THE LIBRARY
OF
THE UNIVERSITY
OF CALIFORNIA
LOS ANGELES
FAMOUS PROBLEMS
ELEMENTARY GEOMETRY
THE DUPLICATION OF THE CUBE
THE TRISECTION OF AN ANGLE
THE QUADRATURE OF THE CIRCLE
AN AUTHORIZED TRANSLATION OF F. KLEIN'S
VORTBAGE UBER AUSGEWAHLTE FRAGEN DER ELEMENTARGEOMETRIE
AUSGEARBEITET VON F. TAGERT
BY
WOOSTER WOODRUFF BEMAX
PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MICHIGAN
AND
DAVID KHiKXE SMITH
;>^>I: OF MAIIU.M in. > i.\ TEAI-HERS COI.LKOK. CuLriiuiA UMVLKSIIV
GINX AND COMPANY
BOSTON NM.\V YoltK CHICACiO I.OXIKN
ATLAXTA DALLAS COLL'-MHUS SAN KUAXCISCO
COPYRIGHT, 1897, BY
WOOSTEB WOODRUFF BEMAN AITO DAVID EUGENE SMITH
ALL RIGHTS RE8ERYED
316.9
atfrtnacum
GINN AND COMPANY- PRO-
PRIETORS BOSTQN U.S.A.
s
i\ f Library
i Jo *3
PREFACE.
THE more precise definitions and more rigorous methods of
demonstration developed by modern mathematics are looked
upon by the mass of gymnasium professors as abstruse and
excessively abstract, and accordingly as of importance only
for the small circle of specialists. With a view to counteract-
ing this tendency it gave me pleasure to set forth last summer
in a brief course of lectures before a larger audience than
\ usual what modern science has to say regarding the possibility
' of elementary geometric constructions. Some time before, I
^ had had occasion to present a sketch of these lectures in an
* Easter vacation course at Gottingen. The audience seemed
*k
to take great interest in them, and this impression has been
confirmed by the experience of the summer semester. I ven-
ture therefore to present a short exposition of my lectures to
the Association for the Advancement of the Teaching of Math-
ematics and the Natural Sciences, for the meeting to be held at
Gottingen. This exposition has been prepared by Oberlehrer
Tagert, of Ems, who attended the vacation course just men-
tioned. He also had at his disposal the lecture notes written
out under my supervision by several of my summer semester
students. I hope that this unpretending little book may con-
tribute to promote the useful work of the association.
F. KLEIN.
GOTTINGEN, Easter, 1895.
TRANSLATORS' PREFACE.
AT the Gottingen meeting of the German Association for
the Advancement of the Teaching of Mathematics and the
Natural Sciences, Professor Felix Klein presented a discus-
sion of the three famous geometric problems of antiquity,
the duplication of the cube, the trisection of an angle,
and the quadrature of the circle, as viewed in the light of
modern research.
This was done with the avowed purpose of bringing the
study of mathematics in the university into closer touch with
the work of the gymnasium. That Professor Klein is likely
to succeed in this effort is shown by the favorable reception
accorded his lectures by the association, the uniform commen-
dation of the educational journals, and the fact that transla-
tions into French and Italian have already appeared.
The treatment of the subject is elementary, not even a
knowledge of the differential and integral calculus being
required. Among the questions answered are such as these :
Under what circumstances is a geometric construction pos-
sible ? By what means can it be effected ? What are tran-
scendental numbers ? How can we prove that e and TT are
transcendental ?
With the belief that an English presentation of so impor-
tant a work would appeal to many unable to read the original,
vi TRANSLATOR'S PREFACE.
Professor Klein's consent to a translation was sought and
readily secured.
In its preparation the authors have also made free use of
the French translation by Professor J. Griess, of Algiers,
following its modifications where it seemed advisable.
They desire further to thank Professor Ziwet for assist-
ance in improving the translation and in reading the proof-
sheets.
August, 1897.
W. W. BEMAN.
D. E. SMITH.
CONTENTS.
INTRODUCTION.
MM
PRACTICAL AND THEORETICAL CONSTRUCTIONS .... 2
STATEMENT OF THE PROBLEM IN ALGEBRAIC FORM ... 3
PART I.
The Possibility of the Construction of Algebraic Expressions.
CHAPTER I. ALGEBRAIC EQUATIONS SOLVABLE BY SQUARE ROOTS.
1^4. Structure of the expression x to be constructed ... 5
5, 6. Normal form of x ........ 6
7, 8. Conjugate values 7
9. The corresponding equation F(x) = o .... 8
10. Other rational equations f(x) = o . . . . . .8
11, 12. The irreducible equation 0(x) = o . . . . 10
13, 14. The degree of the irreducible equation a power of 2 . .11
CHAPTER II. THE DELIAN PROBLEM AND THE TRISECTION OF THE
ANGLE.
1. The impossibility of solving the Delian problem with straight
edge and compasses ....... 13
2. The general equation x 3 = X 13
3. The impossibility of trisecting an angle with straight edge
and compasses 14
CHAPTER III. THE DIVISION OF THE CIRCLE INTO EQUAL PARTS.
1. History of the problem 16
2-4. Gauss's prime numbers ....... 17
5. The cyclotomic equation ....... 19
6. Gauss's Lemma ......... 19
7, 8. The irreducibility of the cyclotomic equation . . .21
viii CONTENTS.
CHAPTER IV. THE CONSTRUCTION OF THE REGULAR POLYGON OF
17 SIDES.
PAGE
1. Algebraic statement of the problem 24
2-4. The periods formed from the roots . . . . .25
6, 6. The quadratic equations satisfied by the periods . . 27
7. Historical account of constructions with straight edge and
compasses ......... 32
8, 9. Von Staudt's construction of the regular polygon of 17 sides 34
CHAPTER V. GENERAL CONSIDERATIONS ON ALGEBRAIC CONSTRUCTIONS.
1. Paper folding ......... 42
2. The conic sections 42
3. The Cissoid of Diocles ........ 44
4. The Conchoid of Nicomedes ...... 45
6. Mechanical devices , 47
PART II.
Transcendental Numbers and the Quadrature of the Circle.
CHAPTER I. CANTOR'S DEMONSTRATION OF THE EXISTENCE OF
TRANSCENDENTAL NUMBERS.
1. Definition of algebraic and of transcendental numbers . 49
2. Arrangement of algebraic numbers according to height . 50
3. Demonstration of the existence of transcendental numbers 53
CHAPTER II. HISTORICAL SURVEY OF THE ATTEMPTS AT THE COM-
PUTATION AND CONSTRUCTION OF TT.
1. The empirical stage 56
2. The Greek mathematicians ...... 56
3. Modern analysis from 1670 to 1770 58
4, 5. Revival of critical rigor since 1770 ..... 59
CHAPTER III. THE TRANSCENDENCE OF THE NUMBER e.
1. Outline of the demonstration 61
2. The symbol hr and the function 0(x) .... 62
3. Hermite's Theorem 65
CONTENTS. lx
CHAPTER IV. THE TRANSCENDENCE OF THE NUMBER TT.
PAGE
1. Outline of the demonstration 68
2. The function ^(x) 70
3. Lindemann's Theorem ....... 73
4. Lindemann's Corollary 74
5. The transcendence of IT . . . . . . 76
6. The transcendence of y = e x . . . . . . .77
7. The transcendence of y = sin J x ..... 77
CHAPTER V. THE INTEGRAPH AND THE GEOMETRIC CONSTRUCTION
OF 7T.
1. The impossibility of the quadrature of the circle with straight
edge and compasses ....... 78
2. Principle of the integraph 78
3. Geometric construction of IT . . . . . . .79
INTRODUCTION.
THIS course of lectures is due to the desire on my part to
bring the study of mathematics in the university into closer
touch with the needs of the secondary schools. Still it is not
intended for beginners, since the matters under discussion are
treated from a higher standpoint than that of the schools.
On the other hand, it presupposes but little preliminary work,
only the elements of analysis being required, as, for example,
in the development of the exponential function into a series.
We propose to treat of geometrical constructions, and our
object will not be so much to find the solution suited to each
case as to determine the possibility or impossibility of a
solution.
Three problems, the object of much research in ancient
times, will prove to be of special interest. They are
1. The problem of the duplication of the cube (also called
the Delian problem).
2. The trisection of an arbitrary angle.
3. The quadrature of the circle, i.e., the construction of TT.
In all these problems the ancients sought in vain for a
solution with straight edge and compasses, and the celebrity
of these problems is due chiefly to the fact that their solution
seemed to demand the use of appliances of a higher order.
In fact, we propose to show that a solution by the use of
straight edge and compasses is impossible.
2 INTRODUCTION.
The impossibility of the solution of the third problem was
demonstrated only very recently. That of the first and second
is implicitly involved in the Galois theory as presented to-day
in treatises on higher algebra. On the other hand, we find
no explicit demonstration in elementary form unless it be in
Petersen's text-books, works which are also noteworthy in
other respects.
At the outset we must insist upon the difference between
practical and theoretical constructions. For example, if we
need a divided circle as a measuring instrument, we construct
it simply on trial. Theoretically, in earlier times, it was
possible (i.e., by the use of straight edge and compasses) only
to divide the circle into a number of parts represented by
2", 3, and 5, and their products. Gauss added other cases
by showing the possibility of the division into parts where
p is a prime number of the form p = 2 2 ^ -(- 1, and the impos-
sibility for all other numbers. No practical advantage is
derived from these results; the significance of Gauss's de-
velopments is purely theoretical. The same is true of all the
discussions of the present course.
