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A PROOF OF THE CONSISTENCY OF GEOMETRY
by
MICHAEL JAY JACOBS
A THESIS
submitted to
OREGON STATE UNIVIRSITY
in partial fulfillment of the requirements for the
degree of
MASTER OP SCIENCE
June 1961
Redacted for Privacy
Professor ,f Nathematics
In Charf.e of ilajor
Redacted for Privacy______________ iimáii öf Dópartment of Natheraati es
Redcted for Privacy
Chairman o 1oo1 G±uate Committee
Redacted for Privacy ean of Graduate Ecboo1
Date thesis is presented
TypeJ by Lilah N. Potter
2n
TABLE OF CONTENTE
i-age
INTRODUCTION. . . . . . . . . . . . . . . . i
I. AXIOItz3 OF CONNECTION. . . . . . . . . . . . 2
II. AXIONS OF ORDER . , . . , 15
III. . . . . . . . . . . . . . . . . . . . . . . 23
IV. PIAYFAIi'S AXIOM. . . . . . . . . . . . . . 44
V. ARCHINEDES' AXIOM . . . . . . . . . . . . . 46
VI. THE CONELETENESS AXIOM. . . . . . . . . . . 48
BIBLIOGRAPHY. . . . . . . . . . . . . . . . 49
A PROOF OF THE CONSISTENCY OF GEOMETRY
INTRODUCTION
It is our purpose to prove the following Theorem:
If the axioms of arithmetic are consistent (i.e.,
don't lead to a contradiction) then so are the axioms of
geometry.
1or the axioms of arithmetic we take the well
known axioms of reano. For the axioms of geometry, we
take the system of axioms given by Hubert.
The proof of the theorem consists of exhibiting
an arithmetic model for geometry. Once such a model is
constructed, then any contradiction in geometry neces-
sarily will involve a contradiction in arithmetic. Thus
the theorem will follow.
The following notations will be used throughout
the text:
Iletadefinition -- A definition which assigns a
geometric name to an arithmetic entity.
Definition -- A definition wbich assigns a name
to a complex of geometric entities, or a definition
which assigns a name to a complex of arithmetic entities.
Theorem -- A deductive result of arithmetic which
has the same wording as an axiom of geometry.
Lemma -- All other deductive results.
2
I. AXIONS OF CONNECTION
Metadef 1. A point is a triple (x,y,z) of real numbers.
Netadef 2. A line is an equivalence class of
pairs of equations equivalent to a pair of equations, A1x+B1y+C1z =0 A2x + B2y + C2z + D2 = O
where A1, 2' B1, B2, C1, C2, D1 and D2 are real con-
stants and x, y and z are real variables and
A B C
rank 1 1 = 2.
2 B2 C2
I'íetadef 3. A plane is an equivalence class of
equations equivalent to an equation,
Ax + By + Cz + D = O
where À, B, C and D are real constants and x, y and z
are real variables and
A2 + B2 + c2 0.
Remark: We will sometimes say that 'a plane is an
equation' or that 'a line is a pair of equations' when
the equation or pair of equations in question are rep- resentatives of the plane or the line. The equations are used as naine for the plane or line.
Netadef 4. A line,
A1x + B1y + C1z + O
+ B2y + C2z +1)2 = O
is incident upon a pair of distinct points, (x1, y1,
z1), (x2, y2, z2) 1ff,
+ B1y1 + C1z1 +
= O
A1x2 + B1y2 + C1z2 + D1= O
2l + B2y1 + C2z1 + D2 = O
À2x2 + B2y2 + C2z2 + D2 = O
Lemnia 1. Two points, P1 = (x1y1z1) and P2 =
(x2y2z2), where y1 , y2, have a unique line incident
upon them.
4e will first show the uniqueness part of the
lemma. Suppose therefore that a line represented by
(ï) . A1x + B1y + C1z + D1 = O
and (2) . . . A2x + B2y + C2z + D2 = O
is incident upon P1 and P2.
Then we have,
(3) . . . + B1y1 + C1z1 + D1 = O
(4) . . . A1x2 + B1y2 + C1z2 + 1)1
(5) . . . A2x1 + B2y1 + 02z1 + D2 = O
(6).. . A2x2+B2y2+C2z2+D2=O From (3) and. (4) and (5) and (6) respectively,
(7) 1(x1-x2) + B1(y1-y2) + C1(z1-z2) = O
(8) . . . A2(x1-x2) + B2(y1-y2) + C2(z1-z2) = O
4
From (7) and (8),
(9) . s . (A1C2-A2C1)(x1-x2) = (B1C2-C1B2)(y2-y1)
and
(lo). . . (A1C2-A2C1)(z1-z2) = (A2B1-A1B2)(y2-y1)
From (3) and (5) and (4) and (6) respectively,
(11). . . (A1C2-C1A2)x1 + (C2B1-C1B2)y1 + (C2D1-C1D2)
=0
(12). . . (A2B1-A1B)y2 + (C1A2-C2A1)z2 + (À2D1-A1D2)
=0
Since y1 y2, from equations (9) and (10),
(l). . . (B1C2-C1B2) = (A1C2-A2C1)(x3-x2)/(y1-y2)
(14). . . (A9B1-A1B2) = (A1C2-A201)(z1-z2)/(y1-y2)
Now we notice using (13) and (14) that if
A1C0-A2C1 = C, the rank of a B1 is equal to zero. A2 B2 02
ience
(15)..... A1C2 - A2C1 o
Hence from (13), (14), (15),
(16). .
