math proof, equivalence relations

[boreal jetty]

could i go about this using a proof by contradiction where i say [a] ≠ [b] and [a]∩[b] ≠ ∅?

[stone rose]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[fleet pelican]

I think its better to reason using double inclusion if the intersection is non empty then it has one element at the very least using this you can prove the equality

use the fact that the equivalence relation is transitive

[boreal jetty]

hmm.. this is my first proofs class and we havent covered double inclusion yet

[fleet pelican]

well here take an element in [a] and prove that its in [b]

then take an element of [b] and prove its in [a]

(using the fact that the intersection is non empty so it has at least one element)

[onyx dagger]

When you have to prove that either P or Q is true, show that !P -> Q. In this case, show that if [a]intersect[b] != O then [a] = [b]

Spoiler:|| since the intersection is not the empty set then there is an element in both [a] and [b], lets call it c.
c is in the class [a] and in the class [b] at the same time. Ok, and?
That is to say, c ~ a and c ~ b at the same time!
But since equivalnce relations are transitive ( if x~ y and y~z then x~ z) then a ~ b, that is to say that a and b are in the same class, that is to say that [a]=[b].
This concludes that when P is not true then Q has to be true||

Think im right

Not sure if this argument actually shows that P or Q without P and Q both true

Oh my bad

I think you just have to prove that if P and Q are both true then it's false

Which should go similar to my proof above I think

[boreal jetty]

so if an element is in a class, that means it is also related to that class? for example can i always say if a ∈ b, then a ~ b ?

also correct me if im wrong but i thought transitivity worked as:
if a~b and b~c, then a~c

can you apply that to c~a and c~b?

[last ore]

Bobby was being handwavy and mixing symmetry in, which is fine since ~ is an equivalence relation