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July 15th, 2016, 09:45 AM  #1 
Newbie Joined: Jul 2016 From: United States Posts: 1 Thanks: 0  Relations and Graphs
1. Let A= {2, 1, 0, 1, 2}. Let r be the relation defined by xry if and only if y=x . Let S be the relation defined by xsy if and only if y= x. (a) Write down as a subset of A X A . (b) Write down as a subset of AXA. (c) With rows and columns labelled in the order (2, 1, 0, 1, 2) , write down the adjacency matrix of r. (d) With rows and columns labelled in the order (2, 1, 0, 1, 2) , write down the adjacency matrix of s. (e) Using your answers to parts (c) and (d) calculate the adjacency matrix of the relation rs. (f) Using your answers to parts (c) and (d) calculate the adjacency matrix of the relation sr. (g) From the solutions to parts (e) and (f) which property, associated with multiplication on numbers, is lacking for the operation of composition for relations? 2. Let A = {(1,2), (2,4), (3,6), (1,4), (2,, (3,12), (1,3), (2,6), (3,4)}. Let r by the relation defined by (a,b)r(c,d) if and only if ad = bc. (a) The relation r is reflexive. Give one example of two elements of r (not A ) that demonstrate the reflexive property. Show clearly that the elements you choose satisfy the reflexive property. (b) The relation r is symmetric. Give one example of two elements of r (not A) that demonstrate the symmetric property. Show clearly that the elements you choose satisfy the symmetric property. (c) The relation r is transitive. Give one example of three elements of r (not A) that demonstrate the transitive property. Show clearly that the elements you choose satisfy the transitive property. (d) As r is reflexive, symmetric and transitive it follows that r is an equivalence relation. What are the equivalence classes of r ? 
July 16th, 2016, 09:56 AM  #2  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
This is ridiculous! You list 11 problems with no attempt to solve them yourself! Why do you want these solved? If you want to learn this material, you have to do them. You don't learn by seeing someone else do them. If you can't do them, show us your attempt, what you do know about these, and where you ran into problem! Quote:
Okay, do you know what "AXA" means? AXA is the set of all ordered pairs with both members from A. Here, since there are only 5 members of A there are $\displaystyle 5^2= 25$ members. AXA is {(2, 2), (2, 1), (2, 0), (2, 1), (2, 2),(1, 2), (1, 1), (1, 0), (1, 1), (1, 2), (0, 2), (0, 1), (0, 0), (0, 1), (0, 2),(1, 2), (1, 1), (1, 0), (1, 1), (1, 2),(2, 2), (2, 1), (2, 0), (2, 1), (2, 2)} For (a) your relation should be the set containing all pairs (x, y) such that y= x that is, (x, x). Which pairs are those? There are 5 such pairs. For (b) your relation should be the set containing all pairs (x, y) such that y= x that is, (x, x). Which pairs are those? There are also 5 such pairs. (Do you see the difference between (a) and (b)? The pair (1, 1) is in (a) because y= 1= x. That pair is NOT in (b) because the absolute value of a number is never negative.) Quote:
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For example, the pair (1, 2) is equivalent to the pair (x, y), according to this definition, if and only if 1+ y= 2+ x. That is the same as y x+ 1, that is, the second member of the pair is 1 larger than the x. I see that (3, 4) is the only other pair here that satisfies that so one equivalence class is {(1, 2), (3, 4)}. (2, 4) is equivalent to (x, y) if and only if 2+ y= 4+ x so that y x= 2. Looking through the list of pairs I see that (1, 3) is the only other pair that has that property. Another equivalence class is {(1, 3), (2, 4)}. Etc. Last edited by Country Boy; July 16th, 2016 at 09:59 AM.  

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