B1C2-C1B2 x1-x2
A1C2-A2C1
(17). A2B1-A1B2 z1-z2
A1CA2C1 = y1-y2
From (11) and (16) and (12) and (17)
5
(18). . . x1 + x1-x2 y1 + C2D1-C1D2 = O
A1C2-C1A2
(19). . . z1-z2 y2 + z2 + A2D1-A1D2 = O
C1A202A1
ie now write a pair of equations which are
equivalent to (1) and (2),
(20). . . (Á102-A2C1)x + (B1C2-B201)y + (D1C2-D2C1) = O
(21). . . (A2B1-A1B2)y + (A2C1-A1C2)z + (D1A2-D2A1) = o
From (16), (17), (18), (19), (20) and (21) we get
another pair of equivalent equations,
(22). .
X y y1 - X1 = O
y1-y2 y1-y2
(23). . . z1-z z -z 2 2
yl2 y2i 2 - = o
This shows uniqueness. To show that such a line
exists we write a pair of equations which are satisfied
by the given triples.
(2'.). . . x(y1-y2) + (x1-x2)y + (x2-x1)y1 + x1(y2-y1) = O
(25). . . y(z1-z2) + (y1-y2)z + (z2-z1)y2 + z2(y2-y1) = O
Since y1 y2 the rank of the equations = 2 and.
thus (24) and (25) are a line incident upon P1 and P2.
Theorem 1. Two distinct points have a unique line
incident upon them.
Proof. If y1 y2 then either z1 z2 or x2. Hence
proceed with a parallel argument to that of the previous
le nun a.
Definition 1. A point F, is a point a line
iff there exists a point , such that the line is mci-
dent upon P and .. e sometimes say the point is on or
the line.
Theorem 2. Two distinct points P and Q of a line
i have the line 1 incident upon them.
Froof'. If P = (x',y',z'), ., = (xu,yu,zhl)
1 = Cz+D=O
o then we have that
Ax' + By' + Cz' + D' = O Ax" + By" + Cz" + D = O A'x' + 13'y' + C'z' + D' = O A'x" + B'y" + C'z" + D' = O
Theorem a. There exist two distinct points on
any line.
Proof. Suppose we are given a line,
A1x + B1y + C1z D1 O
A2x+B2y+C2z D20 (A B c\
In order for the rank of A B to be 2 one of
\ 2 2 2)
the following must be different from zero:
A1B2 - A2B1 A1C1 - A2C1 B1C2 - B2C1
Suppose that A1B2 - A2B1 O. Then we define,
X0 = B1D - B2D1/À1B2 - A2B1
y0 = D1A2 - D2A1/A1B2 - A2B1
i = B1C2 - B2C1
in = C1A - C2A1
n A1B2 A2B1
t = any non-zero real number.
Then the points (x0, y0, o) and (x0-s-lt, y0+int, z0+nt)
are distinct and are on the line.
A parallel proof is required for the assumption
that either i or m is different from zero.
¡'letadef 5. A plane represented by x + By + Cz +
D = O is incident upon three distinct points (x1y1z1)
(x2y2z2) (x3y3z3) iff
Ax1 + By1 + Cz1 + D = O
Ax2 + By2 + Cz2 + 1) = O
Ax3 + By3 + Cz3 + D = O
Definition 2. A point is a point of a plane iff
there exist 2 other distinct points such that the plane
is incident upon the 3 points. e sometimes say that
the point is on or in the plane. e say that a line is
a line of or in a plane iff every point of the line is
in the plane.
roi
Theorem 3b. There exist three distinct points
on every plane such that the three are not in any one
line.
Froof. Given plane, Ax + By + Cz + D = O, where
A2 + B2 + C2 O, then one of À, B and C must be differ-
ent from zero. Assume A O. Then
Ax + By + Cz + D = O y=O
and
Ax + By + Cz + D = O y+l=O
are lines. Cince no triple with y = O is a solution to
the second pair of equations, the lines are distinct.
Call the lines li
and 12s From theorem 3a we have that
there exist two points P1 and P1' on the line 1 and
P ' = (a', o, b') = (a, o, b), 1.
where a' ji( a, or b' b.
There exists a point l2 (c, -1, d) on 12
pl', P1, 2
are points of the plane since they must
satisfy the equation of the plane.
Theorem 3c. There exists a plane.
Froof. x = O is a plane.
Theorem k. Three points which are not in any
one line have a unique plane incident upon them.
rroof. Let the three given points be: (x1, y1, z1),
(x2, y2, z,..), (x, y7, z7)
Case 1.
Assume: X1 y1 z1
X2 y2 z2 = O
X3 y3 z3
therefore
Ax1 + By1 + Cz1 = O
Ax2 + By2 + Cz2 = O
Ax3 + By3 + Cz3 = O
C)
for some values of A, B, C such that A2 + B2 + C2 O.
(for if A2 4- B + C2 =0 then A= B= C = O)
Hence the points are on the plane Ax + By + Cz = O.