Our fundamental problem may be stated : What geometrical
constructions are, and what are not, theoretically possible ? To
define sharply the meaning of the word "construction," we
must designate the instruments which we propose to use in
each case. We shall consider
1. Straight edge and compasses,
2. Compasses alone,
3. Straight edge alone,
4. Other instruments used in connection with straight edge
and compasses.
The singular thing is that elementary geometry furnishes
no answer to the question. We must fall back upon algebra
and the higher analysis. The question then arises : How
INTRODUCTION. 3
shall we use the language of these sciences to express the
employment of straight edge and compasses ? This new
method of attack is rendered necessary because elementary
geometry possesses no general method, no algorithm, as do
the last two sciences.
In analysis we have first rational operations : addition,
subtraction, multiplication, and division. These operations
can be directly effected geometrically upon two given seg-
ments by the aid of proportions, if, in the case of multiplica-
tion and division, we introduce an auxiliary unit-segment.
Further, there are irrational operations, subdivided into
algebraic and transcendental. The simplest algebraic opera-
tions are the extraction of square and higher roots, and the
solution of algebraic equations not solvable by radicals, such
as those of the fifth and higher degrees. As we know how to
construct Vab, rational operations in general, and irrational
operations involving only square roots, can be constructed.
On the other hand, every iii (x) = 0, defined as the equation of lowest degree,
with rational coefficients, admitting the root Xj and conse-
quently all the x t 's (10).
12. PROPERTIES OF THE EQUATION (x) = 0.
I. < (x) is an irreducible equation, i.e., (x) cannot be
resolved into two rational polynomial factors. This irreduci-
bility is due to the hypothesis that < (x) = is the rational
equation of lowest degree satisfied by the x t 's.
For if we had
*(*)=* 00 x(*).
ALGEBRAIC EQUATIONS. 11
then < (Xi) = would require either if/ (x x ) = 0, or % (*i) = 0>
or both. But since these equations are satisfied by all the
conjugate values (10), < (x) =0 would not then be the equa-
tion of lowest degree satisfied by the Xi's.
II. (x) = has no multiple roots. Otherwise < (x) could
be decomposed into rational factors by the well-known meth-
ods of Algebra, and (x) = would not be irreducible.
III. < (x) = has no other roots than the x/s. Otherwise
F (x) and < (x) would admit a highest common divisor, which
could be determined rationally. We could then decompose
(x) into rational factors, and < (x) would not be irreducible.
IV. Let M be the number of x/s which have distinct values,
and let
X l5 X 2 , . . . X M
be these quantities. We shall then have
(x) = C (x x t ) (x . x 2 ) . . . (x X M ).
For < (x) = is satisfied by the quantities x t and it has no
multiple roots. The polynomial < (x) is then determined save
for a constant factor whose value has no effect upon (x) = 0.
V. (x) = is the only irreducible equation with rational
coefficients satisfied by the x t 's. For if f (x) = were another
rational irreducible equation satisfied by x : and consequently
by the Xj's, f (x) would be divisible by (x) and therefore
would not be irreducible.
By reason of the five properties of < (x) = thus estab-
lished, we may designate this equation, in short, as the irre-
ducible equation satisfied by the x/s.
13. Let us now compare F (x) and <(x). These two poly-
nomials have the x/s as their only roots, and (x) has no
multiple roots. F (x) is, then, divisible by < (x) ; that is,
12 FAMOUS PROBLEMS.
F! (x) necessarily has rational coefficients, since it is the quo-
tient obtained by dividing F (x) by (x). If F x (x) is not a
constant it admits roots belonging to F (x) ; and admitting
one it admits all the x/s (10). Hence F! (x) is also divisible
by < (x), and
If F 2 (x) is not a constant the same reasoning still holds, the
degree of the quotient being lowered by each operation.
Hence at the end of a limited number of divisions we reach
an equation of the form
F,, _ ! (x) = Ci 4> ( x )>
and for F (x),
The polynomial F (x) is then a power of the polynomial of
minimum degree (x), except for a constant factor.
14. We can now determine the degree M of <(x). F (x)
is of degree 2 m ; further, it is the vth power of < (x). Hence
2 m = vM.
Therefore M is also a power of 2 and we obtain the following
theorem :
The degree of the irreducible equation satisfied by an expres-
sion composed of square roots only is always a power of 2.
15. Since, on the other hand, there is only one irreducible
equation satisfied by all the x/s (12, V.), we have the converse
theorem :
If an irreducible equation is not of degree 2 A , it cannot be
solved by square roots.
CHAPTER II.
The Delian Problem and the Trisection of the Angle.
1. Let us now apply the general theorem of the preceding
chapter to the Delian problem, i.e., to the problem of the
duplication of the cube. The equation of the problem is
manifestly
x 3 = 2.
3.
This is irreducible, since otherwise V2 would have a
rational value. For an equation of the third degree which is
reducible must have a rational linear factor. Further, the
degree of the equation is not of the form 2 h ; hence it cannot
be solved by means of square roots, and the geometric con-
struction with straight edge and compasses is impossible.
2. Next let us consider the more general equation
x 3 = A.,
X designating a parameter which may be a complex quantity
of the form a + ib. This equation furnishes us the analyt-
ical expressions for the geometrical problems of the multi-
plication of the cube and the trisection of an arbitrary angle.
The question arises whether this equation is reducible, i.e.,
whether one of its roots can be expressed as a rational func-
tion of X. It should be remarked that the irreducibility of
an expression always depends upon the values of the quan-
tities supposed to be known. In the case x 3 = 2, we were
dealing with numerical quantities, and the question was
whether v2 coiild have a rational numerical value. In the
equation x 3 = X we ask whether a root can be represented by
a rational function of A. In the first case, the so-called
14 FAMOUS PROBLEMS.
domain of rationality comprehends the totality of rational
numbers ; in the second, it is made up of the rational func-
tions of a parameter. If no limitation is placed upon this
parameter we see at once that no expression of the form
in which (A) and i/r(A) are polynomials, can satisfy our
equation. Under our hypothesis the equation is therefore
irreducible, and since its degree is not of the form 2 h , it can-
not be solved by square roots.
3. Let us now restrict the variability of A. Assume
A = r (cos < + i sin <) ;
whence VA = Vr Vcos + i sin .
Our problem resolves itself into two, to
~~X extract the cube root of a real number and
also that of a complex number of the form
cos -\- i sin <, both numbers being regarded
as arbitrary. We shall treat these separately.
I. The roots of the equation x 3 = r are
3 / 3 / 3 /
Vr, c Vr, e 2 Vr,
representing by c and e 2 the complex cube roots of unity
- 1 + i V3 . _ - 1 - i V3
C C
Taking for the domain of rationality the totality of rational
functions of r, we know by the previous reasoning that the
equation x 8 = r is irreducible. Hence the problem of the
multiplication of the cube does not admit, in general, of a
construction by means of straight edge and compasses.
II. The roots of the equation
x 3 = cos -\- \ sin
is the analytic expression of the
problem of the trisection of the
angle.
If this equation were reducible,
one, at least, of its roots could be represented as a rational
function of cos and sin <, its value remaining unchanged
on substituting + 2?r for <. But if we effect this change
by a continuous variation of the angle <, we see that the
roots x l5 x 2 , x 3 undergo a cyclic permutation. Hence no root
can be represented as a rational function of cos and sin <.
The equation under consideration is irreducible and therefore
cannot be solved by the aid of a finite number of square roots.
Hence the trisection of the angle cannot be effected with straight
edge and compasses.
This demonstration and the general theorem evidently hold
good only when < is an arbitrary angle ; but for certain spe-
cial values of the construction may prove to be possible,
e.g., when < = .
CHAPTER III.
The Division of the Circle into Equal Parts.
1. The problem of dividing a given circle into n equal
parts has come down from antiquity ; for a long time we
have known the possibility of solving it when n = 2 h , 3, 5, or
the product of any two or three of these numbers. In his
Disquisitiones Arithmeticae, Gauss extended this series of
numbers by showing that the division is possible for every
prime number of the form p = 2^ + 1 but impossible for all
other prime numbers and their powers. If in p = 2^ -J- 1
we make p = and 1, we get p = 3 and 5, cases already
known to the ancients. For p = 2 we get p = 2 22 + 1 = 17,
a case completely discussed by Gauss.
For p. = 3 we get p = 2 ?3 + 1 = 257, likewise a prime num-
ber. The regiilar polygon of 257 sides can be constructed.
Similarly for /A = 4, since 2^ + 1 = 65537 is a prime number.
/* = 5, /A = 6, /i = 7 give no prime numbers. For /u. = 8 no
one has found out whether we have a prime number or not.
The proof that the large numbers corresponding to /x = 5, 6, 7
are not prime has required a large expenditure of labor and
ingenuity. It is, therefore, quite possible that p. = 4 is the
last number for which a solution can be effected.
Upon the regular polygon of 257 sides Eichelot published
an extended investigation in Crelle's Journal, IX, 1832,
pp. 1-26, 146-161, 209-230, 337-356. The title of the
memoir is : De resolutione algebraica aequatlonls x 257 = 1, sine
de divisions circuli per bisectionem anyuli septies repetitam in
paries 257 inter se aequales commentatio coronata.
THE DIVISION OF THE CIRCLE. 17
To the regular polygon of 65537 sides Professor Hermes
of Lingen devoted ten years of his life, examining with care
all the roots furnished by Gauss's method. His MSS. are
preserved in the collection of the mathematical seminary in
Gottingen. (Compare a communication of Professor Hermes
in No. 3 of the Gb'ttinger Nachrichten for 1894.)
