Irrational numbers

25 Pages·2016·0.223 MB·English

by  Hartmut Laue

Irrational numbers
Hartmut Laue
Mathematisches Seminar der Universit¨at Kiel 2016

Some introductory remarks
The following pages present a part of my course in Elementary Number Theory at the
Mathematics Department of the University of Kiel, summer 2016. It contains an intro-
duction to the theory of approximations of real numbers by rational numbers. A large
part is devoted to the theory of continued fractions.
It happens frequently that a close look at classical presentations of mathematical the-
ories reveals surprising disharmonies which could easily be avoided. But it seems that
people are reluctant to rewrite them and rather stick to historically grown habits. Even
small changes which would make the theory work more smoothly and therefore be a
service to its beauty are not taken into consideration. Elementary Number Theory is no
exception from this general observation. If a classical text chooses to write a continued
fraction expansion in the form (a ,a ,a ,...), then this decision is shared by all fol-
0 1 2
lowers, despite the fact that at the point where Fibonacci numbers come in (see p.11),
suddenly a shift of subscripts is required. Writing continued fraction expansions in the
form (a ,a ,a ,...) avoids this crack in the presentation.
1 2 3
“Listening” in this sense to what mathematics wants instead of following historically
grown habits which result in inconsistencies will always pay: This is my firm conviction
after endless experiences of this kind in more than forty years of writing mathematical
texts. The tiny specimen mentioned before has, for example, the unforeseen but agree-
able effect that it is the subscripts divisible by 3 which play the well-known special role
in the continued fraction expansion of e (see p.21), no longer the subscripts which are
congruent to 1 modulo 3 as in the traditional presentation. Now what is closer to the
−
ideal of mathematical beauty, from an unbiased view? Moreover, the first continued
fraction of a number becomes really the first – and not the zeroth. Consequently, the
subscript 0 is still available when you technically need it for the recursive definition of
numerators and denominators of continued fractions (see Lemma3). Clearly, it is possi-
ble to work with the subscript 1 at that place, but again: What is closer to the ideal
−
of mathematical beauty, from an unbiased view?
Perron’s classical text “Die Lehre von den Kettenbru¨chen” is certainly a treasure chest
for its mathematical contents, but certainly not for its mathematical beauty. People
who absolutely want to find the formulae as written in that book should replace our
a , u , v by Perron’s b , A , B , etc., – but these would not be readers which
n n n n 1 n 1 n 1
− − −
the present text was written for. I have frequently been asked why my texts look “so
different” from presentations in certain books. I have taken this occasion – one of the
tiniest of all – to formulate a written answer. Readers are requested to transfer its spirit
to all of my texts. Readers may trust that deviations from the so-called “standard” are
results of careful consideration and should not naively interprete them as a haphazard
or mischievous play with modified terminology or notation. Nothing could be falser. As
frequently said orally to my students, I would be grateful for any suggestion of improve-
ment if it benefits the two outstanding mathematical qualities of elegance and beauty.
It is this the kind of activity which contemporary mathematics deserves. And needs.
3
Apart from basic Linear Algebra and Analysis, only sporadically a little specific know-
ledge from other parts of Number Theory is necessary, like the little Fermat theorem,
Euler’s criterion on quadratic residues, Gauss’s reciprocity law.
The set of positive integers 1, 2, 3, ...will be denoted by N, furthermore N := N 0 ,
0
∪{ }
n := k k N, k n for all n N . Clearly, the notion of greatest common divisor is
0
{ | ∈ ≤ } ∈
defined for all pairs of integers (in particular, also if one or both of them equal 0).
4
Irrational numbers
Rational numbers are well-understood, irrational numbers aren’t. As Q is a dense subset
of R, we find rational numbers as close to a given real number x as we want. For a fixed
positive integer b, however, there is a largest fraction with denominator b which does
not exceed x; i.e., there is a unique integer z such that
z z +1
x < .
b ≤ b
If b = 1, the fraction is an integer, the number z is called the integral part of x and
denoted by x . Then x := x x is called the fractional part of x.
