AN ABSTRACT OF THE THESIS OF
ROBERT WILLIAM ESCHRICH
(Name)
in
for the
presented on
Mathematics
(Major)
Master of Science
(Degree)
November 13, 1967
(Date)
Title: A MODEL OF NON-EUCLIDEAN GEOMETRY IN THREE
DIMENSIONS,
Abstract approved
TT
inature redacted for privacy.
(Harry E. Goheen)
This paper is a continuation of William Zell's thesis, A
Model of Non-Euclidean Geometry in Three Dimensions. The
purpose of that thesis was to show that the axioms of non-Euclidèan
geometry are consistent if Euclidean geometry an& hence arith-
rnetic is consistent. Mr. Zell. discussed the axioms of connection
and order arid the axiom of parallels, and we continue here with the
topic of congruence and the axiom of Archimedes. Thus only consideration of the axiom of completeness remains to complete the
model.
APPROVEE:
Signature redacted for privacy.
Professor of Mathematics
In Charge of Major
Signature redacted for privacy.
of Department of Mathematics
Signature redacted for privacy.
7V7
'
Dean of Graduate School
Date thesis is presented
November 13, 1967
Typed by Carol Baker for Robert William Eschrich
A Model of Non-Euclidean Geometry in Three
Dimensions, II
by
Robert William Eschrich
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Master of Science
June 1968
TABLE OF CONTENTS
Page
I.
CONGRUENCE
IL THE AXIOMOF ARCHIMEDES
BIBLIOGRAPHY
1
28
33
A MODEL OF NON-EUCLIDEAN GEOMETRY IN THREE
DIMENSIONS, II
I.
Definition 7.
CONGRUENCE
Let
be a line with the parametric equations
I
x = f(t)
y = g(t)
z = h(t)
given in lemma 4. If
P
is a point of I, then the points of
I
may be divided into the following three classes by lemma 4:
(1)
{P}
{Q: tQ < t}
{Q: tQ> t}
the ray. If a ray has origin P and if
distinct from P,
is called the origin of
P
A ray is one of the sets (ii) and (iii).
Q
is a point of the ray
-
then the ray is denoted by PQ.
the negative ray of I
(ii) is called
with origin P with respect to the given
parametric equations. (iii) is called the positive ray.
Definition 8.
Given a plane
the inverse with respect to
IT
iT:
D(x2+y2+z2+l) + Ax+ By+ Cz
of the point
P
is defined as
0,
2
p2 + BK1,
(p1 + AK1,
p3 + CK1)
-2(Ap1 + Bp2 + Cp3)
where
K1
-
A2 + B2
+
C2
if
), -+K2(p2+), -+K2(p3+
+ K(p1 +
(-
D=O, and
A2 + B2 + C2
1
an
wheiw
=
(p1 +
A
z
+ (p2 +
B 2
2
+ (p3 +
if DO.
A transformation of points into points which maps every point into its
inverse with respect to
the plane
it.
ir
The mapping
is called the inversion with respect to
(P) = P
defined for all points by
is the only other type of inversion. Henceforth, K1
and
will
K2
refer to the above expressions unless otherwise specified.
Lemma 5.
Proof:
Inversions map points into points.
Suppose P is a point and
to the plane
D(x2
+
I
is the inversion with respect
y2 + z2 + 1) + Ax + By + Cz = 0.
1(P) = (p1 + AK1, p2 + BK1, p3 + CK1).
Since
If
D
0,
3
O<(p1+AK1)2+(p2+BK1)2+(p3+CK1)2p+p+p<1,
is a point.
1(P)
0< (-
11
+
D
0,
1(P)
is a point since
K2(p1+))2+
<1
The proof of this last inequality follows from the inequalities
A2+B2+C2
2
2
2
p1+p2+p3<l
and
2
>1.
4D
Lemma 6;
Proof:
11
I
is an inversion and
It will suffice to prove that
1(1(P))
P
and
is a point,
P
I(I(P))=R
have the same
first coordinate, for their second and third coordinates will then be
equal by symmetry..
Suppose
D(2
+
1(1(P))
y2
+
I
is inversion with respect to the plane
z2 + 1) + Ax + By + Cz = 0.
The first component of
is
A(p1+AK1) + B(p2+BK1) + C(p3+CK1)
p1 + AK1 - 2A
)
A2 + B2 + C2
if
D=O,
and
4
A2
+
B2 + C2
4D2
A
A
A
(Th+K2 (p1+)+
B
B2
c
A2 + ( B-+K2 (p2+)+)
+ ( -+K2
(p3+
X K2 (p1 +
if
0.
