Continued Fractions and the Euclidean Algorithm
Lecture notes prepared for MATH 326, Spring 1997
Department of Mathematics and Statistics
University at Albany
William F. Hammond
Table of Contents
 1 Introduction... *

2 The continued fraction expansion of a real number... *

3 First examples... *

4 The case of a rational number... *

5 The symbol [t_{1}, t_{2}, …, t_{r}]... *

6 Application to Continued Fractions... *

7 Bezout's Identity and the double recursion... *

8 The action of GL_{2}(Z) on the projective line... *

9 Periodic continued fractions... *

References ... *
1. Introduction
Continued fractions offer a means of concrete representation for
arbitrary real numbers. The continued fraction expansion of a real
number is an alternative to the representation of such a number as
a (possibly infinite) decimal.
The reasons for including this topic in the course on Classical Algebra
are:
2. The continued fraction expansion of a real number
Every real number x is represented by a point on the real line
and, as such, falls between two integers. For example, if n is
an integer and
x falls between n and n+1, and there is one and only one
such integer n for any given real x. In the case where x
itself is an integer, one has n = x. The integer n is
sometimes called the floor of x, and one often introduces
a notation for the floor of x such as
Examples:
For any real x with n = [x] the number
u = x  n falls in the unit interval
I consisting of all real numbers u for which 0 <= u < 1.
Thus, for given real x there is a unique decomposition
where n is an integer and u is in the unit interval.
Moreover, u = 0 if and only if x is an integer. This
decomposition is sometimes called the mod one decomposition
of a real number. It is the first step in the process
of expanding x as a continued fraction.
The process of finding the continued fraction expansion of a real
number is a recursive process that procedes one step at a time.
Given x one begins with the mod one decomposition
where n_{1} is an integer and 0 <= u_{1} < 1.
If u_{1} = 0, which happens if and only if x is an integer,
the recursive process terminates with this first step. The idea
is to obtain a sequence of integers that give a precise determination
of x.
If u_{1} > 0, then the reciprocal 1/u_{1} of u_{1} satisfies
1/u_{1} > 1 since u_{1} is in I and, therefore, u_{1} < 1.
In this case the second step in the recursive determination of the
continued fraction expansion of x is to apply the mod one
decomposition to 1/u_{1}. One writes
1/u_{1} = n_{2} + u_{2} ,

where n_{2} is an integer and 0 <= u_{2} < 1. Combining the
equations that represent the first two steps, one may write
x = n_{1} + {1}/{n_{2} + u_{2}} .

Either u_{2} = 0, in which case the process ends with the
expansion
x = n_{1} + {1}/{n_{2}} ,

or u_{2} > 0. In the latter case one does to u_{2} what had just
been done to u_{1} above under the assumption u_{1} > 0.
One writes
1/u_{2} = n_{3} + u_{3} ,

where n_{3} is an integer and 0 <= u_{3} < 1. Then combining
the equations that represent the first three steps, one may write
x = n_{1} + {1}/{n_{2} + {1}/{n_{3} + u_{3}}} .

After k steps, if the process has gone that far, one has integers
n_{1}, n_{2}, …, n_{k} and real numbers
u_{1}, u_{2}, …, u_{k} that are members of the unit interval
I with u_{1}, u_{2}, …, u_{k1} all positive.
One may write
x = n_{1} + {1}/{n_{2} + {1}/{n_{3} + {1}/{… + {1}/{n_{k} + u_{k}}}}} .

Alternatively, one may write
x = [ n_{1}, n_{2}, n_{3}, …, n_{k} + u_{k} ] .

If u_{k} = 0, the process ends after k steps. Otherwise,
the process continues at least one more step with
1/u_{k} = n_{k+1} + u_{k+1} .

In this way one associates with any real number x a sequence,
which could be either finite or infinite, n_{1}, n_{2}, …
of integers. This sequence is called the continued fraction
expansion of x.

