# Relations SoR

1. Mar 1, 2016

### YamiBustamante

1. The problem statement, all variables and given/known data
Suppose that A = { 1, 2, 3} , B = { 4, 5, 6} , R = { (1, 4), (1, 5), (2, 5), (3, 6)} ,
and S = { (4, 5), (4, 6), (5, 4), (6, 6)}. Note that R is a relation from A to B and S is a relation from B to B . Find the following relations:
(a) S ◦ R .

(b) S ◦ S−1 .

2. Relevant equations
S◦R = {(a,c) ∈ (AXC) : ∃ b∈B ((a,b)∈R and (b,c)∈S)}

3. The attempt at a solution
I'm having trouble understanding what a composition relation is. I know you have to the path that connects R and S, but other than that, I don't understand it. This is my first example of a composition relation, so I have little to no prior knowledge in writing S ◦ R . I tried searching for examples online, but I can't find any. My textbook doesn't even cover any examples. I even attempted solving with a picture and "connecting the dots" just like teacher did to demonstrate what S◦R means, but it didn't help as much.
Here's what I have.

a) So I got that S ◦ R = {(1,5),(1,6),(2,4),(3,6)}
The answer is SoR = {(1,5), (1,6), (1,4), (2,4), (3,6)}
But I don't understand how they got (1,4)

b) So for the inverse of S = {(5,4),(6,4),(4,5),(6,6)}
I got that S ◦ S−1 = {(5,5), (4,6), (4,5), (6,6)}
(b)SS1 = {(5,5), (5,6), (6,5), (6,6), (4,4)}
So, I'm completely wrong in this one except for one.

Can someone please walk me through how to get to the answers and why they are correct? Please don't give ambiguous hints.

2. Mar 1, 2016

### YamiBustamante

NEVERMIND I FIGURED IT OUT!

3. Mar 2, 2016

### HallsofIvy

Staff Emeritus
For others who saw this and wondered- R maps 1 to both 4 and 5 (because it contains (1, 4) and (1, 5)) while S maps 4 into 4 and 6 and maps 5 into 4 so SR maps 1 to 4 and 6- it contains (1, 4) and (1, 6). R maps 2 to 5 and S maps 5 to 4. SR maps 2 to 4- it contains (2, 4). Finally, R maps 3 to 6 and S maps 6 to 6 so SR maps 3 to 6. SR= {(1, 4), (1, 6), (2, 4), (3, 6)}.

$S^{-1}$ (not "S- 1") reverses the pairs in S: $S^{-1}$= {(5, 4), (6, 4), (4, 5), (6,6)}. Of course, then, $S^{-1}$ maps 5 to 4 and S maps 4 to 5 so $SS^{-1}$ maps 5 to 5 Similarly, $S^{-1}$ maps 6 to 4 and S maps 4 to 6 so $SS^{-1}$ maps 6 to 6, etc. In fact, $SS^{-1}$ is the identity on B, {(4, 4), (5, 5), (6, 6)}, by definition of "inverse function"!

Last edited by a moderator: Mar 2, 2016