D
Both of these expressions can be simplified to
Inversions are one-to-one and onto.
Lemma 7.
Proof:
Suppose
points such that
lemma 6.
I
is the inversion with respect to the plane
+ 1) + Ax + By + Cz
D(x2 + y2 +
So
p1
1(P) = 1(Q),
If
P and Q are distinct
then P = 1(1(P)) = 1(1(Q))
is one-to-one.
I
0.
I
Q by
is onto, for
P = I((p1 + AK1, p2 + BK1, p3 + CK1 ))
if
D=O,
and
P = I((-+K2(p1+), -+ K2(p2 +), -+ K2(p3+)))
if DO.
Lemma 8.
such that
1(P)
11
8.
P
is a point,
then there is an inversion
I
5
Proof:
P = 0,
If
where
1(P) = e
I=
If
.
P
let I
e,
be inversion with respect to the plane
2
2
2
2
2
2
(x+y+z+1)-p1x-p2y-p3z=O.
2
Then the first coordinate of
1(P)
is
-
A
+ K2(p1 +
A
)
where
A = -p1
p +p
+p
2
and
1
K2-
2
2
2
'-(p1 + p2 + p3
The first coordinate simplifies to zero. By symmetry the second and
third coordinates are also zero, so
Lemma 9.
If
point other than
e,
i
1(P)
e.
is a line throu.gh the origin and P is a
then there are parametric equations
x = at
y = bt
z = ct
of
i
origin
such that P is a point of the positive ray of i
0.
with
6
Proof:
e
If
is a point of the line
2
2
D1(x2+y +z +1)+A1x+ B1y+ C1z = 0
1:
f
(x2 +y2+z 2+1)+A2x+B2y+C2z=O
then D1 = 0 and
= 0.
Case 2 of lemma 4 gives parametric
e quat ions
x = at
y = bt
z
for
.
Since
0,
te
ct
either t > 0 or t < 0.
If
t
> 0,
P is a. point of the positive ray with respect to the above parametric
equations.
If
t
< 0,
p is a point of the positive ray if
x = -at
y
-bt
z
-ct
are used as the parametric equations of
Lemma 10.
If
.
is a line and I is the inversion with
respect to the plane D(x2 +y 2+z +2 1) + Ax+
I(fl
is a line.
y + Cz = 0,
then
7
Proof:
Let
be the line
.
D1
2
2
2
(x+y+z+1)+A1x+B1y+C1z=O
2
2
+y2 +z 2+1)+A2x+B2y+C2z0
First suppose D1 =
parametric equations x
D
y
If
0,
1(1)
Then
= 0.
a0t,
0
y = b0t,
where
so
has
If
-2(Aa0+ Bb0+ Ccc)
K1
-
then
1(L) = L.
A2 + B2
+
such that
If
D
0
(BC' - W C)x + (A' C-AC' )y + (AB' -A' B)z
0
does not exist, then
1(L)
is the line
2
2
2
(x+y+z+1)+A'x+B'y+C'zO
I(fl:
where
.
c0t by lemma 9.
and there exists a real number
(a0, b0, c0)= (k0A, k0B, k0C),
a
z
.,
is the line with parametric equations x = (a0 + AK1)t,
(b0 + BK1)t, z = c0 + CK1,.
D
is a point of
0
A
D
- K2(a0t0 +
B' =
-K2(b0t0+-
=
- K2(ct +
A
A2 + B2 + C2
K2=
4D2
CZ
AZ+ (b0t0+)
B
(a0t0+)
+(c0t0+)
but such
8
-(Aa0 + Bb0 + Cc0)
- 2D(a
+c
+b
Secondly consider the case in which at least one of
is non-zero. Assume
are
The equations defining i
0.
D1
and
D1
equivalent to
x2 +y2 +z 2+1+2ax+2by+2cz0
a'x+b'y+c'z
A
where
A2
Za
B
,
Zb =
2c =
,
a',
B2 = b', and C2 = c'.
