My Math Forum  

Go Back   My Math Forum > Math Forums > Math

Math General Math Forum - For general math related discussion and news


Reply
 
LinkBack Thread Tools Display Modes
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 ?
kendraca is offline  
 
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:
Originally Posted by kendraca View Post
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.
You haven't even copied these correctly! Your (a) and (b) are exactly the same! But "write down" what "as a subset of AXA"? I suspect that (a) asked you to "write down r as a subset of AXA" and (b) asked you to "write down s as a subset of AXA".

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:
(c) With rows and columns labelled in the order (-2, -1, 0, 1, 2) , write down the adjacency matrix of r.
Do you know what "adjacency matrix" of a relation is? The value in "row labeled a" and "column labeled b" is 1 if the pair (a, b) is in the relation, 0 if it is not.

Quote:
(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.
You can show that this is the product of the two previous matrices!

Quote:
(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?
Do you know what the "commutative property" is?


Quote:
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.
A relation that is a subset of UXU is "reflexive" if and only if, for every x in U, (x, x) is in the relation.

Quote:
(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.
A relation is "symmetric" if and only if whenever (x, y) is in the relation so is (y, x).

Quote:
(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.
A relation is "transitive" if and only if whenever (a, b) and (b, c) are in the relation so is (a, c). Notice that the second member of the first pair is the first member of the second pair.

Quote:
(d) As r is reflexive, symmetric and transitive it follows that r is an equivalence relation. What are the equivalence classes of r ?
An "equivalence class" is a subset, P, such that every member of P is "equivalent" to every other member of P. The equivalence classes for a given relation always "partition" the set- that is every member of the set is in one and only one equivalence class.

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.
Country Boy is offline  
Reply

  My Math Forum > Math Forums > Math

Tags
applied math, graphs, relations



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
relations hashom Computer Science 0 May 7th, 2014 04:10 AM
Relations rhymin Number Theory 8 March 30th, 2013 05:18 PM
Question on isomorph Graphs and their complement Graphs MageKnight Applied Math 0 January 17th, 2013 10:38 PM
Relations, R^2 and R^3 leafiann Applied Math 4 March 8th, 2011 02:50 AM
relations parmar Applied Math 2 December 10th, 2010 09:26 PM





Copyright © 2019 My Math Forum. All rights reserved.