Let A = {1, 3, 4, 5} and B = {4, 5, 6, 9} and let S be the "divides" relation. Ordered pairs of s and s-1

Question

Let A = {1, 3, 4, 5} and B = {4, 5, 6, 9} and let S be the "divides" relation. That is,
for all (x, y) is an element of A x B, x S y <->x | y.
State explicitly which ordered pairs are in S and S –1.

Answer 1

Check every element of A to see what it divides in B.

1 divides 4, 5, 6, 9
3 divides 6, 9
4 divides 4
5 divides 5

S = { (1,4), (1,5), (1,6), (1,9), (3,6), (3,9), (4,4), (5,5) }

S-1, the inverse relation, is defined as the set of (y,x) such that xSy. That means take S and reverse all the ordered pairs:

S-1 = { (4,1), (5,1), (6,1), (9,1), (6,3), (9,3), (4,4), (5,5) }
or written in order:
S-1 = { (4,1), (4,4), (5,1), (5,5), (6,1), (6,3), (9,1), (9,3)}

It makes more sense, though, to write S^(-1) instead of S-1.

See:
http://en.wikipedia.org/wiki/Inverse_relation

Answer 2

The elements of S are as follows:

(1,4) (3,6) (4,4) (5,5)
(1,5) (3,9)
(1,6)
(1,9)

1 divides everything, that's the 1st column. 3 only divides 6 and 9, that's the second column and 4 and 5 only divide themselves, that's the 3rd and 4th columns.

Note that S is not a function since 1 and 3 are mapped to multiple outputs.

I am not sure about the "inverse" of S. My algebra is a little rusty, but I think the inverse is a map of B x A s.t. y divides x. If that is correct, then only (4,4) and (5,5) are elements of S inverse.

Any have more info on this part?

Answer 3

1 divides everything, so (1,4),(1,5),(1,6),(1,9) are in S
3 divides 6 and 9 from B add (3,6),(3,9) to S
4 divides only 4 from B: add (4,4) to S
5 divides only 5 from B: add (5,5) to S

Now, I guess (a,b) is in S-1 iff a S b and a<>b ?

Then S-1 = {(1,4), (1,5), (1,6), (1,9), (3,6), (3,9)}