B
B2
1
-
C
,
and
c' =
0
C
and if D2 = 0,
0,
If
then
a
then
'= D1
-
ID2
C
If
-
D
0,
then
I(s)
is the
line
2
(x + y + z
2
+ 1) + 2 (a-AK1 )x+ 2(b-BK )y+ 2(c-CK )z = 0
1(i):
f2
2(Aa + Bb + Cc)
where
*
ntjmber
the line
(a' +AK )+ (b' +BK )y+ (c' +CK )z
If
D
0
and there exists a real
A2 + B2 + C2
stich that
(a,b,c) =(k1A,
k1B, k1C),
then
I(s)
is
0
9
2
2
2
(x+y+z+1)+A"x+B"y+C"zO
1(i):
a'x+ b'y + c'z= 0
where
=
- K21(f(t1) +
)
- K22(f(t2) +
=
- K21(g(t1) +
)
- K22(g(t2) +
=
- K21(h(t1) +
)
- K22(h(t2) +
A2+ B2+C2
-1
4D2
K21 -
)+ (g(t1)+) + (h(t1)+)
(f(t1)+
A2 + B2 +
4D2
K22 (f(t2) +
2
A
)
+ (g(t2) +
B
2
2
+ (h(t2) +
a2 +b2 +c2 -1
(-a) +(--b) +(A
2
t2 = -ti
f(t)
g(t)
= -a + t(- a)
-b + t(- b)
and
h(t) = -c +
c)
B
2
C
2
c)
C
10
If
0
D
but such a
does not exist, then
k1
1(1)
is the line
x2 + y 2 + z 2 + 1) + A"x + B"y + C"z = 0
1(1): ? ( g(t1)h(t2)-g(t2)h(t1)] x+ [f(t2)h(t1)-f(t1)h(t2)J y
+ [f(t1)g(t2)-f(t2)g(t1)J z
Lemma 11.
y = b1t, z = c1t,
and
c2t,
and
y
b2t, z
I
has parametric equations x
a1t,
has parametric equations
x = a2t,
is inversion with respect to the plane
x
z
y
c2
b2
a2
a1
+
+
+
C
1
c2
b2
0
a1
c1
a1
b1
a2
c
a2
b2
2
=0,
then
I(i) =
Proof:
such that
If
P
and
1(12)
is a point of
and
=
1
1'
1(0)
then 1(P)
0.
is the point of I 2
11
2
2
,a1 +b1
+c12
a +b
+c
t1(p) =
1(h)
so
By lemma 6,
=
1
=
1(1(u))
=
1(0)
1(12).
0
=
trivially.
iff
I
fl
there are inversions
o
RS
o
I (P)
21
=
PQ and
Two segments
Metadefinition 7.
I
R
12I
,
and
are congruent, we write
I
,
,
such that
n
o
o
PQ
RS.
I
I
n
are congruent
RS
I (Q)
21
= S.
PQ and
11
Lemma 12. Congruence of segments is an equivalence relation.
Let
Proof:
1(1(P))
=
be any inversion. Then if PQ is a segment,
I
P and
1(1(Q))
=
Q,
so
PQ
is some finite product of inversions
= 4i(4i(P))
Then
i.i(Q)=S.
by lemma 6, so
RS
PQ.
then there is a finite product
nd
c(S)
pQ
Uv.
=
V.
Lemma 13.
I(?)I()I(R).
c2(4i(P))
So
If
I
i4i
PQ.
such that
P and
= U
i4i
RS
(P)
= R
and
of inversions such that
and
V.
*
is an inversion and PQR,
there
RS
4i(4i(Q))
4i(S)
Finally, if PQ
2
PQ
11
and
Q
RS
UV,
c2(R)
U
Hence
then
12
?roof:
4
1
Let
be the line incident upon P and
I
has parametric equations
or t> tQ> tR
<tQ <tR
x = f(t), y = g(t),
Kis_
z = h(t).
Either
without loss of generality.
g1(t) = g(t) + BK1Q
h1(t) = h(t) + CK1R where for any point
-2(As1 +
By lemma
Since these two cases are sym-
metric, we can suppose t <tQ <tR
Let f1(t) = f(t) + AK1
Q.
and
S
+ Cs3)
A2+B2+C2
Also let
A
f2(t) = -
+ K2(f(t) + A
B
g2(t) =
K2Q(g(t) +
-m
B
and
h2(t)=
where for any point
S
A2 + B2 + C2
4D2
A2
(s1+Th)
D = 0,
of
1(1).
x = f1(t),
If
D
0,
y
1(1)
C2
B2 +(S3+)
s2+2D)
g1(t),
z = h1(t)
are parametric equations
is the line with parametric equations
13
f2(t), y = g2(t), z = h2(t).
x
1(Q) = (f.(tQ) g.(tQ), h.(tQ)) and
where
i=1
t1(p) = tp
if
In either case
D L 0.
tI(R) - tR
t1(p) < t1(Q) < t](p)
SO
*
and
I(P)I(Q)I(R).
Theorem 14.
1
and
t1(Q) = tQ
1(R) = (f.(tR). g.(t) hi(tR))
i=2
and
D=0
if
(f.(t) g.(t), h.(t)).
