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Notes on Homework #2
Exercise 1.4.3
We are given real numbers
a < b
and we have to show that there exists an irrational number
t
with
a < t < b
. Let
a
0
=
a
=
√
2 and
b
0
=
b
=
√
2. Since
a
0
< b
0
we can apply Theorem 1.4.3 to justify
saying that there is a rational number
r
with
a
0
< r < b
0
, in other words
a

√
2
< r < b

√
2. Adding
√
2
to both sides we have that
a < r
+
√
2
< b.
Now
t
=
r
+
√
2 is the sum of rational and irrational numbers, so by Exercise 1.4.2
t
is irrational. Thus there
does exist an irrational number
t
with
a < t < b
.
Exercise 2.2.1 b)
First, observe that
±
±
±
±
3
n
+ 1
2
n
+ 5

3
2
±
±
±
±
=
±
±
±
±
2(3
n
+ 1)

3(2
n
+ 5)
2(2
n
+ 5)
±
±
±
±
=
±
±
±
±

13
4
n
+ 10
±
±
±
±
=
+13
4
n
+ 10
.
(The absolute value is important — without it you get an expression which is negative and so will
always
be
< ±
.) From here we see that the
n
we want are those which are
n >
(13
±

1

10)
/
4. So we want to use
an
N
with
N >
(13
±

1

10)
/
4. Here is what the ﬁnished solution should look like:
Consider any
± >
0. By the Archimedian Property there exists a natural number
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This note was uploaded on 11/07/2009 for the course MATH 3224 at Virginia Tech.
 '08
 KBHANNSGEN
 Calculus, Real Numbers

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