In order to show that this is the unique plane we
must only show that the rank of (x1 y1 z1\ is 2.
x2 y2 z2
\x3 Y3 Z3
The rank is less than 3 since the determinant is
zero. The rank is not zero since then
(x1, y1, z1) = (x2, y2, z.b) = (x3, y3, z3)
= (O, O, o)
If the rank is i then we may write without loss
of generality,
(x2, y2 z2) = k (x1, y1, z1),
(x3, y3, z3) = k'(x1, y1, z1),
in the usual vector notation.
Now suppose (x1, y1, z1) and. (x2, y2, z2) satisfy
an equation
then
and
Ax + By + Cz + D = O
Ax1 + By1 + Cz1 + D = O
+Bky1 +Ckz1 +D=O
hence -D Ax1 + By1 + Cz1 = k(Ax1 + By1 + Cz1).
If (x2, y2, z2) = (O, O, O) then D = O. If (x2, y2, z2)
(o, O, O) then k O and then + By1 + Cz1 = O and
hence D = O.
Therefore in any case, D G.
1-lence a line,
Ax + By + Cz = O A'x + B'y + C'z = O
is incident upon (x1 y1 z1) and (x2 y2 z2).
But since (x, y3, z3) = k(x1, y1, z1),
A kx, + B ky1 + C kz1 = O
A'kx1 + B'ky1 + C'kz1 = O
hence (x3, y3, z3) is on the line of (x1, y1, z1) and
(x2, y2, z2) which is a contradiction.
11
Case 2
Then
Xl yl z1
X2 2 Z2 C
X3 Y3 Z3
+By +Cz1 +0=0 + By2 + Cz2 + I) = 0
Ax3 + By3 +0z3 +D=0
has a solution for some D 0.
{ience it must be that
But since
-4_x1 + + ---z1 + i = o
-4--x2 + --2 + + i =
A B C ----x3 + + Z3 + i = O
xl yl zl
X2 y2 z2 AO
X3 y3 z3
this solution is unique for A/D, B/D, C/i).
Theorem 5. Any 3 points of a plane a which are
not in the same straight line have x incident upon them.
Proof. The 3 points (x1, y1, z1) (x2, y2, z2) (x3, y3, z3)
satisfy
1K
+ By1 + Cz1 + i) = O
Ax2 + By2 + Cz2 + D = O
Ax3 + By3 + Cz3 + D = O
where Ax +By +Cz +D=O is the plane in question.
Theorem 6. If a plane contains two points of a
line then it contains every point of the line.
Proof. Suppose a plane Ax + By + Cz + D = O contains
two points, (x1, y1, z1) and (x2, y2, z2), then
Ax1 + By1 + Cz1 + D = O
Ax2 + By2 + Cz2 + D 0.
Now there exists a line,
A'x+B'y+C'z+D' =0
A"x + B"y + C"z + " = O
incident upon (x1 y1 z1) and (x2 y2 z2) and
(A' B' c'\ Rank =2.
\ÀtI B" C")
Therefore
¡A B
Rank A' B' C' 2
A" B" C"
It is impossible that the rank be 3 since the
equations
13
Ax + By + Cz + 1) = O
A'x+B'y+C'z+D' =0 A"x + B"y + Cttz + D" = O
have more than one solution. Jience the rank is 2.
Therefore there exists a k 0 such that either (kA, kB, ko) = (n', B', C')
or
(kA, kB, kC) = (ATI, ti ,
3uppose, without loss of generality, that it is the first of these condition which is true. Then
-D' = kAx1 + kBy1 + kCz1 = -kD
hence
D' = kA)
A'x + B'y + C'z + i)' = O
and
.x + By + C7 + D = O
are equivalent. hence every solution to the line is a
solution to the plane.
Theorem 7. If two distinct planes have a point in common then there exists at least one other point in couimon with the planes. i>roof. Suppose planes
Ax + By + Cz + D = O
anJ
A'x + B'y + C'z + U' = O
have a point (x0, y0, z0) in common.
L$ince the planes have a point in common
(A B Rank
( t =2
A' B' C')
And therefore there exists a line
Ax + By + Cz + i) = O 'x+B'y+C'z+D' =0
such that each point of the line is a point of both
planes. Then by theorem 3a, the present theorem is
proved.
lL.
Theorem 8. There exist 4 points which are not
in any one plane.
froof. Consider the points, Pi = (0, 0, 0), P2 =
(1, 0, 0), P3 = (0, 1, o), P4 = (o, o, 1).
If there were a plane
Ax + By + Cz + D = O
which contained E1, then D = O. If it contained P2,
then A = -D = 0. If it contained P3, then B = -D = O.
If it contained .L4 then C = -D = O. But then
A2 + B2 + C2 o.
15
II. AXIOMS OF ORDER
Lemma 2. The set of all points on a line
x+By+Cz+D =0
A'x+B'y+C'z+D' =0
is the saine as the set of all points (x, y, z) such that
X = X0 + lt
y = y0 + mt
z z + nt, o
where (x0, y0, z0) is any point of the line, 1, in, n are
real constants, i + in + n = i and t takes on all real
values.
Proof. It was shown in the proof of theorem a that
there exists a point (r', y', z') on the line such that
every point (x, y, z) where
X = X' + i't'
y = y' + iiì't'
z = z' + n't'
1', ni', n' are real constants and t' assumes every real
value, is a point of the line and that
k = (lt)2 + (m')2 + (n')2 O
Now let t" = t'/f
i= in =
n = n'/J
16
then x = x' + it"
y = y' + mt"
z = z' + nt"
where t" assumes all real values, is an equivalent set
of equations.