1(P)
Now
a.nd
If
P and Q are distinct points of the line
then on a given ray of
i.s a point of the line
R
with origin R there is a unique point
such that PQ ' RS.
S
A segment is congruent to itself.
Proof:
If
0
is a point of
lemma 8 and let I
2
let
.
Otherwise apply
be inversion with respect to the plane
2
2
2
(x + y + z
Then 11(P)
II =
0.
2
By lemma 10,
+ 1) - p1x-p2y-p3z
11(L j)
0
is some line through
0,
and by lemma 9 this line has parametric equations x = a1t, y = b1t,
z = c 1t
such that
11(Q)
is on the positive ray of
11(L1)
with
origin 0.
Since
x
f(t), y
origin
R
is a line, by lemma 4 it has parametric equations
g(t), z
h(t)
(a <t < b).
The given ray of
with
is either the positive or the negative ray with respect to
these equations. Since these two possibilities are symmetric, we
14
can suppose it is the positive ray without loss of generality. Now
a <tR <
tR+b
<b,
2
so if
tR +b
parameter
then
2
is the point of 12
U
given by the
is a point of the positive ray of
U
2
with origin IL
If
8 and let
let
is a point of 12
0
12 =
Otherwise apply Lemma
.
be inversion with respect to the plane
12
2
2
2
r1+r2+r3
2
(x2 +y 2+z +
1) - r1x-r2y-r3z = 0
2
Then 12(R)
By lemma 10,
0.
Iz(lz)
is a line through
By lemma 9 this line has parametric equations
z = c2t
such that
with origin
and
are the same line and
are the same line and
12(12)
the positive ray of Iz(lz),
let
11(Q)
use lemma 11 and let
Iz(lz),
12(12)
0.
Il(l)
If
y = b2t,
x = a2t,
is a point of the positive ray of
12(U)
0.
13 =
If
11(11)
11(Q)
is on
and
12(12)
is a point of the negative ray of
13
be inversion with respect to the
plane
a1x+b1y+c1z = 0.
'-
let
l
13
and
12(12)
are different lines, using lemma
be inversion with respect to the plane
11,
15
x
a2
a1
222
'Ja1+b1+c1
z
y
222
'.fa1+b1+c1
+
.Ja+b+c
Cl
b2
b1
+
C2
222 Nla+b+c
'.fa2+b2+c2
+
222
N1a2+b2+c2
a1
c1
a1
b1
a2
c2
a2
b2
=0.
Then 13(0) = 0 while
12(12)
with origin 0.
Let S = 121311(Q).
with origin R.
and
is a point of the positive ray of
1311(Q)
121311(Q)
Then
S
is a point of the positive ray of
Furthermore,
PQ
since
RS
121311(P)
R
S.
It remains to show that the point
S
is unique. Suppose there
which is distinct from
S
yet satisfies the hypotheses
is a point S'
ofthetheorem. Either
*
RSS'
or
*
Sincethesetwocases are
RStS.
.1
symmetric, we will consider the case where RS' .
PQ
RS
and PQ
RS'
imply
a finite product of inversions
RS
RS'.
such that
Applying lemma 6 to this last equation gives
13 implies
*
RS'S,
By lemma 12,
This means there is
(B) = B
4i(S') = S.
contradicting theorem 10.
A segment is congruent to itself by lemma 12.
and
(5) = 5'
Now lemma
16
RS and PQ
Theorem 15. If PQ
Proof:
then
UV,
UV.
RS
By lemma 12.
are points of
Theorem 16. If P, Q, and B
Q', and B'
and if
are points of
*
P1,
and
*
P' Q' B'
PQB,
PQP'Q', and QBQ'B', then PBP'B'.
By hypothesis there is a finite product of inversions
Proof:
such that i4i(P) = P'
unique point
R'
(Q) = Q'.
of the ray
is a point of Q'R'
4i(R),
%4J(R)
S
and
Q'B'
and
By theorem 14 there is a
such that
Q'B',
QB
B'.
S
for by lemma 13 PB implies P'
is a point of Q'B'.
(P) = P'
by definition, since
Definition 9.
If
1!
B'
Hence
and
.,
(B),
so
P'B'
B'
is a line of a plane
point of the plane but not a point of
Also
Now PB
4i(B).
(B)
Since
Q' S.
QB
ir
and
Q
then the points of
is a
can
ir
be divided into the following three classes:
{u:u is a point of .}
{u: QU has no points of
.
and
U
is not in
(i), or
UQ}
{U: U
is not in
QU contains a point of
(i) but
The set (ii) is called the side of
.
containing
Q.
