Download 10 Towards a Natural Axiomatization of Geometry

10
10.1
Towards a Natural Axiomatization of Geometry
The Uniformity Theorem
Main Theorem 15 (Uniformity Theorem). Any Hilbert plane is of either one of
the following three types:
semi-Euclidean The angle sum of every triangle is two right angle, and every Lambert
or Saccheri quadrilateral is a rectangle.
semi-hyperbolic The angle sum of every triangle is less than two right angle, and
every Lambert or Saccheri quadrilateral has one respectively two acute angles.
semi-elliptic The angle sum of every triangle is larger than two right angle, and every
Lambert or Saccheri quadrilateral has one respectively two obtuse angles.
Figure 10.1: Across the longer side BC is the larger angle δ.
Lemma 10.1. Given is any quadrilateral ABCD with right angles at vertices A and
B, vertices C and D lying on the same side of line AB. The angles γ at vertex C and
δ at vertex D satisfy
γ < δ if and only if BC > AD
γ∼
= δ if and only if BC ∼
= AD
γ > δ if and only if BC < AD
Proof. As shown in the figure, we assume BC > AD. We have to check whether γ < δ.
By transferring segment AD, we produce the congruent segment BE ∼
= AD such that
ABED is a Saccheri quadrilateral. Let its top angles be congruent to ε.
In the triangle DEC, the exterior angle theorem yields γ < ε. Since points A and
C lie on different sides of line DE, angle comparison at vertex D implies ε < δ. Hence
γ < ε < δ, and transitivity of angle comparison yields γ < δ, as to be shown.
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By a similar argument, we prove that BC < AD implies γ > δ. Finally BC ∼
= AD
∼
implies γ = δ, since the top angles of a Saccheri quadrilateral are congruent.
Figure 10.2: For an acute top angle, the perpendicular dropped from a point P inside the
top segment CD is shorter than the opposite sides.
Lemma 10.2. Given is a Saccheri quadrilateral ABCD, with right angles at vertices
A and B. From any point P inside the top segment CD, the perpendicular is dropped
onto the segment AB, with foot point Q. The following three cases can occur:
γ < R and P Q < AD
γ = R and P Q ∼
= AD
γ > R and P Q > AD
Proof. At vertex P there occur the supplementary angles α = ∠DP Q and β = ∠CP Q.
We begin by assuming P Q < AD, and look for a result about the top angle γ. Using
the previous Lemma 10.1 for the quadrilateral AQP D, we conclude δ < α. Using the
Lemma 10.1 once more for the quadrilateral QBCP , we conclude γ < β. Hence angle
addition yields
2γ = γ + δ < β + α = 2R
and hence γ < R.
By a similar argument, the assumption P Q > AD implies γ > R, and finally, indeed,
the assumption P Q ∼
= AD implies γ = R.
Lemma 10.3. Given is a Saccheri quadrilateral ABCD, with right angles at vertices
−−→
A and B. From any point P on the ray DC outside of the top segment CD, the
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Figure 10.3: For an acute top angle γ, the perpendicular P Q dropped from a point outside
the top segment is longer than the opposite sides BC.
perpendicular is dropped onto the segment AB, with foot point Q. The following three
cases can occur for the top angle:
γ < R and P Q > AD
γ = R and P Q ∼
= AD
γ > R and P Q < AD
Proof. By transferring segment AD, we produce the congruent segment QE ∼
= AD such
that AQED is a Saccheri quadrilateral. Let its top angles be congruent to ε. We get
the third Saccheri quadrilaterals BQEC, denote its top angles by β. The line BC
intersects segment DE (Why?). Let F be the intersection point.
At first, we assume QP > AD, and check that γ < R. The three points A, B and Q
on the base line lie on the same side of top line DE, since baseline and top line of the
Saccheri quadrilateral AQED have the middle line as their common perpendicular,
and hence are parallel. Since QP > AD ∼
= QE, points Q and P lie on different sides
of line DE. Hence points P and C lie on the (upper) side of DE, whereas A, B and Q
lie on the lower side. The intersection point F of line DE with line BC lies between B
and C.
Angle addition at vertex C yields γ + β + χ = 2R. In the triangle DEC, the
exterior angle χ > δ − ε = γ − ε. Finally, comparison of angles at vertex E shows that
β > ε. Put together, we get
2R = γ + β + χ > γ + β + γ − ε > 2γ
and hence γ < R, as to be shown.