2. We may restrict the problem of the division of the
circle into n equal parts to the cases where n is a prime num-
ber p or a power p a of such a number. For if n is a com-
posite number and if /* and v are factors of n, prime to each
other, we can always find integers a and b, positive or nega-
tive. such that -, ii
1 = a /A + bv ;
1 a . b
whence = | .
fLV V fl
To divide the circle into p.v = n equal parts it is sufficient to
know how to divide it into p. and v equal parts respectively.
Thus, for n = 15, we have
1 _2_3
15~3 5'
3. As will appear, the division into p equal parts (p being
a prime number) is possible only when p is of the form
p = 2 h + 1. We shall next show that a prime number can
be of this form only when h = 2 M . For this we shall make
use of Fermat's Theorem :
If p is a prime number and a an integer not divisible by p,
these numbers satisfy the congruence
p 1 is not necessarily the lowest exponent which, for a
given value of a, satisfies the congruence. If s is the lowest
exponent it may be shown that s is a divisor of p 1. In
particular, if s = p 1 we say that a is a primitive root of p,
18 FAMOUS PROBLEMS.
and notice that for every prime number p there is a primitive
root. We shall make use of this notion further on.
Suppose, then, p a prime number such that
(1) P = 2 h + 1,
and s the least integer satisfying
(2) 2 8 = + 1 (mod. p).
From (1) 2 h < p ; from (2) 2"> p.
.-. s>h.
(1) shows that h is the least integer satisfying the congruence
(3) 2 h = 1 (mod. p).
From (2) and (3), by division,
2 8 ~ h = 1 (mod. p).
.-.(4) s-h^h, s<2h.
From (3), by squaring,
2 a = 1 (mod. p).
Comparing with (2) and observing that s is the least expo-
nent satisfying congruences of the form
2* = 1 (mod. p),
.-. s = 2h.
We have observed that s is a divisor of p 1 = 2 h ; the same
is true of h, which is, therefore, a power of 2. Hence prime
numbers of the form 2 h -}- 1 are necessarily of the form
4. This conclusion may be established otherwise. Sup-
pose that h is divisible by an odd number, so that
h = h'(2n + l);
then, by reason of the formula
THE CYCLOTOMIC EQUATION. 19
p = 2 h ' (2n + 1) + l is divisible by 2 h> + 1, and hence is not a
prime number.
5. We now reach our fundamental proposition :
p being a prime number, the division of the circle into p equal
parts by the straight edge and compasses is impossible unless p
is of the form
p = 2 h + 1 = 2^ + 1.
Let us trace in the z-plane (z = x + iy) a circle of radius 1.
To divide this circle into n equal parts, beginning at z = 1, is
the same as to solve the equation
z" 1 = 0.
This equation admits the root z = 1 ; let us suppress this root
by dividing by z 1, which is the same geometrically as to
disregard the initial point of the division. We thus obtain
the equation
z n-l_j_ z n-2 +> .. +2 + 1=0,
which may be called the cyclotomic equation. As noticed
above, we may confine our attention to the cases where n is
a prime number or a power of a prime number. We shall
first investigate the case when n = p. The essential point of
the proof is to show that the above equation is irreducible.
For since, as we have seen, irreducible equations can only be
solved by means of square roots in finite number when their
degree is a power of 2, a division into p parts is always im-
possible when p 1 is not equal to a power of 2, i.e., when
p gfe 2 h + 1 = 2^ + 1.
Thus we see why Gauss's prime numbers occupy such an
exceptional position.
6. At this point we introduce a lemma known as Gauss's
Lemma. If
F(z) = z m + kz m ~ l + Bz m ~ 2 +. . . + Lz+ M,
20 FAMOUS PROBLEMS.
where A, B, . . . are integers, and F(z) can be resolved into
two rational factors f (z) and (z), so that
F (z) = f (z) ^ (z) = (z'"' + ai z m - 1 + a 2 Z ra '- 2 + . . .)
then must the a's and /3's also be integers. In other
words :
If an integral expression can be resolved into rational factors
these factors must be integral expressions.
Let us suppose the a's and /3's to be fractional. In each
factor reduce all the coefficients to the least common denom-
inator. Let a and b be these common denominators.
Finally multiply both members of our equation by a b . It
takes the form
a b F(z) = fi (z) fa (z) = (a z m ' + a^"'- 1 + . . .)
(boz^' + biz- 11 - 1 +...).
The a's are integral and prime to one another, as also the b's,
since a and b are the least common denominators.
Suppose a and b different from unity and let q be a prime
divisor of a b . Further, let a { be the first coefficient of f a (z)
and b k the first coefficient of fa (z) not divisible by q. Let
us develop the product f a (z) ! (z) and consider the coefficient
of z m ' + m " - * - k . It will be
1+2k
b k _ 2 + . . .
According to our hypotheses, all the terms after the first are
divisible by q, but the first is not. Hence this coefficient is not
divisible by q. Now the coefficient of z m'+m"-i-k j n ^ g rst
member is divisible by a b , i.e., by q. Hence if the identity
is true it is impossible for a coefficient not divisible by q to
occur in each polynomial. The coefficients of one at least of
the polynomials are then all divisible by q. Here is another
absurdity, since we have seen that all the coefficients are
THE CTCLOTOMIC EQUATION. 21
prime to one another. Hence we cannot suppose a and b
different from 1, and consequently the a's and /3's are in-
tegral.
7. In order to show that the cyclotomic equation is irre-
ducible, it is sufficient to show by Gauss's Lemma that the
first member cannot be resolved into factors with integral
coefficients. To this end we shall employ the simple method
due to Eisenstein, in Crelle's Journal, XXXIX, p. 167, which
depends upon the substitution
z = x + l.
We obtain
All the coefficients of the expanded member except the first
are divisible by p ; the last coefficient is always p itself, by
hypothesis a prime number. An expression of this class is
always irreducible.
For if this were not the case we should have
f (x + 1) = (x m + a^- 1 + . . . + a_, x + a m )
where the a's and b's are integers.
Since the term of zero degree in the above expression of
f (z) is p, we have a m b m - = p. p being prime, one of the fac-
tors of a m b m ' must be unity. Suppose, then,
Equating the coefficients of the terms in x, we have
22 FAMOUS PROBLEMS.
The first member and the second term of the second being
divisible by p, a m _ib m - must be so also. Since b m - 1,
a m _x is divisible by p. Equating the coefficients of the terms
in x 2 we may show that a m _ 2 is divisible by p. Similarly
we show that all of the remaining coefficients of the factor
x m -f- a 1 x m ~ 1 + . . . + a m _i x -f- a m are divisible by p. But
this cannot be true of the coefficient of x m , which is 1.
The assumed equality is impossible and hence the cyclo-
tomic equation is irreducible when p is a prime.
8. We now consider the case where n is a power of a
prime number, say n = p a . We propose to show that when
p > 2 the division of the circle into p 2 equal parts is impos-
sible. The general problem will then be solved, since the
division into p a equal parts evidently includes the division
into p 2 equal parts.
The cyclotomic equation is now
z-1
It admits as roots extraneous to the problem those which
come from the division into p. equal parts, i.e., the roots of
the equation p _ ..
Suppressing these roots by division we obtain
as the cyclotomic equation. This may be written
Z P(P-1) _|_ Z P(P~2) _|_ _j_ 2 p _J_ J __ Q
Transforming by the substitution
z = x + l,
we have
(x + 1)PCP-D + ( X + l)p 2.
CHAPTER IV.
The Construction of the Regular Polygon of 17 Sides.
1. We have just seen that the division of the circle into
equal parts by the straight edge and compasses is possible
only for the prime numbers studied by Gauss. It will now
be of interest to learn how the construction can actually be
effected.
The purpose of this chapter, then, will be to show in an
elementary way how to inscribe in the circle the regular poly-
gon of 17 sides.
Since we possess as yet no method of construction based
upon considerations purely geometrical, we must follow the
path indicated by our general discussions. We consider, first
of all, the roots of the cyclotomic equation
x 16 +x ls +. . . + x 2 +x+l = 0,
and construct geometrically the expression, formed of square
roots, deduced from it.
We know that the roots can be put into the transcendental
form
< = cos - + i sin j- (K = 1, 2, . . . 16) ;
and if
2-n- . . . 2-n-
= cos +i sin,
that
Geometrically, these roots are represented in the complex
plane by the vertices, different from 1, of the regular polygon
of 17 sides inscribed in a circle of radius 1, having the origin
THE REGULAR POLYGON OF 17 SIDES. 25
as center. The selection of e x is arbitrary, but for the con-
struction it is essential to indicate some e as the point of
departure. Having fixed upon e^ the angle corresponding to
K is K times the angle corresponding to e^ which completely
determines C K .
2. The fundamental idea of the solution is the following :
Forming a primitive root to the modulus 17 we may arrange
the 16 roots of the equation in a cycle in a determinate order.
As already stated, a number a is said to be a primitive root
to the modulus 17 when the congruence
a 8 = + 1 (mod. 17)
has for least solution s = 17 1 = 16. The number 3 pos-
sesses this property ; for we have
3 13 = 12
3 14 = 2 ,
3 15 EEE 6
3 16 = 1J
Let us then arrange the roots e K so that their indices are
the preceding remainders in order
e 8> *9> e 10j C 13) e 55 e !5> c ll> C 16j e !4? C 8) C 75 C 4) 12f 2? C 6 , Cj.