.
}
17
Definition 10.
P, Q, and B are noncollinear or if
If
-I-
P, Q, andB arecollinearand PB,
is called an angle with vertex
P.
called a straight angle. PQ '.jPB
or by
L
(PQ,
then
PQL1PBL) {P}
In the latter case the angle is
is denoted by L QPB
{P}
PB).
Metadefinition 8.
are called
L P' Q' B
and
L PQB
congruent angles iff there is a finite product of inversions
such that
4i(Q) = Q'
point of
ji(Q) 4i(R).
and
LPQB
If
this is denoted by
r2<O,
is a
B'
are congruent,
C,
and if
p2 > 0,
q1 > 0
-'-
then PB.
Using lemma 5 let
Proof:
LP' Q'B'
and
P = (p1,p2,O), Q = (q1,O,O), and
(r1,r2, 0) are points of a line
and
4i(Q)i4i(P),
LP' Q'B'.
LPQR
Lemma 14. If
B
is a point of
P'
,
iji
I
be inversion with respect to the
2
plane
i-- (x 2 + y 2
2
+ z + 1) -q1x
is a line through e.
y = bt, z
ct.
So
Since
I(.C)
0.
t1(Q) = 0.
1(Q) = e,
1
2
q1
1----:
q1
z
PZ3
2
1(Q)
0
and
ICC)
has parametric equations x = at,
The second coordinate of the point
q1
Then
2
p2
1(P)
is
18
0 < q1 < 1
which is greater than zero since
second coordinate of
and p2> 0.
The
is
1(R)
1
1
2
r1
q1
(r1 - -
r2
2
+ r22 + r32
)
q1
which is less than zero since
btl(R) < 0.
or t1(p) > 0 > t1(flr
we have either t1(p) < t1(Q) <t1(fl)
t1(p)> t1(Q) > tI(R)
and 13
bt1(p) > 0 while
Thus
So either t1(p) < 0 <t1(fl)
t1(Q) = 0,
Since
r2 < 0.
In either case
I(P)I(à)I(R).
or
By lemma 6
PR.
Theorem l7a. Any two straight angles are congruent.
and LSQ' U are straight angles.
Proof: Suppose LPQB
theorem 14, there is a point P'
of
Q' S
So there is a finite product of inversions
and
4i(Q)
Q'.
Q' U by lemma 13.
point of
4i (Q)
such that
ti(R)
By theorem 1 and lemma 10,
of the line incident upon
is a point of
4i
such that
(R).
U,
4i(Q) = Q',
Now
4i(Q) 4i(P),
and
Q'
and
Hence
and
4i (B)
ti(Q)4i(P)
ti(Q) ti(R) = Q'R'
LPQR
L.SQ? U.
By
P' Q - PQ.
P'
i (P)
is a point
is a point of
Q' pt
so
U
o
is a
S
19
Any angle which is congruent to a straight angle
Theorem 17b.
is itself a straight angle.
LP'QtR'
is a straight angle and
LPQR
LPQR.
Proof:
Suppose
Let
be the finite product of inversions giving the congruence of
4'
LP'QtR'
and
lemma 10 implies that
*
implies
PQR
Since
LPQR.
t4i (P),
*
i(P)Q"Ii(R)
P, Q, and R are collinear,
Q' , and
t4i
by lemma 13.
congruence of angles, either Pt
are collinear, and
(R)
By the definition of
is a point of
Qtij(P)
and
R'
is a point of Q' 4' (R) or P' is a point of Qi (R) and R' is a point of
and R' arecollinear and P'Q'R'.
Ineithercase
Q'i(P)
Theorem 17c.
is a line of the plane
V
r
and
there is in
L(r,$),
is not a straight angle and
is a given ray of
with origin
then
is a side of
and
LPQR
LPQR
11
a ray
s
with origin
V
such that
and
-; is unique.
Proof: (1) Since
equations
is a line, by lemma 4 it has parametric
x = f(t), y = g(t), z = h(t)
(a <t <b).
The given ray
with origin V is either the positive or the negative ray. We will
consider the case where it is the positive ray, the other case being
20
let I
Q = 0,
Otherwise let
=
2
we may write
is a point of thepositive ray with origin V,
If
for t0 -
U: (f(t0), g(t0), h(t0))
Since the point
symmetric
b+tv
be inversion with
I
respect to the plane
2
2
2
q1 + q2 + q3
2
(x + y
2
Then
Q and
distinct lines through
R
Thus
i
not a straight angle,
Rt.
