Fractions - School of Mathematics

Fractions
Author(s): L. R. Ford
Source: The American Mathematical Monthly, Vol. 45, No. 9 (Nov., 1938), pp. 586-601
Published by: Mathematical Association of America
Stable URL: http://www.jstor.org/stable/2302799 .
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586
[November,
FRACTIONS
yk(S)
=
(Cs--
()kCn{cs
r7r
~~a
-
ank)
+
Rk}
then fork sufficiently
large Rk is uniformlysmall on the interval (-a, a). Now
we may write
k
+ 1)7rs
Cn+ 1
1(n +ls
?
a
(1)COS(
Rk=
cn nr+ IJ
Cn,+2
+
(n
Cn
x
k
)((n + 2
an?lk)
nl
2),7rs\
(nt +
)
-
n
?+2k)
+
Since the Fourier series (1) is convergent,the C's have a maximum absolute
value C, and hence throughoutthe interval (-a, a)
Cnk
IRk<
where
T=
1
(n+
+1
1)k
+
_
j
Cn
T
rX
1
dx
JO (n +
(n + 2)k
1
x) k
(k
-
)nk-1
It followsthat uniformlyon the interval (-a, a)
limI Rk I _ lim
kn
R
k?
and our theoremis a consequence.
oo I Cn
Cn
(k
- 1)
= 0,
FRACTIONS*
L. R. FORD,
Armour Institute of Technology
Perhaps the author owes an apology to the reader forasking him to lend his
attention to so elementarya subject, for the fractionsto be discussed in this
paper are, forthe most part, the halves, quarters,and thirdsof arithmetic.But
the fact is that the writerhas, forsome years,been lookingon these entitiesin a
somewhat new way. Here will be founda geometricpicturizationwhich will be
novel to the readerand whichwill supply a visual representationof arithmetical
resultsof diversekinds.
The idea of representinga fractionby a circleis one at which the author arrived by an exceedinglycircuitousjourney. It began with the Group of Picard.
In the geometrictreatmentofthis groupas carriedout by Bianchi in accordance
with the general ideas of Poincare certain invariant familiesof spheres appear.
These spheres,whichare foundat the complex rational fractions,are mentioned
later in this paper. They suggest analogous known invariant familiesof circles
at real rational points in the theoryof the Elliptic Modular Group in the com* Some of the material of this article was presentedin an address beforethe American Mathematical Society at Lawrence, Kansas, November 28, 1936. Other parts have been given in lectures
at the Rice Institute, the Universityof Texas, NorthwesternUniversity,and the Armour Institute.
1938]
587
FRACTIONS
plex plane. Finally, it became apparent that this intricatescaffoldingof group
theory could be dispensed with and the whole subject be built up in a completely elementaryfashion. It is this treatmentthat is undertakenhere.
1. The geometric representation. We begin with real fractions.These are
usually represented,along with real irrationals,by points on a line. Let this line
be the x-axisin an xy-planeof rectangularco6rdinates.
Through each rational point x = p/q, wherep and q are integersand the fraction is in its lowest terms,we constructa circle of radius 1/(2q2) tangent to the
x-axisand lyingin the upper half-plane.It is this circle,whichtouches the x-axis
at the point usually taken to representthe fraction,whichwill be the geometric
picture of the fraction.The integersare representedby circlesof radius 1/2; the
fractions1/3, 2/3, 4/3, etc., by circles of radius 1/18; and so on. Every small
interval of the x-axis contains points of tangency of infinitelymany of these
circles.*
A
~~~E
D
)
P
q
P
Q'
1
FIG.
Let p/q and P/Q be two differentfractionsin theirlowest terms. Consider
the distance betweenthe centersof theirrepresentativecircles. In the figurethe
horizontaldistance A C is I (P/Q) - (p/q) and the vertical distance CB is the
difference of the radii, I (1/2Q2) - ( /2q2) . We have
AB2
(P
AB-Q
-
( 21
\2Q2
P+(
q-
+
$2
2q28
IQ2q2
(AD+EB)2 +
)
2--21)
2Q2(Pq-
(Pq
-
pQ)2pQ)2-1
Q2q2
* Families of circles tangent to the real axis at the rational points are intimatelyinvolved in
the geometryof the modular group and in the theory of quadratic forms. In these connections
they have been used forsome time. Circles definedas in the text except forthe more general radius
1/(2hq2) were used, with various values of h, by the presentauthor in papers in the Proceedings
of the Edinburgh Mathematical Society in 1917. They were probably used earlier by others.
