CS 381 Solutions to Homework 11

Textbook pp. 615 - 616:
2
a) and b) are equivalence relations, but
c) - e) are not because they are not transitive.

10.
Since R is an equivalence relation, equivalence classes exist among the elements of A. Let f be a mapping from A to the set of equivalence classes, that is f(x) = [x], where [x] is the equivalece class that has x as a member. Then f is a function because for every element x of A, a unique [x] exists. Also (x,y) is in R if and only if y is in [x], and y is in [x] if and only if [x] = [y]. Hence (x,y) is in R if and only if f(x) = f(y).

16.
R is reflexive, because ((a,b), (a,b)) is in R since ab = ba.
It is symmetric because ((a,b), (c,d)) is in R if and only if ((c,d), (a,b)) is in R since ad = bc if and only if cb = da.
It is transitive. For if ((a,b), (c,d)) is in R and ((c,d), (e,f)) is in R, then ad = bc and cf = de. Hence c = de/f. Hence ad = bc = bde/f. Hence a = be/f. Hence af = be. Hence ((a,b), (e,f)) is in R.

40 a) { (x, 5x/3) | x is a positive integer multiple of 3}

44
b) No, because 0 is excluded.
e) No, because the set of integers not divisible 3 and the set of even integers are not disjoint. For example 2 is in both sets.

Textbook pp. 630 - 631:
10.
Not a partial order because it is not transitive. The arrow (c,b) is missing.

22 c) The vertices are 2, 3, 5, 10, 11, 15, 25 and the arcs are (2,10), (3,15), (5,10), (5,15) and (5,25).

34. a) 27, 48, 60 and 72
b) 2 and 9
c) No greatest element
d) No least element
e) 18, 36 and 72
f) 18
g) 2, 4, 6 and 12
h) 12