2
+ z + 1) -q1x-q2y-q3z
Denote
0.
i
by
I1(.2)
11(P)
by
0
P and Q by
Denote the line incident upon
11(Q) = 0.
and the line incident upon
by
2
Since
and
P'
LPQR
11(i3)
and
2
is
are
11(R)
The plane
x
z
y
=0
rr2 =
r
is incident upon
r
r
P'. 0, and R' , and by theorem 4 it is the only
such plane.
11
V=0
let
12 =
.
Otherwise let
12
be inversion with
respect to the plane
2
2
2
v1+v2+v3
2
2
2
(x + y + z
2
+ 1)-v1x-v2y-v3z = 0
21
Then 12(V)
If
9
aiid
is a point of
S
write 12(S) = S'
is a line through
12(1 j)
,
and
collinear by lemma 10.
then
is a point of
12(S)
then S'
U' ,
12(U)
9
by lemma 10.
12(ir1).
0, and U'
,
If we
are non-
The plane
x
z
y
=0
113 =
is the unique plane incident upon
S ,0
and 15'.
Let
A=
r2
frz
p3
p31
p1
r3'
r3'
r1
S3t
'
p37
c=1P2
r3'
Izt
and
s2,
A'
B=
13
Then
r2
is the plane Ax + By + Cz = 0 and
A' x + B' y + C' z
let
13
Now if
0.
112 = 113
let
be inversion with respect to the plane
x
n1
y
z
m2
m3
n2
n3
=0
i-s the plane
ir3
13 =
.
Otherwise
22
where M is the point other than
common to
0
rr2
and
rr3
given by theorem 7, and where
A
A'
+
B
'.JA+B2+C2 i.JA+BI2+Ct2
.JA2+B2+C2
C
and
if3
,.JAI2+B!2+Ct2
Ct
+
)
,.JAt2+Bt2+Cl2
'.JA2+B2+C2
13(112)
B'
+
1311(Q) = 12(V) = 0.
Use lemma 9 to get parametric equations x = a1t, .y
z
c1t for
I(
ray with origin
l
0.
X - a2t, y = b2t, z
such that
12(U)
c2t
1311(QP) = 12(VtJ)
1311(QP)
12(VU),
a1x + b1y + c1z = 0.
14
is a point of the positive
Likewise obtain parametric equations
for
I3Il(2) such that
a point of the positive ray with origin
and
b1t,
let
If
let
14
14
.
1311(P)
0.
If
If
I3I1(2) = Iz(
1311 (
is
Iz( 0
i
but
be inversion with respect to the plane
I3I1(2)
1z
i'
be inversion with respect to the plane
use lemma 11 and let
23
x
a2
a1
b2
b1
c1
C2
+
+
+
2
2
.ia+b+c
z
y
2 2 2
2
2
2
2 2 2
'ja1+b1+c1 'ja2+b2+c2 'ja1+b1+c1 i.ja+b+c22
2
'ia2+b2+c2
b1
C1
a1
c1
a1
b1
b2
C2
a2
c2
a2
b2
=
In any case 14(ir3) =
12141311(R)
Now
point of
If
and
1'
12141311(R)
12(VU).
141311(QP)
Furthermore, it is not a
is a point of
for this would imply the collinearity of P, Q, and B.
is a point of
,
Otherwise let
let 15 =
15
be inversion with respect to the plane
x
y
z
a1
b1
c1
At
B?
C!
Denote the product of inversions
let W
Let
s
Then
YW.
since qi(Q) = V,
point of
1512141311
W is a point of
&i(R),
ti(Q)iIj(R),
LI
s
=0.
by
Then if we
.
is in
and has origin V,
is a point of qi(Q)qi(P),
LPQR '
&i.
LUVW.
and
qi(B)
and
is a
24
(ii) To show the uniqueness of
s'
with origin V which is distinct from
in
LPQR
s')
L(
Otherwise let
is the plane
If
n
and the plane
A'
line
m
m
n
n
x
C'
jA'2+B'2+C'2
iA '2+B'2+C'2
1612(12) =
If
but
I6I2(.2)
=
u
If
.
be the positive ray of
and
I6I2(VU)u,
be the
Let
1612(VU) = u,
let
17
let
17 =
be inversion with
1612(12)
use lemma 9 to get
x = a3t, y = b3t, z = 0
for I6I2(. i such
is the positive ray with origin U with respect to
parametric equations
1612 (VU)
common
U
and where
t, y = 0, z = 0, and let
U.