⌊ ⌋ h i −⌊ ⌋
z 1
0.1. Let x R, b N . Then there exists at most one integer z such that x < .
∈ ∈ >1 − b b2
Proof. Let z,z Z such that x z , x z′ < 1. Then (cid:12)(cid:12) (cid:12)(cid:12)
′ ∈ | − b| | − b| b2
z z z z 2 1
′ ′
− x + x < ,
b ≤ b − − b b2 ≤ b
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
hence z = z′. (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
The following remark will make use of the so-called pidgeon-hole principle: If you have
to put n+1 objects into n boxes, one of the boxes will necessarily have to take in at least
two objects. Despite its disarming simplicity,1, it has numerous interesting applications
as it may be used to obtain non-constructive existence statements. It is attributed to
Dirichletwhointroduceditbyproviding thelineofreasoningintheproofofthefollowing
assertion:
0.2. Let x RrQ, n N. Then there exist elements b n, z Z such that
∈ ∈ ∈ ∈
z 1
0 < x < ,
− b bn
(cid:12) (cid:12)
(equivalently: 0 < bx z < 1). (cid:12) (cid:12)
| − | n
Proof. The fractional parts 0 x , 1 x ,..., n x are mutually distinct, because
h · i h · i h · i
ix = jx (where i,j are distinct integers) implies that
h i h i
(j i)x = jx ix Z+ jx ix = Z,
− − ∈ h i−h i
a contradiction as x / Q. Thus they are n + 1 elements of the interval [0,1[ which
∈
we may write as the disjoint union of the n intervals [k 1, k[ where k n. It follows
−n n ∈
1The pidgeon-hole principle is just a formulation in everyday terms of the simple fact that a function
of a set of n+ 1 elements into a set of n elements cannot be injective. – The German term is
“Schubfachprinzip”.
5
that (at least) one of these intervals contains (at least) two of those fractional parts.
Hence there exist integers i,j such that 0 i < j n and 0 < jx ix < 1.
≤ ≤ |h i − h i| n
Put b := j i, z := jx ix . Then b n and bx = z + jx ix , hence
− ⌊ ⌋ − ⌊ ⌋ ∈ h i − h i
z jx ix 1
x = |h i−h i| ]0, [.
− b b ∈ bn
(cid:12) (cid:12)
T(cid:12) heor(cid:12)em 1. For every x R, the following are equivalent: 2
∈
(i) x is irrational.
z 1
(ii) There exist infinitely many pairs (z,b) Z N such that 0 < x < .
∈ × − b b2
z 1
(iii) There exist infinitely many pairs (z,b) Z N such that 0 < x < .
∈ × b − b2
z 1
(iv) There exist infinitely many pairs (z,b) Z N such that 0 = x < .
∈ × 6 − b b2
(cid:12) (cid:12)
(v) There exist a real number M and infinitely many pairs (z,b)(cid:12) Z (cid:12)N such that
∈ ×
z M
0 = x < .
6 − b b2
(cid:12) (cid:12)
Proof. Th(cid:12)e impl(cid:12)ications (ii) (iv), (iii) (iv), (iv) (v) are trivial, while (i) (ii), (iii)
⇒ ⇒ ⇒ ⇒
will follow immediately from results at a later stage (see 9.3). Thus it would be enough
to show at this point that (v) implies (i). Still we will also give Dirichlet’s short proof,
with the aid of 0.2, that (i) implies (iv): Let k N , b ,...,b N, z ,...,z Z such
0 1 k 1 k
∈ ∈ ∈
z 1 1 z
i i
that 0 = x < for all i k. Choose n N such that < x for all i k.
6 − b b2 ∈ ∈ n − b ∈
i i i
By 0.2, (i(cid:12)) implie(cid:12)s that (cid:12) (cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
z 1 1 1 z
i
0 < x < min , < x for all i k
− b bn ≤ b2 n − b ∈
i
(cid:12) (cid:12) (cid:8) (cid:9) (cid:12) (cid:12)
for suitably chosen (cid:12)integer(cid:12)s b n, z Z. Hence ((cid:12)z,b) =(cid:12)(z ,b ) for all i k and (iv)
i i
∈ ∈ 6 ∈
follows.