Convention.
When [n_{1}, n_{2}, ...] is called a
continued fraction, it is understood that all of the numbers
n_{j} are integers and that n_{j} >= 1 for j >= 2.
3. First examples











1 + {1}/{2 + {1}/{{4}/{3}}} 



1 + {1}/{2 + {1}/{1 + {1}/{3}}} 






3 + {1}/{{1}/{SQRT{10}3}} 






3 + {1}/{6 + {1}/{{1}/{SQRT{10}3}}} 



3 + {1}/{6 + {1}/{SQRT{10}+3}} 



3 + {1}/{6 + {1}/{6 + {1}/{…}}} 









2 + {1}/{3 + {1}/{[5, 2]}} 



2 + {1}/{3 + {1}/{5 + {1}/{2}}} 



2 + {1}/{3 + {1}/{{11}/{2}}} 













Let
x = 1+{1}/{2+{1}/{3+{1}/{2+{1}/{3+{1}/{2+…}}}}} .

In this case one finds that
where
y = 2+{1}/{3+{1}/{2+{1}/{3+{1}/{2+…}}}} .

Further reflection shows that the continued fraction structure for
y is selfsimilar:
This simplifies to
and leads to the quadratic equation
with discriminant 60. Since y > 2, one of the two
roots of the quadratic equation cannot be y, and, therefore,
Finally,
The idea of the calculation above leads to the conclusion that any
continued fraction [n_{1}, n_{2}, …] that eventually repeats
is the solution of a quadratic equation with positive discriminant
and integer coefficients. The converse of this statement is also
true, but a proof requires further consideration.
4. The case of a rational number
The process of finding the continued fraction expansion of a rational
number is essentially identical to the process of applying the
Euclidean algorithm to the pair of integers given by its numerator and
denominator.
Let x = a/b, b > 0, be a representation of a rational number
x as a quotient of integers a and b. The mod one decomposition
{a}/{b} = n_{1} + u_{1} , u_{1} = {a  n_{1} b}/{b}

shows that u_{1} = r_{1}/b, where r_{1} is the remainder for
division of a by b. The case where u_{1} = 0 is the case
where x is an integer. Otherwise u_{1} > 0, and the mod one
decomposition of 1/u_{1} gives
{b}/{r_{1}} = n_{2} + u_{2} , u_{2} = {b  n_{2} r_{1}}/{r_{1}} .

This shows that u_{2} = r_{2}/r_{1}, where r_{2} is the remainder for
division of b by r_{1}. Thus, the successive quotients in Euclid's
algorithm are the integers n_{1}, n_{2}, … occurring in the
continued fraction. Euclid's algorithm terminates after a finite
number of steps with the appearance of a zero remainder. Hence, the
continued fraction expansion of every rational number is finite.

Theorem 1.
The continued fraction expansion of a real number is
finite if and only if the real number is rational.
Proof. It has just been shown that if x is rational, then
the continued fraction expansion of x is finite because its calculation
is given by application of the Euclidean algorithm to the numerator
and denominator of x. The converse statement is the statement
that every finite continued fraction represents a rational number.
That statement will be demonstrated in the following section.
5. The symbol [t_{1}, t_{2}, …, t_{r}]
For arbitrary real numbers t_{1}, t_{2}, …, t_{r}
with each t_{j} >= 1 for j >= 2
the symbol [ t_{1}, t_{2}, …, t_{r} ] is defined recursively
by:
(1) [t_{1},t_{2},…,t_{r} ] = t_{1}+{1}/{[t_{2},…,t_{r} ]} .

