](7,8)\in\mathcal{R}_1\cup\mathcal{R}_2\Large\bf{?}[/imath]
Now please, please tell us what you do or don't understand about that.
[imath][/imath][imath][/imath]