implies that
x
≤
0
by Theorem 11.7 (set
y
= 0
in
the theorem).
x
≥
0
and
x
≤
0
means
x
= 0.
6. (a)

x

=

x

y
+
y
 ≤ 
x

y

+

y

. Therefore

x
  
y
 ≤ 
x

y


y

=

y

x
+
x
 ≤ 
y

x

+

x

=

x

y

+

x

. Therefore

y

x
 ≤ 
x

y

and

x

y
 ≥ 
x

y

.
That is

x

y
 ≤ 
x
  
y
 ≤ 
x

y

which implies that
vextendsingle
vextendsingle

x
  
y

vextendsingle
vextendsingle
≤ 
x

y

.
(b) By (a)

x
  
y
 ≤ 
x

y

. Therefore, if

x

y

< c
,
then

x
  
y

< c
so

x

< c
+

y

.
(c)

x

y
 ≥
0. If

x

y

< epsilon1
for every
epsilon1 >
0,
then

x

y

= 0
by Pb. 4.

x

y

= 0
implies
x
=
y
.
3
7. Use induction. We know

x
1
 ≤ 
x
1

and

x
1
+
x
2
 ≤ 
x
1

+

x
2

so the inequality holds for
n
= 1
, n
= 2. Assume that the inequality holds for any set of
k
real numbers. Then, for any
set of
k
+ 1
real numbers,

x
1
+
x
2
+
· · ·
+
x
k
+
x
k
+1

=

(
x
1
+
x
2
+
· · ·
+
x
k
)+
x
k
+1
 ≤ 
x
1
+
x
2
+
· · ·
+
x
k

+

x
k
+1
 ≤ 
x
1

+
x
2

+
· · ·
+

x
k

+

x
k
+1

R
B. (a) Assume that the positive integer
k
∈
S
. That is, assume
1 + 2 + 3 +
· · ·
+
k
=
(
k
+ 2)(
k

1)
2
Prove that
k
+ 1
∈
S
.
1 + 2 + 3 +
· · ·
+
k
+ (
k
+ 1)
=
(
k
+ 2)(
k

1)
2
+ (
k
+ 1)
=
(
k
2
+
k

2 + 2(
k
+ 1)
2
=
k
2
+ 3
k
2
=
(
k
+ 3)(
k
)
2
Therefore,
k
+ 1
∈
S
.
(b)
Since
1 + 2 + 3 +
· · ·
+
n
=
n
(
n
+ 1)
2
negationslash
=
(
n
+ 2)(
n

1)
2
for all
n
,
there is n
o positive
integer for which this equation holds.
C. (a) Show
S
is closed with respect to addition and multiplication.
(b)
1
2

5
√
2
=
1
2

5
√
2
·
2 + 5
√
2
2 + 5
√
2
=
2 + 5
√
2

46
=

1
23

5
46
√
2
4
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 Fall '08
 Staff
 Integers, Natural number, Prime number