Download 4 Arithmetic of Segments—Hilbert`s Road from Ge

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Arithmetic of Segments—Hilbert’s Road from Geometry to Algebra
In this section, we explain Hilbert’s procedure to construct an arithmetic of segments,
also called Streckenrechnung. Hilbert constructs a field from a given model of geometry
and thus opens up a quiet general road from geometry to algebra.
The approach is explained in Hilbert’s foundations. It works for an arbitrary affine
plane for which the theorem of Pappus holds. Since existence and uniqueness of the
parallel is assumed, one can take advantage of similar triangles in the construction of
the field of segment arithmetic. Neither one of the axioms of continuity shall be assumed.
Depending on the given model of geometry, we obtain a variety of different fields, rather
than the real numbers exclusively.
One can even generalize the approach, and assume only the weaker theorem of Desargues instead of Pappus. In this case, one obtains on the side of algebra only a skew
field—thus only the associative law still holds, but the commutative law does no longer
hold. These important results have already been stated in the Main Theorems 2 and 3.
In the context of projective geometry, the commutative law turns out to be equivalent
to the theorem of Pappus and the associative law to the theorem of Desargues.
As an alternative, we may avoid the theorem of Pappus and get the segment arithmetic more directly from angles in a circle via Euclid’s III.32. This approach is used in
Hartshorne’s book and presented here in an own variant. It works in every Pythagorean
plane and breaks any connection with ideas from projective geometry. Recall the definitions:
Definition (1.2 Pythagorean plane). A Pythagorean plane is a Hilbert plane for
which the Euclidean parallel postulate holds.
Definition 4.1 (Pythagorean field). A Pythagorean field is an ordered field with the
property
√
(Pyth) If a, b ∈ F, then a2 + b2 ∈ F.
Main Theorem 18. Any affine plane for which the Theorem of Pappus holds is isomorphic to the Cartesian plane over a field. Conversely, the Theorem of Pappus holds
in a Cartesian plane over a field.
The first result is elaborated in this section in detail. The second statement repeats
part of the Main Theorem 3, and can be checked directly with some rather tedious
calculations which we leave out. Furthermore, this section gives the detailed direct
proof of the following result:
Theorem 4.1. Any Pythagorean plane is isomorphic to the Cartesian plane F2 over its
field F of segment lengths. Furthermore, this is a Pythagorean field.
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4.1
Fields
Definition 4.2 (Field). A field is a system of undefined elements (symbols, numbers),
two binary relation symbols + and · and two constants 0 and 1 with the following
properties:
(1) (i) For any two numbers a and b exists the sum a + b.
(ii) (a + b) + c = a + (b + c) for any a, b, c.
(iii) a + b = b + a for any a, b.
(iv) a + 0 = a for any a.
(v) for each number a there exists a number −a such that a + (−a) = 0.
(2) (i) For any two numbers a and b exists the product a · b.
(ii) (a · b) · c = a · (b · c) for any a, b, c.
(iv) a · 1 = 1 · a = a for any a.
(v) for each number a = 0 there exists a number a−1 such that a·a−1 = a−1 ·a = 1.
(3) (i) 0 = 1.
(ii) (a + b) · c = a · c + b · c for any a, b, c.
(iii) c · (a + b) = c · a + c · b for any a, b, c.
(4) a · b = b · a for any a, b.
A system of elements satisfying only (1)(2)(3) is called a skew field.
Any model for the field axioms is simply called a field, too. In this context, we
denote by F the set of elements. Clearly 0, 1 ∈ F.
4.2
Construction of the field of segment arithmetic
To relate Euclidean geometry to algebra, we need to construct a field from the congruence classes of segments.
One can define the sum of any two segments by putting them together along a
line. Thus, if point B lies between A and C, we define AC = AB + BC. Because
of Hilbert’s postulate III.3, the addition of segments defines even an addition for the
congruence classes of segments. As already shown in Proposition 5.7 in the section about
congruence of segments, angles and triangles, we get from these congruence classes an
ordered Abelian group.
In the theorem 8 about the measurement of segments, we have already used a unit
segment. To this unit segment, say OE, we assign the length 1. The construction of the
length for any segment as an (finite or infinite) binary fraction done in the measurement
theorem 8 depends on the Archimedean axiom.
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Although this ”old segment arithmetic” is very practical, it depends essentially on the
Archimedean axiom, and the possibility to approximate irrational numbers by rational
numbers. Only to some very special segments, a finite binary or proper fraction can
be assigned as length. Hence the old segment arithmetic is not always suitable for a
constructive definition of the multiplication of the field.
The sum of segments is defined similarly in the old and the new segment arithmetic.