(i) (v): Let r Z, s N such that x = r. Let M R. If b N, z Z and
¬ ⇒ ¬ ∈ ∈ s ∈ ∈ ∈
0 6= |x− zb| < Mb2, then b1s ≤ |brb−szs| < Mb2, hence b < Ms. This means that the number of
all b N for which there exists an integer z such that 0 = x z < M has the upper
∈ 6 | − b| b2
bound Ms and is hence finite.
Theconcept tobeintroducednowisoffundamentalimportanceforthetheoryofrational
approximations of irrational numbers.
2The condition in (ii) – (v) that there exist infinitely many pairs (z,b) Z N could be replaced by
∈ ×
the condition that there exist infinitely many b N such that the respective inequalities hold for
∈
some z Z: For every b N the number of suitable z is finite.
∈ ∈
6
Definition 2. For every n N, a R, a ,...,a R we define
1 2 n >0
∈ ∈ ∈
K( ) := 0, K(a ) := a ,
1 1
∅
and recursively,
1
K(a ,...,a ) := a +
1 n 1
K(a ,...,a )
2 n
which is called the (regular) continued fraction for (a ,...,a ).3 Then K is a real-valued
1 n
function defined on the set of all tuples over R the components of which are positive
except possibly the first one. We have
1 1
K(a ,a ) = a + , K(a ,a ,a ) = a + , ...
1 2 1 a 1 2 3 1 a + 1
2 2 a3
Even in the case where a = a = ... = 1, the corresponding continued fractions are
1 2
of a non-trivial nature: We have K(1) = 1, K(1,1) = 2, K(1,1,1) = 3, K(1,1,1,1) = 5,
2 3
K(1,1,1,1,1) = 8, ...We will see later that this special sequence of continued fractions
5
plays a distinguished role in the theory of rational approximations (see Proposition10(3)
and comment on p.18).
2.1. K(a ,...,a ) = K a ,...,a ,K(a ,...,a ) for all m n.
1 n 1 m 1 m n
− ∈
1
(cid:0) (cid:1)
In particular: K(a ,...,a ) = K(a ,...,a ,a + ) if n > 1.
1 n 1 n 2 n 1
− − an
Proof. If m = 1, the claim is trivial. We proceed by induction on n. If n = 1, then
m = 1 and the claim holds. For the inductive step, let n > 1, m n. Then, by the
∈
definition of K,
1
K a ,...,a ,K(a ,...,a ) = a +
1 m 1 m n 1
− K a2,...,am 1,K(am,...,an)
−
(cid:0) (cid:1) 1
=(cid:0)a + = K(a ,(cid:1)...,a ),
1 1 n
K(a ,...,a )
2 n
where the second equality makes use of the inductive hypothesis.
2.2. The mapping R R, x K(a ,...,a ,x), is injective and differentiable.
>0 1 n 1
→ 7→ −
Proof. The claim is trivial if n = 1. Let n > 1. If the mapping R R , x
>0 >0
→ 7→
K(a ,...,a ,x), is injective and differentiable, then so is the mapping R R,
2 n 1 >0
x a + − 1 . Now the claim follows inductively. →
7→ 1 K(a2,...,an−1,x)
3There is a more general notion of continued fraction which allows also numerators = 1. This is the
6
reason for the adjective “regular” for the special kind of continued fraction defined in the text. We
will only consider regular continued fractions and therefore omit the adjective. – It would be more
precise to write “K (a1,...,an) ” but we prefer the notation with a single pair of parentheses.
(cid:0) (cid:1)
7
2.3. Let (x ) be a converging sequence over R with a positive limit x . Then
k k N >0 ∗
∈
K(a ,...,a ,x ) K(a ,...,a ,x ),
1 n 1 k 1 n 1 ∗
− k−→ −
→∞
(cid:3)
because the mapping considered in 2.2 is continuous.