These circles are called "Speiser circles" by some writers.This name appears to be due to a
note of a dozen lines by A. Speiser in the Actes de la Socifte Helvetique des Sciences Naturelles in
1923. Mention should also be made of a very interestingninety-page booklet by Ziillig, GeometrischeDeutung unendlicherKettenbruache(1928).
588
[November,
FRACTIONS
If lPq-pQl >1, then AB>AD+EB,
and the two circles are wholly external to one another.
If lPq-PQ|
=1, then AB=AD+EB,
and the two circles are tangent.
If lPq-PQI <1 we have, since Pq-pQ is an integer, that Pq-pQ=O.
From this P/Q = p/q, which is contraryto the assumptionthat the fractionsare
different.So this last inequality is not possible. We may state the followingresult:
THEOREM 1. The representative
circlesoftwodistinctfractionsare eithertangent
or whollyexternalto one another.
2. Adjacent fractions.We shall call two fractionsp/q and P/Q adjacent if
theirrepresentativecirclesare tangent. The conditionforthis is that Pq - pQ
= 1. (We assume here and henceforththat fractions appear in their lowest
terms.) We shall prove
THEOREM
2. Each fractionp/q possessesan adjacentfraction.
That integersP and Q exist satisfyingthe equation IPq - PQ =1 (or, arrangingsigns suitably, satisfyingPq -pQ = 1) is a fundamentalpropositionof
the theoryof numbers.A great many proofshave been given. For the sake of
completenesswe prove the theoremhere. The proofgiven is so arranged as to
cover the case of complexfractionswhichwill appear subsequentlyin this paper.
The proofis by induction. Clearly the theoremholds for IqI = 1, since p/I
then has the adjacent fraction(p+1)/1. To prove the theoremin general we
assume that all fractionswhose denominatorsare less in absolute value than
qI possess adjacent fractionsand prove it forthe fractionp/q.
Let n be the integernearest to p/q, whence we can write,with m integral,
nq-+m
= nm
yO<
+-=
p _n
q
q
q
Iml<Iq
Since jmj < I q ,then q/m has an adjacent fractionr/s, so Isq - rm1-1. Then
the fraction
P
s
nr+s
= n + -=
is adjacent to p/q; for
IPq
-
PQ =
Q
r
r
(nr+ s)q - (nq + m)r|
=
sq - rm = 1.
This establishes the theorem.
We can now give a formulaforall fractionsadjacent to p/q.
THEOREM
3. If P/Q is adjacent to p/q thenall fractionsadjacent to p/q are
P?z
P + np
Qn
Q + nq
wheren takeson all integralvalues.
1938]
589
FRACTIONS
We findreadily that the fractionsgiven are adjacent to p/q, for
I (P + np)q
-
p(Q + nq) I =
Pq
-
pQ
= 1.
are adjacent to one another, since
We findthat Pn,/Qnand P?+1/Qn+1
[P +np][Q+(n+1)q]-
[P+(n+1)p][Q+nq]
I = IPq-PQI
= 1.
The circles correspondingto these fractionsforma ring around the circle of
p/q, all tangent to the circle of p/q, and each tangent to the circleswhich precede and follow it in the sequence (Figure 2). That this ring completelysur-
p
FIG.
2
rounds the circle of p/q we see from
(a)
Qn
==p
q
Pq-pQ
q(Q + nq)
q
+
1
q2(n + Q/q)
When n-->+ oo, Pn/Qn approaches p/q from one side; when n-o>- c, Pn/Qn approaches p/q fromthe other side.
It is obvious fromthe geometricpicture that it is not possible to draw a
cilcle lying in the upper half-plane,touching the x-axis, and tangent to the
circle of p/q but not intersectingthe circles of the ring surroundingthe circle
of p/q. It followsthat there are no furtherfractionsadjacent to p/q.
THEOREM 4. Of thefractionsadjacent to p/q (1q I > 1) exactlytwohavedenominatorsnumericallysmallerthan q.
That two circlesof the precedingringabout the circle of p/q are largerthan
that circle is fairlyevident fromthe geometricalpicture. We see also that
I Q + nq| < I ql,
or
I n + Q/q| < 1
for exactly two values of n; namely, those integersbetween which -Q/q lies.
