MAT 337 Problem Set 1 Sample Solutions
Ian Zwiers  February 10, 2009
This document is for demonstration purposes; it is not a marking key.
2.4 A(f)
Compute the limit
lim
n
→∞
√
n
2
+
n

n
; then use
= 10

6
to find integer
N
that satisfies
the defition.
Multiplying by the conjugate,
√
n
2
+
n

n
=
n
√
n
2
+
n
+
n
=
1
√
1+
1
n
+1
. Thus,
1
2
=
1
√
1 + 1
=
1
q
1 + lim
n
→∞
1
n
+ 1
=
1
lim
n
→∞
q
1 +
1
n
+ 1
= lim
n
→∞
p
n
2
+
n

n.
Since
1
n
>
0,
p
n
2
+
n

n

1
2
=
1
2

1
q
1 +
1
n
+ 1
<
1
2

1
(
1 +
1
n
)
+ 1
=
1
2 (2
n
+ 1)
.
Note that
1
2(2
n
+1)
<
10

6
whenever
10
6
4

1
2
< n
. Thus, for
= 10
6
,
N
=
10
6
4
satisfies the
definition.
2.5 (I)
Show the set
S
=
n
+
m
√
2 :
m, n
∈
Z
is dense in
R
.
Suppose that
n
+
m
√
2 =
n
0
+
m
0
√
2.
Then if
m
6
=
m
0
,
√
2 =
n

n
0
m
0

m
∈
Q
, which is false.
Therefore
m
=
m
0
⇒
n
=
n
0
. This proves that,
n
b
m
√
2
c
+
m
√
2
m
∈
Z
+
} ⊂
S
∩
[0
,
1]
has an infinite number of elements.
Let
>
0 be arbitrary.
We may divide [0
,
1] into
N
=
d
1
e
subintervals
I
1
, I
2
, . . . I
N
, each
with length no more than
. We have proven
S
∩
[0
,
1] is infinite, so there are at least
N
+ 1
elements of
S
in [0
,
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 Spring '09
 subinterval Ij, Ian Zwiers

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