Download M3P14 LECTURE NOTES 11: CONTINUED FRACTIONS 1

M3P14 LECTURE NOTES 11: CONTINUED FRACTIONS
1. Rational Continued Fractions
Given a rational number pq , we can write pq = a0 + r0 , where a0 is an
integer and 0 ≤ r0 < 1 is between 0 and 1. If r0 is not zero, then we can
write r10 = a1 + r1 , with a1 and integer and 0 ≤ r1 < 1. We then have:
p
1
.
= a0 + r0 = a0 +
q
a1 + r1
Continuing in this way, as long as ri is nonzero we set r1i = ai+1 + ri+1 ,
with ai an integer and 0 ≤ ri+1 < 1. The denominators of the ri are strictly
decreasing, so eventually some rn = 0 and we have:
p
1
.
= a0 +
1
q
a1 + a +···+
1
2
an
This expression is called the continued fraction expansion of pq . It is closely
related to Euclid’s algorithm; indeed, it’s not hard to see that the ai are the
quotients qi from Euclid’s algorithm applied to the pair p, q. Note that each
ai is an integer and for i ≥ 1, ai ≥ 1.
2. Infinite Continued Fractions
Now let α be an irrational real number. As above, we can write α =
a0 + r0 , where a0 is an integer and 0 ≤ r0 < 1, and then for each i set
1
ri = ai+1 + ri+1 with ai+1 an integer and 0 ≤ ri+1 < 1. Note that unlike in
the rational case, this sequence will never terminate, as ri is always irrational
and thus never zero. We write:
1
α = a0 +
1
a1 + a2 +...
and call this the continued fraction expansion of α. Note that so far this is
just a formal expression! In fact, we can make mathematical sense of this
expression, but it requires some justification.
First, we introduce some useful notation: For a0 , . . . an real numbers, we
define [a0 ; a1 , . . . , an ] to be the real number:
a0 +
1
a1 +
1
a2 +···+ a1
n
when this expression is well-defined (that is, when it does not involve a
division by zero.)
1
2
M3P14 LECTURE NOTES 11: CONTINUED FRACTIONS
The value of this expression can computed by a recurrence relation, as
follows:
Lemma 2.1. Given a0 , . . . , an real numbers, define pi , qi for 0 ≤ i ≤ n by
p0 = a0 , q0 = 1, p1 = a0 a1 + a, q1 = a1 , together with the recurrences:
pi = ai pi−1 + pi−2
qi = ai qi−1 + qi−2 .
Then [a0 ; a1 , . . . , an ] =
pn
qn ,
assuming no qi is zero.
Proof. We prove this by induction on n; the cases n = 0 and n = 1 are
clear. Let p0i and qi0 be the numbers defined by the recurrence attached to
the sequence a0 , a1 , . . . , an−2 , an−1 + a1n . The induction hypothesis tells us
that [a0 ; a1 , . . . , an−2 , an−1 +
1
an ]
=
and q 0 , we have
p0n−1
.
0
qn−1
By the recurrence defining the p0
(an−1 + a1n )p0n−2 + p0n−3
p0n−1
=
0
0
0
qn−1
+ qn−3
(an−1 + a1n )qn−2
Since the sequences defining p0i ,qi0 and pi ,qi agree for i ≤ n − 2, the latter is
equal to:
(an−1 + a1n )pn−2 + pn−3
.
(an−1 + a1n )qn−2 + qn−3
Multiplying both numerator and denominator by an , and using pn−1 =
an−1 pn−2 + pn−3 , pn = an pn−1 + pn−2 (and similarly for qn ) we find that
this expression is equal to pqnn .
Thus we have:
pn
1
[a0 ; a1 , . . . , an ] = [a0 ; a1 , . . . , an−2 , an−1 + ] =
an
qn
as claimed.
Note that if ai ≥ 1 for all i (as will be the case, for instance, if the ai come
from taking a continued fraction expansion of some real number), then the
qi are all nonzero and form a strictly increasing sequence. (Indeed, if all ai
are at least one, then qi ≥ qi−1 + qi−2 ≥ 2qi−2 , so this sequence increases
exponentially fast.)
Now suppose we have an infinite sequence a0 , a1 , a2 , . . . , of real numbers,
and assume that ai ≥ 1 for i ≥ 1. Define pi and qi by the recurrence given
above. We call pqii the ith convergent to the continued fraction:
a0 +
1
a1 +
1
a2 +...
.
Lemma 2.2. We have pn qn−1 − qn pn−1 = (−1)n−1 .
M3P14 LECTURE NOTES 11: CONTINUED FRACTIONS
3
Proof. This is again by induction on n; the base case n = 1 is clear. Assume
this is true for n − 1. Then by the defining recurrence for the pi and qi , we
have:
pn qn−1 −qn pn−1 = (an pn−1 +pn−2 )qn−1 −(an qn−1 +qn−2 )pn−1 = pn−2 qn−1 −qn−2 pn−1 = −(−1)n−2
and the claim follows.
n−1
It follows that | pqnn − pqn−1
| < qn q1n−1 . Since the qn increase exponentially,
pi
it follows that the qi form a Cauchy sequence, and hence converge to a real
number.
A natural question to ask at this point is: “if the sequence a0 , a1 , . . .
arises from the continued fraction expansion of an irrational number α, is
the limit of the convergents pqnn equal to α?”. This is indeed the case:
Lemma 2.3. Let α be an irrational real number, and let a0 , a1 , . . . be the
sequence of integers arising from its continued fraction expansion. Let pqnn be
the nth convergent. Then pqnn < α if n is even, and pqnn > α if n is odd.