590
FRACTIONS
[November,
For one of these two values of n, n+Q/q is positive,forthe other,negative.
We see thenfrom(a) that one of the fractionsof Theorem 4 is greaterthan p/q,
the other is less.
We shall add one furthercircleto our collection.The fraction1/0 is formally
adjacent to the integersp/1, since 1 1-p- O = 1. We take as its circle the line
y 1, which touches the circles of all the integers.We consider as the interior
of the circle that part of the plane above the line.
3. The mesh triangles. The parts of the upper half plane exteriorto all the
circles of the system consist of an infinitenumber of circular arc triangles to
which the name of "mesh triangles"will be given. Any two sides of a mesh triangle lie on circles belongingto adjacent fractions.Its angles are zero.
A furtherstudy of the mesh trianglewill be made in a later section,whereit,,
propertieswill be used in the theoryof approximation.
4. Farey's series. Let a curve be drawn across the set of circlesthat we have
just defined.We start with a point Ao of the upper half-planeand trace a continuous curve L which remains in the upper half plane except that its terminal
point, if any, may possibly lie on the x-axis. We consider the fractionswhose
circles are passed throughin succession by L. Each circle K is surroundedby
mesh triangles.If L issues fromK into one of these trianglesand if it does not
returnto K it will on leaving the trianglepass in generalinto a circletangentto
K. We agree that K shall not be counted twice ifL passes out of K and then returnsto K again withoutenteringanother circle. We also make the convention
that if L touches two circles at theirpoint of tangencywithoutenteringeither
then one or the othershall be consideredas crossed by L (the largercircle,say,
or the one on the left,or the one whose fractionis the greater).We can thenstate
as an established propositionthe following:
PRINCIPLE.
If twocirclesof thesystemare penetratedin succsssionbyL thenthe
correspondingtwofractionsare adjacent.
As a firstillustrationlet L be a line, y = k, parallel to the x-axis,startingsay
at a point on the positive y-axis and stopping at x= 1. The points of tangency
with the x-axis of the circlesthroughwhich L passes are arrangedin orderfrom
left to right; that is, the correspondingfractionsare arranged in numerical
L intersects the circles of all fractions in the
order. If 1/(n+1)2<k<1/n2,
interval O<x<1 whose denominators do not exceed n and the circles of no
other fractions.These fractionsarranged in numericalorder constitutewhat is
known as a Farey's series of ordern forthe interval.
Certain theorems concerning Farey's series are now readily established.*
Let p/q <p'/q' be successive fractionsof the series,all the numbersbeing positive.
* See, for example, Landau, Vorlesungen iuberZahlentheorie, 1927, Band 1, pp. 98-100.
Farey discussed these fractionsmore than a centuryago.
1938]
(1) P'q-pq'=
591
FRACTIONS
1. This results from the Principle stated earlier in this section.
(2) q+q'_n+l.
For (p+p')/(q+q')
is a fractionbetween p/q and p'/q';
since it does not belong to the series its denominatoris greaterthan n.
(3) For any number co of the interval there is a fractionp/q of Farey's
series of order n such that
p
1
+ l)q
q
|(n
The numbercolies in an interval formedby a Farey's fractionp/q and an
adjacent fraction(p+pl)/(q+ql) not of the series (pi/qi being the Farey's fraction precedingor followingp/q). The lengthof this intervalis t/(q+qi)q, where
q+ql_n+lt.
If we take other simple formsfor the curve L we get series analogous to
Farey's. Thus if we take a line with a positive slope startingabove y = 1 and
terminatingat the originwe have all non-negativefractionssuch that
pq < n,
wheren is suitably chosen. If these be arrangedin orderof numericalmagnitude
then successive fractionssatisfy (1) above; (2) is replaced by (p+p')(q+q')
>n?l .
The seriessuch that 0 <p ? n may be got by takingforL the arc of a suitable
circle tangent to the x-axis at the origin.
5. The problem of approximation. Dirichlet showed by elementarymeans
that if cois a real irrationalnumber then the inequality
|q
(1)
p
qq
C
k
<-
is satisfiedby infinitelymany fractionsp/q if k = 1. If, however,co= r/s, a rational, then the inequality is satisfiedby only a finitenumber of rationals p/q
however large k be chosen. For
q
r
p
rq-ps
s
q
sq
1
s
k
sql
q2
q2
except for a finitenumber of values of q for which qI
I ks|; also for each q
thereis clearly only a finitenumberof fractionssatisfyingthe inequality.
