how to prove this? (a book of set theory Charles. C. Pinter exercise 3.3 number3-b
)
Given ,
H = {[(a,b) , (c,d)] : (b-a) = (d-c)}
We know that a relation R={(a,b): a is related to b} is said to be an equivalence relation if ,
(i) R is reflexive. i.e. a€R => (a,a)€R
(ii) R is symmetric. i.e. (a,b)€R => (b,a)€R
(iii) R is transitive i.e. (a,b)€R and (b,c)€R => (a,c)€R.
Here, we have,
H={[(a,b) , (c,d)] : (b-a)=(d-c)}
1. REFLEXIVITY:
Let (a,b)€R
we can write ,
(b-a) = (b-a)
=> [(a,b) , (a,b)] € H
i.e. (a,b) € H => [(a,b), (a,b)] € H
Thus , H is reflexive.
2. SYMMETRY:
Let [(a,b) , (c,d)] € H
=> (b-a) = (d-c)
=> (d-c) = (b-a)
=> [(c,d) , (a,b)] € H
i.e. [(a,b) , (c,d)] € H => [(c,d) , (a,b)] € H.
Thus, H is symmetric.
3. TRANSITIVITY:
Let [(a,b) , (c,d)] € H and [(c,d) , (e,f)] € H.
=> (b-a) = (d-c) ........... (i)
and (d-c) = (f-e) ............. (ii)
So, from (i) and (ii) , we get ,
( b-a) = (f-e)
=> [(a,b) , (e,f)] € H
i.e. [(a,b) , (c,d)] € H and [(c,d) , (e,f)] € H => [(a,b) , (e,f)] € H.
Thus, H is transitive.
Hence , From (1) , (2) and (3) , we can say that H is an equivalence relation.
. PROVED!
how to prove this? (a book of set theory Charles. C. Pinter exercise 3.3 number3-b ) Find the equivalence lelatio a)...
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