Also i2 + + 1.
Now since (x0, y0, z0) is on the line, there exists a
t such that,
1ow if
then
X = X' + its, o o
yo = Y' + mt
z =z' +nt" o o
xl = X' + lt:'Ji
= y' + mt
z1 = y' + nt
x = x + l(t -t) 1 o
y1 y0 + m(t -t) z1 = z + n(t -t)
o
Hence if we let T = t "-T " i i o
xl = Xo + ltl
y1 = y + mt1
z1 = z0 + nt1
Now assume that (x9 y2 z2) is on the line. Then
Ax2+By2+Cz2+D =0 + B'y2 + C'z2 + D' O
and since
we have
Also
Hence
Ax +By +Cz +D =0 o o o
A'x0 + B'y0 + C'z0 + D' = O
ì(x -x ) + B(y0-y2) + C(z0-z2) = O 02 A'(x0-x2) +B'(y0-y2) +C'(z0-z2) = O
+ B(y1-y0) + C(z1-z) = o
-x ) + B'(y1-y0) + C'(z1-z0) = O lo
Al + Bin + Cn = O
i.'l + B'm + C'ri = O
fA B c\ Now since
( has rank 2, there exists a
IA' B' C'
t2 O such that
Hence
xo-x2 = t2l
yo-y2 = -t2m
z0-z2 = -t2n
X =X +tl 2 o 2
y2 y0 + t2in
z2 = z0 + t2n
17
Since the steps can be reversed, the lemma is proved in
both directions.
Remark: These equations, which are equivalent
tOILX+ By +Cz D= O, A'x+B'y+C'z +D' =Oare referred to as the parametric equations of the line with
respect to the point (x0, y0, z0).
Jietadef 6. Let
X = X0 + lt
y = y0 + mt
z = z0 + nt
be the parametric equations of a line with respect to
(x0 y0 z0). e say that three distinct points on the
line, A, B and C, are such that B is between A and C
(written ABC) 1ff one of the following is true,
t\> tB'> t or tK tB< ta,
where tA tB t are the values of t corresponding to
A, B and C in the above equations.
Note that if (x1, y1, z1) (x0, y0, z0) is on
the line, we will have
X = X1 + it'
y = y1 + mt'
z = z1 + nt'
is another set of parametric equations for the line.
1f t], t1, t are the values corresponding to A, B and
19
C then it is obvious that if tB is between tA and t0 in
the real ordering then t is between tk and t in the
real ordering.
Theorem 9. If A, B and C are points, then BC
implies OBA.
Proof. By symmetry.
Theorem 10. If A, B and C are points, then at
most one of ABC, BAC and ACB hold.
iroof. At most one of three real numbers is between
the other two in the ordering of the real numbers.
Theorem 11. If A and B are distinct points of
a line, then there exists a point C of the line such
that B is between A and C.
Proof. By lemma 2, the parametric equations in the
metadefinition of 'between' exist. uppose tA corre-
sponds to point A, tB corresponds to point t3 (with
respect to some point (x0, y0, z0)), then
tA < tß implies that there exists t such
that tA < tB t0
tA > tB implies that there exists t such
that tA > tB > to.
20
Definition 3. A line segment AB is the set of
all points between A and B, where A and B are distinct
points.
Definition 4. A closed line segment is the
union of AB and A y B , where A and B are distinct
points.
Definition 5. i. triangle L. ABC is defined for
3 points A, B, C such that they are not all in any one
line and is the set y y
Theorem 12. If a line 1, in the plane a, of the
points P, R (P, Q and R are not in the same straight
line) has a point of PQ y QR y RF on it, then it has
another distinct point of the triangle fR on it. Proof. Assume that the equation of the plane a is
A'x + B'y + C'z + D' = 0. Since line i has all its
points in the plane a, it may be represented by
A'x+B'y+C'z+ L)' =0
Ax + By + Cz + D = O
(A' B' '
where rank Ç
= 2, as was shown in the proof A B C
of theorem 6.
Now let
21
P = (x , y, z ) P p
= (Xq Yq z ) q
R = (Xr r' Zr
Then by lemma 2 there exists a representation of the
line of f and as follows:
X = X + lt
y = y + mt
Z = Z + nt p
3ince i has a point in common with the line of
P<,, the function
f(t) = i.(x0 + it) + B(y0 + mt) + C(z0 + nt)
+D
must have the value zero at some point of PQ. We have
two cases, since f(t) is a linear function of t.
(1) f(t) is identically zero.
Then the theorem is proved since there must exist
another point on PQ in common with 1.
(2) f(t1) = o for exactly one value of t. This
value of t corresponds to a point of P. and therefore
o=t (t <'dt p l\ q
where tq is the value of t for point .
Therefore, since f is linear on t, f(t) and
f(tq) have opposite sign.
22
Now in exactly symmetrical fashion we define
g(t') and h(t") to be the corresponding functions with
respect to the lines of and RF respectively.