Various suites of fractionsapproximatingto co satisfy (1) with suitable k.
The convergentsin the ordinarycontinued fractionforcosatisfythe inequality
with k = 1. A suite of Hermite admits the smaller value k= 1/V/3.
The determinationof the best value of k is due to Hurwitz,*who proved the
followingtheorem.
* Mathematische Annalen, vol. 39, 1891, pp. 279-284. See also Borel, Journal de Mathematiques, 5th ser., vol. 9, 1903, pp. 329 ff;L. R. Ford, Proceedings of the Edinburgh Mathematical Society, vol. 35, 1917.
592
[November,
FRACTIONS
THEOREM 5. If k = 1/V\, thenfor each irrationalX thereare infinitely
many,
fractionsp/q satisfyingtheinequality(1).
If k < 1 /V, thenthereare irrationalscosuch thatonlya finitenumberoffractionsp/q satisfy(1).
Let the curve L of the previoussection be a vertical line whichterminatesat
the irrationalco.Then L intersectsinfinitelymany circles of the system we are
consideringand is tangent to none. If L cuts the circle of p/q then the distance
fromcoto p/q is less than the radius:
p_
1
q
2q2
The curve L thus provides us with an infinitesuite of all those fractionssatisfying (1) when k= 1/2. If p/q and p'/q' are successive fractionsof this suite then,
fromthe Principle of the preceding section, we have P'q-pq'= + 1. Also, we
ofthesuiteto X is thenimthat Iq'| > Iqf. The convergence
see geometrically
mediately proved.
We propose now to prove Hurwitz' theoremby a study of the circles of out
system. This elementaryproof will be free of continued fractionson the one
hand and of the theoryof the Modular Group on the other.
The line L cuts across infinitelymany mesh triangles.The firstpart of the
theoremof Hurwitz is a consequence of the followingresult which will now be
proved:
THEOREM 6. Of thethree
fractionswhosecirclesformtheboundaryofa meshtriangle whichL crosses,at least one satisfies(1) withk==/d5.
We consider the mesh triangle in some detail. Let P/Q, p/q, pl/ql be the
fractionswhose circles bound the area, where
O < Q < q < qi = q+Q,
Pl=P+P.
We suppose, to fix the picture, that the largest of the three circles is on the
right;that is, P/Q > p/q, whence Pq-pQ = + 1. If not, we could make suitable
changes of sign in the following;or, more simply, we could reflectthe entire
figurein the y-axis by changingthe signs of the numeratorsand of X and at the
conclusion of our analysis we could reflectagain.
The vertexA (see Figure 3) is the point of contact of the circlesof p/q and
P/Q. It divides the line of centers of these circles in the ratio of their radii,
1/2q2:1/2Q2,or Q2:q2. The abscissa of this point is found by an early formula
of analytic geometry,or by the methods of high school plane geometry,to be
a=
P
P
_
q2. + Q2-.
q
Q
q2 +
Q2
_
-
q+PQ
q2 _+
Q2
1938]
593
FRACTIONS
The abscissas of B and C, the othervertices,may be writtendown by an interchange of letters,
pq + piql
PQ + piqi
q2 +
Q2 +
q2
q?2
In order that L cross the mesh triangle under considerationit is necessary
and sufficientthat colie in the intervalwhose rightend is c and whose leftend
A
ab
b
,
Pl c
c
cafe I
CseXf
FIG. 3
is the lesser of the two quantities a and b. We find
a---
p
Q
q
q(q2+Q2)
b-=-
P
q_
q
q(q2+qq2)
whence subtractingand rememberingthat qi = q + Q, we have
q2 - qQ - Q2
b-a=
(q2 + Q2)(q2+ q?2)
We put q/Q = s > 1 The sign of b - a is (dividing the preceding numerator
by Q2) the same as the sign of s2- s -1. Now, factoring,
s2
- s-
1 =
(s
)
+
-
21),
and the square root of 5 firstenters the picture. Here the firstfactoris positive
and the sign is determinedby the second factor.We have two cases to consider.
