CS 381 Solutions to Homework 11

Textbook pp. 413 - 414:

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

6. 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 $f(x) = f(y)$.

8. 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.

24 a) $\{ (x,3x) \mid$ $x$ is a positive integer}

26
a) is a partition because an integer is even or odd and because no integer is even and odd at the same time.
c) is a partition because an integer is divisible by 3 or leaves 1 or 2 as the remainder when divided by 3, and because no integer has any two of those properties at the same time.

Textbook pp. 428 - 429:

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

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

24 a) l and m
b) a, b and c
c) No
d) No
e) k, l and m
f) k
g) None
h) None