d
)
2
+
b
2
.
(2.1.3)
Let
y
=
b
d
. Dividing (2.1.2) and (2.1.3) by
d
2
, we obtain
p
5
y
(
y
+
1
)
≥
(
y
+
1
)
2
+
y
2
and
p
5
y
≥
y
2
+
1.
(2.1.4)
The first of these inequalities simplifies to
p
5
(
y
2
+
y
)
≥
2
y
2
+
2
y
+
1, which yields
(
2

p
5
)
y
2
+
(
2

p
5
)
y
+
1
≤
0. Dividing by 2

p
5, which is negative, we obtain
y
2
+
y

(
2
+
p
5
)
≥
0.
Solving for the roots, we obtain
y
≤

3

p
5
2
or
y
≥
1
+
p
5
2
. On the other hand, the second
inequality in (2.1.4) can be written
y
2

p
5
y
+
1
≤
0, which yields
p
5

1
2
≤
y
≤
p
5
+
1
2
. Therefore
the only possibility is
y
=
1
+
p
5
2
, which is a contradiction since
y
is rational.
Case 2:
Similar to Case 1 and left to the reader.
Theorem 2.1.9.
p
5
is the best (i.e. largest) constant possible in Hurwitz Theorem.
In other
words, for every c
>
p
5
there exists an irrational number x such that there exist only finitely
many rational numbers
h
k
such that
x

h
k
<
1
ck
2
.
In fact x
=
1
+
p
5
2
satisfies this property for all
such c.
Proof.
Suppose there exists infinitely many rational
h
k
with
x
=
1
+
p
5
2
such that
h
k

x
<
1
ck
2
.
(2.1.5)
19
We will show that
c
≤
p
5. By the triangle inequality,
h
k

1

p
5
2
=
h
k

1
+
p
5
2
+
p
5
≤
h
k

1
+
p
5
2
+
p
5
<
1
ck
2
+
p
5.
(2.1.6)
If we multiply (2.1.5) and (2.1.6), then the left hand side is:
h
k

1
+
p
5
2
·
h
k

1

p
5
2
=
h
2
k
2

h
k

1
=
1
k
2
h
2

hk

k
2
≥
1
k
2
.
Therefore we obtain
1
k
2
<
1
ck
2
1
ck
2
+
p
5
. Multiplying by
ck
2
we obtain
c
<
1
ck
2
+
p
5. As
k
goes to infinity, the right side approaches
p
5. Therefore
c
≤
p
5 as desired.
2.2
Continued Fractions
Let
a
0
,
a
1
,... be real numbers. Assume that
a
i
>
0 for
i
>
0. Define
⟨
a
0
,
a
1
,...,
a
n
⟩
=
a
0
+
1
a
1
+
1
a
2
+
1
.
.
.
+
a
n

1
+
1
a
n
.
Example
3
.
⟨
2,3,4
⟩
=
2
+
1
3
+
1
4
=
2
+
4
13
=
30
13
.
Definition 2.2.1.
A
simple continued fraction
is a continued fraction
⟨
a
0
,
a
1
,...,
a
n
⟩
where
each
a
i
is an integer.
Clearly, a simple continued fraction is a rational number. The converse is also true:
Proposition 2.2.2.
Any rational number can be written as a simple continued fraction.
Proof.
We are given a rational number
p
q
in reduced form. If
q
=
±
1, then
p
/
q
=
±
p
=
⟨±
p
⟩
,
and we are done. Therefore suppose that

q
 6
=
1. We assume without loss of generality that
20
p
>
0. We let
u
0
=
p
and
u
1
=
q
. By the Euclidean Algorithm
u
0
=
u
1
a
0
+
u
2
with 0
≤
u
2
<

u
1

u
1
=
u
2
a
1
+
u
3
with 0
≤
u
3
<

u
2

.
.
.
u
j
=
u
j
+
1
a
j
.
Claim:
⟨
a
0
,
a
1
,...,
a
j
⟩
=
p
q
. Define
r
i
=
u
i
u
i
+
1
. Dividing the equation
u
1
=
u
i
+
1
a
i
+
u
i
+
2
by
u
i
+
1
,
we see that
r
i
=
a
i
+
1
r
i
+
1
. Starting from
r
0
=
a
0
+
1
/
r
1
, and applying this equation recursively
for
r
1
,
r
2
,..., until
r
j
=
a
j
, we see that So
r
0
=
⟨
a
0
,
a
1
,
a
2
,...,
a
j
⟩
. Since
r
0
=
p
/
q
, this proves
the desired result.
Note that there is a simple way to alter a given simple continued fraction representation
of a given rational number. Namely, suppose we have
x
=
⟨
a
0
,
a
1
,...,
a
j
⟩
with
a
j
>
1. Then we
also have
x
=
⟨
a
0
,
a
1
,...,
a
j

1,1
⟩
, since
(
a
j

1
) +
1
/
1
=
a
j
. To eliminate this ambiguity, we
impose the condition on our simple continued fractions that they do not end with 1 except for
the case 1
=
⟨
1
⟩
. We formalize this with the following definition:
Definition 2.2.3.
We say that a simple continued fraction
x
=
⟨
a
0
,
a
1
,...,
a
j
⟩
is in
simplest form
if
j
=
0 or if
j
>
0 and
a
You've reached the end of your free preview.
Want to read all 95 pages?
 Spring '14
 Math, Number Theory, Prime number, Rational number, kN, Continued fraction