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/sci/ - Science & Math

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4923612 No.4923612 [Reply] [Original]

I'm working on teaching myself set theory with a free textbook and the part on equivalence relations is very short and I'm still a little confused.

Unfortunately it only gives solutions to odd problems, and even those I'm a little unsure of. But here, for this one:

If I understand it right, an equivalence class a is the set of all Xs such that xRa, and two are equal if their Xs match.

So here, if I'm doing it right, is the set R = {(a,a), (b,b), (c,c), (d,d) (e,e), (a,d), (b,c), (e,a), (c,e)}?

If so, how do I get the equivalence classes from that?

>> No.4923635

Dude!!!!!!!!!!!!!!!!!!!!!! don't even go down this "teach yourself" route...........I've tried doing it and it gets difficult and tedious. You need people to talk about it with and these little
things add up. if you had a professor, you could ask him these questions.............plus if you're looking to do anything school related, there is no credit if you taught it to yourself.

>> No.4923638

There is only one.

>> No.4923650

I'm not sure what you were doing with the X's, but equivalence classes are subsets with equivalent members.

Because R is an equivalence relation, (i) aRa, (ii) aRb implies bRa, and (iii) aRb and bRc implies aRc. Using these basic axioms we can construct the equivalence classes.

aRd, so there is an equivalence class, call is S, containing a and d. eRa, so e is also in S (because eRaRc). cRe, so c is in S. Finally, bRc, so b is in S. This is the entire set, so there is only one equivalence class.

>> No.4923662
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4923662

I tried sketching the diagrams, is the top one correct or is the bottom one?

I don't know if aRb means that a has an arrow going to b AND b has an arrow going to a.

If the direction goes both ways, then they're part of the same equivalence class, right? So then is all of this one equivalence class?

>> No.4923667
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4923667

>>4923650
oh, sweet. So I am on the right track. Thank you so much!

>> No.4923675

>>4923662
Oh shit nigga, what are you doing?
xRy means x and y are in the same equivalence class and elements of the same equivalence class are indistinguible by the relation.

>> No.4923693

>>4923635
I have a professor but it's a mostly self-directed course. I'm not a math major and I'm maxed out in credits, I'm just taking some courses out of interest.

>> No.4923706

You have a set A and you put the elements in boxes (The equivalence classes).
If x and y are in the same box, you write xRy.

>> No.4923753

So then if I have to write all the partitions of {a,b,c}, the answer is

{{a,b,c}}, {{a,b},{c}}, {{a},{b,c}}, {{a},{b},{c}}

>> No.4923757

>>4923753
right?

>> No.4923767

>>4923757
You're missing {{a,c},{b}}

>> No.4923850

>>4923767
Awesome, thanks.

If I have ln : (0, inf) -> R, obviously it's injective.

For subjectivity, I can rewrite ln(x) = y as x=(e^y). Does that mean it's subjective, as e^y is positive so x can always be found in (0, inf)?

>> No.4923895

>>4923850
Surjective, and yes, it's a bijection.