Notice that if r is the remainder of 3* (mod. 17), we have
3*= 3
3 5 = 5
3 9 =14
02 Q
O - i7
3 6 =15
3 10 EE 8
3 3 = 10
3 r =ll
3 U = 7
3 4 = 13
3 8 = 16
3^= 4
whence c r = e =
If r r is the next remainder, we have similarly
, __ , 8 K+1 - /, 3 K \3 - / \3
e r i ( e l ) (*r)
Hence in this series of roots each root is the cube of the preceding.
Gauss's method consists in decomposing this cycle into
sums containing 8, 4, 2, 1 roots respectively, corresponding
to the divisors of 16. Each of these sums is called a period.
26 FAMOUS PROBLEMS.
The periods thus obtained may be calculated successively as
roots of certain quadratic equations.
The process just outlined is only a particular case of that
employed in the general case of the division into p equal
parts. The p 1 roots of the cyclotomic equation are cyclic-
ally arranged by means of a primitive root of p, and the
periods may be calculated as roots of certain auxiliary equa-
tions. The degree of these last depends upon the prime fac-
tors of p 1. They are not necessarily equations of the
second degree.
The general case has, of course, been treated in detail by
Gauss in his Disquisitiones, and also by Bachmann in his
work, Die Lehre von der Kreisteilung (Leipzig, 1872).
3. In our case of the 16 roots the periods may be formed
in the following manner : Form two periods of 8 roots by
taking in the cycle, first, the roots of even order, then those
of odd order. Designate these periods by x x and x 2 , and
replace each root by its index. We may then write symbol-
ically
x : = 9 + 13 + 15 + 16 + 8 + 4+ 2 + 1,
x 2 = 3 + 10 + 5 + 11 + 14 + 7+12 + 6.
Operating upon x x and x 2 in the same way, we form 4 periods
of 4 terms :
yi = 13 + 16+ 4+ 1,
y 2 = 9 + 15+ 8+ 2,
y 8 = 10 + ll+ 7+ 6,
y 4 = 3+ 5 + 14 + 12.
Operating in the same way upon the y's, we obtain 8 periods
of 2 terms :
2 1 = 16+1, 2 E = H+ 6,
z 2 = 13 + 4, 2 6 = 10+ 7,
2 S = 15 + 2, z 7 = 5 + 12,
z 4 = 9+8, Zg = 3 + 14.
THE EEGULAR POLYGON OF 17 SIDES. 27
It now remains to show that these periods can be calculated
successively by the aid of square roots.
4. It is readily seen that the sum of the remainders corre-
sponding to the roots forming a period z is always equal to 17.
These roots are then e r and ei 7 _ r ;
2-7T 2-7T
e r = cos r + i sin r ,
2ir 2ir
V = 17 _ r = cos (17 - r) ^ + i sm (17 r) ,
2?r . . 2ir
= cos r i sm r .
Hence
27T
e r + e r - = 2 COS r .
Therefore all the periods z are real, and we readily obtain
2,r
= 2 cos-,
z s = 2 cos
4r
17'
z 6 = 2 cos
'
-2 2 2?r
17'
z 7 = 2 cos
5 rr
-2 8 27r
cos 17 ,
z s = 2cos
2^
17"
Moreover, by definition,
*i = zi + z 2 + z 3 + z 4 , x 2 = z 5 + z 6 + z 7 + z 8 ,
Yl = Zl + Z 25 Y2 = Z-3 + Z4> Y3 = Z5 + Z6> Y4 = Z7 + Z 8-
5. It will be necessary to determine the relative magnitude
of the different periods. For this purpose we shall employ
the following artifice : We divide the semicircle of unit radius
into 17 equal parts and denote by Si, S 2 , S 17 the distances
28
FAMOUS PROBLEMS.
of the consecutive points of division A t , A 2 , . A 17 from the
initial point of the semicircle, S 17 being equal to the diam-
eter, i.e., equal to 2. The angle
A K Ai 7 has the same measure as the
half of the arc A K 0, which equals
2/<7T TT
. Hence
34
(17 K)T
= 2c S 34
That this may be identical with
2 cos h , we must have
4h = 17 K,
K = 17 4h.
Giving to h the values 1, 2, 3, 4, 5, 6, 7, 8, we find for K the
values 13, 9, 5, 1, 3, 7, 11, 15. Hence
z i Sis,
Z 2 = Si,
Z S = S 9 ,
Z 4 ~
Z 5 = S 7 ,
Z 6 == Sn,
Z 7 == S 3 ,
z 8 = S 5 .
The figure shows that S K increases with the index ; hence the
order of increasing magnitude of the periods z is
Z 6>
Z 8)
Moreover, the chord A^A^+p subtends p divisions of the semi-
circumference and is equal to S p ; the triangle OA K A K + p shows
that
and a fortiori
Calculating the differences two and two of the periods y, we
easily find
THE REGULAR POLYGON OF 17 SIDES. 29
Yi y 2 = Sis + Si S 9 + S 15 > 0,
yi y 3 = S 13 + $! + S 7 + S n > 0,
yi y4 = Sis + Si + S 3 Si > 0,
y 2 y s = S 9 Sis -f S 7 + S u > 0,
y 2 y* = S 9 Sis + S 3 S 5 < 0,
y s y* = S 7 Sn + S 3 S 5 < 0.
Hence
y 3 < y 2 < y* < yi-
Finally we obtain in a similar way
x 2 < x t .
2rr
6. We now propose to calculate z l = 2 cos . After mak-
ing this calculation and constructing z 1? we can easily deduce
the side of the regular polygon of 17 sides. In order to find
the quadratic equation satisfied by the periods, we proceed to
determine symmetric functions of the periods.
Associating Zi with the period z 2 and thus forming the
period y l5 we have, first,
zi + z 2 = yi-
Let us now determine z^. . We have
where the symbolic product xp represents
C K ' C P K + p'
Hence it should be represented symbolically by K + p, remem-
bering to subtract 17 from K -J- p as often as possible. Thus,
z 1 z 2 = 12 + 3 + 14 + 5 = y 4 .
Therefore Z! and z 2 are the roots of the quadratic equation
(0 z 2 - yi z + y 4 = 0,
whence, since z l > z 2 ,
_ yi yi
2 ~ 2
30 FAMOUS PROBLEMS.
We must now determine y x and y 4 . Associating y x with the
period y 2 , thus forming the period x 1? and y s with the period
y 4 , thus forming the period x 2 , we have, first,
Yi -f y 2 = x lt
Then,
yi y 2 =(13 + 16 + 4 + l) (9 + 15 + 8 + 2).
Expanding symbolically, the second member becomes equal
to the sum of all the roots ; that is, to 1. Therefore y A
and y 2 are the roots of the equation
(,) y-x iy -l = 0,
whence, since y t > y 2 ,
Similarly,
y
and
y s y 4 = - 1.
Hence y 8 and y 4 are the roots of the equation
(V) y 2 -x 2 y-l = 0;
whence, since y 4 > y 8 ,
_x 2 + Vx 2 2 + 4 x 2 -Vx 2 2
Y4- ~ -, y* = -
It now remains to determine X! and x 2 . Since x x + x 2 is
equal to the sum of all the roots,
Xl + X 2 = 1.
Further,
xix 2 = (13 + 16 + 4 + 1 + 9 + 15 +8 + 2)
(10 + 11 + 7 + 6 + 3 + 5 + 14 + 12).
Expanding symbolically, each root occurs 4 times, and thus
= 4.
THE REGULAR POLYGON OF 17 SIDES. 31
Therefore x t and x 2 are the roots of the quadratic
() x 2 +x-4 = 0;
whence, since x t > x 2 ,
_-l + Vl7 _ - 1 - Vl7
xi- 2 ' * 2 ~ 2
Solving equations , 17, i/, in succession, z x is determined
by a series of square roots.
Effecting the calculations, we see that z t depends upon the
four square roots
Vl7, V Xl 2 + 4, Vx 2 2 + 4, V yi ' 2 -4y 4 .
If we wish to reduce z l to the normal form we must see
whether any one of these square roots can be expressed
rationally in terms of the others.
Now, from the roots of (rj),
Vx t 2 + 4 = y! y 2 ,
Vx 2 2 + 4 = y 4 y,.
Expanding symbolically, we verify that
(yi Ys) (Y4 Ys) = 2 (xj x a ),*
* (yi - y 2 ) (y 4 - y) = (13 + 16 + 4 + 1 - 9 - 15 - 8 - 2) (3 + 6 + 14
+ 12-10-11-7-6)
= 16+ 1 + 10+ 8- 6- 7- 3- 2
+ 2+ 4+13+11- 9-10- 6- 5
+ 7 + 9 + 1 + 16-14-15-11-10
+ 4+ 6 + 15+13-11-12- 8- 7
-12-14- 6- 4+ 2+ 3+16+15
- 1- 3-12-10+ 8+ 9+ 5+ 4
- 11 - 13 - 5 - 3 + 1 + 2 + 15 + 14
- 5 - 7 - 16 - 14 + 12 + 13 + 9 + 8
= 2(16 + 1 + 8 + 2 + 4 + 13 + 15+9-10-6-7-3-11-5-14
-12)
= 2(Xi - x g ).
32 FAMOUS PROBLEMS.
that is, _ _
Vx t 2 + 4 Vx 2 + 4 = 2 Vl7.
Hence Vx 2 3 -f- 4 can be expressed rationally in terms of the
other two square roots. This equation shows that if two of
the three differences y,. y 2 , y 4 y 3 , *i *2 are positive, the
same is true of the third, which agrees with the results ob-
tained directly.
