purely periodic continued fractions

purely periodic continued fractions

We know that

periodic continued fractions represent quadratic irrationals

; this article characterizes

purely periodic continued fractions

. We will use freely the results on

convergents

to a

continued fraction

.

Theorem 1

.

(Galois) A quadratic

irrational

t

is represented by a purely periodic

simple continued fraction

if and only if

t

>

1

and its

conjugate

s

under the

transformation

d

↦

-

d

satisfies

-

1

<

s

<

0

.

Proof.

Suppose first that

t

is represented by a purely periodic continued fraction

t

=

[

a

0

,

a

1

,

…

,

a

r

-

1

¯

]

.

Note that

a

0

≥

1

since it appears again in the continued fraction. Thus

t

>

1

. The

r

th

complete convergent is again

t

, so that we have

t

=

p

r

-

2

+

t

⁢

p

r

-

1

q

r

-

2

+

t

⁢

q

r

-

1

so that

q

r

-

1

⁢

t

2

+

(

q

r

-

2

-

p

r

-

1

)

⁢

t

-

p

r

-

2

=

0

Consider the

polynomial

f

⁢

(

x

)

=

q

r

-

1

⁢

x

2

+

(

q

r

-

2

-

p

r

-

1

)

⁢

x

-

p

r

-

2

.

f

⁢

(

t

)

=

0

, so the other root of

f

⁢

(

x

)

is the conjugate

s

of

t

. But

f

⁢

(

-

1

)

=

(

p

r

-

1

-

p

r

-

2

)

+

(

q

r

-

1

-

q

r

-

2

)

>

0

since the

p

i

and the

q

i

are both

strictly increasing

sequences

, while

f

⁢

(

0

)

=

-

p

r

-

2

<

0

. Thus

s

lies between

-

1

and

0

and we are done.

Now suppose that

t

>

1

and

-

1

<

s

<

0

, and let the continued fraction for

t

be

[

a

0

,

a

1

,

…

]

. Let

t

n

be the

n

th

complete convergent of

t

, and

s

n

=

t

n

¯

. Thus

s

0

=

s

. Then

t

=

t

0

=

a

0

+

1

t

1

so that

s

0

=

t

0

¯

=

a

0

+

1

t

1

¯

=

a

0

+

1

s

1

and thus

1

s

1

=

-

a

0

+

s

0

<

-

a

0

≤

-

1

so that

-

1

<

s

1

<

0

. Inductively, we have

-

1

<

s

n

=

t

n

¯

<

0

for all

n

≥

0

. Suppose now that the continued fraction for

t

is not purely periodic, but rather has the form

t

=

[

a

0

,

a

1

,

…

,

a

k

-

1

,

a

k

,

a

k

+

1

,

…

,

a

k

+

j

-

1

¯

]

for

k

≥

1

. Then

t

k

=

t

k

+

j

and so

t

k

-

1

-

t

k

+

j

-

1

=

(

a

k

-

1

+

1

t

k

)

-

(

a

k

+

j

-

1

+

1

t

k

+

j

)

=

a

k

-

1

-

a

k

+

j

-

1

But

a

k

-

1

≠

a

k

+

j

-

1

, otherwise

a

k

-

1

would have been the first element of the repeating period. Thus

t

k

-

1

-

t

k

+

j

-

1

is a nonzero

integer

and thus

s

k

-

1

-

s

k

+

j

-

1

is as well. But

-

1

<

s

k

-

1

-

s

k

+

j

-

1

<

1

, which is a

contradiction

. Thus

k

=

0

and the continued fraction is purely periodic.
∎

References

1

A.M. Rockett & P. Szüsz,

Continued Fractions

, World Scientific Publishing, 1992.

Title

purely periodic continued fractions

Canonical name

PurelyPeriodicContinuedFractions

Date of creation

2013-03-22 18:04:44

Last modified on

2013-03-22 18:04:44

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 11Y65

Classification

msc 11A55

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