Under the assumption P Q < AD, several modifications occur, as can be seen in the
figure on page 293. Indeed, because of QP < AD ∼
= QE, points Q and P lie on the
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Figure 10.4: For an obtuse top angle γ, the perpendicular P Q dropped from a point outside
the top segment is shorter than the opposite sides BC and AD.
same side of line CE. Hence all five points A, B, Q, E and D lie on the same (lower) side
of top line DE. The intersection point F of line DE with line BC lies outside segment
BC.
Angle addition at vertex C yields γ + β − χ = 2R. In the triangle DEC, the
exterior angle χ > −δ + ε = −γ + ε. Finally, comparison of angles at vertex E shows
that β < ε. Put together, we get
2R = γ + β − χ < γ + β + γ − ε < 2γ
and hence γ > R, as to be shown.
Figure 10.5: Two Saccheri quadrilaterals with a common middle line M N have either both
acute, both obtuse, or both right top angles.
Lemma 10.4. Given are two Saccheri quadrilaterals ABCD and A B C D with
a common middle segment M N . Then there top angles γ and γ are either both acute,
both right, or both obtuse.
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Proof. Without loss of generality we may assume A ∗ A ∗ M ∗ B ∗ B as order of the
vertices on the base line. Using the previous Lemma 10.2 for the Saccheri quadrilateral
ABCD, we get the three equivalences
γ
γ
γ
< R if and only if A D
∼
= R if and only if A D
> R if and only if A D
Using Lemma 10.3 for the Saccheri quadrilateral
lences
γ < R if and only if
γ ∼
= R if and only if
γ > R if and only if
< BC
∼
= BC
< BC
A B C D , we get the three equivaA D A D A D < BC
∼
= BC
< BC
Put together, we see that angles γ and γ are either both acute, both right, or both
obtuse, as to be shown.
Figure 10.6: Any two Saccheri quadrilaterals can be put into a position such that the middle
line of the first one is the base line of the second one, and vice versa.
End of the proof of the Uniformity Theorem. Given a Saccheri quadrilateral ABCD
with middle line M N , and any second Saccheri quadrilateral. We transfer the middle line
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−−→
of the second Saccheri quadrilateral onto the ray M B. By means of congruent triangles,
it is straightforward to verify that we can produce a Saccheri quadrilateral QP RS
which is congruent to the second given one, and has the bottom line SQ = M N and
the middle line M K = AB.
The two lines CD and P R do intersect (Why?). With the intersection point L, we
get a Lambert quadrilateral M KLN . Let λ = ∠KLN be its top angle.
By Lemma 10.4, we conclude that the top angles γ and λ of the quadrilaterals
ABCD and M KLN are either both acute, right, or obtuse. Similarly, we conclude
that the top angles ϕ and λ of the quadrilaterals QP RS and M KLN are either
both acute, right, or obtuse. Hence the top angles γ and ϕ of the two given Saccheri
quadrilaterals are either both acute, right, or obtuse.
10.2
A hierarchy of planes
As we have learned in definition 1.1:
Definition (1.1). A Hilbert plane is any model for two-dimensional geometry where
Hilbert’s axioms of incidence (I.1)(I.2)(I.3a)(I.3b), order (II.1) through (II.4), and congruence (III.1) through (III.5) hold.
Neither the axioms of continuity—Archimedean axiom and the axiom of completeness—
nor the parallel axiom need to hold for an arbitrary Hilbert plane. The Uniformity
Theorem brings confidence that these few axioms are a correct start towards a natural
axiomatization of full-blown Euclidean geometry.
Which postulates do we want to be added to the Hilbert plane axioms? This section
gives an account of some recent achievements, mainly of M.J. Greenberg in the axiomatization of geometry. We begin with the axioms of a Hilbert plane, and successively
postulate few additional axioms with the effort to arrive at a system as close as possible
to Euclid’s system. By further axioms—mainly of continuity—the system is narrowed
to a system that has the real Cartesian plane as its unique model.
The Uniformity Theorem motivates the following definition:
Definition 10.1 (Three basic types of Hilbert planes). According to the three
cases occurring in the Uniformity Theorem—
A semi-Euclidean plane is a Hilbert plane for which the angle sum of every triangle
is two right angles.
A semi-hyperbolic plane is a Hilbert plane for which the angle sum of every triangle
is less than two right angles.
A semi-elliptic plane is a Hilbert plane for which the angle sum of every triangle is
larger than two right angles.
As the next step towards a natural axiomatization of Euclidean geometry, we investigate the Euclidean parallel property.