Case I. a<b, and * s> (V/5+1). We shall show that P/Q satisfies the
inequality (1) with k= 1/\/5.We have
P
Q
-co
P
<--a
Q
=
q
Q(q2 + Q2)
-
s
1
s2 + 1
Q2
* Hereand inmanysubsequent
relationsthesignofequalityis notpossible,foronememberis
rationaland theotheris irrational.
594
[November,
FRACTIONS
If now
we have
s
1
s2+ 1
V5
(5 -
I ?)
s + l<
s2 -
<
vX)
(
whichis impossiblesince both factorsin the firstmemberare positive. It follows
that s/(s2+ 1) < 1/\/5,and the inequality is satisfied.
Case II. b<a, and s< 2(\5+ 1). We shall show that pl/ql satisfiesthe inequality (1) with k = 1/A/5.It is clear that c is nearer pil/ql than b is, since C
is higheron the circle than B. Then
Pi
--wc
If
<-
q,
Pi
q
b =
qi
ql(q2+
s(s+
s2 +
q2)
(s +
1)
1)2
1
q?2
1
s(s+1)
s2 + (S + 1)2
then
> 0,
(V5- 2)s2? (/5-2>-1
or, dividing,
s2 + s -V5-2
> 0.
Factoring,
(s
?
d
) (s + v
> O
?)
which is impossible since the firstfactoris negative and the second is positive.
We have proved that one, at least, of the threefractionswhose circlesbound
the mesh trianglesatisfies(1) with k = 1/\/S. Since L passes throughinfinitely
many mesh trianglesit followsthat infinitelymany rational fractionssatisfythe
inequality with k = 1/ 5.
It remains to exhibit an irrationalco for which (1) holds for only a finite
and any
We take, in fact, co==(\5+1)
number of fractions for k<l/v5.
h < 1 and show that thereis a finitenumberof fractionsforwhich
(2)
q
_
S+1
2
<
<\Vq2
For a fractionsatisfyingthis we may write
p _\v5 +t1
q
2
0
\1,5q22
1938]
595
FRACTIONS
where I0 | < h < 1. Writingthis
p__
1
2
q
VH5
=
?
2
0
\V5q2
squaring, and rearranging,we get
pq
-
5q2[(p2
-
]
-
q2)
2
=
The expressionin square brackets must be positive, whence the integerp2-pq
-q 2, whichcannot be zero (since then p/q would turnout irrational)must equal
or exceed 1. We have
h2
ft2
pq
5[(p2_
- q2)
-
5(1-h)
]
This limits q to a finitenumber of values. For each q, the inequality (2) then
limits p; and the number of fractionsis finite.
Other irrationals might have been used here; for example, co= (rV\5?s)/t,
where r, s, t are integers.These last irrationalnumbersare found in every interval of the x-axis.
6. Continued fractions. One of the most successfulinterpretationssupplied
by our system of circles is the geometricpicture of the continued fraction
(3)
ao +
1
a, +
a2 4
a3 +...
or, as oftenwritten,
aO +
1
1
1
a, + a2?
a3 + * *
Here the quantities a,,,are integers.If all the integersare positive, except possibly a0, the continued fractionis simple.*
The nth convergent,
pI/qn, is the quantity that remains when all that part
of the expressionfollowinga,,,.-,is erased. We have
Pi
aO
P2
1
aoa +1
qi
1
a2
a,
a,
1
p3
-
q3
=a+
(aoa, + 1)a2 + aO
1
a, + a
=.
a1a2 +1
=
a2p2 + pI
a2q2
+
ql
a2
* So called by Chrystal in whose Algebra, Part II, pp. 396ff,will be found one of the best
elementarytreatmentsof continued fractionsin English.
596
[November,
FRACTIONS
Here and in the followingwe are using the p's and q's to representthe actual
numeratorsand denominatorsof the fractionsin the second membersand not
merelynumbers proportionalthereto.
The last equation exhibits forn=2 the recurrenceformulafromwhich we
calculate the convergentsstep by step,
(4)
Pn+1
anpn+ pn-1
qn+1
anqn +
qn-i
This formulais readily established by induction. It holds, we have just seen,
forn = 2; assuming it up to any later n we prove it forn +1. We get Pn+2/qn+2
fromPn+l/qn+iby replacingan by an+l/an+l;
Pn+2
(an + 1/an+l)pn+ Pn-1
anl+(anPn+ Pn-1)+ pn
an+lPn+l+ pn
qn+2
(an + 1/an+i)qn +
an+l(anqn + qn-1) + qn
an+lqn+l + qn
qn-1
This is the required formulawith n replaced by n+1 and (4) holds forn >2. It
holds also forn = 1 if we take po/qo= 1/0.