In order for this definition to make sense one needs to know that
the denominator in the righthand side of (1) is nonzero.
The condition t_{j} >= 1 for j >= 2 guarantees, in
fact, that [t_{2},…,t_{r} ] > 0, as one may prove using
induction.
It is an easy consequence of mathematical induction that the symbol
[t_{1}, t_{2}, …, t_{r}] is a rational number if each t_{j} is
rational. In particular, each finite continued fraction is a
rational number. (Note that the symbol [t_{1}, t_{2}, …, t_{r}]
is to be called a continued fraction, according to the convention of
the first section, only when each t_{j} is an integer.)
Observe that the recursive nature of the symbol [t_{1}, …, t_{r}]
suggests that the symbol should be computed in a particular case working
from right to left. Consider again, for example, the computation above
showing that [2, 3, 5, 2] = 81/35. Working from right to left one
has:





5+{1}/{[2]} = 5+{1}/{2} = {11}/{2} 



3+{1}/{[5, 2]} = 3+{2}/{11} = {35}/{11} 



2+{1}/{[3, 5, 2]} = 2+{11}/{35} = {81}/{35} 

There is, however, another approach to computing
[t_{1}, t_{2}, …, t_{r}].
Let, in fact, t_{1}, t_{2}, … be any (finite or infinite)
sequence of real numbers. One uses the double recursion
(2) p_{j} = t_{j} p_{j1} + p_{j2} , j >= 1 , p_{0} = 1 , p_{1} = 0

to define the sequence {p_{j}} , j >= 1. The double
recursion, differently initialized,
(3) q_{j} = t_{j} q_{j1} + q_{j2} , j >= 1 , q_{0} = 0 , q_{1} = 1

defines the sequence {q_{j}} , j >= 1.
Note that p_{1} = t_{1}, p_{2} = t_{1}t_{2} + 1, …
and q_{1} = 1, q_{2} = t_{2}, q_{3} = t_{2}t_{3} + 1,
… .
One now forms the matrix
(4) M_{j} =  (  
p_{j} 
q_{j} 
p_{j1} 
q_{j1} 
  ) 
 for j >= 0 .

Thus, for example,
It is easy to see that the matrices M_{j} satisfy the double recursion
(5) M_{j} = 
 M_{j1} , j >= 1

as a consequence of the double recursion formulas for the p_{j} and q_{j}.
Hence, a simple argument by mathematical induction shows that
This is summarized by:

Proposition 1.
For any sequence {t_{j}}, j >= 1 of
real numbers, if {p_{j}} and {q_{j}} are the sequences
defined by the double recursions (2) and (3), then one
has the matrix identity
(7)  (  
p_{r} 
q_{r} 
p_{r1} 
q_{r1} 
  ) 
 = 
 … 
 


for each integer r >= 1.

Corollary 1.
One has the identity
p_{r} q_{r1}  q_{r} p_{r1} = (1)^{r} for each
integer r >= 1.
Proof. The number p_{r} q_{r1}  q_{r} p_{r1} is the
determinant of the matrix M_{r}. From the formula (6)
the matrix M_{r} is the product of r matrix factors, each of
which has determinant 1. Since the determinant of the product
of matrices is the product of the determinants of the factors, it is
clear that det(M_{r}) = (1)^{r}.

Corollary 2.
One has the vector identity
for each integer r >= 1.
Proof. First recall (i) that the product of a matrix and a (column)
vector is defined by the relation
(ii) that the transpose of a matrix is the matrix whose rows
are the columns of the given matrix, and (iii) that the transpose
operation reverses matrix multiplication. One tranposes both sides of
the relation (7) to obtain:
(9)  (  
p_{r} 
p_{r1} 
q_{r} 
q_{r1} 
  ) 
 = 
 
 … 
 .

To this relation one applies the principle that the first column of
any 2 \times 2 matrix is the product of that matrix with the
column
in order to obtain the column identity (8).

Theorem 2.
For any sequence {t_{j}}, j >= 1 of
real numbers, if {p_{j}} and {q_{j}} are the sequences
defined by the double recursions (2) and (3), and
if t_{j} >= 1 for j >= 2, then the value of the symbol
[t_{1}, …, t_{r}] is given by the formula
(10) [t_{1}, t_{2}, …, t_{r}] = {p_{r}}/{q_{r}} for r >= 1 .