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10.3
Wallis’ axiom
Already in the section on Legendre’s theorem, we have addressed the question to find
a clear cut, suggestive or self-evident postulate that would better replace the Euclidean
postulate.
John Wallis (1616-1703), in a treatise on Euclid published in 1693, was astute enough
to propose a new postulate that he believed to be more plausible than Euclid’s parallel
postulate. He phrased it as follows:
Finally (supposing the nature of ratio and of the science of similar figures
already known), I take the following as a common notion: to every figure
there exists a similar figure of arbitrary magnitude.
We want to make Wallis’ idea precise in the context of axiomatic geometry, beginning
with Hilbert’s axiom. Euclid’s theorems about similar triangles depends on the Archimedean axiom—via his definition of equality of ratios. Hence it is better to restrict
attention to equiangular triangles instead of similar figures. Recall that equiangular
triangles have three pairs of congruent angles. Both Greenberg and Hartshorne suggest
a modification of Wallis’ axiom along these lines.
(Wallis-Greenberg Postulate). Given any triangle ABC and segment DE there
exists a triangle with DE as one side such that the triangles ABC and DEF are
equiangular.
(Wallis-Hartshorne Postulate). Given any triangle ABC and segment DE there
exists a triangle A B C have side A B ≥ DE such that the triangles ABC and
A B C are equiangular.
In this paragraph, we show the following theorem:
Theorem 10.1. The following three postulates are equivalent in any Hilbert plane:
(a) the Euclidean parallel postulate
(b) the Wallis-Greenberg postulate
(c) the Wallis-Hartshorne postulate
Proposition 10.1. The Euclidean parallel postulate implies the Wallis-Greenberg postulate.
Proof. Given is triangle ABC and segment DE. We transfer segment AB onto the
−−→
ray DE and get the congruent segments AB ∼
= DE .
By the extended ASA-theorem, we construct the congruent triangles ABC ∼
=
DE F . Indeed, as already explained in the proof of the extended ASA-theorem, the
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Figure 10.7: Uniqueness of parallels implies Wallis’ postulate.
−−→
ray rD forming the angle ∠BAC with ray DE and the ray rE forming the angle ∠ABC
−−→
with ray E D, both constructed in the same half-plane of line DE, intersect in point F .
−−→
Additionally, we construct the ray rE forming the same angle ∠ABC with ray ED,
again in the same half-plane of line DE. The rays rE and rE are parallel by [Euclid I.27],
see 5.37. The ray rD intersects one of these parallel rays. Hence, as already explained
in Problem 2.1, it intersects the line of the second ray rE , too. Indeed, otherwise, the
line of rE would have the two different parallels rD and rE both through point F .
Let now F be the intersection point of rD and rE . The triangles ABC and DEF
have two pairs of congruent angles at vertices A and D, as well as vertices B and E, by
construction.
We have seen in Proposition 7.1 above that uniqueness of parallels implies that every
triangle has angle sum 2R—every Pythagorean plane is semi-Euclidean.
Hence the triangles ABC and DEF have a third pair of congruent angles at
vertices C and F and are equiangular, as to be shown.
Proposition 10.2. The Wallis-Hartshorne postulate implies the Euclidean parallel postulate.
Proof. Given is line l and a point P not on l. One parallel can be obtained as the ”double
perpendicular” as already explained in Proposition 5.38. One drops the perpendicular
p from point P onto line l and denotes the foot point by F . Next, one erects at point
P the perpendicular to line P F . Thus, one gets a line m parallel to the given line l.
We now suppose towards a contradiction that there exists a second line t through
→
−
point P which is parallel to line l, too. Let t be the ray on line t with vertex P and
lying between the two parallels l and m. We choose any point Q = P on this ray, and
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Figure 10.8: Wallis’ postulate implies uniqueness of parallels.
drop the perpendicular onto the line P F . The foot point R lies in the segment P F ,
since lines t and l do not intersect.
We now apply the Wallis-Hartshorn axiom to the triangle P QR and the segment
P F . Hence there exists an equiangular triangle P Q R with side P R ≥ P F .
−→
We transfer this segment and then the triangle onto the ray P F and get a triangle
P Q R , which is congruent to P Q R and hence equiangular to triangle P QR.
We can put points Q and Q on the same side of line P F . Because of congruence of the
angles α = ∠R P Q ∼
= ∠RP Q, the uniqueness of angle transfer imply that these rays
−→ −−→
are equal: P Q = P Q . Hence point Q lies on the line t = P Q.
