Equivalence relation Puns

As the title says, what is the significance of equivalence relations.

Is it still true that the functional state of the art for the union/find problem lags behind the well known imperative approach? The best I can find (2007) are techniques using internal mutation with an external functional interface. Are these in practical use, and has anything else appeared?

I know the following theorem:

Theorem: Let S be a set, and ~ an equivalence relation on S. Then the equivalence classes with respect to ~ form a partition of S. In particular, if a is in S, the a is in [a] and furthermore, a is in [b] if and only if [a]=[b].

But I'm completely stuck as to how to apply it.

Any sort of tips or help would be appreciated.

Hi guys, i need help with this relation :

∀a, b ∈ Q* (a, b) ∈ R ⇔ (∃h ∈ Z)(a = 5^h b)

Q*=(x ∈ Q| x != 0)

I have to show that it's an equivalence relation (So it must be Reflexive, Symmetric and Transitive). Can someone help me by showing to me all the steps to solve this problem?

In particular i need help to show that the relation is symmetric (∀a, b ∈ A)((a, b) ∈ R ⇒ (b,a) ∈ R)

I've tried this: In this case (a) is equal to (5^h b), so (b) has to be equal to (5^h a). To show that it's Symmetric 5^h b must be equal to 5^h a and if we isolate the unknown terms (a and b) the solution is b = a so (b,a) ∈ R. Is this right?

PS: I really apologize for my english XD, there are some mathematical technical terms in this post that i don't know how to write correctly (I'm italian)

Hi math students. I made a practice quiz about equivalence relations and turned it into a YouTube video. Here's a link:

Equivalence Relations Practice Quiz with Solutions

It could be useful to you if you are studying equivalence relations.

Have a nice day!

It's ex 18.a from function chapter,How to prove it, Velleman. I've taken the solution from Inchmeal. The author says he's not confident with it.

My problem is with the existencial part, since I haven't understood the other yet.

Solution Suppose R is compatible with f. Suppose h= {(X,y)∈A/RxB s.t ∃x∈X(f(x)=y)}

Existence. Suppose X∈A/R and y∈B s.t (X,y)∈h. Thus there exist an x s.t x∈X and y=f(x). Since x∈X and X∈A/R, it follows that X=[x]R. Thus ([x]R,f(x))∈h. Or h([x]R)=f(x). Thus such a relation exists.

Questions:

****What's his flow of thought?

I can see that h={ ([x]R,f(x))∈A/RxB s.t x∈A} (roster notation of h) because of the definition of h.

I understood that he assumes this set written in set builder notation, then takes an arbitrary element from it, explores the definitions and concludes that if (X,y)∈h where h is the assumed relation, then (X,y)=([x]R,f(x)) for any (X,y)∈h.

Why does he do that? Why is he allowed to assume h in SBN and he's not allowed to do that by enumeration? Couldn't he just assume h by roster notation as a relation and then prove it's a function?

****Why doesn't he prove that h is a function? **** Strangeness: he concludes the existencial part saying h is a relation.In the uniqueness part he says "where h is the function we defined"

OT:

Uniqueness part: (I don't care of it for now) Suppose there is another relation k. Suppose (X,y)∈k s.t X=[x]R, y=f(x) and h(X)=f(x). Suppose p∈[x]R. Thus there exists a p s.t p∈[x]R and p=f(x). It follows that ([x]p,f(x))∈h where h is the function we defined. Or (X,y)∈h. Since (X,y) are arbitrary, it follows that k=h.

public int compareTo(Node that) {
        if(this.x < that.x && this.y < that.y) return -1;
        if(this.x > that.x && this.y > that.y) return 1;
        if(approxEquals(this.x, that.x) && this.y > that.y) return 1;
        if(approxEquals(this.x, that.x) && this.y < that.y) return -1;
        if(approxEquals(this.y, that.y) && this.x > that.x) return 1;
        if(approxEquals(this.y, that.y) && this.x < that.x) return -1;
        if(approxEquals(this.x, that.x) && approxEquals(this.y, that.y)) return 0;
        return 0;
    }

I use floats for x,y which is why the approxEquals method exist.

The problem is the first two lines of code, which give the error "Comparison method violates its general contract!". I don't understand why they contradict eachother

EDIT: The Exception is of type java.lang.IllegalArgumentException. I read that my error here means that the code violates the rules of transitivy, symmetry and/or reflexivity

https://github.com/Ahajha/cpp-equiv

I created this out of necessity for some of my research work, but just recently polished it up, gave it a licence, and put it into its own repository for others to use.

Would love any feedback; Code reviews, suggestions for improvements (especially from those that may have a use for this), or anything else!

The question says that for a,b in R, aRb if b-a is in Q, and asks us to show this is indeed an equivance relation in the real number.

I don't quite understand this question as sqrt(2) and 1 are both in R but 1-sqrt(2) is not in Q. How is this an equivalent relation on R?

Directions:

Suppose R1 and R2 are equivalence relations on a set A. Let R be defined by the condition that aRb if and only if aR1b ∧ aR2b. (Note: I've already proven that R is an equivalence relation on a separate proof). If [a], [a]1 and [a]2 denote the equivalence class of a with respect to R, R1 and R2, respectively, prove that [a] = [a]1 ∩ [a]2.

What I've come up so far:

Since there's this theorem that says [a]1 ∩ [a]2 = ∅, I think I should prove that [a] = ∅. But I don't know how to do that.

A screenshot of the problem.

Thank you so much

Hi can someone pleaseee help me with how to prove this?

A relation ~ is defined on Q by: x~y if and only if there exists m in Z such that x = y + 2m

Prove it’s an equivalence relation

Thanksss :)

does the union form one of the partitions of X or the entire original set X ?

Suppose L is the language represented by the regular expression ((O U 1)(O U 1)(0 U 1))*.

I have that 0 ≈𝐿 00 is False, but I do not understand why. Doesn't ≈𝐿 mean that the two strings are either both in L or both not in L? Isn't 0 and 00 both unable to be in L because you need to have at least 3 characters?

I am confused for the others as well - why is 1 ≈𝐿 1001 true? Is it because neither is in L, or both in L?

As with 11 ≈𝐿 1101. This is false, but how?

Suppose I want to prove that for every a in R and k in N, there exists some b in R, s.t

b ~ a (b~a if a-b is in Q) and 1/k > b and b >= 0.

My attempt would just pick b=0(Can I do this?) since b=0 is in R and in Q, therefore proving the three properties of equivalence is easy. And obviously, since k is in N, 1/k > 0 always and 0>=0 always.

Would this approach work or it's flawed for whatever reason(maybe for irrational number)? I have a feeling this is way too simple

I'm sorry for the stupid question, but this has bothered me for a while. Why is "≡" necessary when "=" already represents equality? Or maybe I'm confusing the difference between equivalence and equality. Can somebody explain why "≡" is different from "="?

Where a path from c to d is a continuous function in a topological space X, f :[a,b] -> X, where f(a) = c and f(b) = d

Let ~ be the relation on R^* = R - { 0 } given by the following rule:

x ~ y if and only if x/y ∈ Q

Show that ~ is an equivalence relation on R^*.