Proof. What is slightly strange about this important
result is that while the {p_{r}} and the {q_{r}} are
defined by the front end recursions, albeit double recursions,
(2) and (3), the symbol [t_{1}, …, t_{r}]
is defined by the back end recursion (1). The proof
begins with the comment that the righthand side of (10)
does not make sense unless one can be sure that the denominator
q_{r} ≠ 0. One can show easily by induction on r that
q_{r} >= 1 for r >= 1 under the hypothesis t_{j} >= 1
for j >= 2.
The proof proceeds by induction on r. If r = 1, the
assertion of the theorem is simply the statement t_{1} = p_{1}/q_{1},
and, as noted above, p_{1} = t_{1} and q_{1} = 1. Assume now
that r >= 2. By induction we may assume the correctness of
the statement (10) for symbols of length r1, and,
therefore, for the symbol [t_{2}, …, t_{r}]. That case of the
statement says that [t_{2}, …, t_{r}] must be equal to a/c ,
where by corollary 2
with
Now by (1)
[t_{1}, t_{2}, …, t_{r}] = t_{1} + {1}/{a/c} = t_{1} + {c}/{a} = {at_{1} + c}/{a} .

But by corollary 2 again

 = 
 
 
 =  (  
at_{1} + c 
bt_{1} + d 
a 
b 
  ) 
 
 = 
 .

Hence,
{p_{r}}/{q_{r}} = {at_{1} + c}/{a} = [t_{1}, t_{2}, …, t_{r}] .

6. Application to Continued Fractions
Recall that [n_{1}, n_{2}, … ] is called a continued fraction
only when each n_{j} is an integer and n_{j} >= 1 for j >=
2. The sequence n_{1}, n_{2}, … may be finite or
infinite. The symbol c_{r} = [n_{1}, n_{2}, …, n_{r}]
formed with the first r terms of the sequence, is called the
r^{th} convergent of the continued fraction.
Associated with a given sequence n_{1}, n_{2}, … are two sequences
p_{1}, p_{2}, … and q_{1}, q_{2}, … that are given,
according to the double recursions (2), (3) of the
previous section with t_{j} = n_{j}.

Proposition 2.
If [n_{1}, n_{2}, …] is a continued
fraction, then the integers p_{r} and q_{r} are coprime for each
r >= 1.
Proof. By Corollary 1 of the previous section
p_{r} q_{r1}  q_{r} p_{r1} = (1)^{r}. Hence, any positive
divisor of both p_{r} and q_{r} must divide the lefthand side of
this relation, and, therefore, must also divide (1)^{r}.

Proposition 3.
The difference between successive convergents
of the continued fraction [n_{1}, n_{2}, …] is given by
the formula
(11) c_{r}  c_{r1} = {(1)^{r}}/{q_{r} q_{r1}} for r >= 2 .

Proof. According to the theorem (formula 10) at the
end of the last section the convergent c_{r} is given by
c_{r} = {p_{r}}/{q_{r}} .

Hence,


{p_{r}}/{q_{r}}  {p_{r1}}/{q_{r1}} 



{p_{r} q_{r1}  p_{r1} q_{r}}/{q_{r} q_{r1}} 



{(1)^{r}}/{q_{r} q_{r1}} . 

(The last step is by Corollary 1 above.)

Remark 1.
The formula (11) remains true if
c_{r} = [t_{1}, …, t_{r}] where the t_{j} are real numbers
subject to the assumption t_{j} >= 1 for j >= 1.

Lemma.
The sequence {q_{j}} is a strictly
increasing sequence for j >= 2.
Proof. This is easily proved by induction from the recursive
definition (3) of the sequence.

Theorem 3.
If [n_{1}, n_{2}, …] is an infinite
continued fraction, then the limit
lim_{r > INFTY} {p_{r}}/{q_{r}}

always exists.
Proof. As one plots the convergents c_{r} on the line of real
numbers, one moves alternately right and left. The formula
(11) for the difference between successive convergents
elucidates not only the fact of alternate right and left movement but
also the fact that each successive movement is smaller than the one
preceding. Therefore, one has
c_{1} < c_{3} < c_{5} < … < c_{6} < c_{4} < c_{2} .

