This preview shows pages 1–2. Sign up to view the full content.
MATH 290 – HOMEWORK 4 – SOLUTIONS
1. For each natural number
n
, let
M
n
=
{
k
∈
Z
:
n

k
}
. (In other words, for each natu
ral number
n
,
M
n
=
{
...,

3
n,

2
n,

n,
0
, n,
2
n,
3
n, .
..
}
.) Prove that
∞
[
n
=1
M
n
=
Z
and
∞
\
n
=1
M
n
=
{
0
}
. (Hint: It is not suﬃcient to just explain why the given identities are true,
you have to supply an actual proof.)
Solution:
(a). To see that
∞
[
n
=1
M
n
=
Z
we must show inclusion both ways. To see that
∞
[
n
=1
M
n
⊆
Z
note that each
M
n
⊆
Z
so that if
x
∈
M
n
for some
n
, then
x
∈
Z
.
To see that
Z
⊆
∞
[
n
=1
M
n
, note that since 1 divides every integer,
Z
⊆
M
1
(in fact
Z
=
M
1
but we don’t need that here), and so
Z
⊆
M
1
⊆
∞
[
n
=1
M
n
. Therefore
∞
[
n
=1
M
n
=
Z
.
(b). To see that
∞
\
n
=1
M
n
=
{
0
}
we again must show inclusion both ways. To see
that
{
0
} ⊆
∞
\
n
=1
M
n
, note that since every natural number divides 0, that 0
∈
M
n
for all natural numbers
n
. Hence
{
0
} ⊆
∞
\
n
=1
M
n
. To see that
∞
\
n
=1
M
n
⊆ {
0
}
, let
k
be a nonzero integer. We know that if
m
is a natural number such that
m

k
then
m
≤ 
k

. Therefore letting the natural number
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.
This note was uploaded on 03/05/2011 for the course MATH 290 taught by Professor Alligood,k during the Fall '08 term at George Mason.
 Fall '08
 Alligood,K
 Math, Advanced Math

Click to edit the document details