If the continuedfractionterminates,its value is the last convergent.If there
are infinitelymany a's the value to be assigned to the fractionis lim pn/qnif
this limit exists.
We investigatethe suite of convergents
THEOREM
Po
1
P
ao
p2
p3
qo
Oi
q,1\
1
q2
q3
7. Successiveconvergents,
pn/qnand Pn+l/qn+l,are adjacentfractions.
From (4)
pr+lqn- qn+lPn = (anPn +
pn-i)qn
-
(anqn +
qn-l)pn
=
-
(pnqn-1
-
qnpn-1).
= -1 we have that
Since piqo-poqi= ao O-1
Pn+lqn - qnFlPn =
(-
1)n+l,
which establishes the theorem.
Given the tangent circles of pn_1/qn_i
and pn/qnand the integeran, how do
we findthe circleofPn+l/qn+l?For all integralan we get from(4) (see Theorem 3,
Section 2) the ringof tangentcirclesaround the circleof pn/qn.For an= 0 we get
foran= 1 we get a circlenext to this; foran = 2 we get the
the circleof Pn_l/qn-1;
next around the ring;and so on. For an -1, -2, etc.we pass around in the opposite direction.
To determinethe directionconsider
pn+1
Pn
q1) n+q
qn+1
qn
qn+lqn
(nq
(anqn
?
q+l
+
qn-l)qn
qn (
2n
(an+
1)+
1
1938]
597
FRACTIONS
For a. large and positive the last denominatoris positive. If n+1 is even then
Pn+l/qn+lis to the right of pn/qnand we have counted around the circle of
and we
is less than pn/qn
Pn/q2in a clockwisedirection;if n +1 is odd pn+1/qn+1
have counted in a counter-clockwisedirection.
2
ao
P3P2
q3
q2
FIG. 4
We may liken the circle of pn/qn to the face of a clock (Figure 4). Set the
hand to point toward the centerof the circle of Pn_i/qn_.
It is now in the zero
position. If we seek an even convergent turn clockwise to the anth point of
tangency; if we seek an odd convergent turn counter-clockwiseto the anth
point of tangency. If an is negative the turningis in the opposite directionin
each case. The hand now points to the centerof Pn+l/qn+li
Initially (n = 1) we set the hand to point verticallyupward to the point of
tangency of y= 1 with the circleof ao/1, as in the figure.The positive direction
is clockwise. On the next clock the positive directionis counter-clockwise;on
the next, clockwise; and so on. In the figurea,=2, a2=2, and if ao0=O the first
fourconvergents,startingwith po/qo,are
1
0
1
2
0
1
2
5
We see then that thecirclesoftheconvergents,
in order,of a continuedfraction
forma sequence,or chain,of circles.The chain beginswiththecircley = 1 and each
circleis tangentto thecirclesprecedingand followingit in thesequence.
Conversely,any such chain has a unique correspondingcontinued fraction.
We determine ao as the integer whose circle touches the firstcircle, y = 1, in
the chain; and a2, a3, etc., are got uniquely by countingpoints of tangency,as
already explained.
A chain can be definedin many ways and with much arbitrariness.A simple
598
FRACTIONS
[November,
way is to use a curve L, startingabove y= 1, as explained in Section 4. We shall
= Pn-i/qn_-i
nor to qn= 0, which means
not object to an=0, whichgives Pn+l/qn+l
that y= 1 reappears as a circle of the chain.
The preceding picture answers various questions about convergence. We
see that it is possible so to choose the chain that pn/qnapproaches a prescribed
rational or irrationallimit in a great variety of ways, or so that pn/qnbecomes
positivelyor negativelyinfinite,or so that Pn/qnwanders moreor less arbitrarily
over the x-axis. We may even choose the chain so that each interval of the xaxis contains infinitelymany convergents.Such a chain is the set of circles
passed over by the broken line L formedby joining in succession the points
A1, A2, * * whereAn has the coordinates [(-1)8n, 1/n]. In this example each
real rational numberappears infinitelyoftenamong the convergents.
