MATH 290 – HOMEWORK 6 – SOLUTIONS
1. Define the relation
R
on the integers
Z
by
xRy
if and only if either
x

y
or
y

x
.
(a) Prove that the relation
R
is not transitive. (Hint: To show it is not transitive, you must find
three integers
x
,
y
, and
z
such that
xRy
and
yRz
but
x
6
Rz
.)
(b) For each
n
∈
Z
, define
A
n
to be the set of integers related to
n
, that is,
A
n
=
{
x
∈
Z
:
xRn
}
.
Since
R
is symmetric we also have
A
n
=
{
x
∈
Z
:
nRx
}
. Describe explicitly the sets
A
20
and
A
0
. This description can be done by listing the set, or by giving a precise description of the
set in words.
(c) Prove that
\
n
∈
Z
A
n
=
{
1
,
0
,
1
}
.
(Hint: To prove the inclusion “
⊆
” you might want to try proof by contrapositive, that is,
show that if
m /
∈ {
1
,
0
,
1
}
then
m /
∈
T
n
∈
Z
A
n
.)
Solution:
(a). Let
x
= 5,
y
= 10, and
z
= 2. Since 5

10,
xRy
and since 2

10,
yRz
. But since 2
does not divide 5 and 5 does not divide 2, 2
6
R
5 and hence
R
is not transitive.
(b).
The set
A
20
consists of all multiples of 20, that is, all
m
∈
Z
such that 20

m
,
together with the factors of 20 and their negatives, that is, all
m
∈
Z
such that
m

20.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Alligood,K
 Advanced Math, Integers, Rational number, equivalence class, xRy

Click to edit the document details