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I am looking for a reference for the claim that the Pasch axiom is equivalent to the Pythagorean field condition, and with respect to what base theory this should be true.

Since posting the question, somebody directed me to

Szmielew, Wanda. The role of the Pasch axiom in the foundations of Euclidean geometry. Proceedings of the Tarski Symposium (Proc. Sympos. Pure Math., Vol. XXV, Univ. California, Berkeley, Calif., 1971), Amer. Math. Soc., Providence, R.I., 1974, pp. 123–132.

as being the source, but I don't have a copy of this yet.

Note. The reason the rationals are not enough is because the hypothenuse of a right-angle triangle with sides $1$ and $a$ needs to be in the base field in order to be able to do Euclidean geometry over it.

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  • $\begingroup$ Is this clear? Plane geometry over $\mathbb Q$ also satisfies Pasch's axiom, but $\mathbb Q$ is not Pythagorean. Maybe instead of a Pythagorean field, you want an ordered field? $\endgroup$ – Sebastian Goette May 4 '16 at 11:35
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    $\begingroup$ This depends on what other axioms you assume. You are not going to be able to do much work with circles over the rationals :-) $\endgroup$ – Mikhail Katz May 4 '16 at 11:48
  • $\begingroup$ What is the base theory over which the equivalence is claimed? $\endgroup$ – Todd Trimble May 4 '16 at 11:58
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    $\begingroup$ It doesn't really answer your question (in fact, it casts some doubt upon it), but you might like to read this account I wrote on Reddit(!) a few years ago. $\endgroup$ – Gro-Tsen May 4 '16 at 12:10
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    $\begingroup$ I'm inclined to think that "Pythagorean" is indeed needed for working with circles, but the Pasch axiom looks entirely linear to me and so should not require a Pythagorean field. $\endgroup$ – Andreas Blass May 4 '16 at 21:52
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WANDA SZMIELEW's article consider Tarski's axiom system for geometry, exposed into: What is elementary geometry ? (1957/58).

[Tarski] axiom system consists only of twelve short first order $\Pi \Sigma$ sentences and the continuity axiom, in case of the second order geometry, or the continuity axiom schema, in case of the first order (i.e., elementary) geometry.

The first version from 1930, published by Tarski only in 1948 as a note in the monograph A decision method for elementary algebra and geometry, still consists of eighteen elementary axioms, but six of them turned out step by step to be superfluous before Tarski's paper was published. After that two more axioms (identity and connectivity axioms for betweenness) were proved to depend on the remaining ones, and then the only ten left are: transitivity axiom for betweenness; Pasch's axiom:

$P: \ \ \exists a'[B(apa') \land B(ba'c)] \to \exists b'[B(bpb') \land B(ab'c)];$

identity, reflexivity, and transitivity axioms for equidistance; axiom of segment construction; five segments axiom; two dimension axioms (lower and upper); and Euclid's axiom. [...] After such a change let us denote the ten axioms by $A_1,\ldots, A10$.

The Pasch axiom is independent of all the remaining axioms together with the continuity axiom. [...] The independence model is constructed as follows: Let

$$\mathcal R=(R,+,\times,R^+)$$

be the ordered field of real numbers, $R^+$ being the positive cone of $\mathcal R$. Let the function $f$ be any one of the discontinuous one-to-one mappings of $R$ onto itself which satisfies the condition

$$f(x + y)=f(x)+f(y).$$

Then the Cartesian plane constructed over the so-called semi-ordered field

$$\mathcal R_f=(R,+,\times,f(R^+))$$

of reals is the independence model for the Pasch axiom.

Let us notice that the set $P = f(R^+)$, called the semipositive cone of $\mathcal R_f$ satisfies the conditions

(i) $x \in P \lor -x \in P$,

(ii) $x \in P \land -x \in P \to x = 0$,

(iii) $x \in P \land y \in P \to x + y \in P$,

but does not satisfy the condition

(iv) $x \in P \land y \in P \to x \times y \in P$.

In general, given a field $(F, +, \times)$ and a subset $P$ of $F$, we refer to the algebraic system

$$\mathcal F=(F,+,\times,P)$$

as a semi-ordered field iff the set $P$ (the semi-positive cone of $\mathcal F$) satisfies the conditions (i), (ii), (iii) (but not necessarily (iv)). In other words, a semi-ordered field is a field with the additive group ordered.

Since the Pasch axiom turns out to be independent, the next question arises: What is Euclidean geometry without the Pasch axiom? The answer is given by Szczerba and Szmielew in [On the Euclidean geometry without the Pasch axiom, Bull.Acad.Polon.Sci.Sér.Sci.Math.Astronom.Phys.18 (1970), 659-666, MR43#2596.].

