suppose p is an equivalence relaiton.
i) (x,y) E p --> [x]p = [y]p
ii) (x,y) not E p --> [x]p intersection [y] = null
sorry but hw to use mathematical symbols here...
I presume that's what you mean (see my quote)...I can't quite rememeber all the formalitites surrounding the notation of equivalence relations and sets...anyway, for the code, simply click on the LaTeX image. There are two of them, one for i) and one for ii). If you ever get stuck by LaTeX, just click on someone elses code that contains the characters you want to use.
For i), use transitivity to show that , and that .
For ii) you are quite right, use contradiction. How far have you got with your proof?
thanks im already done both.by Swlabr;
For i), use transitivity to show that
, and that .
For ii) you are quite right, use contradiction. How far have you got with your proof?
for the first one its the same way as you told. first assume (x,y) E p and let z be an arbitary elment belongs to [x]p then using first assumption and the transitivity we can show that z E [y]p and the other way round.
for the second one its quite simple when use p --> q = ~q --> ~p
thats all....
i was lately thinking of this....(x,y) E p --> [x]p = [y]p
use transitivity to show that , and that .
why cant i do it in one proof? like this
suppose (x,y) E p
let z be arbitrary and suppose
z E [x]p <--> (x,z) E p
<--> .........
like wise by using iff then u dont have to split it into 2 parts,
because i dont see any wrong in it.
thanks in advance..