The essential step for the ”new segment arithmetic” claimed by Theorem 4.1 is the correct definition of the product of two segments. This is done using the legs of equiangular
right triangles. Thus all dependence on the axioms of continuity can be avoided.
Figure 4.1: Construction of the product ab of given segments |OA | = a, |OB| = b and
|OE| = 1.
To be remembered. For the definition of the product ab, we use two similar right triangles. The first triangle has the legs the first factor a and 1, the second one has as its
legs ab and the second factor b.
Look at the figure on page 401. On the perpendicular ray through O, we have put
the segment with prescribed length
|OA | = a
On the horizontal ray emanating from vertex O to the right, there are put points O, E
and B such that the segments measured from the origin O have the prescribed lengths:
|OE| = 1 , |OB| = b
Now we draw the parallel to line EA through point B. Why does this parallel does
intersect the vertical line? Indeed, by Proclus’ Lemma 2.1, a line intersecting one of two
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parallel lines intersects the other one, too. Since the vertical line intersects line EA ,
it intersects its parallel through B, too. Hence we get the unique intersection point P and define the product ab as length of the segment OP :
ab := |OP |
Indeed, we have used the similar right triangles OEA and OBP .
Question. How does the order of the factors a and b become relevant for the construction?
Answer. The first factor a = |OA | and the unit segment |OE| = 1 appear together as
legs of one triangle OEA . The second factor b determine the segment OB through
one end of which one draws the parallels EA BP .
Remark. Indeed, the product of segments a and b turns out to be such that the rectangle
with sides a and b has the same area as the rectangle with sides 1 and ab. But this
simple definition of the area of the rectangle is only possible because of commutativity.
The definition of area does not automatically imply commutativity. 37
The first obvious question is whether our definition given above makes the product
commutative. Furthermore, we need to go on and check whether the remaining axiom
required for a field are satisfied, and whether we get an ordered field.
4.2.1
Commutativity
To check the commutative law, we repeat the above construction, with the roles of a
and b exchanged. We construct point A on the horizontal line such that
|OA| = a
and a point on the perpendicular line such that
|OB | = b
Now we draw the parallel to line EB through point A, and get the intersection point
Q . By definition ba is length of the segment OQ :
ba = |OQ |
As you can see from the figure on page 403, it is now really a question from geometry,
whether ab = ba!
Proposition 4.1. The segment arithmetic is commutative.
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It would be interesting to find out in a psychological study who really knows that!
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Figure 4.2: Construction of the products ab and ba, with an empirical error exaggerated.
Proof. The product ab is constructed as in the figure on page 403, by the pair of parallels
EA BP Because of the isosceles and equiangular triangles OAA and OBB , we get another
pair of parallel lines AA BB . The hexagon AP BB EA and the horizontal and
vertical lines h and v are a Pappus configuration.
Remark. To get the three pairs of opposite sides of the hexagon, it is convenient to shift
the points cyclically by three:
A P B B E A
B E A A P B
The three pairs of opposite sides of the hexagon appear underneath each other:
AP P B BB B E EA A A
The Pappus configuration contains the two pairs of parallel opposite sides
EA BP AA BB 403
Hence, by the theorem of Pappus, there is a third pair of parallels
EB AP The product ba is constructed as in the figure on page 403, by the pair of parallels
EB AQ
Both EB AQ and EB AP , and both P and Q lie on the vertical line v. The
uniqueness of the parallels and the uniqueness of intersection point implies P = Q .
Hence ab = |OP | = |OQ | = ba, as to be shown.
Remark. Conversely, the commutative law of multiplication, together with the theorem
that equiangular triangles have proportional sides, can be used to get Pappus theorem
in the simplified version.
Remark. In this and the following proofs, we consider only the case of positive quantities
a, b, c. It is left to the reader to check that the corresponding results hold for positive
as well as negative quantities.
4.2.2
The distributive law
Figure 4.3: Proof of the distributive law ab + ac = a(b + c).
Proposition 4.2. The segment arithmetic is distributive.
Proof. We get the segment [c, c + b] by transferring the segment [O, b] onto the ray with
vertex c on the horizontal axis. The construction of the products ab, ac, and a(b + c)
is shown in the figure on page 404. In all three cases, we use parallels to the line EA ,
through points b, c and (b + c) on the horizontal axis, respectively. On the vertical axis,
we get the intersection points ab, ac and a(b + c), the distance of which to the origin
define the respective products.
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Finally, we draw the parallel to the vertical axis through point c, and get intersection
point P with the line through points b+c and a(b+c). By ASA congruence, the triangles
Ob(ab) and c(b + c)P are congruent, since they are equiangular and have one pair
of congruent sides. Hence the segments [O, ab] and [c, P ] are congruent.