7. Simple continuedfractions.*The situationis quite different
ifall an(n > 0)
are positive integers.The firstclock runs right (see Figure 4) and makes P2/q2
greaterthan pl/ql; the next runs leftand places p3/qa between pi/ql and P2/q2.
Each convergentfalls between the two immediatelyprecedingit. Certain facts
then appear:
(a) The fractionconvergesto a value w.
(b) The odd convergentsincrease monotonicallyto w,the even convergents
decrease monotonicallyto co.
(c) colies between each pair of successive convergents.
(d)
qn+l
>
qn
1).
(>
co 1<
<2
(e) O ltn_
qn
1
qnqn+l
1
anqn
c1
qn
Here the second expressionis the distance between the convergentsPn/qnand
? anqn+qn-1 > anqn (n> 1).
Pn+l/qn+i, and qn+l
-pn
qn
,
an>
qnqn+2
forthe second memberis the distance between pn/qnand pn+2/qn+2.
We see geometricallyhow to findthe simple continued fractionfora given
co.We select for ao the integernext below X (see Figure 4). We then turn the
clock to the rightuntil p2/q2 lies as far to the left as possible without passing
co,thus findinga,. We turnthe second clock to the leftuntil p3/q3 is as farto the
rightas possible withoutpassing w,whence we have a2; and so on.
That the developmentis unique is apparent. The firstan's that differin two
developments throw the values of the two continued fractionsinto different
intervalsthereafter.There is an unimportantexception in the case of a fraction
which terminates.A last an > 1 may be written (an-1) + 1/1, thus insertingan
additional convergent.
* Zullig,loc. cit.,has thechainofcirclesgivenherewithfigures
formanynumericalcases.
1938]
599
FRACTIONS
It is not difficult
to identifythis process of formingthe continuedfractionfor
co with the usual arithmeticprocess: (1) taking ao as the largest integerin W;
(2) formingw1,the reciprocalof the remainderw - ao, and takinga, as the largest
integerin wi; (3) taking a2 as the largest integerin (2, the reciprocal of wl-a1;
and so on.
We see geometricallythat pn/qn is a good approximationto w if an is very
large, for our clock has then turned far toward the bottom. This is clear, of
course, from(e) above. Thus in the continued fraction
7r=
3 +-
1
1
1
-
1
--
1
1
-
7 + 15 + I + 292+ I + I +
*
the second convergent22/7 and the fourth355/113 (here a4= 292) are especially
good and are, in fact, much used.
Suppose we take an,= 1 forall n in order to avoid a good approximationas
far as we can. We have
co=1?--
1
1
I
I
+
1?
+-
1
-,
(A)
From this, since w> 1,
or w2-w-1=0.
co= 21(5 + 1).
This is the irrationalquantitywe used in the latterpart ofthe proofofHurwitz's
theorem.
8. Complex fractions.*Complex numbersx+iy, where x and y are real and
i=
-, are commonlyrepresentedby points (x, y) in an xy-plane using rectangular coordinates. The complex integers,n=n'+in",
where n' and n" are
real integers,forma square lattice in the plane. A fractionis the quotient of two
integers
p
pi
q
qI +
+?ip"l
iq"
p?+ip"l
q' +
iq"
l
q'
-iq"
-
iq"
p'q' + p"lq"
q2 -4- q/2
/t-pl
q'2 +
q"12
and is representedby the point with the co6rdinatesshown in the last member.
The complex fractionsthus consist of the numbersx+iy, wherex and y are real
fractions.We thinkof the xy-planeas being horizontal,and we introducea third
or z-axis perpendicularto the x- and y-axes.
The analogue of the circle of Section 1 is a sphere. Through the point in the
plane which representsthe fractionp/q (in its lowest terms) we construct a
sphere touching the complex plane there, lying in the upper half-space, and
having the radius 1/(2qq). Here q is the conjugate of q, q=q'-iq";
whence
* For the matterstreated in this section see L. R. Ford, Transactions of the American Mathematical Society, vol. 19, 1918, pp. 1-42.
600
qq=q
[November,
FRACTIONS
12 +q'
12=q|
2.
This spherewill be the geometricrepresentation
of p/q.
Every small area in the complex plane contains points of tangencyof infinitely
many of these spheres.
