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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. Tags equivalence, relations Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Jet1045 Abstract Algebra 5 May 19th, 2013 01:03 PM bvh Advanced Statistics 2 April 10th, 2013 09:18 PM remeday86 Applied Math 1 June 13th, 2010 12:10 AM shine123 Number Theory 1 December 31st, 1969 04:00 PM Jet1045 Number Theory 0 December 31st, 1969 04:00 PM

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