Furthermore, the segments [c, P ] and [ac, a(b + c)] are congruent since they are
opposite sides of a parallelogram. Hence [O, ab] ∼
= [ac, a(b + c)], which shows that
ab + ac = a(b + c).
4.2.3
The associative law
Figure 4.4: The proof of the associative law yields at first c(ab) = a(cb).
Proposition 4.3. The segment arithmetic is associative.
Proof. From given segments a and c on the vertical line, and e = 1 and b on the
horizontal line, we construct at first segments ab and cb on the vertical line. These
two segments are transferred onto the horizontal axis, using 45◦ -lines. Finally, one gets,
both on the vertical line, segment c(ab) from the parallels [e, c] [ab, c(ab)] and segments
a(cb) from the parallels [e, a] [cb, a(cb)].
The three points ab, b and cb on the horizontal axis and the three points ab, c(ab) and
cb on the vertical axis give a Pappus configuration with two, and hence three pairs of
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parallels. We conclude that a(cb) = c(ab). By applying the already proved commutative
law twice, we get
a(bc) = a(cb) = c(ab) = (ab)c
4.2.4
A direct proof of commutativity of segment arithmetic
Figure 4.5: Prove commutativity ab = ba.
The following proof of commutativity of segment arithmetic does not refer to Pappus’ theorem. Instead, we use angles in a circle directly. Thus we get a proof that
commutativity of segment arithmetic holds in every Hilbert plane were the parallel to
a given line through a given point is unique.
Indeed, only one circle is used for the proof. Once more, we are using measurements
along the horizontal and vertical axes crossing at point O perpendicularly. The given
positive segment |OA | = a > 0 is transferred to the vertical axis. The segments
|OE| = 1 and |OB| = b > 0 are transferred to the horizontal axis.
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Question. Recall how the product ab has been constructed.
Answer. We draw the line through B parallel to EA , and get its intersection point with
the vertical axis at point P with distance |OP | = ab from the origin.
As next step, the triangle OA E is reflected across the −45◦ line to produce the
triangle OA− E − .
Question. Explain why, because of these constructions, there appear three congruent
angles.
Answer. The angles ∠OA E ∼
= ∠OP B are congruent because the vertical axis traverses
a pair of parallel lines. The angles ∠OA E ∼
= ∠OA− E − are congruent by SAS congruence of the corresponding triangles. We have marked the three congruent angles by
α.
We now draw a circle through points A− , E − and B.
Question. Explain why this circle goes through the point P .
Answer. The angles ∠E − A− B ∼
= ∠E − P B lie both on the same side of E − B. They are
circumference angles of the arc E − B in the circle drawn and hence congruent. Hence
point P lies on this circle.
The triangle OBE − is reflected across the +45◦ line, as needed in the next step.
Question. Explain how the product ba has to be constructed, in the left upper quadrant.
Answer. We draw the line through A− parallel to E − B , and get its intersection point
with the vertical axis at point ba with distance |Oba| = ba from the origin.
Question. Explain why, because of these constructions, there appear three congruent
angles.
Answer. The angles ∠OB E − ∼
= ∠ObaA− are congruent because the vertical axis traverses a pair of parallel lines. The angles ∠OBE − ∼
= ∠OB E − are congruent because
of SAS congruence of the corresponding triangles. We have marked the three congruent
angles by β.
Question. Explain why the circle through A , E and B—as already drawn—goes through
the point ba.
Answer. The angles ∠E − BA− ∼
= ∠E − baA− are circumference angles of the same arc,
and lie both on the same side of its chord E − A− . Hence they are congruent. Hence ba
lies on this circle.
−→
Since the circle has only one intersection point with the vertical ray OA, and we
have obtained that both ab and ba is this intersection point, we conclude that ab = ba.
Remark. In the drawing from a former version—which was not from Leibniz’ ”best of all
worlds”—the construction did not come out right. In such a case, would Leibniz have
trusted the proof or the drawing?
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Figure 4.6: Commutativity ab = ba follows from both lying on the circle.
4.2.5
Alternative proof of associativity
The given positive segments |OA| = a > 0 and |OB| = b > 0 are transferred to the
vertical axis. The segments |OE| = 1 and |OC| = c > 0 are transferred to the horizontal
axis. The products ac and bc are constructed on the vertical axis.
Problem 4.1. One goes on, almost as above, with E, A, B replaced by C, AC and BC.
The angles α and β each appear four time. Explain in detail how the figure on page 409
yields a proof of a(bc) = b(ac) with any three positive a, b, c > 0, and thus confirms the
associative law.
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Figure 4.7: a(bc) = b(ac) both lie on the circle.
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