16 =
=0
B'
+1,
respect to the plane x = 0.
that
z
is the plane z = 0 and 1612(V) = U.
with origin
If
y
z = 0,
JA'2+B'2+C '
16(Tr3)
let
z = 0,
is the point given by theorem 7 other than
where M'
N'=(
and that
s
be inversion with respect to the plane
16
x
to
suppose there is a ray
5,
this representation. Then use lemma 11 and let
with respect to the plane
17
be inversion
25
x
y
a3
b3
+1
222
22
Now
line
-b3
0
L71612(S)
Otherwise let
let
has a positive second coordinate,
18 =
be inversion with respect to the plane y = 0.
18
18L71612(s)
is the positive ray with origin
x = a4t, y = b4t, z = 0 , and
18171612(s' )
with origin e of some line x = a5t,
a4
happen that
=0.
0
ja3 + b3+c3
0
If
z
22
ja4+b4
y
a5
-
22
ja5+b5
,
0
of some
is the positive ray
b5t, z
It cannot
0.
for then we would have
18171612(5) = I8I7I6I(s1). Lemma 6 would then imply
So either
a4
22
Ja4+b4
a5
22
or
22
ja4+b4
ja5+b5
a5
a4
>
22
Since these cases are symmetric, we will consider only the former
case.
Again apply lemma 11, letting
to the plane
19
be inversion with respect
26
x
y
a5
b5
z
+1
0
1918171612 (s) =,
Now
origin
and
x
9
1918171612 (VJ)
5
is the positiye ray with
of some line x = a6t, y = b6t, z =
18171612 (
X =(x1, x2, 0)
(z1,z2,0) is a point of
b6 > 0,
where
0
is the positive ray with origin
9
of some line
b7 < 0.
a7t, y = b7t, z = 0 where
Sippose
Z
-b
0
0
=0.
I
is a point of
18171612 (s)
1918171612 (VU)
Then x2> 0
and
and
z2<O.
By theorem 1, the line
222
22
22
22
22
+[z1(x1+x2+1)_x1(z1+z2+l)]y0
(x +y +z + 1)+[ x2(z +z2 + 1)-z2(x1 +x2 + 1)] x
1.
:IS
incident Ipon X and
Z.
The point
Y = (y1, 0,0), where
is the smaller of the two roots of
(z2x1-x2z1)(x +l)+[x2(z1 +z2+1)-z2(x1+x2+1)]x = 0,
-
is a point of u
and of
By lemma 14,
*
XYZ.
27
Since L(r, s) L(r, s'),
LZOX
By the definition
LZOY.
of congruence of angles, there is a finite product of inversions
such that
Z
(0) = 0,
O2(Z),
is a point of
X
2
is a point of
and Y is a point of O(X). By Lemma 14,
O(Y),
*
(X)(Z)2(Y).
*
implies
By lemma 13, XYZ
*
(X)2(Y)2(Z).
This
contradicts theorem 10.
Theorem 18.
PQ
such that
PR
LP'Q'R'
are triangles
P' Q' R'
and
PQR
Pt Q',
tPQR
then
If
P1 Rt,
and
LQPR
LQ' P' R',
tPRQtP'R'Q'.
and
Proof: By the definition of congruence of angles, there is a product
of inversions
and
R'
and
4i(R)
point of
PR
Lu
such that
Pi(R).
is a point of
is a point of
P'Q'
P'R",
= 4'(Q)
immediate.
and
and
then
R'
Q'
Thus 4i(Q)
is a point of P' 4s(Q),
is a point of
P'R'. By theorem 14, if
R't
of
QI = Q!I
=
P',
4s(P)
4s(R).
PtR',
and
R'
and if
R"
PQ
Q"
P' Q"
P'Qt
is a
and
This means that
The conclusion of the theorem is now
28
IL THE AXIOM OF ARCHIMEDES
PQ and
Theorem 19. If
R1,R2,
there exist points
P1R2 and
of the ray
,
:E3
such that
R12R3 and R1R2RS,
RS,
PR1
are line segments, then (i)
RS
and R.i-i R. = RS,
general R.1-1 R.R.
1+1
1
and in
and (ii) there exists
1
a natural ntzmber
*
such that
n
PQR
n
Proof: (i) Denote the line incident upon P and
the line incIdnt upon
and
R
If not, use lemma 5 and let
222
p1+p2+p3
2
Then 11(P) = 0
c1t
IU l
such that
a1 < 0
the plane
let I
and
x
0.
b1
0
If
c1,
P= 0,
If
and
1
let
Ii
be inversion with respect to the plane
and by lemma
with origin 0.
If
by i.
