CS 381 Solutions to Homework 11

Textbook pp. 563 - 564:

2
a) Equivalence relation
c) Not transitive. Hence not an equivalence relation

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 $f(x) = [x]$. Then $f$ is a function because for every element $x$ of $A$, a unique $f(x)$ exists. Also $(x,y) \in R$ if and only if $[x] = [y]$. Hence $(x,y) \in R$if and only if $f(x) = f(y)$.

12. It is reflexive because any bit string completely agrees with itself.
It is symmetric because if a string x agrees with a string y everywhere except at the first three bits, then y agreew with x everywhere except at the first three bits.
It is transitive because if x agrees with y and y agrees wtih z everywhere except at the first three bits, respectively, then x agrees with z everywhere except at the first three bits.

40 a) { (x, 3x/2) | x is a positive even integer}


44
b) Not a partition because 0 is not included.
d) Yes, it is a partition.

Textbook pp. 578 - 580:

10. Not a partial order because it it not transitive. $(c,b)$ is missing.

20. The vertices are 0, 1, 2, 3, 4, 5 from the top and the arcs are (5,4), (4,3), (3,2), (2,1) and (1,0).

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