Then we have,
h(ttI) = f(t) ß(t'q) = f(tq)
and
h(t") =
Hence h(tI') and g(t'q) have opposite signs. Now
since h(t") = g(t') it must be that there is a zero of
h(t") over the segment lIP or a zero of g(t') over the
segment ¿R or that h(t"r) = (t') = O. In any case the
theorem follows.
Remark:
X,
(1)... y' =
z'
III.
A transformation
¡a11 a12 a13\
(
&)] a22 a23 y
\a31 a32 a33 z
is denoted
(aij), (ai)
where
(a11 a12 a13
(aij) = f
a a22 a23
\a31 a32 a33
a1
(ai) = a2
a3
a1
a
a7
23
The statement of equation (1) is written
() =
(aij), (ai)1
() The product of two transformations (aij), (ai)' and
(bij), (bi) is written
(bij), (bi) (aij), (ai)
= (bij)(aij), (bij)(ai) + (bi)
24
Lemma 5.
(bij), (bi) ' (aij), (ai) ()
= (bij), (bi) (ai), (ai) (y)]
Proof. Follows from the above definition.
Remark: (aij), (ai)1 is a riRid motion if f
(1) all the a's in (aij) and (ai) are real.
(2) (aij)(aijì I
where (aij)T = the transpose of (aij)
and I = the 3 x 3 identity matrix.
() Iaii\ = i
Lemma 4. The set of all rigid motions with the
prouct mapping forms a group, G.
:roof. (1) The operation is associative.
(2) To establish closure we must show
(B)(BA)T i
where A = (aij) B = (bij)
and (aij),(ai)' , (bij), (bi) are rigid motions.
If we denote the right inverse of a matrix X by X1, we
have
(BA) = A1B1 ÀTBT = (BÂ)T
25
iso we note that
det (BA) = det Bdet A = 11 1.
(3) The identity is
" (
(4) (aij), (ai) -1 = (aij)(aij)T(ai)
where -1 stands for the right inverse of a rigid motion
with respect to the ;roup operation and identity.
Lemma 5. A transformation (aij), (ai)
is a member of the group G only if aij\ = i and
(1)
= i for i = 1, 2, 3
(2)
= O for all i Ai'
,j=l
(3)
a12 = i for i = 1, 2, 3
(4)
= O for all j A j'
j j;
all hold. ither conditions i and 2 or conditions 3 and
4 are sufficieLt for (aij), (ai) to be in G if
f
aij =
26
Proof. (aij)(ai)1 conditions i and 2
(aid), (ai) G (aij)T - (alj)T(ai) G
T ..TT (aij) (aij) = (aij) (aij) =
conditions 3 and 4.
The sufficiency statement is obtained by rever-
sing the steps in the above arguments.
Netadef 7. An ordered pair of distinct points
(F], 12) is congruent to an ordered pair of distinct
points (1Di, £2 ) , (written i] 2) E (.E ' ,
P2' ) where
Ii = (x1 y1 z1) P2 = (x2 y2 z2)
'ji' = (x1' y1' z1 ) P2' = (x2' y2' z2')
iff that there exists a rigid motion (aij), (ai)5
such that
and
I xi'
I ±
y1'1=
{(aij), (ai) (
y\
z1,) z])
1X2\ f X\
1y2
(aij), (ai) ( yj
z2'
Remark. We will later change the above nietadef.
Lemnia 6.
1. (i:i, I2) is congruent to (iIi' , 12 (' )
27
is congruent to (, k2). 2. (p1, P2) E (P1', Pj)
and
implies (P1, P2) (P1', 12)
l' Fi' P2)
Proof. Follows from the fact that G is a group.
Lemma 7. (Fi, (P1',p2t)irr
2 2 + + (z1-z2) = (x1'-x2')2 + (y1'-y2')2
+ (z1'-z2')2
2 2 Proof. Let u = x1-x2 , r = u + u X X y
uy =
uz = zl_z2 = r2 + 2
z
then we have that at least one of u, u, u is not
zero since the points are distinct. Assume either
uy O or u O and hence r O and b O. 'e define
a transformation, A, as follows:
( r/ O u/\ o\
( ui'r u/r O
A = O i O , O . u/r u/r O
u/ O -r/L o )
o o i
(X2\
, I2 Z2/)
It is easy to verify that is a rigid motion and
o X2'\
O A y2
o z2
= fi
o z1
If both x1 = X2 and y1 = y2 then the same result
is easily obtained by interchanging the second and third
rows of each matrix in A, and writing r = u2 + u2
and interchanging u and u in each matrix.
Hence by repeated application of the previous
lemma,
°l' '2 ((o, O, O), (c, O, O)
Similarly we can show that
(P1', F21)a ((o, o, O), (c', O, O)
where b' = (x1'-x2) + (y1'-y2')2 + (z1'-z2')2
But we have by hypothesis
(J' = b.
Hence by the previous lemma,
(F1, k2) (E1' , F2' )
The steps are reversible so the implication holds in the
other direction.
29
Leinìa 8. (Pi, k2) = 2' r1) where P1 and k2
are arbitrary distinct points.
Proof. Follows from the previous lemma since
+ (y1-y2)2 + (z1-z2)2
+ (y2-y1)2 + (z2-z1)2
where P (x, y, z) P2 = (x2, y2, z2)
ietadef 7'. The metadef 7 for congruence is now
changed by striking the word 'ordered' wherever it
appears.