Since any strictly increasing sequence of positive integers
must have infinite limit, the seqence q_{j} q_{j1} has infinite
limit, and so the sequence of reciprocals 1/q_{j} q_{j1} must
converge to zero. Hence, the sequences of odd and evenindexed
convergents must have the same limit, which is the limit of the
sequence of all convergents.

Definition 1.
The limit of the sequence of convergents of an
infinite continued fraction is called the value of that
continued fraction.

Theorem 4.
If [n_{1}, n_{2}, …] is the continued
fraction expansion of an irrational number x, then
x = lim_{r > INFTY} {p_{r}}/{q_{r}} ;

that is, the value of the continued fraction expansion of a real
number is that real number.
Proof. For each r >= 1 the continued fraction expansion
[n_{1}, n_{2}, …] of x is characterized by the identity
(12) x = [n_{1}, n_{2}, …, n_{r} + u_{r}] ,

where u_{r} is a real number with 0 <= u_{r} < 1.
The sequences of p's and q's for the symbol
[n_{1}, n_{2}, …, n_{r} + u_{r}] agree with those for the symbol
[n_{1}, n_{2}, …, n_{r}] except for the r^{th} terms.
One has by (10)
[n_{1}, n_{2}, …, n_{r} + u_{r}] = {P_{r}}/{Q_{r}} ,

where by (3)





(n_{r} + u_{r}) q_{r1} + q_{r2} 

Hence,
Q_{r} = q_{r} + u_{r} q_{r1} .

Therefore, the displacement from c_{r1} to x is by (11)
{(1)^{r}}/{Q_{r} q_{r1}} = {(1)^{r}}/{(q_{r} q_{r1} + u_{r} q_{r1}^{2})} ,

which is in the same direction but of smaller magnitude than the
displacement from c_{r1} to c_{r}. Therefore, x must be larger
than every oddindexed convergent and smaller than every evenindexed
convergent. But since all convergents have the same limit, that limit
must be x.
7. Bezout's Identity and the double recursion
It has already been observed that the process of finding the continued
fraction expansion of a rational number a/b (b > 0), involves
the same series of long divisions that are used in the application of
the Euclidean algorithm to the pair of integers a and b. Recall
that at each stage in the Euclidean algorithm the divisor for the current
stage is the remainder from the previous stage and the dividend for the
current stage is the divisor from the previous stage, or, equivalently,
the dividend for the current stage is the remainder from the second
previous stage. The Euclidean algorithm may thus be viewed as a double
recursion that is used to construct the sequence of remainders. One
starts the double recursion with
r_{1} = a and r_{0} = b .

At the j^{th} stage one performs long division of r_{j2}
by r_{j1} to obtain the integer quotient n_{j} and the integer
remainder r_{j} that satisfies 0 <= r_{j} < r_{j1}. Thus,
(13) r_{j} = r_{j2}  n_{j} r_{j1} .

The Euclidean algorithm admits an additional stage if r_{j} > 0.
Since
0 <= r_{j} < r_{j1} < … < r_{2} < r_{1} < r_{0} = b ,

there can be at most b stages.
One may use the sequence of successive quotients n_{j} (j >= 1)
to form sequences {p_{j}} and {q_{j}}, as in the
previous section, according to the double recursions:
(14) p_{j} = n_{j} p_{j1}+p_{j2} , j >= 1 ; p_{0} = 1 , p_{1} = 0 .

(15) q_{j} = n_{j} q_{j1}+q_{j2} , j >= 1 ; q_{0} = 0 , q_{1} = 1 .

It has already been observed that q_{j} >= 1 for j >= 1 and
[n_{1}, n_{2}, …, n_{j}] = {p_{j}}/{q_{j}} , j >= 1 .

