Download MTH6128 Number Theory 5 Periodic continued fractions

MTH6128
Number Theory
Notes 5
5
Spring 2017
Periodic continued fractions
Continued fractions give a very useful way of representing real numbers if you
are interested in finding good approximations; also we will soon see uses of them
in solving diophantine equations. But for ordinary arithmetic, they are hopeless:
there is no simple rule even for adding two continued fractions. Here is an
example.
√
We have
√ that 2 + 1 is the value of [2; 2, 2, 2, . . .] = [2;]. If we add it to itself,
we get 2( 2 + 1) which is the value of [4; 1, 4, 1, . . .] = [4; 1].√Could you imagine
a rule to produce this? Maybe. But adding again, we get 3( 2 + 1) which is the
value of [7; 4, 8, 4, 8, . . .] = [7; 4, 8], and it is dificult to see how any rule would
produce this.
√
is the value of [4; 4, 4, 4, . . .] =
But worse is to come.√We have that √5 + 2 √
[4;], but adding this to 2 + 1 we get 2 + 5 + 3 which is the value of
[6; 1, 1, 1, 6, 8, 1, 2, 1, 2, 2, . . .]. We see no pattern at all, and it really stretches
the imagination to see how any simple rule could produce this from [2;] + [4;].
√ In√fact, we are going to show in this chapter that the continued fraction for
2+ 5+3 is non-periodic: it never repeats. We will decide just which irrational
numbers have periodic continued fractions:
√ they are precisely the quadratic num√
bers. Now convince yourself that 2 + 5 + 3 does not satisfy a quadratic with
rational coefficients. (Its minimal polynomial has degree 4.)
Definition The infinite continued fraction
[a0 ; a1 , a2 , . . .]
is periodic if there exist integers k, l with k > 0 such that
an+k = an for all n ≥ l.
1
If an+k = an for all n ≥ l, we write the continued fraction as
[a0 ; a1 , a2 , . . . , al−1 , al , al+1 , . . . , al+k−1 ].
It is purely periodic if there exists k > 0 such that
an+k = an for all n ≥ 0.
The notation is similar to the one used for periodic decimals. For example,
[2; 1, 2, 1, 2, 1, 2, 1, . . .] = [2; 1]
[3; 5, 2, 1, 2, 1, 2, 1, . . .] = [3; 5, 2, 1].
We now calculate these two continued fractions. Let c be the value of [2; 1].
Then
c = [2; 1, c]
[2, 1, c]
(by Proposition 3.3(b))
=
[1, c]
3c + 2
=
.
c+1
√
2
So c2 + c = 3c + 2, so that c√
− 2c − 2 = 0, or c = 1 ± 3. But c > 2, so we must
take the plus sign; c = 1 + 3.
Now let d be the value of [3; 5, 2, 1]. Then
d = [3; 5, c]
[3, 5, c]
=
[5, c]
16c + 3
=
5c + 1 √
19 + 16 3
√
=
6+5 3
√
√
(19 + 16 3)(6 − 5 3)
√
√
=
(6 + 5 3)(6 − 5 3)
√
126 − 3
=
.
39
Note that d, like c, is a “quadratic irrational”, an algebraic integer satisfying a
quadratic equation. (We saw this already for c; and d satisfies (39x − 126)2 = 3.)
2
In this chapter we are going to show that the result suggested by these examples is true in general. A real number has a periodic continued fraction if and
only if it is a quadratic irrational. We will also find which numbers have purely
periodic continued fractions. We will apply these results to sums of squares and
to a diophantine equation called Pell’s equation in the next chapter.
Recall
√ from Chapter 2 that a quadratic irrational x ∈ R is of the form x =
u + v d, where u and v are rational numbers, v 6= 0, and d is a squarefree
integer greater than 1. (We require v 6= 0 and d > 1 since, if either of these
conditions failed, y would be rational.) Quadratic irrationals are precisely the
roots of irreducible quadratics with rational coefficients.
√
Definition √
If x = u + v d is a quadratic irrational, we define its conjugate to
be x0 = u − v d.
Note that x and x0 are the two roots of the same irreducible quadratic.
Definition A reduced quadratic irrational is a quadratic irreducible x such that
x and its conjugate x0 satisfy
and
− 1 < x0 < 0.