Replacing now x x , y x , y* by their numerical values, we
obtain in succession
1 4- Vl7 + \/34 - 2 Vl7
~~~
- 1 - Vl7 + V34 + 2 V17
" 4
1 + Vl7 + V34 2 Vl7
Zi =
8
68 fl2Vl7 16V34+2V17 2(1 Vl7)V34 2V17
The algebraic part of the solution of our problem is now
completed. We have already remarked that there is no known
construction of the regular polygon of 17 sides based upon
purely geometric considerations. There remains, then, only
the geometric translation of the individual algebraic steps.
7. We may be allowed to introduce here a brief historical
account of geometric constructions with straight edge and
compasses.
In the geometry of the ancients the straight edge and com-
passes were always used together ; the difficulty lay merely in
bringing together the different parts of the figure so as not to
THE REGULAR POLYGON OF 17 SIDES. 33
draw any unnecessary lines. Whether the several steps in
the construction were made with straight edge or with com-
passes was a matter of indifference.
On the contrary, in 1797, the Italian Mascheroni succeeded
in effecting all these constructions with the compasses alone ;
he set forth his methods in his Geometria del compasso, and
claimed that constructions with compasses were practically
more exact than those with the straight edge. As he ex-
pressly stated, he wrote for mechanics, and therefore with a
practical end in view. Mascheroni's original work is difficult
to read, and we are under obligations to Hutt for furnishing
a brief resume in German, Die Mascheroni' schen Constructionen
(Halle, 1880).
Soon after, the French, especially the disciples of Carnot,
the author of the Geometrie de position, strove, on the other
hand, to effect their constructions as far as possible with
the straight edge. (See also Lambert, Freie Perspective,
1774.)
Here we may ask a question which algebra enables us to
answer immediately : In what cases can the solution of an
algebraic problem be constructed with the straight edge alone?
The answer is not given with sufficient explicitness by the
authors mentioned. We shall say :
With the straight edge alone we can construct all algebraic
expressions ivhose form is rational.
With a similar view Brianchon published in 1818 a paper,
Les applications de la theorie des transver sales, in which he shows
how his constructions can be effected in many cases with the
straight edge alone. He likewise insists upon the practical
value of his methods, which are especially adapted to field
work in surveying.
Poncelet was the first, in his Traite des proprietes projectives
(Vol. I, Nos. 351-357), to conceive the idea that it is sufficient
to use a single fixed circle in connection with the straight lines
34
FAMOUS PROBLEMS.
of the plane in order to construct all expressions depending
upon square roots, the center of the fixed circle being given.
This thought was developed by Steiner in 1833 in a cele-
brated paper entitled Die geometrischen Constructionen, ausge-
fuhrt mittels der geraden Linie und eines festen Kreises, als
Lehrgegenstand fur hohere Unterrichtsanstalten und zum Selbst-
unterricht.
8. To construct the regular polygon of 17 sides we shall
follow the method indicated by von Staudt (Crelle's Journal,
Vol. XXIV, 1842), modified later by Schroter (Crelle's Jour-
nal, Vol. LXXV, 1872). The construction of the regular
polygon of 17 sides is made in accordance with the methods
indicated by Poncelet and Steiner, inasmuch as besides the
straight edge but one fixed circle is used.*
First, we will show how with the straight edge and one fixed
circle we can solve every quadratic equation.
At the extremities of a diameter of the fixed unit circle
(Fig. 4) we draw two tangents, and select the lower as the
4 axis of X, and the diameter
perpendicular to it as the
axis of Y. Then the equa-
tion of the circle is
Let
y(y-2)=0.
px + q=
FIG. 4.
be any quadratic equation
with real roots x t and x 2 . Required to construct the roots /!
and x 2 upon the axis of X.
Lay off upon the upper tangent from A to the right, a seg-
4
ment measured by - ; upon the axis of X from 0, a segment
* A Mascheroni construction of the regular polygon of 17 sides by
L. Ge'rard is given in Math, Annalen, Vol. XL VIII, 1896, pp. 390-392.
THE REGULAR POLYGON OF 17 SIDES. 35
measured by ; connect the extremities of these segments by
P
the line 3 and project the intersections of this line with the
circle from A, by the lines 1 and 2, upon the axis of X. The
segments thus cut off upon the axis of X are measured by X!
and x 2 .
Proof. Calling the intercepts upon the axis of X, X! and x 2 ,
we have the equation of the line 1,
2x + x 1 (y-2)=0;
of the line 2,
2x + x 2 (y-2) = 0.
If we multiply the first members of these two equations we
get
x 2 + ^- 2 x (y-2) + x -^ (y -2)* =
as the equation of the line pair formed by 1 and 2. Subtract-
ing from this the equation of the circle, we obtain
x(y-2) + ^ (y-2)'-y(y-2) = 0.
This is the equation of a conic passing through the four
intersections of the lines 1 and 2 with the circle. From
this equation we can remove the factor y 2, correspond-
ing to the tangent, and we have left
x + X -f (y-2)-y = 0,
which is the equation of the line 3. If we now make
*i H~ x a P an d X i x 2 = q> we get
and the transversal 3 cuts off from the line y = 2 the seg-
36
FAMOUS PROBLEMS.
Thus the
merit - , and from the line y = the segment -
correctness of the construction is established.
9. In accordance with the method just explained, we shall
now construct the roots of our four quadratic equations.
They are (see pp. 29-31)
() x 2 + x 4 = 0, with roots Xx and x 2 ; x x > x 2 ,
0?) y 2 xjy 1 =0, with roots yi and y 2 ; yi > y 2 ,
(V) y 2 X 2y 1 = 0, with roots y 3 and y 4 ; y 4 > y 3 ,
() z 2 y t z + y 4 = 0, with roots z and z 2 ; z > z 2 .
These will furnish
z: = 2cos^,
whence it is easy to construct the polygon desired. We
notice further that to construct z v it is sufficient to construct
We then lay off the following segments : upon the upper
tangent, y = 2,
444
~ 4 ' 7' 7' 7 ;
AI Ag jri
upon the axis of X,
This may all be done in the following manner : The
straight line connecting the point -f- 4 upon the axis of X
with the point 4 upon the tangent y = 2 cuts the circle in
JL
o
FIG. 5.
THE REGULAR POLYGON OF 17 SIDES. 37
two points, the projection of which from the point A (0, 2),
the upper vertex of the circle, gives the two roots x x , x 2 of the
first quadratic equation as intercepts upon the axis of X.
4
To solve the second equation we have to lay off above
x i
and -- below.
*1
To determine the first point we connect K! upon the axis of
X with A, the upper vertex, and from 0, the lower vertex,
draw another straight line through the intersection of this
line with the circle. This cuts off upon the upper tangent
4
the intercept . This can easily be shown analytically.
x i
The equation of the line from A to ^ (Fig. 5),
2x + Xl y = 2 Xl ,
and that of the circle,
x 2 + y(y-2) = 0,
give as the coordinates of their intersection
4 Xl 2 Xl 2
The equation of the line from through this point becomes
4
cutting off upon y == 2 the intercept .
x i
We reach the same conclusion still more simply by the use
of some elementary notions of projective geometry. By our
construction we have obviously associated with every point x
of the lower range one, and only one, point of the upper, so
that to the point x = oo corresponds the point x' = 0, and con-
versely. Since in such a correspondence there must exist a
38 FAMOUS PROBLEMS.
linear relation, the abscissa x' of the upper point must satisfy
the equation f const.
x
X
Since x' = 2 when x = 2, as is obvious from the figure, the
constant = 4.
__L o +1 y<
FIG. 6.
To determine upon the axis of X we connect the point
x i
4 upon the upper with the point + 1 upon the lower tan-
gent (Fig. 6). The point thus determined upon the vertical
4
diameter we connect with the point above. This line
cuts off upon the axis of X the intercept . For the
x i
line from 4 to + 1,
intersects the vertical diameter in the point (0, ). Hence
the equation of the line from to this point is
*i
5y 2x lX = 2,
and its intersection with the lower tangent gives .
x i
The projection from A of the intersections of the line from
1 4
to with the circle determines upon the axis of X the
x i *i
two roots of the second quadratic equation, of which, as
THE REGULAR POLYGON OF 17 SIDES.
39
already noted, we need only the greater, y x . This corres-
ponds, as shown by the figure, to the projection of the upper
intersection of our transversal with the circle.
Similarly, we obtain the roots of the third quadratic equa-
tion. Upon the upper tangent we project from the inter-
section of the circle with the straight line which gave upon
the axis of X the root + x 2 . This immediately gives the
4
intercept , by reason of the correspondence just explained.
-4
FIG. 7.
If we connect this point with the point where the vertical
diameter intersects the line joining 4 above and -f- 1 below,
we cut off upon the axis of X the segment , as desired.
x 2
If we project that intersection of this transversal with the
circle which lies in the positive quadrant from A upon the
axis of X, we have constructed the required root y 4 of the third
quadratic equation.
We have finally to determine the root z x of the fourth quad-
4 y
ratic equation and for this purpose to lay off above and -
below. We solve the first problem in the usual way, by pro-
jecting the intersection of the circle with the line connecting
A with + y x below, from upon the upper tangent, thus
4
obtaining . For the other segment we connect the point
-f- 4 above with y 4 below, and then the point thus determined
40
FAMOUS PROBLEMS.
