Register to reply 
Composition of Relations 
Share this thread: 
#1
Apr3012, 11:35 PM

P: 1,302

1. The problem statement, all variables and given/known data
R = { (1,2), (3,5), (2,2), (2,5) } S = { (2,1), (5,3), (5,1), (5,5) } Explicitly find the relation R^1 o S^1 2. Relevant equations 3. The attempt at a solution This was on my test. First I just wrote down the inverses: R^1 = { (2,1), (5,3), (2,2), (5,2) } S^1 = { (1,2), (3,5), (1,5), (5,5) } I didn't know what to do because the definition we learned defines 3 other sets, and all of the exercises in my test book has those 3 other sets defined. For example, there are usually sets A, B, and C along with the sets R and S. So I have no idea how I can apply the definition to do this. 


#2
May112, 07:25 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,565

[itex]R^{1}[/itex] also maps 2 to 2 so [itex]R^{1}oS^{1}[/itex] also maps 1 to 2 and contains the pair (1, 2). 


#4
May112, 09:55 AM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819

Composition of Relations



#5
May112, 10:02 AM

P: 1,302

I think the other three sets in my definition are A, B, and C and are dupposed to be the domain of R, the Range of R/domain of S, and the range of S.
Sound reasonable? 


#6
May112, 12:16 PM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819

The domain of R is {1,2,3}. The domain of S is {2,5}. etc. 


Register to reply 
Related Discussions  
Can you name this composition?  Materials & Chemical Engineering  6  
Gauss Composition? and a naive composition law  Linear & Abstract Algebra  3  
Relations bet. Groups, from Relations between Resp. Presentations.  Linear & Abstract Algebra  1  
Phase flow is the oneparameter group of transformations  Differential Geometry  6  
Beginner's mathematical proof / composition of relations  Calculus & Beyond Homework  11 