√
√
In our worked example, c = 1 + 3 > 0 and −1 < c0 = 1 −√ 3 < 0, so c is
a reduced quadratic
irrational. On the other hand, d = (126 − 3)/39 > 1 but
√
0
d = (126 + 3)/39 is greater than d. So d0 is not reduced.
x>1
We are going to prove the following result:
Theorem 5.1 (a) A real number has periodic continued fraction if and only
if it is a quadratic irrational.
(b) A real number has purely periodic continued fraction if and only if it is a
reduced quadratic irrational.
There are four things to prove here, and the proofs will take the rest of this
chapter.
Claim 1 We begin by showing that a number with a purely periodic continued
fraction is a reduced quadratic irrational.
Let x be the value of [a0 ; a1 , . . . , ak−1 ]. We will suppose that k ≥ 3; the
argument for k = 1, 2 is easy to do directly, or we can simply pretend that the
period is longer than it is (for example, [2; 1] = [2; 1, 2, 1]).
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We know that x is irrational, since the continued fraction for a rational number terminates. Also, just as we argued for the value c of [2; 1], we have
x = [a0 ; a1 , . . . , ak−1 , x]
xpk−1 + pk−2
,
=
xqk−1 + qk−2
where ci = pi /qi is the ith convergent of [a0 ; a1 , . . . , ak−1 ].
Hence x2 qk−1 + x(qk−2 − pk−1 ) − pk−2 = 0, so that x is a quadratic irrational.
Also, a0 = ak ≥ 1 (remember that all terms except possibly the first in a
continued fraction are positive), so x > a0 ≥ 1. It remains to show that the
algebraic conjugate x0 of x satisfies −1 < x0 < 0.
From the properties of quadratic equations, we have xx0 = −pk−2 /qk−1 < 0.
Also,
pk−1
ck−1
pk−2
,
<
=
−x0 =
xqk−1
xqk−1
x
because pk−2 < pk−1 ; and also
−x0 =
pk−2
pk−2
ck−2
<
=
,
xqk−1
xqk−2
x
because qk−2 < qk−1 . One of k − 1 and k − 2, say j, is even; and we know from
Corollary 3.8 that cj < x. So −x0 < 1, or x0 > −1. Now we have verified all
parts of the definition of a reduced quadratic irrational.
Claim 2 Next we show that any number with periodic continued fraction is a
quadratic irrational.
Let y be the value of [a0 ; a1 , . . . , am , am+1 , . . . , am+k ]. Let z be the value of
[am+1√
; . . . , am+k ]. By Proof 1, z is a (reduced) quadratic irrational, say z =
u + v d, where u and v are rational numbers and d is a squarefree integer. We
have
[a0 , . . . , am , z]
[a1 , . . . , am , z]
[a0 , . . . , am ]z + [a0 , . . . , am−1 ]
=
.
[a1 , . . . , am ]z + [a1 , . . . , am−1 ]
y = [a0 ; a1 , . . . , am , z] =
Let [a0 , . . . , am ] = A, [a0 , . . . , am−1 ] = B, [a1 , . . . , am ] = C, [a1 , . . . , am−1 ] = D
(these are all positive integers). Then
y =
Az + B
Cz + D
4
√
Au + B + Av d
√
=
Cu + D + Cv d
√
√
(Au + B + Av d)(Cu + D − Cv d)
=
,
(Cu + D)2 − (Cv)2 d
√
which is a quadratic irrational since it has the form U + V d for some rational
numbers U and V .
Now our goal is to prove the converse of the last two results: if x is a [reduced]
quadratic irrational, then its continued fraction is [purely] periodic. Let us begin
with an example.
√
√
Example Find the continued fraction of 2 + 7. Note
that
2
+
7 is reduced:
√
it is greater than 1, and its algebraic conjugate 2 − 7 lies between −1 and 0.
x0
x1
x2
x3
x4
√
= 2 + 7,
√
√
= 1/(2 + 7 − 4) = (2 + 7)/3,
√
√
= 3/(2 + 7 − 3) = (1 + 7)/2,
√
√
= 2/(1 + 7 − 2) = (1 + 7)/3
√
√
= 3/(1 + 7 − 3) = 2 + 7 = x0
√
a0 = b2 + 7c = 4
√
a1 = b(2 + 7)/3c = 1
√
a2 = b(1 + 7)/2c = 1
√
a3 = b(1 + 7)/3c = 1
√
So 2 + 7 is the value of [4; 1, 1, 1].