4.
upon the vertical diameter produced with . This line cuts
y .
off upon the axis of X exactly the segment desired, . For
the line a (Fig. 8) has the equation
Z, 0,
FIG. 8.
It cuts off upon the vertical diameter the segment
The equation of the line b is then
_T7'
and its intersection with the axis of X has the abscissa .
yi
If we project the upper intersection of the line b with the
2ir
circle from A upon the axis of X, we obtain z l = 2 cos .
If we desire the simple cosine itself we have only to draw a
diameter parallel to the axis of X, on which our last projecting
o
ray cuts off directly cos . A perpendicular erected at this
point gives immediately the first and sixteenth vertices of the
regular polygon of 17 sides.
The period Zj was chosen arbitrarily ; we might construct
in the same way every other period of two terms and so find
the remaining cosines. These constructions, made on separate
figures so as to be followed more easily, have been combined
in a single figure (Fig. 9), which gives the complete construc-
tion of the regular polygon of 17 sides.
THE REGULAR POLYGON OF 17 SIDES.
41
FIG. 9.
CHAPTER V.
General Considerations on Algebraic Constructions.
1 . We shall now lay aside the matter of construction with
straight edge and compasses. Before quitting the subject we
may mention a new and very simple method of effecting cer-
tain constructions, paper foldiny. Hermann Wiener has
shown how by paper folding we may obtain the network of
the regular polyhedra. Singularly, about the same time a
Hindu mathematician, Sundara Kow, of Madras, published a
little book, Geometrical Exercises in Paper Foldimj* (Madras,
Addison & Co., 1893), in which the same idea is consider-
ably developed. The author shows how by paper folding we
may construct by points such curves as the ellipse, cissoid, etc.
2. Let us now inquire how to solve geometrically prob-
lems whose analytic form is an equation of the third or of
higher degree, and in particular, let us see how the ancients
succeeded. The most natural method is by means of the
conies, of which the ancients made much use. For example,
they found that by means of these curves they were enabled
to solve the problems of the duplication of the cube and the
trisection of the angle. We shall in this place give only a
general sketch of the process, making use of the language
of modern mathematics for greater simplicity.
Let it be required, for instance, to solve graphically the
cubic equation
x 8 + ax 2 + bx + c = 0,
or the biquadratic,
x 4 +ax 3 +
* See American edition, revised by Beman and Smith, The Open Court
Publishing Co., Chicago, 1901.
ALGEBRAIC CONSTRUCTIONS.
Put x 2 = y ; our equations become
43
and
xy + ay + bx + c =
y 2 + axy + by + ex + d = 0.
The roots of the equations proposed are thus the abscissas
of the points of intersection of the two conies.
The equation
x 2 = y
represents a parabola with axis vertical. The second equa-
tion,
xy + ay + bx + c = 0,
represents an hyperbola whose asymptotes are parallel to the
axes of reference (Fig. 10). One of the four points of inter-
FIG. 10.
FIG. 11.
section is at infinity upon the axis of Y, the other three at a
finite distance, and their abscissas are the roots of the equa-
tion of the third degree.
In the second case the parabola is the same. The hyper-
bola (Fig. 11) has again one asymptote parallel to the axis of
X while the other is no longer perpendicular to this axis.
The curves now have four points of intersection at a finite
distance.
44
FAMOUS PROBLEMS.
The methods of the ancient mathematicians are given in
detail in the elaborate work of M. Cantor, Geschichte der
Mathematik (Leipzig, 1894, 2d ed.). Especially interesting is
Zeuthen, Die Kegelschnitte im Altertum (Kopenhagen, 1886,
in German edition). As a general compendium we may men-
tion Baltzer, Analytische Geometrie (Leipzig, 1882).
3. Beside the conies, the ancients used for the solution of
the above-mentioned problems, higher
curves constructed for this very pur-
pose. We shall mention here only
the Cissoid and the Conchoid.
The cissoid of Diodes (c. 150 B.C.)
may be constructed as follows (Fig.
12) : To a circle draw a tangent (in the
figure the vertical tangent on the right)
and the diameter perpendicular to it.
Draw lines from 0, the vertex of the
circle thus determined, to points upon
the tangent, and lay off from upon
each the segment lying between its
intersection with the circle and the
tangent. The locus of points so deter-
mined is the cissoid.
To derive the equation, let r be the
FIG. 12. radius vector, the angle it makes with
the axis of X. If we produce r to the
tangent on the right, and call the diameter of the circle 1,
the total segment equals
cos 0'
circle is cos 0. The difference of the two segments is
hence
The portion cut off by the
r, and
r =
cos Q
cos =
sin 2 6
cos 0'
ALGEBRAIC CONSTRUCTIONS. 45
By transformation of coordinates we obtain the Cartesian
equation,
(x 2 -f y 2 ) x y 2 = 0.
The curve is of the third order, lias a cusp at the origin,
and is symmetric to the axis of X. The vertical tangent to
the circle with which we began our construction is an asymp-
tote. Finally the cissoid cuts the line at infinity in the cir-
cular points.
To show how to solve the Delian problem by the use of
this curve, we write its equation in the following form :
We now construct the straight line,
This cuts off upon the tangent x = 1 the segment A, and
intersects the cissoid in a point for which
y _ A 3
l-x~
This is the equation of a straight line passing through the
point y = 0, x = 1, and hence of the line joining this point
to the point of the cissoid.
This line cuts off upon the axis of Y the intercept \ 3 .
We now see how ^2 may be constructed. Lay off upon
the axis of Y the intercept 2, join this point to the point
x = 1, y = 0, and through its intersection with the cissoid
draw a line from the origin to the tangent x = 1. The inter-
cept on this tangent equals ^2.
4. The conchoid of Nicomedes (c. 150 B.C.) is constructed
as follows : Let be a fixed point, a its distance from a fixed
46
FAMOUS PROBLEMS.
line. If we pass a pencil of rays through and lay off on
each ray from its intersection with the fixed line in both
directions a segment b, the locus of the points so determined
is the conchoid. According as b is greater or less than a,
the origin is a node or a con-
jugate point ; for b = a it is
a cusp (Fig. 13).
Taking for axes of X and Y
the perpendicular and paral-
lel through to the fix-, as is
easily seen from the figure.
Our previous investigations have shown us that the prob-
lem of the trisection of the angle is a problem of the third
degree. It admits the three solutions
Every algebraic construction which solves this problem by
the aid of a curve of higher degree must obviously furnish all
the solutions. Otherwise the equation of the problem would
not be irreducible. These different solutions are shown in
the figure. The circle and the conchoid intersect in eight
points. Two of them coincide with the origin, two others
with the circular points at infinity. None of these can give
a solution of the problem. There remain, then, four points
of intersection, so that we seem to have one too many. This
is due to the fact that among the four points we necessarily
find the point B such that M B = 2 b, a point which may be
determined without the aid of the curve. There actually
remain then only three points corresponding to the three
roots furnished by the algebraic solution.
5. In all these constructions with the aid of higher alge-
braic curves, we must consider the practical execution. We
need an instrument which shall trace the curve by a con-
tinuous movement, for a construction by points is simply a
method of approximation. Several instruments of this sort
have been constructed ; some were known to the ancients.
Nicomedes invented a simple device for tracing the conchoid.
It is the oldest of the kind besides the straight edge and
compasses. (Cantor, I, p. 302.) A list of instruments of
more recent construction may be found in Dyck's Katalog,
pp. 227-230, 340, and Nachtrag, pp. 42, 43.
PART II.
TRANSCENDENTAL NUMBERS AND THE QUADRATURE OF THE
CIRCLE.
CHAPTEK I.
Cantor's Demonstration of the Existence of
Transcendental Numbers.
1. Let us represent numbers as usual by points upon the
axis of abscissas. If we restrict ourselves to rational numbers
the corresponding points will fill the axis of abscissas densely
throughout (iiberall dicht), i.e., in any interval no matter how
small there is an infinite number of such points. Neverthe-
less, as the ancients had already discovered, the continuum
of points upon the axis is not exhausted in this way ; between
the rational numbers come in the irrational numbers, and the
question arises whether there are not distinctions to be made
among the irrational numbers.
Let us define first what we mean by algebraic numbers.
Every root of an algebraic equation
a co n + a^ 11 - 1 + + a n _ lW + a n =
with integral coefficients is called an algebraic number. Of
course we consider only the real roots. Rational numbers
occur as a special case in equations of the form
a w -f- a t = 0.
50 FAMOUS PROBLEMS.
We now ask the question : Does the totality of real
algebraic numbers form a continuum, or a discrete series
such that other numbers may be inserted in the intervals?
These new numbers, the so-called transcendental numbers,
would then be characterized by this property, that they cannot
be roots of an algebraic equation with integral coefficients.
This question was answered first by Liouville (Comptes
rendus, 1844, and Liouville's Journal, Vol. XVI, 1851), and
in fact the existence of transcendental numbers was demon-
strated by him. But his demonstration, which rests upon the
theory of continued fractions, is rather complicated. The
investigation is notably simplified by using the developments
given by Georg Cantor in a memoir of fundamental impor-
tance, Ueber eine Eigenschaft des Inbegriffes reeller algebra-
ischer Zahlen (Crelle's Journal, Vol. LXXVII, 1873). We
shall give his demonstration, making use of a more simple
notion which Cantor, under a different form, it is true, sug-
gested at the meeting of naturalists in Halle, 1891.
2. The demonstration rests upon the fact that algebraic
numbers form a countable mass, while transcendental numbers
do not. By this Cantor means that the former can be arranged
in a certain order so that each of them occupies a definite
place, is numbered, so to speak. This proposition may be
stated as follows :
The manifoldness of real algebraic numbers and the mani-
foldness of positive integers can be brought into a one-to-one
correspondence.