Bezout's Identity says not only that the greatest common divisor of a
and b is an integer linear combination of them but that the coefficents
in that integer linear combination may be taken, up to a sign, as q
and p.

Theorem 5.
If the application of the Euclidean algorithm to
a and b (b > 0) ends with the m^{th} long division,
i.e., r_{m} = 0, then
(16) r_{j} = (1)^{j1} 
 , 1 <= j <= m .

Proof. One uses induction on j. For j = 1 the statement
is r_{1} = q_{1} a  p_{1} b. Since by (14, 15)
q_{1} = 1 and p_{1} = n_{1}, this statement is simply the
case j = 1 in (13). Assume j >= 2, and that
the formula (16) has been established for indices smaller than
j. By (13) one has
r_{j} = r_{j2}  n_{j} r_{j1} .

In this equation one may use (16) to expand the terms
r_{j2} and r_{j1} to obtain:


 {  (1)^{j3}(q_{j2}a  p_{j2}b)  } 
  n_{j}  {  (1)^{j2}(q_{j1}a  p_{j1}b)  } 
 



 {  (1)^{j1}(q_{j2}a  p_{j2}b)  } 
 + n_{j}  {  (1)^{j1}(q_{j1}a  p_{j1}b)  } 
 



(1)^{j1}  {  (q_{j2}a  p_{j2}b) + n_{j} (q_{j1}a  p_{j1}b)  } 
 



(1)^{j1}  {  (q_{j2} + n_{j} q_{j1})a  (p_{j2} + n_{j} p_{j1})b  } 
 





Corollary 3.
The greatest common divisor d of a and b
is given by the formula
(17) d = (1)^{m} (q_{m1}a  p_{m1} b) ,

where m is the number of divisions required to obtain zero
remainder in the Euclidean algorithm.
Proof. One knows that d is the last nonzero remainder
r_{m1} in the Euclidean algorithm. This formula for d is the
case j = m1 in (16).

Corollary 4.
(18) p_{m} = {a}/{d} , q_{m} = {b}/{d} .

Proof. The last remainder r_{m} = 0. The case j = m
in (16) shows that a/b = p_{m}/q_{m}. Since, by the first
proposition of the preceding section, p_{m} and q_{m} have no common
factor, this corollary is evident.
8. The action of GL_{2}(Z) on the projective line
If a, b, c, d are real
numbers with ad  bc ≠ 0 and
is the matrix with entries a, b, c, and d,
then M · z, for z real, will denote the expression
(19) M · z = {a z + b}/{c z + d} .

One calls M · z the action of M on z.
M · z is a perfectly good function of z except for
the case z =  d/c where the denominator cz + d vanishes.
If it were also true that az + b = 0 for the same z, then
one would have  b/a =  d/c, in contradiction of the assumption
ad  bc ≠ 0. Thus, when z =  d/c, the
value of M · w increases beyond all bounds as w approaches z,
and it is convenient to say that
where INFTY is regarded as large and signless. If further it is agreed
to define
which is the limiting value of M · w as w increases
without bound, then one may regard the expression M · z
as being defined always for all real z and for INFTY. The set
consisting of all real numbers and also the object (not a number)
INFTY is called the projective line. The projective line is
therefore the union of the (ordinary) affine line with a single point
INFTY.

Proposition 4.
If [n_{1}, n_{2}, …] is any continued
fraction, then
(20) [n_{1}, n_{2}, …, n_{r}, n_{r+1}, …] = M · [n_{r+1}, …] .

where
Proof. Let z = [n_{r+1}, …]. Then
[n_{1}, n_{2}, …, n_{r}, n_{r+1}, … ] = [n_{1}, n_{2}, …, n_{r}, z ] .

The statement of the proposition now becomes
[n_{1}, n_{2}, …, n_{r}, z ] = M · z .