Since P R ≥ P F , we conclude that either R = F or R and P lie on different
sides of F . We consider the second case. The foot points R and R lies on different
sides of F and hence the line l. The lines RQ, l and R Q are parallel, being all three
perpendicular to P F . Hence points R and Q lie on one side of line l and points R and
Q lie on the other side. Hence the segment QQ intersects the line l, say in point T .
In the first case, we conclude that point Q =: T lies on line l.
Hence, in both cases, the lines l and t do intersect. We have confirmed no second
parallel t to line l through point P can exist.
10.4
Proclus’ Theorem
We shall repeatedly use
Lemma 10.5 (Proclus lemma). If one of two parallel lines is intersected by a third
line, the other one is intersected, too.
What is the missing link leading from the semi-Euclidean plane to a plane with
uniqueness of parallels? Based on Proclus commentaries to Euclid, Greenberg has suggested the following angle unboundedness axiom. This axiom goes back to Aristole’s
book I of the treatise De Caelo (”On the heavens”).
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Definition 10.2 (Aristole’s Angle Unboundedness Axiom). For any acute angle
θ and any segment P Q, there exists a point X on one side of the angle such that the
perpendicular XY dropped onto the other side of the angle is longer than the given
segment: XY > P Q.
Main Theorem 16 (Proclus’ Theorem). A Hilbert plane is Pythagorean if and only
if it is semi-Euclidean and Aristole’s axiom holds.
Corollary 26. In any Hilbert plane, the uniqueness of parallels is equivalent to the
following two requirements:
• Every triangle has angle sum two right angles.
• For every acute angle and every given segment, there exist a right triangle with
the given angle and the leg across from it congruent to the given segment.
We have seen in Proposition 7.1 above that uniqueness of parallels implies that every
triangle has angle sum 2R. Hence every Pythagorean plane is semi-Euclidean. Indeed,
uniqueness of parallels implies Aristole’s axiom, too.
Proposition 10.3. Hilbert’s parallel postulate implies Aristole’s axiom.
Figure 10.9: Uniqueness of parallels implies unbounded opening of an angle.
Proof. Let the acute angle θ = ∠(m, n) and a segment be given. We transfer the segment
onto the perpendicular erected at the vertex of the angle onto the side m and get the
segment P Q. Next we erect the perpendicular l onto line P Q at Q. The lines l and m
are parallel since they are both perpendicular to line P Q.
Because uniqueness of parallels has been assumed, the lines n and l are not parallel.
Let S be their intersection point. Take any point X on the line n such that S is between
P and X. We drop the perpendicular from X onto line m and let Y be the foot point.
Since X and Y are on different sides of line l, the segment XY intersects the line l. Call
T the intersection point.
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The Lambert quadrilateral QT Y P is indeed a rectangle, as shown by the Uniformity Theorem. Its opposite sides are congruent by Lemma 10.1. Hence we get
PQ ∼
= YT <YX
as claimed by Aristole’s angle unboundedness axiom.
By Proposition 10.3 and Proposition 7.1, we know that Aristole’s Axiom and the
angle sum 2R for triangles are both necessary for uniqueness of parallels to occur. To
complete the proof of Proclus’ Theorem, we now show that these two assumptions are
sufficient.
Proposition 10.4. A semi-Euclidean Hilbert plane, for which the Aristole’s Axiom
holds is Pythagorean.
Figure 10.10: The second line n = m, different to the double perpendicular m, drawn
through point P is along the side of a triangle P XR, and hence by Pasch’s axiom intersects
line l.
Proof. Given is a line l and a point P not on this line. As explained in Proposition 5.38,
we use the standard ”double perpendicular” m to get a parallel to line l through point
P . Let Q be the foot point of the perpendicular dropped from point P onto line l, and
let m be the perpendicular erected at point P onto the first perpendicular P Q.
We need to show that m is the unique parallel through point P to line l. Take any
other line n through point P . Let θ = ∠(rm , rn ) be the acute angle between two rays
on the lines n and m from vertex P . We can choose these rays such that the angle θ is
firstly acute, and secondly, the interior of the angle lies in the same half plane of line m
as line l.
We now explicitly prove existence of an intersection point of ray rn and line l. Let
XY > P Q be the segment between the sides of angle θ = ∠XAY , as postulated by
Aristole’s Axiom. We can choose point X to lie on the ray rn , and drop the perpendicular
from point X onto the line P Q, obtaining the foot point R. We have obtained a Lambert
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quadrilateral RXY P . In a semi-Euclidean plane, this is a rectangle. Its opposite sides
are congruent by Lemma 10.1. Hence we conclude
PR ∼
= XY > P Q
and points P and R lie on different sides of point Q.