Lemma 3. Let (a ) be a sequence4 of real numbers, a > 0 for all n N . Define
n n N n >1
∈ ∈
u := 1, u := a , u := a u +u
0 1 1 n+1 n+1 n n 1
−
v := 0, v := 1, v := a v +v
0 1 n+1 n+1 n n 1
−
for all n N. Then, for all n N, v > 0 and
n
∈ ∈
u
n
(1) K(a ,...,a ) = ,
1 n
v
n
(2) u v u v = ( 1)n,
n n 1 n 1 n
− − − −
(3) u v u v = ( 1)na .
n+1 n 1 n 1 n+1 n+1
− − − −
Note that (2) and (3) may also be expressed in the following form if n > 1:
u u ( 1)n u u ( 1)na
n n 1 n+1 n 1 n+1
(2′) − = − , (3′) − = −
v − v v v v − v v v
n n 1 n 1 n n+1 n 1 n 1 n+1
− − − −
Proof. Clearly, v > 0 for all n N. The claim in (1) is trivial for n 2. Let n 2 and
n
∈ u ≤ ≥
n
assume that, inductively, K(a ,...,a ) = for every sequence (a ) of real numbers
1 n n n N
vn ∈
with a > 0 for all n N . Given such a sequence, we consider the modified sequence
n >1
∈
1
a ,...,a ,a + ,a ,...
1 n 1 n n+2
− an+1
Applying first 2.1, then our inductive hypothesis, we obtain
K(a ,...,a ) = K(a ,...,a ,a + 1 ) = (an + an1+1)un−1 +un−2
1 n+1 1 n−1 n an+1 (an + an1+1)vn−1 +vn−2
a (a u +u )+u a u +u u
n+1 n n 1 n 2 n 1 n+1 n n 1 n+1
= − − − = − = .
a (a v +v )+v a v +v v
n+1 n n 1 n 2 n 1 n+1 n n 1 n+1
− − − −
(2) Let n N. For all m N we have
>1
∈ ∈
u v u v = (a u +u )v u (a v +v )
m m 1 m 1 m m m 1 m 2 m 1 m 1 m m 1 m 2
− − − − − − − − − −
= (u v u v ),
m 1 m 2 m 2 m 1
− − − − − −
hence, by repeated application, u v u v = ( 1)n 1(u v u v ) = ( 1)n.
n n 1 n 1 n − 1 0 0 1
− − − − − −
4 Note that the subsequent settings and assertions of Lemma3 remain valid if referred to a given
l-tuple(a1,...,al)insteadofasequence,lettingnrangeoverl, lr 1 resp.,insteadofN, N>1 resp.
{ }
8
(3) Let n N. Then
∈
u v u v = (a u +u )v u (a v +v )
n+1 n 1 n 1 n+1 n+1 n n 1 n 1 n 1 n+1 n n 1
− − − − − − − −
= (u v u v )a = ( 1)na ,
n n 1 n 1 n n+1 n+1
− − − −
by (2).
In the sequel, we will use the variables u , v as defined in Lemma 3 throughout without
n n
further notice when a sequence (a ) as in Lemma 3 is given.
n n N
∈
Corollary 4. Suppose the hypotheses of Lemma 3. Then we have
(1) K(a ,...,a ) is strictly monotonically increasing, K(a ,...,a ) is
strictl1y mono2tmo−n1icamll∈yNdecreasing, and 1 2m m∈N
(cid:0) (cid:1) (cid:0) (cid:1)
K(a ,...,a ) < K(a ,...,a ) for all l,m N.
1 2m 1 1 2l
− ∈
(2) Both sequences considered in (1) are convergent, and for all m N,
∈
K(a ,...,a ) < lim K(a ,...,a ) lim K(a ,...,a ) < K(a ,...,a )
1 2m 1 1 2l 1 1 2l 1 2m
− l − ≤ l
→∞ →∞
(3) If the sequence (a ) has a positive lower bound, then v v . If
n n∈N>1 n n+1 n−→ ∞
→∞
v v , then K(a ,...,a ) is convergent. If K(a ,...,a ) con-
n n+1 n−→ ∞ 1 n n N 1 n n N
verges to→y∞ R, then ∈ ∈
(cid:0) (cid:1) (cid:0) (cid:1)
∈
u 1 u 1
2m 1 2m
0 < y − < , 0 < y <
− v v v v − v v
2m 1 2m 1 2m 2m 2m 2m+1
− −
u 1
n
for all m N. In particular, y < for all n N.
∈ − v v v ∈
n n n+1
(cid:12) (cid:12)
Proof. (1) The first two claims fol(cid:12)low from(cid:12) Lemma3(1),(3), choosing first n to be even,
′
then odd. Furthermore, Lemma3(1),(2) imply, for all m N, that
′
∈
u u 1
2m 1 2m
K(a1,...,a2m 1) = − = < K(a1,...,a2m).