InterpretingFigure 1 as a pair of spheres at complex fractions,we have
AB2=++
P
p
Q
q
2
1
\2
1
1
1
2QQ
2qq/
2QQ
2qq
Pq
-pQ121
QQqq
If I Pq-pQ I > 1, thenAB >AD +EB, and the spheres are whollyexternalto
one another. If IPq -pQ = 1 the spheresare tangent. In no case do the spheres
intersect.We call the fractionsadjacentifthe spheresare tangent.
FIG.
5
That p/q has an adjacent fractionP/Q we already know, for we have designed the proofof Theorem 2 to cover this case. Then the fraction
Pn
P + np
Qn
Q + nq
of Theorem 3, wheren is any complex integer,is adjacent to p/q. We note that
the four fractionsPm/Qm,where m=rn+l, n-1, n+i, n-i, are adjacent to
= |m-rnl | *Pq-PQI
=1.
Pn/Qn; for I (P+?np)(Q+mq)-(P+mp)(Q+?nq)|
That is, each sphere tangent to a given sphere touches four other spheres
tangent to the given sphere. It is not difficultto show that the fractionsPn/Qn
constitutethe complete set of fractionsadjacent to p/q.
How many of the spheres touching the sphere of p/q are larger than that
sphere? We require the numberof solutions of
fQ+nql
<Iql,
or
I n+Q/qI
< 1.
Now n + Q/qI is the distance fromthe integern to the point -Q/q. The problem takes the form: If a unit circle is drawn in the complex plane, how many
complex integersdoes it contain? For various positions of the center (-Q/q)
the answer is 2, 3, or 4, as is readily seen. We put aside the case I q = 1, which
puts the centerat an integer,when clearly thereare no largertangent spheres.
We introduce the plane z= 1 to touch the spheres of the integersand be the
sphere of theiradjacent fraction1/0.
1938]
MATHEMATICAL
601
EDUCATION
The use of complex integersin the continuedfraction(3) offersno difficulty.
We are led to the geometricpicture of the continued fractionas a chain of
spheres,beginningwith the plane z= 1, and proceedingthence fromsphere to
tangent sphere. We have here the same possibilitiesof convergenceor divergence as before. At the worst we can set up continued fractionswhose convergents are present in every small region of the complex plane. There is no
immediate generalizationof the simple continuedfractionto the complex case.
There are, however,schemes fordeveloping a complex number in a continued
fraction,the best processes being due to Hurwitz.*
The analogue of Hurwitz's theoremon rational approximationsto an irrational numberwas firstdiscoveredby the presentauthor.t The inequality (1) is
replaced by
k
p
|
_
q
co
< -1
qq
and the minimumk for which infinitelymany fractionsalways satisfythe inequality is k = 1/VX. The proofof this,however,is not elementary.
MATHEMATICAL
EDITED BY
EDUCATION
of Colorado
C. A. HUTCHINSON, University
This Departmentof the MONTHLY has been createdas an experimentto afforda place for the
discussion of theplace of mathematicsin education. Withthis topic will naturallybe associated other
mattersemphasizingthe educational interestsof thosewho teach mathematics.It is not intendedto
critical
take up minutedetails ofteachingtechnique.The columnsare open to thosewho havethoughtful
commentto make, be itfavorableor adverseto thecause of mathematics.The successofthisdepartment
to
obviouslywill depend upon thecooperationof thereadersofthe MONTHLY. Address correspondence
ProfessorC. A. Hutchinson, Universityof Colorado,Boulder, Colorado.
MATHEMATICAL
EDUCATION
IN GERMANY BEFORE
1933
RICHARD COURANT, New York University
At the suggestionof the editors I shall try to give a briefaccount of trends
in the teaching of mathematics in Germany,particularlyat German universities, in the period fromthe World War until 1933.
The situation of mathematicsin Germany,since the early part of the 19th
century,has been pivoted around a definiteconnectionbetween universityand
high school. In a sweepingreformfollowingthe French Revolution, the institution of the German "humanistischesGymnasium"was established. Teachers at
these institutionswere required to undergoa very thoroughacademic preparation,and the task of preparingthemwas entrustedto the philosophicalfaculties
of the universities.The teachers' trainingwas in no way of an elementarychar-
154.
* Acta Mathematica, vol. 11, 1887, pp. 187-200.
t L. R. Ford, Transactions of the American Mathematical Society, vol. 27, 1925, pp. 146-