Theorem 1. Congruence of pairs of points is
transitive and symmetric.
Proof. Follows from the preceeding lemmas.
Definition 6. A ray from a point P is a set con-
taming a given point B and all points Y such that YPB
is not true and Y is on the line of IB.
Lemma 9. t set of points X is a ray from a point
(x0, y0, z0) iff there exists a line 1 represented by
X = X1 + lt (where x1, y1, z1 is
y = y1 4 mt any point of the line)
z = z1 + nt
and
30
X = X + lt o i o
yo = Yi + mt
z o =z
i +nt
and (x, y,, z), (XB, B' ZB) are members of X imply
that eì.ther
t > to and tB > to
tA Kt0 and tBK t o
where XA = X1 + itA
YA=Yl+mtA
ZA = Z1 + fltA
and XB=Xl+ltB
= yl + mt3
ZB = Z1 + fltB
i'roof. Follows from lemma 2, metadef 6 and definition 6.
Theorem 14. In a ray from a point I there exists
a unique point such that F E RS where R and are
arbitrary distinct points.
iroof. Let F = (x y, z, = (Xq Yq Zq) R =
(Xr r' zr) S = (x5, '
za). !e have for any point
(x1, y1, z1) on a ray from P that
X = X + lt 1 p 1
y1 = y + mt1
z1 = z + nt1
Then if (x2, y2, z2) is another point on this ray
31
X = X + lt 2 p 2
y2 = y + mt2
z = z + nt 2 p 2
and it follows from lemma 9 that t1 and t2 have the
same non-zero sign. hence
(x1-x)2 + (y1-y)2 + (z1z0) =
+ z2-ç2 = \t2\
since i2 + m2 + = 1. Hence if
it follows that there exists a uni'ue point on the ray
such that
x = x ± 1(t\
y = y +
z = z ± n\t
where either the + or the - sign applies according to
whether the ray is characterized by positive or negative
values of t.
Theorem 15. If ABC on a line i and if A'B'C' on
a line 1' and AB A'B', BC B'C' then AC A'C'.
Proof. Let
= a ) '' (a', a' a' it =(a,
B =(b, y) B' = (b', b' b's)
=(c, c, c) ' = (c'y, c'i, c's)
Then the line i is represented by
X = b + lt X
=b+mt z=b +nt
z
and line 1' by
X' =b' +l't' X
y' =b' +m't' y
z' =b' +n't' z
Let ta tb t and ta tb t' represent the
values of t and t' corresponding to ., B, C and .', B',
C' on lines 1 and 1' respecLively. Then
t =t' =0 b
and we may suppose that ta O and hence t O. Now,
in this case, if t'a O we let ni" = -ni', n" = -n' and
1" = -1'. iLvidently this new set of equations if equi-
valent to the old set. So in any case we have
ta > O , t'a> O
tc<o , t'c<o
ïlso, it is evident that
Then
AB A'B' implies t'a = ta
BC B'C' implies t' = t
(a-c) + (a_c)2 + (a-c)2 = ta -
= t' - t' - J(a' -c' )2
+ (a'-c')2+(a'-a')2 a c x x
Jefinition 7. If we have two distinct rays r1 and
r2 from a point A we define an angle 4 (ri, A, r2) as
the ordered set (r1, A, r2).
hemark. e will later change the definition of
angle.
Definition 8. A straight angle is an angle 3 (r1, A, r2) such that the points of r1 and r2 are on the
same line.
rietadef 8. 4 (r1, ¡, r2) 2. (r1',A', r2')
(read congruent) iff there exists a rigid motion which
takes points of r1 to points of r1' and points of r, to
points of r2' and to '.
Lemma 10. A rigid motion takes points of lines
into points of lines. Therefore if a rigid motion takes
two distinct points of a line into two image points, the
rigid motion takes the line of the points into the line
3L.
of the image points.
Proof. A rigid motion (aij), (ai) takes points of
the line
A1 B1 Cl'\ x D1
O J
A2 B2 02/ y
z
into points of the line
(o A1 B1 C1 Ç(aij), (ai) y +
A B2 02 )
The 1econd part of the lemma follows from theorem 2.
Lemma 11. (r1 P1 r2) 4 (r2 r1)
Proof. Let P1 = (x1 y1 z1) and P2 be a point of r1 and
let k2 = (x2, y2, z2).
Now the rigid motion 'A' in the proof of lemma 7
takes P3 into (O, O, O) and P2 into (ci, O, C). Hence
by the preceding lemma, it takes r1 into a ray r1' from
(O, O, o) and containing all points of the form (x, O O)
where x as is obviously verifiable.
Now on r2 there is a point Z such that
(P1, z) ( (O, O, o), (O, o, l)
hence the transformation A takes Z into a point Z' =
(1, m, n) such that i2 + in2 + n2 = i. Let the image of
r2 under be r21. Then (1, m, n)E r2'.
35
If we can find a rigid motion B, sucí that B
takes r1' into r2' and. vice versa, the present lemma
would follow since ABÀ will take r into r1 and vice
versa.
íe want to find a rigid motion B such that
I/o o
(c B O
\; O
i
m = B O
n O
i
= B O m
O n
Such a rigid motion is,
n
B -1-n2
= In
n\ O
rim u
-l-m2 O 1+1
Definition 7'. henceforth definition 7 should
omit the word 'ordered'.