We now use Pasch’s axiom for line l and triangle P XR. Line l intersects the
side P R in point Q and is parallel to side RX, since the lines l and RX are both
perpendicular to P Q. Hence by Pasch’s axiom, the line l intersects the third side P X
of the triangle, say in point S.
Thus we have explicitly checked existence of an intersection point lines n = P X with
line l, as to be shown.
10.5
More about Aristole’s axiom
Proposition 10.5. In a semi-elliptic Hilbert plane, Aristole’s Axiom does not hold.
Especially, it does not hold for the angle between any two parallels to the same line, nor
for any angle given by the excess of the angle sum of a triangle over two right angles.
Figure 10.11: For the angle between two parallels m and n to line l, intersecting at P , the
distance from a point on one side to the other side of angle ∠(m, n) is always smaller than the
perpendicular P Q.
Proof. Given is a line l and a point P not on this line. Again, let Q be the foot point
of the perpendicular dropped from point P onto line l, and let m be the perpendicular
erected at point P onto the first perpendicular P Q. As explained in Proposition 5.38,
”double perpendicular” m is a parallel to line l through point P .
Let n be a second parallel to line l through point P . Take any point X on the line
n such that X and Q lie on the same side of line m. We drop the perpendicular from
point X onto the line P Q and let Z be its foot point. All three line m, ZX and l are
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perpendicular to P Q, and hence parallel. Hence points X and Z both lie between the
parallels l and m. Hence point Z lies between P and Q, and hence.
PQ > PZ
The perpendicular from point X onto the line m has foot point Y . Thus we have
obtained the Lambert quadrilateral ZXY P . In a semi-elliptic plane, its fourth angle
χ = ∠ZXY is obtuse. From Lemma 10.1 we conclude
P Z > XY
Thus we have obtained that P Q > P Z > XY holds for any point X on one of the
rays from P on line n. Hence the distance from any point of line n to line m is strictly
smaller than P Q, and Aristole’s angle unboundedness axiom does not hold.
Proposition 10.6. In a semi-Euclidean as well as in a semi-hyperbolic Hilbert plane,
Archimedes’ Axiom implies Aristole’s Axiom.
Figure 10.12: Doubling the distance from a point on one side to the other side of angle θ.
Proof. Let the acute angle θ = ∠(m, n) be given. We choose any point C on side
−→
m = AC, and drop the perpendicular onto n to get the foot point D. In order to double
the segment CD, we construct point E such that C is the midpoint of segment DE and
point F on ray m such that C is the midpoint of segment AF . The triangles
ACD ∼
= F CE
are congruent by SAS-congruence, because of the vertical angles at point C and two
pairs of congruent sides AC ∼
= CF and DC ∼
= CE.
Because of the triangle congruence, ∠CDA ∼
= ∠CEF = ∠DEF are both right
angles. We drop the perpendicular from point F onto the other side n and get foot
point G.
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Now DEF G is a Lambert quadrilateral, with an acute or right angle ∠EF G.
By Lemma 10.1 either side adjacent to the acute angle is longer than the respectively
opposite side. Hence we conclude
F G ≥ ED = 2 · CD
We need to assure that doubling the original segment CD is sufficient to get another segment XY longer than any arbitrarily segment P Q. To this end, we use the Archimedean
Axiom. Let F G := C1 D1 . By induction, we can construct segments
Cn Dn ≥ 2n · CD > n · CD
for all natural numbers n.
Given any segment P Q, the Archimedean axiom tells there exists a natural number
n such that
n · CD > P Q
Define XY := Cn Dn for such a number n. Both inequalities together imply
XY > n · CD > P Q
as required for Aristole’s angle unboundedness axiom to hold.
From Proposition 10.5, the first Legendre Theorem and the Uniformity Theorem we
can recapitulate:
Corollary 27. In a semi-elliptic Hilbert plane, neither Aristole’s Axiom nor Archimedes
Axiom does hold.
Together with Proposition 10.6, we get the remarkable result—for which it would be
nice to have a direct proof:
Corollary 28. In every Hilbert plane, the Archimedean Axiom implies Aristole’s Axiom.
Too, from Proposition 10.6 and Proposition 10.4, we get once more the third Legendre Theorem—the direct proof of which we have given earlier.
Corollary 29. A semi-Euclidean plane for which the Archimedean axiom holds is
Pythagorean.
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