Let us denote by $\mathcal E^P$ and call weak Euclidean geometry the theory based upon the axioms $A_1,\ldots, A10$. Now, let us drop the Pasch axiom ($A10$), but instead let us add to the remaining nine axioms two linear statements:

$A10_1: \ \ B(abc) \to B(cba)$

$A10_2: \ \ \exists b \ [B(abc) \land D(abbc)]$,

in the proofs of which, in Tarski's system, the Pasch axiom is involved. $A10_1$ is the commutativity axiom for betweenness; $A10_2$ is the axiom of midpoint existence.

Let us denote by $\mathcal E$ and call weak Euclidean Pasch-free geometry the theory based upon the eleven axioms just listed.

It is well known that $\mathcal E^P$ is the theory of Cartesian planes over arbitrary ordered (thus formally real as well) Pythagorean fields; in other words, the models of $\mathcal E^P$ coincide up to isomorphism with the Cartesian planes over such fields. Now $\mathcal E$ turns out to be the theory of Cartesian planes over arbitrary semi-ordered formally real Pythagorean fields.

Moreover, in theory $\mathcal E$ the Pasch axiom occurs to be just equivalent to the condition (iv), stating that in the semi-ordered scalar field, constructed on an arbitrarily chosen line, the semi-positive cone is closed under multiplication.

In turn, let us extend both the theories $\mathcal E^P$ and $\mathcal E$ by the continuity axiom $Co$. In this way we get full Euclidean geometry $(\mathcal E^P)^{Co}$ and Euclidean Pasch-free geometry $\mathcal E^{Co}$ both second order theories. As is shown by Szczerba in [The Pasch-free geometry with the continuity axiom, Bull.Acad.Polon.Sci.Sér.Sci.Math.Astronom.Phys.19 (1971), 613-616], the continuity axiom loses a lot of its power when the Pasch axiom is lacking. In fact, $(\mathcal E^P)^{Co}$ is known to be the theory of the Cartesian plane over the field $\mathcal R$ of real numbers, but $\mathcal E^{Co}$ turns out to be the theory of Cartesian planes not only over continuously semi-ordered fields of real numbers, but over continuously semi-ordered arbitrary formally real and Pythagorean fields. Any such field can be obtained up to isomorphism from the field $\mathcal R$ by means of a suitable modification $\circ$ of multiplication, and thus has the power of continuum.

Conversely, every field of characteristic $0$ and power of the continuum may be continuously semi-ordered (characteristic $0$ is a necessary condition for a field to be semi-orderable). In conclusion, with regards to the scalar field (formally real and Pythagorean) treated as a reduct of the semi-ordered scalar field (constructed on a fixed line), the continuity axiom without the Pasch axiom determines only that its power is that of the continuum.

Finally, let us extend the theories $\mathcal E^P$ and $\mathcal E$ , instead by the continuity axiom, only by the circle axiom

$C: \ \ B(abc) \to \exists c'[B(pbc') \land D(acac')]$

This axiom says that a half-line with the origin $b$ inside or on a circle cuts this circle in a point $c'$.

$(\mathcal E^P)^C$, called geometry of elementary constructions, is known to be the theory of Cartesian planes over arbitrary ordered Euclidean fields. What is the theory $\mathcal E^C$ ? One might expect that with $\mathcal E^C$ the class of the semi-ordered formally real Euclidean fields is connected. No; it is not the case. The theories $(\mathcal E^P)^C$ and $\mathcal E^C$ turn out just to coincide with each other

$$(\mathcal E^P)^C=\mathcal E^C$$

since, as is shown by Szmielew in [The Pasch axiom as a consequence of the circle axiom, Bull.Acad.Polon.Sci.Sér.Sci.Math.Astronom.Phys.18 (1970), 751-758], the Pasch axiom is a theorem of $\mathcal E^C$. In other words, the Pasch axiom is dependent (thus superfluous) in the axiom system of geometry of elementary constructions. At first glance the situation seems to be paradoxical, since it is common to say that the circle axiom is a particular
consequence of the continuity axiom, but, at the same time,

$$P \in \mathcal E^C \ \text {and} \ P \notin \mathcal E^{Co}.$$

In fact, in $(\mathcal E^P)^{Co}$, $C$ can not be proved without the use of $Co$, however in the proof of $C$ besides $Co$ also some other axioms are involved ; now it is clear that one of them must coincide with $P$.

In conclusion, the circle axiom, known to restrict the class of fields underlying the models from Pythagorean to Euclidean, in fact restricts also the semi-ordered fields to ordered ones.

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