)
Theorem 16. An angle is congruent to itself.
Proof. Follows from the previous lemma and the fact
r
t}iat , O transforms rays and points into them-
O) selves.
Definition 9. A side of a line in a plane is a
set of points containing a given point A not on the line,
and all points X in the plane of A and the line, such
that there does not exist a point of the line such that
XQA.
Remark. Je will use the expression "A rigid mo-
tion takes planes into planest' to mean that all the points
of a plane are taken into all the points of an image plane
by the rigid motion. Similarly we will sa,y that 'lines
are taken into lines' to mean that points of lines are
taken into points of lines.
Lemma li. A rigid motion takes a plane to an
image plane.
Proof. If A + B + C + D = O is a plane, and
(aii)(ai) is a rigid motion and
Ax0+By+Cz+f)0 and
() = {(ai)(ai)
37
then we have
where
A'x1 + B'y1 + C'z1 + D' = O
(aij)
=
and D' = -A'a1 - B'a2 - C'a3 + D.
Lemma 12. A set of points X is a side of a line
A'x+B'y+C'z+D' =0 Ax + By + Cz + i) = O
in a plane
Â'x+B'y+C'z +D' =0
if f
(x1, y1, z1) X and (x2, y2, z2) ( X
imp li es
Ax1 + By1 + Cz1 + 1)
has the same non-zero sign as does
+ By2 + Cz2 + D.
froof. If the points (x1, y1, z1) and (x2, y2, z2) have
no point of 1 between them and the line of (x1, y1, z1)
and (x2, y2, z2) is represented by
X = X0 + lt
y = y0 + mt
z = z0 + nt
we define
f(t) = .(x0+lt) + í3(y0-i-mt) + C(z0+nt) +
then in the range t1 t t2 or t2
ever applies), f(t) has no zero values.
a linear function of t, f(t1) and .f(t2)
same sign. Reversing the steps we see
follows.
j)
t t1 (which
.5ince f(t) is
must have the
that the converse
Lemma 13. A rigid motion takes a side of a
plane with respect to a line into a side of the image
plane with respect to the image of the line.
Iroof. The proof follows from the fact that lines are
transformed into lines and hence the line determined by
two points on a given side of the plane goes into an
image line. This image line does not intersect the image
of the line defining the side of the plane because
Uistances' are preserved and hence a non-zero 'distance'
cannot go into a zero 'distance' as would be the case if
the image lines had a common point.
Theorem 17. If 1 is a line in a
point of 1, r1 is a ray from P on 1, an
1 in a, there exists a ray r2 in X such
4 (r1, A, r2) is congruent to a given
B, r2') in plane .
plane cx, P is a
X is a side of
that the angle
angle (r1',
39
Froof. Let the equation of cx. be
Ax + By + Cz + Ii = O
where
A2 + B2 + C2 O.
ì'ow consider the matrix,
O O A/k
A = i O B/k
o i C/k
where k = A2 + B2 + C2 O.
Obviously the columns of A are linearly independent and
(A/k)2 + (B/k)2 + (C/k)2 = i.
Hence by the Gram-Schmidt process we can find a matrix
(a11
B
i
a12 /k
a22 B/k
a32 C/k
where the columns are mutually orthogonal and are normal.
Je notice that
B O B
i C
Further the determinant of B is either ±1 and (if
it is -1) can be made +1 by multiplying one of the first
two columns by -1. Hence by lemma 11, since
o
i
the rigid motion
into the plane
z = K
J.
= B
B' (\
40
takes tx + By + Cz + D
where K is some constant. Further, the rigid motion
( I, (-K\ will take z = K into the plane z = 0.
\-KI)
vie also note that sides of the plane are taken into sides
of the image plane, z = 0, by lemma l. Let 1' be the
image of the line 1 in the plane z = O and let the image
of the point i' be (x', y', 0). Then the ri!id motion
I, (:;
takes points of z = O into points of z = O and the point
(x', y', 0) into (0, 0, o). Now the image of 1' under
this last transformation is a line 1". There exists a
point (x", y", O) on i" such that + y"2 = i.
Now the rigid motion
¡x' Y' O (o
(-i' X' o
\ O i O
14.1
takes points of z = O to points of z = O, (O, O, O)
into itself and (x", y", O) into (1, 0, 0).
Hence in the proof of the theorem we will assume
that these transformations have ail been applied to the
orií"inal plane and that the image of line i is y = O
z=O
(by lemnia 10) and that the image of F is (O, O, O).
Then if we can prove the theorem for this particular line
and this particular point on the line and in the plane
(z = 0), we can then apply the inverse transformations
one at a time and prove the theorem in general.
Je first notice that there exist exactly two
sides of the plane with respect to the line y = O
z =0
The one side is given by the points (x, y, o) where
y > O and the other by the points (x, y, O) where
y 'Z.° That these sets of points each constitutes a
side of the plane is obvious. To show that there cannot
exist another side of the plane with respect to the given
line we note that two distinct sides of a plane with
respect to a given line cannot have a non-empty inter-
section. This is obvious when we consider the definitions
involved.
hence let us suppose that the side of the plane
in question is given by the set (x, y, 0) y > O
\e may also suppose that the ray in question is the ray
(x, O, O): x > O since the aruinents will be sym-
metric for all four cases.