Proof. Once again we use induction on n; the case n = 0 is clear. Note that
1
[a1 ; a2 , . . . , an ] is the (n − 1)st convergent to α−a
. Thus, by the induction
0
hypothesis, if n is odd we have:
1
[a1 ; a2 , . . . , an ] <
α − a0
and thus
1
= [a0 ; a1 , a2 , . . . , an ].
α < a0 +
[a1 ; a2 . . . , an ]
If n is even the same argument works with the inequalities reversed.
Corollary 2.4. Let α be an irrational real number, and let a0 , a1 , . . . be the
sequence of integers arising from its continued fraction expansion. Let pqnn be
the nth convergent. Then |α − pqnn | < qn q1n+1 . In particular the limit pqnn as n
approaches infinity is α.
Proof. Since exactly one of
|α −
pn
qn
and
pn+1
qn+1
is less than α, we have
pn
pn+1 pn
1
|<|
− |=
.
qn
qn+1
qn
qn qn+1
Since the qn increase exponentially quickly the second claim is clear.
Note that since qn q1n+1 < q12 , this gives a second, completely constructive
n
proof of Dirichlet’s theorem on rational approximations!
3. Best Approximations
In fact, there is a sense in which the convergents to the continued fraction
expansion of α are the best rational approximations to α. We will make this
precise below.
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M3P14 LECTURE NOTES 11: CONTINUED FRACTIONS
Fix α real and irrational, and as above define ai and ri by setting α =
a0 + r0 , with a0 an integer and 0 ≤ r0 < 1, and r1i = ai+1 + ri+1 with ai+1
an integer and 0 ≤ ri+1 < 1. Define pn and qn from the sequence a0 , a1 , . . .
by the recurrences of the previous section.
Lemma 3.1. For all n, we have:
α=
pn + pn−1 rn
.
qn + qn−1 rn
Proof. For n = 1 this is easy. Assume it is true for n − 1; that is, that we
have:
pn−1 + pn−2 rn−1
α=
.
qn−1 + qn−2 rn−1
1
1
We have rn−1
= an + rn ; substituting an +r
for rn−1 in the above and using
n
the relations pn = an pn−1 + pn−2 , qn = an qn−1 + qn−2 yields the claimed
result.
Corollary 3.2. For all n, |α −
pn
qn |
< |α −
pn−1
qn−1 |.
Proof. We have:
|
Substituting α =
α−
α−
pn
qn
pn−1
qn−1
pn +pn−1 rn
qn +qn−1 rn ,
|=
qn α − p n
qn−1
|
|.
qn qn−1 α − pn−1
we find that the right hand side is equal to:
n−1 rn
qn pqnn +p
qn−1
+qn−1 rn − pn
|
|.
n−1 rn
qn qn−1 pqn +p
− pn−1
n +qn−1 rn
Simplifying, we find that this is equal to:
qn−1 qn pn−1 rn − pn qn−1 rn
qn−1
|
|=
rn < 1.
qn
qn−1 pn − pn−1 qn
qn
The claim follows.
Theorem 3.3. Let h, k be integers with |k| ≤ qn . Then | hk − α| ≥ | pqnn − α|.
In other words, pqnn is the best rational approximation to α with denominator
at most qn .
Proof. Suppose that | hk − α| < | pqnn − α|. Then by the corollary we also have
n−1
| hk − α| < | pqn−1
− α|. Suppose that n is even. Then we have:
pn
pn−1
<α<
,
qn
qn−1
and our inequalities then imply that we also have
pn
h
pn−1
< <
.
qn
k
qn−1
M3P14 LECTURE NOTES 11: CONTINUED FRACTIONS
5
If n is odd the same holds with the inequalities reversed, by a similar argument. Either way we have
h pn−1
pn pn−1
1
| −
|<| −
=
.
k
qn−1
qn
qn−1
qn qn−1
On the other hand, the left hand side of this is a positive rational number
1
with denominator kqn−1 , so must be at least kqn−1
. This is a contradiction
since k ≤ qn .
4. Periodic Continued Fractions
In this section we single out certain irrationals with particularly nice
continued fraction expansions:
Definition 4.1. The continued fraction expansion [a0 ; a1 , a2 , . . . ] of an irrational number α is eventually periodic if there exist positive integers N and
d such that an = an+d for all n ≥ N . It is periodic if there exists a positive
integer d such that an = an+d for all n.
Definition 4.2. An irrational number α is a quadratic irrational if there is
a polynomial ax2 + bx + c, with a, b, c integers, that has α as a root.
We then have:
Proposition 4.3. Suppose that α has an eventually periodic continued fraction expansion. Then α is a quadratic irrational.
Proof. We first show this when α is periodic. We then have a d such that
1
.
α = a0 +
a1 + a +···+1 1
2
1
ad−1 + α
Simplifying the right hand side, we find integers x, y, z, w such that
xα + y
α=
.
zα + w
Then zα2 + (w − x)α − y = 0; since α is irrational z cannot be zero, so α is
a quadratic irrational.
If α is only eventually periodic there is a β with periodic continued fraction
expansion such that we have:
1
.
α = a0 +
a1 + a +···+ 1 1
2
aN −1 + 1
β
Then β is quadratic irrational. It thus suffices to show that if β is quadratic
irrational, so is β1 and n + β for integer n.
Note that if β is a root of ax2 + bx + c, then β1 is a root of a + bx + cx2 ,
and n + β is a root of a(x − n)2 + b(x − n) + c, so these claims are clear. In fact, all quadratic irrationals have eventually periodic continued fraction expansions, but we will not prove this.