9
9
7
(x+y+z+1)p1xp2yp3z -
origin. By lemma
y = b1t, z
S
Q by
1
j)
0
is a line through the
has parametric equations x
11(Q)
is a point of the positive ray of
a1 > 0 and b1
let
a1t,
0
c1,
let
12
be inversion with respect to
If neither of these is the case, use lemma 11 and
be inversion with respect to the plane
29
x
y
z
a1
b1
ci
+1
222
222
J a+b1+c1
ci
0
Denote
by
1211
let
R = 0,
-b1
1(P) = 1(e) = 8
where
0,0)
Q' =
is some point
If
Then
.
=0.
13 =
.
q
by lemma 11, and
> 0.
If not, use lemma 5 and let
13
inversion with respect to the plane
222
r1 +r2+r3
(x
Then 13(R)
origin. By lemma 9
with origin 0.
a2 < 0
plane
14
such that
If
I3(.
=
13(S)
14
x = a2t,
is on the positive ray of I3(.2)
a2 > 0 and b2
let
0.
is a line through the
has parametric equations
I3(.2)
and b2 = c2 = 0,
x = 0.
+ y2 + z 2 +i)-r1x-r2y-r3z
and by lemma 10
0,
y = b2t, z = c2t
2
0 = c2,
let
14 =
.
If
be inversion with respect to the
If neither of these is the case, use lemma 11 and let
be inversion with respect to the plane
30
x
y
z
a2
b2
C2
222
.Ja2 +b2+c2
c2
-b2
0
Let
Then
1413 =
222
222
'.Ja2 +b2 +C2
= 0.
by lemma 11, and
4i2(R) = 14(0) = 0
4i2(S)
is somepoint S1 z(s1,0,O) where O<$i< 1.
If
q1'
< s,
and
= R1,
14 since
eQ'S1 and
In genera],
c2k1
PR
Furthermore, PR
RS.
1
qi1(0) = P,
k> 1,
for
Since
= qi1qi2(S).
and since qi1qi2(R) = P
is a point of
R1
,
and let
n=1
let
let
by lemma
qi1(S1) =R1.
and
= Q,
1
where
=
is inversion with respect to the plane
2
(x2+y2+z2+1)
i+$kl
x0 and
s0=e,
and let
=1
1
If
are computed, we find that
s1,s2,
i = 2,3,4,
0< si_i < 1,
Since
*
*
ls23
Hence, 5O5152'
*
*
PR1R, R1R2R3
,
,
s.1 - (
L-iSjSi+i
S.
, R.i-i RiR i+i
o
,
oc2c2qi(S)
1+s
2
2
1+s.1-i
,",
*
,
2s 1- 1
s.i -
lk
for
2
1-i
) s i-i > s.i-i
and by lemma 13,
Also, if
k
isodd,then
31
and
Rki = l2k-1 °
2i2 (R),
°
and if k is evex,then
ik-i
°
Rki = ik-1
°
Rk
and
(Note that
whether
c2
for
(S.) = S.
11
1
°
i
i,2,.) So RkRkl _ RS
is even or odd.
k
The first conclusion of the theorem has now been proved.
remains to show that there exists
define
zsi-i
s
=
i+s
i- 1
2
*
can be written
s.
1
then
,
n
0<s
*
such that
PQRn.
< s. < 1
since
1
1
If we
0 < s.
i-i
i-i
s
1(2i11)
.
We want to find
It
i0
< 1.
such that
jo_i
2
S.
1
5
1
i-i
-
>q11.
_i)si
1+(2
Both
q1' (1_s l
and
s1(l-q1') are positive and finite. So if
iog[2(
D-
-
then
D
i.s finite.
q(l-s1)
s1(i-q)
log 2
By the Archirnedean principle for real numbers,
32
there exists a natural number i0 such that i0> D.
Now i0> D
implies
i0
q(l-s1)
>2 s1(l-q)
implies
(1-s
i0- 1
>
2
s1(l-q
implies
S.
10
Thus
*
PQR.-
*
9Q' S.
.
Since
-
>
i -1
l+(2
0
1)si
ljJ
1(Qt) = Q,
and
10
by lemma 13.
1.0
namely n = i0.
So there exists n such that
) = R.
ijj
10
*
PQR
n
10
33
BIBLIOGRAPHY
Hille, Einar. Analytic function theory. Vol.
1.
New York,
Blai.sdell, 1965. 308 p.
Jacobs, Michael J. A proof of the consistency of geometry.
Master' s thesis. Corvallis, Oregon State University, 1961.
49 numb. leaves.
Zell, William Lee. A model of non-Euclidean geometry in three
dimensions. Master's thesis. Corvallis, Oregon State
University, 1967. 34 numb. leaves.