This may be seen to follow by multiplying both sides in formula
(9), after replacing t_{j} with n_{j}, by the column
The matrix M in the preceding proposition is an integer matrix with
determinant ± 1. The notation GL_{2}(Z) denotes the set
of all such matrices. (The 2 indicates the size of the matrices,
and the Z indicates that the entries in such matrices are numbers
in the set Z of integers.) It is easy to check that the product of
two members of GL_{2}(Z) is a member of GL_{2}(Z) and
that the matrix inverse of a member of GL_{2}(Z) is a member
of GL_{2}(Z). Thus, GL_{2}(Z) forms what is called
a group. The formula (19) defines what is called the
action of GL_{2}(Z) on the projective line.
One says that two points z and w of the projective line are
rationally equivalent if there is a matrix M in GL_{2}(Z)
for which w = M · z. Since (i) GL_{2}(Z) is a group,
(ii) M_{1} · (M_{2} · z) = (M_{1} M_{2}) · z, and (iii)
w = M · z if and only z = M^{1} · w, it is easy to
see that every point of the projective line belongs to one and only one
rational equivalence class and that two points rationally equivalent to
a third must be rationally equivalent to each other.

Terminology.
The rational equivalence of points
on the projective line is said to be the equivalence relation on the
projective line defined by the action of GL_{2}(Z).

Example 1.
The set of real numbers rationally equivalent to
the point INFTY is precisely the set of rational numbers.

Example 2.
The proposition above shows that any continued
fraction is rationally equivalent to each of its tails. It follows that
all tails of a continued fraction are rationally equivalent to each
other.
9. Periodic continued fractions
In one of the first examples of a continued fraction expansion, it was
shown that SQRT{10} = [3,6,6,6,…]. This is an example of
a periodic continued fraction. After a finite number of terms
the sequence of integers repeats cyclically. If a cyclic pattern is
present from the very first term, then the continued fraction is called
purely periodic. Evidently, [6,6,6,…] = SQRT{10}3
is an example of a purely periodic continued fraction.
Note that a periodic continued fraction cannot represent a rational
number since the continued fraction expansion of a rational number is
finite.

Theorem 6.
Every periodic continued fraction is the continued
fraction expansion of a real quadratic irrational number.
Proof. For clarity: it is being asserted that every periodic
continued fraction represent a number of the form
where a, b, c, and m are all integers with m > 0,
c ≠ 0, and m not a perfect square.
Numbers of this form with fixed m but varying integers a,
b, and c ≠ 0 may be added, subtracted, multiplied, and
divided without leaving the class of such numbers. (The statement
here about division becomes clear if one remembers always to
rationalize denominators.) Consequently, for M in
GL_{2}(Z) the number M · z will be a number of this
form or INFTY if and only if z is in the same class.
Since a periodic continued fraction is rationally equivalent to a
purely periodic continued fraction, the question of whether any
periodic continued fraction is a quadratic irrationality reduces to
the question of whether a purely periodic continued fraction is such.
Let
x = [n_{1}, …, n_{r}, n_{1}, …, n_{r}, n_{1}, …, n_{r}, …]

be a purely periodic continued fraction. By the proposition of the
preceding section, x = M · x where M is notationally
identical to the M in (20). Ignoring the computation
(9) of M in terms of convergents, let
Then
or, otherwise said, x is a solution of the quadratic equation
cx^{2}  (ad) x  b = 0 .


Remark 2.
It is conversely true that the continued fraction expansion of every
real quadratic irrationality is periodic.
This converse will not be proved here.
REFERENCES
 [1]
G. Chrystal, Algebra: An Elementary Textbook (2 vols.),
Chelsea.
 [2]
G. Hardy & E. Wright, An Introduction to the Theory
of Numbers, Oxford Univ. Press.
 [3]
S. Lang, Introduction to Diophantine Approximations,
AddisonWesley.
 [4]
O. Perron, Die Lehre von den Kettenbrüchen, 2nd ed.,
Chelsea.