Now it is obvious that the plane 1 may be trans-
formed by rigid motions into the plane z = O and the
ray r1' into the ray (x, O, O) x> O . Now if the
image of r2' is in the side (x, y, O): y> O we
can pause for a moment. If the image ray of r2' is in
the other side of the plane, the rigid motion
ç(i o o o
(o -i o o
\ \O o -i O
will take it into (x, y, O) : y O and leave the
other ray and the plane of the two rays unchanged.
Hence the existence part of the theorem is proved since
an angle is congruent to itself.
The uniqueness part is next proved. 3uppose
there is another distinct ray r3 from (O, O, O) in the
given side of the plane such that the two angles are
congruent. Then there exists a point (x, y, O) where
x2 + y2 i on this ray. It is obvious that any rigid
motion which leaves the ray (x, O, O) x > O fixed
must leave ______________
+ »2
fixed where x' and yt are the imace points of x and y
4.3
and also we have x + y = 1. It is ap:arent that
these two conditions completely determine the point
(x', y', O) and further that (x', y', O) = (x, y, O).
This proves the theorem since the line of r3 must be
the line of r2.
Derinition. In a triangle ABC we define A3C
as the angle 4 rBr where r1 contains A and r2 con-
tains B.
that
Theorem 18. If in A's ABC and A'B'C' we have
ABC A'B'C'
AB E A'B'
and BC B'C'
then BAC E ß'A'C'.
Iroof. B,y the same series of transformations used in
theorem 17, we can transform 4 's ABC and A'B'C' into
the angle 3 (r1, P, r2) where i = (O, O, O), r1 =
(- (x, O, O)t X > O and r2 is a ray from i in the
plane z = O. Under this set of transformations it is
obvious that the image points of A and A', of B and 13'
will coincide. Hence BaC and - B'A'C' coincide.
IV. PLAYFAIR'S AXIOM
Theorem 19. If
and a point I- not on 1
containing P such that
of' 1. (Flayfair's xi
.iroof. Let Ax + By +
x + By +
'x + B'y
represent 1. ¡ny line
by
44
in a plane cx we have a line 1
then there exists a unique line
it is in cx and contains no point
)Ifl)
Cz + J = O represent cx and 1
Cz + J = O
+ C'z + D' = O
1' which is in a can be represented
Ax + By + Cz + t) = O
A"x+B"y+C"z+D" =0.
Now is 1' does not have a point in common witi 1, then
(A B C
Rank A' B' C' 2.
A B" CT
Hence since Rank (A B = 2, the rank
\A' B' c'i
of the first matrix must be exactly 2. Therefore (since
Rank (A B C = 2) we have that (A', B", C") and
" c"l
(A', B', C') must be linearly dependent. iience
( , I, ' ' I '\ i ( t -i: T I
b , = , i. ,
where k 0. Low if 1' is to contain (x0, y0, z0) then
Ll.5
DU = - - 13??y - =
o o o
-k(Á'x0 + B'y0 + C'z).
Therefore a unique line 1' exists which satisfies the
stated conditions.
L46
V. ARCHIMED.ES' AXIOM
Definition. A set of points B1, B2
on a line is liiearly ordered iff i < i Ç k is equi-
valent to B.. Bk.
Theorem 20. For any seent XI there exists a
sequence B1, B2, B3 .... of points on a line 1, such
that the sequence is linearly ordered and BB+i XY
for all i, and such that for any point Z on i there
exists a B such that B1 'Z1 B.
roof. For any line i with parametric equations
X = X0 + lt
y = y0 + mt
z = z0 + nt,
let B1 = (x0, y0, z0). Then there exists a t1 O
such that the points (x0 ± it1, y0 mt1, z0 ± nt1)
and (x0, y0, z0) are congruent to X and Y. It follows
that the sequences
= X0 ± lit1, y0 ± lit1, z0 ± lit1
are each linearly ordered. Now if the point ¿ is
(xi, y, z) then there exists at Z such that
4.7
X z
= X o
+ lt z
yz = Y0 + lflt
zz = Yo + fltz
If te.) O then the sequence contains
a point Bk = (x0 + lkt1, y0 + inkt1, z0 + nkt1) where
kt1 ) t) O. If t < O a symmetric proof is required.
VI. THE CO1PLETENESS AXION
The completeness axiom of geometry reads as
follows:
If we have a system of points, lines and planes
which satisfy the previous axioms, then it is impossible
to add additional points to the set of points on a line
and still have a system which satisfies the axioms.
It then follows as a theorem of geometry that
the set of points on a line is the set of real numbers.
since the set of points on a line in the present model
is the set of real numbers, it follows that we cannot
add points to this set without contradicting the above
theorem of geometry. Hence the completeness axiom is
satisfied.
4.9
BIBLIOGRAIHY
1. Hubert, David. Grundlagender Geometrie 8. Auf 1, mit Revisionen and rganzungen von iaul Bernays, L;tuttgart, Teubner, 1956. 271 p. (Followed in lecture notes of Dr. Harry Gohoen, OSU Nathematics Dept., 1961.)
2. Landau, Edmund Georg Hermann. Foundations of analysis. Trans. by Steinhar3t. New York, Chelsea, 1951. 134 p.