We seem here to meet a contradiction. The positive inte-
gers form only a portion of the algebraic numbers ; since
each number of the first can be associated with one and one
only of the second, the part would be equal to the whole.
This objection rests upon a false analogy. The proposition
that the part is always less than the whole is not true for
TRANSCENDENTAL NUMBERS. 51
infinite masses. It is evident, for example, that we may
establish a one-to-one correspondence between the aggregate
of positive integers and the aggregate of positive even num-
bers, thus :
1 2 3 n
2 4 6 2n
In dealing with infinite masses, the words great and small are
inappropriate. As a substitute, Cantor has introduced the
word power (Mdchtigkeit), and says : Two infinite masses have
the same power when they can be brought into a one-to-one cor-
respondence with each other. The theorem which we have to
prove then takes the following form : The aggregrate of real
algebraic numbers has the same power as the aggregate of
positive integers.
We obtain the aggregate of real algebraic numbers by seek-
ing the real roots of all algebraic equations of the form
a o) n + a^"- 1 + + a n _ lW -f a n = ;
all the a's are supposed prime to one another, a positive,
and the equation irreducible. To arrange the numbers thus
obtained in a definite order, we consider their height N as
defined by
(aj representing the absolute value of a^ as usual. To a
given number N corresponds a finite number of algebraic
equations. For, N being given, the number n has certainly
an upper limit, since N is equal to n 1 increased by positive
numbers ; moreover, the difference N (n 1) is a sum of
positive numbers prime to one another, whose number is
obviously finite.
52
FAMOUS PROBLEMS.
N
n
la,!
tail
Kl
I..I
w
EQUATION.
* (x) we con-
struct the curve
y = (x) we will now establish three important
properties :
1. x being supposed given and p increasing without limit,
<(x) tends toward zero, as does also the sum of the absolute
values of its terms.
Put u = x (1 x) (2 x) (n x) ; we may then write
which for p infinite tends toward zero.
To have the sum of the absolute values of < (x) it is suffi-
cient to replace x by |x| in the undeveloped form of <(x).
The second part is then demonstrated like the first.
2. h being an integer, (h) is an integer not divisible by p
and therefore different from zero.
Develop <(x) in increasing powers of x, noticing that the
terms of lowest and highest degree respectively are of degree
p 1 and np + p 1. We have
r=np+p 1
,,,_V c'x"-' , c"x" , HP+P-I
= 2, v "5=iji+gp55i + *fr=3)i'
r=p 1
Hence
r=np+p 1
*(h) = 2 c r h'.
r=p-l
Leaving out of account the denominator (p 1) !, which
occurs in all the terms, the coefficients c r are integers. This
denominator disappears as soon as we replace h r by r!, since
the factorial of least degree is h 1 *" 1 = (p 1) !. All the terms
of the development after the first will contain the factor p.
As to the first, it may be written
(1-2-8... n)p.(p-l)!
(p-1)!
and is certainly not divisible by p since p > n.
Therefore < (h) = (n !) (mod. p),
and hence h0,
TRANSCENDENCE OF THE NUMBER e. 65
Moreover, < (h) is a very large number ; even its last term
alone is very large, viz.:
3. h being an integer, and k one of the numbers 1, 2
< (h -f- k) is an integer divisible by p.
We have <(h + k) = 2c r (h + k) r =^c' r h r ,
a formula in which we are to replace h r by r ! only after hav-
ing arranged the development in increasing powers of h.
According to the rules of the symbolic calculus, we have
first
_ -
(p-1)!
One of the factors in the brackets reduces to h ; hence the
term of lowest degree in h in the development is of degree p.
We may then write
r=np+p 1
The coefficients still have for numerators integers and for
denominator (p 1)!. As already explained, this denomi-
nator disappears when we replace h r by r ! . But now all the
terms of the development are divisible by p; for the first
may be written
( l) k P kP- 1 [(k 1) ! (n k) !]" p !
(p-1)!
= ( l) kp k^ 1 [(k - 1) ! (n k) !]" p.
< (h -f- k) is then divisible by p.
3. We can now show that the equation
F(e)=C +C 1 e + C 2 e 2 +- +C n e" =
is impossible.
66 FAMOUS PROBLEMS.
For the number M, by which we multiply the members of
this equation, we select < (h), so that
* (h) F (e) = C < (h) + d4> (h) e + C 2 < (h) e 2 + + C D (h)e n .
Let us try to decompose any term, such as C k < (h) e k , into an
integer and a fraction. We have
e* 4 (h) = e k c r h r .
r
Considering the series development of e k , any term of this
sum, omitting the constant coefficient, has the form
h r k h r k 2 h r k r h r k r+1
Replacing h r by r !, or what amounts to the same thing, by one
of the quantities
rh- 1 , r(r l)h*-- -, r(r !) 3 h 2 , r(r-l)- 2 h,
and simplifying the successive fractions,
h r =
,
~i
J
The first line has the same form as the development of
(h + k) r 5 in the parenthesis of the second line we have the
series
k k 2
f 7+T + (r + l)(r + 2)~ l
whose terms are respectively less than those of the series
..
The second line in the expansion of e k h r may therefore be
represented by an expression of the form
q r , k ' e k k r ,
q r>k being a proper fraction.
TRANSCENDENCE OF THE NUMBER e. 67
Effecting the same decomposition for each term of the sum
e k 2 c r h r
r
it takes the form
e k 2 c r h r = c r (h + k) r + e* q r , k c r k r .
r r r
The first part of this sum is simply (h -|- k) ; this is a
number divisible by p (2, 3). Further (2, 1),
tends toward zero when p becomes infinite : the same is true
a fortiori of q r , k c r k r , and also, since e k is a finite quantity,
r
of e k q r?k c r k r , which we may represent by e k .
r
The term under consideration, C k e k < (h), has then been put
under the form of an integer C k <(h + k) and a quantity C k e k
which, by a suitable choice of p, may be made as small as we
please.
Proceeding similarly with all the terms, we get finally
+ C n e n .
It is now easy to complete the demonstration. All the
terms of the first line after the first are divisible by p ; for
the first, | C 1 is less than p ; < (h) is not divisible by p; hence
C <(h) is not divisible by the prime number p. Consequently
the sum of the numbers of the first line is not zero.
The numbers of the second line are finite in number ; each
of them can be made smaller than any given number by a
suitable choice of p ; and therefore the same is true of their
sum.
Since an integer not zero and a fraction cannot have zero
for a sum, the assumed equation is impossible.
Thus, the transcendence of e, or Hermite's Theorem, is
demonstrated.
CHAPTER IV.
The Transcendence of the Number it.
1. The demonstration of the transcendence of the number
TT given by Lindemann is an extension of Hermite's proof in
the case of e. While Hermite shows that an integral equa-
tion of the form
C + C 1 e + C 2 e 1 '+- + C n e" =
cannot exist, Lindemann generalizes this by introducing in
place of the powers e, e 2 sums of the form
1 + e 2 + ' ' ' + e v
e
where the k's are associated algebraic numbers, i.e., roots of
an algebraic equation, with integral coefficients, of the degree
N ; the I's roots of an equation of degree N', etc. Moreover,
some or all of these roots may be imaginary.
Lindemann's general theorem may be stated as follows :
The number e cannot satisfy an equation of the form
(1) C +C 1 (e kl + e k2 +- - + e k ")
+ C,(e !l + e 2 + - + e v )+- - =
where the coefficients C t are integers and the exponents k u li, * '
are respectively associated algebraic numbers.
The theorem may also be stated :
The number e is not only not an algebraic number and there-
fore a transcendental number simply, but it is also not an
interscendental * number and therefore a transcendental number
of higher order.
* Leibnitz calls a function x\ where X is an algebraic irrational, an
interscendental function.
TRANSCENDENCE OF THE NUMBER it. 69
Let
ax" + a^"- 1 + + a N =
be the equation having for roots the exponents k } ;
bx N ' + biX"'- 1 + - + b N , =
that having for roots the exponents Ij, etc. These equations
are not necessarily irreducible, nor the coefficients of the first
terms equal to 1. It follows that the symmetric functions of
the roots which alone occur in our later developments need
not be integers.
In order to obtain integral numbers it will be sufficient to
consider symmetric functions of the quantities
aki, ak 2 , ak N ,
bl b b! 2 , bl N ., etc.
These numbers are roots of the equations
y" + atf*- 1 + a 2 ay N - 2 + + a H a*" 1 = 0,
Y N/ + biy"'- 1 + b 2 by N '- 2 + + b^b 1 *'" 1 = 0, etc.
These quantities are integral associated algebraic numbers,
and their rational symmetric functions real integers.
We shall now follow the same course as in the demonstra-
tion of Hermite's theorem.
We assume equation (1) to be true ; we multiply both
members by an integer M ; and we decompose each sum,
such as
M(e kl + e k2 +- - + e k >0,
into an integral part and a fraction, thus
Our equation then becomes
C M + C 1 M 1 + C 2 M 2 + -
+ C 2 e 2 + ' = 0.
70 FAMOUS PROBLEMS.
We shall show that with a suitable choice of M the sum of
the quantities in the first line represents an integer not
divisible by a certain prime number p, and consequently
different from zero ; that the fractional part can be made as
small as we please, and thus we come upon the same contra-
diction as before.