− v2m 1 v2m − v2m 1v2m
− −
Let l,m N. If l < m we apply the first statement in this number and obtain
∈
K(a ,...,a ) < K(a ,...,a ) < K(a ,...,a ). If l > m we apply the second
1 2l 1 1 2m 1 1 2m
− −
statementinthisnumberandobtainK(a ,...,a ) < K(a ,...,a ) < K(a ,...,a ).
1 2l 1 1 2l 1 2m
−
(2) is an immediate consequence of (1).
(3) We show first, by induction on n, that v v n s for all n N if s R is a
n n+1 ≥ ⌈2⌉ ∈ 0 ∈
lower bound of the sequence (a ) . We have v v = 0 = 0 s which settles the case
n = 0. Let n > 0 and assume thnant∈vN>1 v n 1 0s.1 ⌈2⌉
n−1 n ≥ ⌈ −2 ⌉
If n is odd, then v v = v v +a v2 n 1 s+s = n s as v 1 by (1).
n n+1 n−1 n n+1 n ≥ −2 · 2 n ≥
(cid:6) (cid:7)
9
If n is even, then v v v v n 1 s = n s by (1), completing the inductive
n n+1 ≥ n−1 n ≥ −2 2
step.
(cid:6) (cid:7) (cid:6) (cid:7)
Ifv v ,then 1 0,hence lim K(a ,...,a ) = lim K(a ,...,a ),
n n+1 n−→ ∞ vnvn+1 n−→ m 1 2m−1 m 1 2m
→∞ →∞ →∞ →∞
by (2) and Lemma3(2). Thus K(a ,...,a ) is convergent.
′ 1 n n N
∈
Suppose K(a1,...,an) y. T(cid:0)hen (2) and L(cid:1)emma3(1) imply for every m N that
n−→ ∈
→∞
u u
2m 1 2m
− < y < ,
v v
2m 1 2m
−
u u u 1
2m 1 2m 2m 1
hence 0 < y − < − = , by Lemma3(2′). The second chain
− v v − v v v
2m 1 2m 2m 1 2m 1 2m
of inequalities follo−ws analogously,−and the−assertion about the absolute value is an
immediate consequence.
It is not difficult to show that for every strictly monotonically decreasing sequence
(r ) over R there exists a sequence (a ) over R such that K(a ,...,a )
n n N >0 n n N >0 1 n+1
∈ ∈ | −
K(a ,...,a ) = r for all n N. It follows that the distance between the limits of the
1 n n
| ∈
two sequences considered in Corollary 4(2) can take every non-negative value. We now
analyze the case of a constant sequence (a ) .
n n N
∈
Proposition 5. Let a R . Let y be the positive, y˜ the negative zero of the polynomial
>0
∈
t2 at 1 R[t], i.e., y = 1 a+√a2 +4 , y˜= 1 a √a2 +4 . Then
− − ∈ 2 2 −
(1) K(a,...,a) y, (cid:0) (cid:1) (cid:0) (cid:1)
n n−→
→∞
yn y˜n
(2) v = − for all n N.
n
y y˜ ∈
−
Proof. (1) By Corollary4(3), K(a,...,a) is convergent. Let yˆ be its limit. By 2.3,
n n N
∈
(cid:0) (cid:1)
1 1
yˆ:= lim K(a,...,a) = a+ = a+ ,
n n limn K(a,...,a) yˆ
→∞ →∞ n
hence yˆ2 ayˆ 1 = 0. As yˆ> 0, it follows that yˆ= y.
− −
yn y˜n
(2) Set w := − for all n N . Then w = 0, w = 1. For every n N, the
n 0 0 1
y y˜ ∈ ∈
definition of y, y˜ i−mplies yn+1 = ayn +yn 1, y˜n+1 = ay˜n +y˜n 1, hence
− −
yn+1 y˜n+1 yn y˜n yn 1 y˜n 1
− −
w = − = a − + − = aw +w .
n+1 n n 1
y y˜ · y y˜ y y˜ −
− − −
Since w satisfies the same recursive condition as v , it coincides with v .
n n n
10