MATH 235 – Algebra 1
Solutions to Assignment 1
September 18, 2007
Solution of 1.
1. If
N
>
1:
N
[
n
=1
[

n,n
] = [

N,N
]
.
This is true because if
x
∈
S
N
n
=1
[

n,n
], then
x
∈
[

n,n
] for some 1
6
n
6
N
, and since
[

n,n
]
⊂
[

N,N
] we have
x
∈
[

N,N
]. To show the other inclusion, if
x
∈
[

N,N
] then
x
∈
[

n,n
] for
n
=
N
, so
x
∈
S
N
n
=1
[

n,n
].
N
\
n
=1
[

n,n
] = [

1
,
1]
.
This is true because if
x
∈
T
N
n
=1
[

n,n
], then
x
∈
[

n,n
] for all 1
6
n
6
N
, so, in particular,
x
∈
[

1
,
1]. To show the other inequality, if
x
∈
[

1
,
1],
x
∈
[

n,n
] for all 1
6
n
6
N
(since
[

1
,
1]
⊂
[

n,n
] for all such
n
), so
x
∈
T
N
n
=1
[

n,n
].
2.
∞
[
n
=1
[
n,n
+ 1] = [1
,
∞
)
.
This is true because if
x
∈
S
∞
n
=1
[
n,n
+ 1], then
x
∈
[
n,n
+ 1] for some
n
>
1, so
x
>
n
>
1,
and thus
x
∈
[1
,
∞
). To show the other inclusion, if
x
∈
[1
,
∞
), and we let
n
be the greatest
integer such that
n
6
x
, then
x
∈
[
n,n
+ 1], and
n
>
1 (since
x
>
1), so
x
∈
S
∞
n
=1
[
n,n
+ 1].
∞
[
n
=1
(
n,n
+ 2) = (1
,
∞
)
.
This is true because if
x
∈
S
∞
n
=1
(
n,n
+ 2), then
x
∈
(
n,n
+ 2) for some
n
>
1, so
x > n
>
1,
i.e.
x
∈
(1
,
∞
). To show the other inclusion, if
x
∈
(1
,
∞
), and we let
n
be the greatest integer
such that
n < x
, then
x
∈
(
n,n
+ 2), and
n
>
1 (since
x >
1), so
x
∈
S
∞
n
=1
(
n,n
+ 2).
3.
∞
[
n
=1
A
n
=
A
1
=
N
.
Clearly
S
∞
n
=1
A
n
⊂
N
because every
A
i
is, by deﬁnition, a subset of
N
. To show the other
1
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x
∈
N
can be written as
x
n
for
n
= 1, i.e.
x
∈
A
1
, so
x
∈
S
∞
n
=1
A
n
.
∞
\
n
=1
A
n
=
{
0
,
1
}
.
Clearly 0 and 1 are in every
A
n
since 0 = 0
n
and 1 = 1
n
for every
n
>
1. So
{
0
,
1
} ⊂
T
∞
n
=1
A
n
.
We will prove the other direction by contradiction. Suppose that
x
∈
T
∞
n
=1
A
n
,
x
6∈ {
0
,
1
}
.
Then, in particular,
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 Fall '07
 Goren
 Algebra, Inverse function

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