Solutions to Assignment
Section 2.1
#4 Solution 1
Consider the real number
a
°
(
a
+ (
±
1))
:
We have
a
°
(
a
+ (
±
1))
=
(
a
°
a
) + (
a
°
(
±
1))
;
by distributive law
=
a
+ (
±
a
)
;
by our assumption
a
°
a
=
a
=
0
Hence by Theorem 2.1.3 (b), we have
a
= 0
or
a
+ (
±
1) = 0
:
In other words,
a
= 0
or
a
= 1
:
#4 Solution 2
Suppose
a
6
= 0
:
Then
9
1
a
2
R
such that
a
°
1
a
= 1
:
Hence
1 =
a
°
1
a
= (
a
°
a
)
°
1
a
=
a
°
(
a
°
1
a
) =
a
°
1 =
a:
In other words,
a
= 1
:
(Technically we don°t need to verify that
0
and
1
are solutions to the equation
a
°
a
=
a
, although this is very easy to do.)
#8 (a)
We can write
x
and
y
in terms of fractions
x
=
a
b
;
and
y
=
c
d
;
such that
a; b; c; d
2
Z
;
and
bd
6
= 0
:
then using the properties of real numbers and the fact that
Z
is closed under
+
and
°
;
we see that
x
+
y
=
ad
+
bc
bd
2
Q
xy
=
ac
bd
2
Q
as well. Hence
Q
is closed under
+
and
°
:
#8 (b)
Suppose
y
is irrational. Note by the same argument as above,
Q
is closed under taking additive
inverse, and
Q
nf
0
g
is closed under taking multiplicative inverse. So if
x
2
Q
and
(
x
+
y
)
2
Q
;
then
(
±
x
)
2
Q
and
y
= (
x
+
y
) + (
±
x
)
2
Q
;
which is a contradiction. If
x
6
= 0
; x
2
Q
and
xy
2
Q
;
then
1
x
2
Q
;
and
y
= (
xy
)
°
(
1
x
)
2
Q
;
which is a contradiction.
#20 (a)
By Theorem 2.1.7 (c), the condition
0
< c <
1
;
if the inequalities are multiplied by
c;
means
0
< c
2
< c:
Hence
0
< c
2
< c <
1
by Theorem 2.1.7 (a).
#20 (b)
Again,
1
< c
leads to
c < c
2
;
and hence
1
< c < c
2
:
Section 2.2
#4
Here it is implicit that
" >
0
;
otherwise two statements are trivially true and hence they imply each
other. By Theorem 2.2.2 (c),
j
x
±
a
j
< "
if and only if
±
" < x
±
a < ":
By Theorem 2.1.7 (b), this is equivalent to
a
±
" < x < a
+
";
by adding
a
or adding
(
±
a
)
to go back
and forth between the two statements.
#15
Let
"
=
j
a
°
b
j
2
:
If
x
2
V
"
(
a
)
\
V
"
(
b
)
;
then
j
x
±
a
j
< "
and
j
x
±
b
j
< ":
Hence
j
a
±
b
j
=
j
a
±
x
+
x
±
b
j ² j
a
±
x
j
+
j
b
±
x
j
<
j
a
±
b
j
2
+
j
a
±
b
j
2
=
j
a
±
b
j
;
which is a contradiction.
Hence
V
"
(
a
)
\
V
"
(
b
) =
;
:
QED. (Warning:
assuming
U
\
V
6
=
;
for all
neighborhoods of
a
and
b
does
not
mean a single element
x
belongs to all of them!)
1
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Section 2.3
#4
Note
1
=
2
²
1
±
(
°
1)
n
n
²
2
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 Spring '08
 BERTRANDGUILLOU
 Linear Algebra, Algebra, Order theory, Natural number, lower bound, upper bound, Archimedean

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