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 October 27th, 2014, 02:10 PM #1 Newbie   Joined: Feb 2014 Posts: 15 Thanks: 0 Help with Equivalence Relations Let R be the relation on Z×Z, that is elements of this relation are pairs of pairs of integers, such that ((a,b),(c,d))∈R if and only if a+d=b+c. Show that R is an equivalence relation. So I need to show that the relation is reflexive, symmetric and transitive. For reflexive I need to show that when (a,b) E Z x Z. (a,b) = (a.b). How do I do this? I assume I have to somehow use a +d = b + c, but I'm not sure how to do it. Last edited by extreme112; October 27th, 2014 at 02:14 PM. Reason: Accidently Hit Enter
October 27th, 2014, 02:35 PM   #2
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Quote:
 Originally Posted by extreme112 Let R be the relation on Z×Z, that is elements of this relation are pairs of pairs of integers, such that ((a,b),(c,d))∈R if and only if a+d=b+c. Show that R is an equivalence relation. So I need to show that the relation is reflexive, symmetric and transitive. For reflexive I need to show that when (a,b) E Z x Z. (a,b) = (a.b). How do I do this? I assume I have to somehow use a +d = b + c, but I'm not sure how to do it.
You want to see if (a,b)R(a,b). So, c = a, d = b. Thus a + d = b + c becomes a + b = b + a. Is this true?

-Dan

October 27th, 2014, 03:10 PM   #3
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 Originally Posted by topsquark You want to see if (a,b)R(a,b). So, c = a, d = b. Thus a + d = b + c becomes a + b = b + a. Is this true? -Dan
Thanks, I think I worked it out for reflexive and symmetric, but how about transitive?

Last edited by skipjack; November 11th, 2014 at 11:59 PM.

October 27th, 2014, 03:16 PM   #4
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 Originally Posted by extreme112 Thanks, I think I worked it out for reflexive and symmetric, but how about transitive?
Given (a, b)R(c, d) and (c, d)R(e, f) we know that
(1) a + d = b + c
(2) c + f = d + e

From (2) we get d = c + f - e

Does this give you any ideas?

-Dan

Last edited by skipjack; November 11th, 2014 at 11:59 PM.

October 27th, 2014, 04:02 PM   #5
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Quote:
 Originally Posted by topsquark Given (a, b)R(c, d) and (c, d)R(e, f) we know that (1) a + d = b + c (2) c + f = d + e From (2) we get d = c + f - e Does this give you any ideas? -Dan
Well the only thing I can think of is that by definition of transitive if (a,b)R(c,d) and (c,d)R(e,f) then (a,b)R(e,f) to make an equation (3) a + f = b + e.

Edit: I think I got the answer.
If I plug in d = c+f-e into d for equation (1). Then I get equation (3). Thanks for the help.

Last edited by extreme112; October 27th, 2014 at 04:38 PM. Reason: I think I found the answer.

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