Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (curry):

Question regarding finding the equivalence class.

OpenStudy (curry):

OpenStudy (curry):

I am having trouble finding the correspoinding equivalence class. :/ How do i find it for this problem?

OpenStudy (curry):

@dan815 @ganeshie8

OpenStudy (anonymous):

you showed it was an equivalence relation right?

OpenStudy (anonymous):

so what, for example, is in the equivalence class of \(2\)?

OpenStudy (curry):

and when I'm proving that it's a equivalence relation, here was my work. can i have some validation here?

OpenStudy (curry):

So wouldn't the equivalence class for 2 be -2 and +2?

OpenStudy (anonymous):

yes

OpenStudy (curry):

so for any integer n, it'd be, -n and +n?

OpenStudy (anonymous):

in fact all equivalence classes have to elements, except 0 which is in its own class

OpenStudy (curry):

what would be the correct notation to write that all?

OpenStudy (anonymous):

idk depends on how you write them

OpenStudy (anonymous):

\[\{a,-a\}\] maybe

OpenStudy (curry):

well usually, in class they said, {elements} << that was generally how they expressed it. but for htis case, i'm not too sure how to write it.

OpenStudy (curry):

and a = N?

OpenStudy (anonymous):

i do no like your proof however, you have the if and then in the wrong place

OpenStudy (curry):

oo! kk, i'll go back and edit that. How should i make it better?

OpenStudy (anonymous):

you want to show it is reflexive meaning \(aRa\) you do not assume \(aRa\) you prove \(aRa\) i.e by saying "because |a|=|a|, it is true that \(aRa\) so R is reflexive

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!