Note that all of x0 , x1 , x2 , x3 are reduced quadratic irrationals, and we can
read off their continued fractions: for example, x2 is the value of [1; 1, 4, 1].
Other observations
which will be important in the proof are that, in each case,
√
xi = (Pi + 7)/Qi , where Pi and Q√
i are integers (the
√ Pi are 2, 2, 1, 1, . . . and the
Qi are 1, 3, 2, 3, . . .); and 0 < Pi < 7, 0 < Qi < 2 7.
We will see that all these properties hold quite generally.
Before we start the proofs, we introduce a slightly different way of writing
quadratic irrationals.
√
Lemma 5.2 (a) A real quadratic irrational can be written as x = (P + D)/Q,
where P, Q are integers, D is a positive integer which is not a square, and
Q divides D − P 2 .
√
√
(b) If x is reduced, then 0 < P < D and 0 < Q < 2 D.
√
Proof (a) We know that x = u + v d where u and v are rationals and d is
squarefree.
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Suppose first that v is positive. Let q be the least common multiple of the
denominators of u and v, and u = p/q, v = r/q. Then
p
√
√
pq + q 2 r2 d
p + r2 d
p+r d
=
=
x=
.
q
q
q2
Put P = pq, Q = q 2 , and D = q 2 r2 d, and note that Q divides P 2 − D.
If v < 0, then write −x in the specified form and then replace Q by −Q.
(b) Now suppose that x is reduced; recall that this
√ means x > 1 and −1 <
0
0
x < 0, where x is the conjugate of x (so x = (P − D)/Q). Then
√
√
• x > 0 > x0 , so (P + D)/Q > (P − D)/Q. Hence Q > 0.
√
√
• x > 1 > −x0 , so (P + D)/Q > (−P + D)/Q. Hence P > 0.
√
√
• x0 < 0, so P − D < 0. Hence P < D.
√
√
√
• x > 1, so (P + D)/Q > 1. Hence Q < P + D < 2 D.
0
Claim 3 We show that a reduced quadratic irrational has a purely periodic
continued fraction.
The proof is quite long, so I begin by outlining the steps. Let x be a reduced
quadratic irrational.
• We show that, if we run one step of the continued fraction algorithm, that
is, with x0 = 0 and a0 = bx0 c, x1 = 1/(x0 − a0 ), that x1 is also a reduced
quadratic irrational.
• It follows that all the numbers xn that come up in the algorithm are reduced
quadratic irrationals.
√
√
D)/Q
,
where
0
<
P
<
D
• So all these numbers
x
have
the
form
(P
+
n
n
n
n
√
and 0 < Qn < 2 D.
• Since there are only finitely many possibilities for Pn and Qn , we must
eventually return to a value we saw before; the algorithm is periodic from
this point.
• Finally we show it is periodic right from the start, that is, purely periodic.
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Now for the details!
√
Suppose that x = P +Q D is a reduced quadratic irrational (P, Q ∈ Z, D ∈ Z,
√
√
D > 0, D not a square, Q | D − P 2 , 0 < P < D, 0 < Q < 2 D). We start
building the continued fraction for x0 = x. The first step is
a = bxc,
y=
1
.
x−a
∗
√
+ D
Claim 3: y is a reduced quadratic number of the form P Q
.
∗
Certainly y > 1, since y = 1/(x − a) and x − a < 1. We have to show that
−1 < y 0 < 0.
Let P ∗ = Qa − P . Then
y =
=
=
=
=
1
√
(P + D)/Q − a
1
√
(P + D − Qa)/Q
1
√
∗
(−P + D)/Q
√
P∗ + D
(D − (P ∗ )2 )/Q
√
P∗ + D
,
Q∗
where Q∗ = (D − (P ∗ )2 )/Q. Now
Q∗ =
D − (Qa − P )2
D − P2
= −Qa2 + 2P a +
,
Q
Q
so Q∗ is an integer. Moreover, Q∗ divides D − (P ∗ )2 (the quotient is just Q). So
we have written y in the same form as x, with the same D, but maybe different
P and Q.
