1. ## relations proofs

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...

2. i guess the second one has to be proved by contradicton?

3. Originally Posted by doresa
sorry but hw to use mathematical symbols here...

CB

4. thanks

5. Originally Posted by doresa
suppose p is an equivalence relaiton.
i) $(x,y) \in \rho \Rightarrow [x]\rho = [y]\rho$
ii) $(x,y) \notin \rho \Rightarrow [x]\rho \cap [y]\rho = \{\phi\}$

sorry but how 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 $[x]\rho \subseteq [y]\rho$, and that $[x]\rho \supseteq [y]\rho$.

For ii) you are quite right, use contradiction. How far have you got with your proof?

6. by Swlabr;

For i), use transitivity to show that
$[x]\rho \subseteq [y]\rho$, and that $[x]\rho \supseteq [y]\rho$.

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....

7. (x,y) E p --> [x]p = [y]p

use transitivity to show that $[x]\rho \subseteq [y]\rho$, and that $[x]\rho \supseteq [y]\rho$.
i was lately thinking of this....
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.