2. We shall again use the symbol h r =r! and select as
the multiplier the quantity M =^(h), where ^(x) is a gene-
ralization of <(x) used in the preceding chapter, formed as
follows :
[(li - x) (I, - x) (I N - - x)] b* b*'* - b"""
where p is a prime number greater than the absolute value of
each of the numbers
C , a, b, , a K , b w ,,
and later will be assumed to increase without limit. As to
the factors a* p , b N ' p , , they have been introduced so as to
have in the development of i/r(x) symmetric functions of the
quantities
ak!, ak 2 , -, ak K ,
HI, b! 2 , , bl,,,
that is, rational integral numbers. Later on we shall have
to develop the expressions
The presence of these same factors will still be necessary if
we wish the coefficients of these developments to be integers
each divided by (p 1) !.
1. .) p a N ' p a s " p
(bli b! 2 bl x ,) p b Np b*" p -
( l) lf P +N 'p + '"(a N a N - 1 ) p a I '' p a N ' p - (b N ,b N '- 1 ) p b Np b N "P
If in this term we replace \\*~ l by its value (p 1) ! the
denominator disappears. According to the hypotheses made
regarding the prime number p, no factor of the product is
divisible by p and hence the product is not.
The second term c p h p becomes likewise an integer when
we replace h p by p! but the factor p remains, and so for all
of the following terms. Hence i/^(h) is an integer not divis-
ible by p.
2. For x, a given finite quantity, and p increasing without
limit, i/f (x) = c r x r tends toward zero, as does also the sum
We may write
ki/ + C 2 *f 2 el', c r l',.q r , 1( ,+ =0.
v=l r v =l r
By 2, 2 we can make 2|c r k r | as small as we please by taking
r
p sufficiently great. Since | q r _ k | < 1, this will be true a fortiori
of
2 c r k r q r , k
r
and hence also of
Since the coefficients C are finite in value and in number, the
sum which occurs in the second line of (2) can, by increasing
p, be made as small as we please.
The numbers of the first line are, after the first, all divis-
ible by p (3), but the first number, C $(h), is not (1).
Therefore the sum of the numbers in the first line is not
divisible by p and hence is different from zero. The sum of
an integer and a fraction cannot be zero. Hence equation (2)
is impossible and consequently also equation (1).*
4. We now come to a proposition more general than the
preceding, but whose demonstration is an immediate conse-
* The proof for the more general case where Co = may be reduced
to this by multiplication by a suitable factor, or may be obtained directly
by a proper modification of f (h).
TRANSCENDENCE OF THE NUMBER it. 75
quence of the latter. For this reason we shall call it Linde-
mann's corollary.
The number e cannot satisfy an equation of the form
(3) C' +C' 1 e kl + C' s e I '+- = <),
in which the coefficients are integers even when the exponents
k u Ij, are unrelated algebraic numbers.
To demonstrate this, let k 2 , k 3 , , k N be the other roots of
the equation satisfied by k x ; similarly for I 2 , I 3 , , I M -, etc.
Form all the polynomials which may be deduced from (3)
by replacing ^ in succession by the associated roots k 2 , ,
l t by the associated roots I 2 , Multiplying the expres-
sions thus formed we have the product
a = 1, 2, , N
= Q, + d (e kl + e k2 H ----- h e k ") + C 2 (e kl+kj + e k2+ks -\
In each parenthesis the exponents are formed symmetrically
from the quantities ki, l t , , and are therefore roots of an
algebraic equation with integral coefficients. Our product
comes under Lindemann's theorem ; hence it cannot be zero.
Consequently none of its factors can be zero and the corollary
is demonstrated.
We may now deduce a still more general theorem.
The number e cannot satisfy an equation of the form
CW + C ( 1 1) e k +C ( 2 >e 1 + - - =
where the coefficients as well as the exponents are unrelated
algebraic numbers.
For, let us form all the polynomials which we can deduce
from the preceding when for each of the expressions C a \ we
substitute one of the associated algebraic numbers
C ( f, C, C^.
76 FAMOUS PROBLEMS.
If we multiply the polynomials thus formed together we get
the product
" a = 1, 2, , N
0=1,2, -,,
7 = 1,2,
n
*, V.
c _i_ r P k -i- r p 1 -i-
^->0 \^ ^'k*' i^ v^jC ^^
+ C k , k e k+k +C k ,,e k+1 +
where the coefficients C are integral symmetric functions of
the quantities
, C< 2 >, -, CW,
and hence are rational. By the previous proof such an
expression cannot vanish, and we have accordingly Linde-
mann's corollary in its most general form :
The number e cannot satisfy an equation of the form
C +C 1 e k +C a e 1 +- - =
where the exponents k, I, -as well as the coefficients C , Ci,
are algebraic numbers.
This may also be stated as follows :
In an equation of the form
C +C 1 e k +C 2 e 1 +- - =
the exponents and coefficients cannot all be algebraic numbers.
5. From Lindemann's corollary we may deduce a number
of interesting results. First, the transcendence of TT is an
immediate consequence. For consider the remarkable equa-
tion
1 4- e iir = 0.
TRAJfSCSNDEXCX OF THE NUMBER it. 77
The coefficients of this equation are algebraic ; hence the
exponent \TT is not. Therefore, TT is transcendental.
6. Again consider the function y = e 1 . We know that
1 = e. This seenis to be contrary to our theorems about the
transcendence of e. This is not the case, however. We
must notice that the case of the exponent was implicitly
excluded. For the exponent the function ^r(x) would lose
its essential properties and obviously our conclusions would
not hold.
Excluding then the special case (x =0, y = 1). Lindemann's
corollary shows that in the equation y = e x or x = log e y, y and
x. i.e., the number and its natural logarithm, cannot be alge-
braic simultaneously. To an algebraic value of x corresponds
a transcendental value of y, and conversely. This is certainly
a very remarkable property.
If we construct the curve y = e x and mark all the algebraic
points of the plane, i.e., all points whose coordinates are alge-
braic numbers, the curve passes among them without meeting
a single one except the point x = 0. y = 1. The theorem still
holds even when x and y take arbitrary complex values. The
exponential curve is then transcendental in a far higher sense
than ordinarily supposed.
7. A further consequence of Lindemann's corollary is the
transcendence, in the same higher sense, of the function
y = sin" 1 x and similar functions.
The function y = sin" 1 x is defined by the equation
We see, therefore, that here also x and y cannot be algebraic
simultaneously, excluding, of course, the values x = <>. y 0.
We may then enunciate the proposition in geometric form :
The curre y = sin" 1 x, like the curve y = e x .
no algebraic point of the plane, except x = 0. y = 0.
CHAPTER V.
The Integraph and the Geometric Construction of n.
1. Lindemann's theorem demonstrates the transcendence
of TT, and thus is shown the impossibility of solving the old
problem of the quadrature of the circle, not only in the sense
understood by the ancients but in a far more general manner.
It is not only impossible to construct TT with straight edge
and compasses, but there is not even a curve of higher order
defined by an integral algebraic equation for which TT is the
ordinate corresponding to a rational value of the abscissa.
An actual construction of TT can then be effected only by the
aid of a transcendental curve. If such a construction is
desired, we must use besides straight edge and compasses
a " transcendental " apparatus which shall trace the curve by
continuous motion.
2. Such an apparatus is the integraph, recently invented
and described by a Russian engineer, Abdank-Abakanowicz,
and constructed by Coradi of Zurich.
This instrument enables us to trace the integral curve
Y = F(x)=/f(x)dx
when we have given the differential curve
y = f(x).
For this purpose, we move the linkwork of the integraph
so that the guiding point follows the differential curve ; the
tracing point will then trace the integral curve. For a fuller
description of this ingenious instrument we refer to the
original memoir (in German, Teubner, 1889 ; in French,
Gauthier-Villars, 1889).
GEOMETRIC CONSTRUCTION OF it.
79
We shall simply indicate the principles of its working.
For any point (x, y) of the differential curve construct the
auxiliary triangle having for vertices the points (x, y), (x, 0),
(x 1, 0) ; the hypotenuse of this right-angled triangle makes
with the axis of X an angle whose tangent = y.
Hence, this hypotenuse is parallel to the tangent to the inte-
gral curve at the point (X, Y) corresponding to the point (x, y).
FIG. 16.
The apparatus should be so constructed then that the trac-
ing point shall move parallel to the variable direction of this
hypotenuse, while the guiding point describes the differential
curve. This is effected by connecting the tracing point with
a sharp-edged roller whose plane is vertical and moves so as to
be always parallel to this hypotenuse. A weight presses this
roller firmly upon the paper so that its point of contact can
advance only in the plane of the roller.
The practical object of the integraph is the approximate
evaluation of definite integrals ; for us its application to the
construction of ir is of especial interest.
3. Take for differential curve the circle
30 FAMOUS PROBLEMS.
the integral curve is- then
Y =f Vr* x 2 dx = ~ sin" 1 - + o ^ ~ * 2 -
*. T Zi
This curve consists of a series of congruent branches. The
points where it meets the axis of Y have for ordinates
+^, -..
- 2 '
Upon the lines X = r the intersections have for ordinates
'V 4 '
If we make r = 1, the ordinates of these intersections will
determine the number IT or its multiples.
It is worthy of notice that our apparatus enables us to
trace the curve not in a tedious and inaccurate manner, but
with ease and sharpness, especially if we use a tracing pen
instead of a pencil.
Thus we have an actual constructive quadrature of the
circle along the lines laid down by the ancients, for our
curve is only a modification of the quadratrix considered
by them.
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ces
ihrary
ACK
JUL72