Also, we have y = 1/(x − a), and so y 0 = 1/(x0 − a). Now x0 < 0, so
x0 − a < −a ≤ −1; so −1 < y 0 < 0 (since x0 − a < 0), as required.
Our claim is proved.
Proof of Theorem 5.1 (b) ⇒√ is Claim 1. Let us prove ⇐: Let x be a
1
reduced quadratic number, x = P +Q D . Put x0 = x, ..., an = bxn c, xn+1 = xn −a
.
n
By Claim 3 (induction), we know that xn =
7
√
Pn + D
.
Qn
Pn and Qn are integers
√
√
satisfying 0 < Pn < D and 0 < Qn < 2 D. There are only finitely many
possible values of Pn and Qn , so after some number of steps, we must return to
values we have seen before. Suppose this first happens when xn = xm . Clearly
the sequence repeats after this point, that is, xi+k = xi for all i ≥ m, where
k = n − m.
We have to show that the repetition starts with m = 0. Let m > 0 be such
that
1
1
= xn =
,
xm =
xm−1 − am−1
xn−1 − an−1
so
1
1
x0m = 0
= x0n = 0
,
xm−1 − am−1
xn−1 − an−1
So x0m−1 − am−1 = x0n−1 − an−1 . Thus x0m−1 and x0n−1 differ by an integer. But
they both lie between −1 and 0, so they are equal, whence xm−1 = xn−1 . This
allows us to go back to m = 0.
So, finally, we have proved the converse implication in part (b) of the big
theorem: that is, the irrational number x has a purely periodic continued fraction
if and only if it is a reduced quadratic irrational.
Proof of Theorem 5.1 (a) The final part is to prove part (a) of the theorem:
the irrational number x has a periodic continued fraction if and only if it is a
quadratic irrational.
We have proved ⇒ in Claim 2. So suppose that x is a quadratic irrational;
we have to show that its continued fraction is periodic (from some point on).
Calculate its continued fraction: that is, put x0 = x, and then
an = bxn c,
xn+1 =
1
x n − an
for n ≥ 0. It is clear that all the xn are quadratic irrationals, since subtracting
an integer from a quadratic irrational, and taking the reciprocal of one, gives
again a quadratic irrational. We have to prove that, for some value of n, the
number xn is a reduced quadratic irrational. Then by part (b) of the Theorem,
the continued fraction is periodic from that point on. So we have to prove that,
if n is large enough, then xn > 1 and −1 < x0n < 0.
By construction, we have xn > 1 for all n > 0, since xn = 1/(xn−1 − an−1 )
and xn−1 − an−1 < 1. So this part is easy!
Also, we have for any n > 0
x = [a0 ; a1 , . . . , an−1 , xn ] =
8
xn pn−1 + pn−2
,
xn qn−1 + qn−2
by standard results about continued fractions.
Hence
x0 pn−1 + pn−2
x0 = n0
,
xn qn−1 + qn−2
so rearranging we obtain
x0n
−x0 qn−2 + pn−2
= 0
.
x qn−1 − pn−1
So, taking factors qn−2 and qn−1 out of numerator and denominator,
x0n = −
qn−2 x0 − cn−2
,
qn−1 x0 − cn−1
where cn is the nth convergent. (Remember that cn = pn /qn .)
Since cn → x as n → ∞, we have
−
qn−1 0
x0 − x
x0 − cn−2
xn = 0
→ 0
=1
qn−2
x − cn−1
x −x
as n → ∞.
So for large enough n, we have |(x0 − cn−2 )/(x0 − cn−1 ) − 1| < 1, and so this
fraction is positive. Thus −(qn−1 /qn−2 )x0n > 0, so that x0n < 0.
Also, we can ensure that n is also large enough so that |cn − x| < |x0 − x|.
If x0 < x, we use the fact that even-numbered convergents are smaller than
x and odd-numbered convergents are greater; choosing n even, we have x0 <
cn−2 < x < cn−1 . If x < x0 , then choosing n odd we have cn−1 < x < cn−2 < x0 .
In either case, we have
qn−2 x0 − cn−2 0
< 1,
−xn =
qn−1 x0 − cn−1 so finally we conclude that, for large enough n,
−1 < x0n < 0
and xn is a reduced quadratic irrational, as required.
The proof is complete.
9