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In the book of Foundations of Geometry by Hilbert, at page 2, it is stated that

I, 1. Two distinct points A and B always completely determine a straight line a. We write AB = a or BA = a.

I, 2. Any two distinct points of a straight line completely determine that line; that is, if AB = a and AC = a, where $B \not = C$, then is also BC = a.

However, what I understand from "Any two distinct points of a straight line completely determine that line" is that given a line $a$, and points $A,B,C$ on that line, we have $AB = AC=BC = a$, but the logical formulation states, i.e the part after "that is", that given any 3 point $A,B,C$ s.t $AB = AC = a$ implies $BC = a$ (of course provided that $B \not = C$ in both cases), so is there any explanation how should we understand

Any two distinct points of a straight line completely determine that line

as

if AB = a and AC = a, where $B \not = C$, then is also BC = a.

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  • This is indeed very odd. Assuming you are using the translation at https://math.berkeley.edu/~wodzicki/160/Hilbert.pdf, it never clarifies the exact relationship between "$AB=a$" (a relation between two points and a line) and "$A$ lies on $a$" (a relation between one point and a line). If $AB=a$ is supposed to mean that $a$ is the unique line on which $A$ and $B$ lie (which is a perfectly reasonable interpretation of the very imprecisely stated Axiom I,1) then Axiom I,2 would be entirely redundant... – Eric Wofsey May 13 '18 at 08:01
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    Note that other presentations of Hilbert's axioms (e.g., Wikipedia) do not use the words "completely determine" in Axiom I,1, so that Axiom I,1 says there exists a line which $A$ and $B$ both lie on but says nothing about uniqueness. – Eric Wofsey May 13 '18 at 08:03
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    The translator seems to have taken liberties there. The german version that I have (13th ed.) says: I 1. Zu zwei Punkten $A, B$ gibt es stets eine Gerade $a$, die mit jedem der beiden Punkte $A, B$ zusammengehört. I 2. Zu zwei Punkten $A, B$ gibt es nicht mehr als eine Gerade, die mit jedem der beiden Punkte $A, B$ zusammengehört. Nothing like $AB=a$. I concur with Eric Wofsey's answer. – ccorn May 13 '18 at 10:19
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    In the 1903 edition it says: I 1. Zwei von einander verschiedene Punkte $A, B$ bestimmen stets eine Gerade $a$. I 2. Irgend zwei voneinander verschiedene Punkte einer Geraden bestimmen diese Gerade. Obviously, the phrasing has evolved since, but the first axiom clearly states existence, and the second axiom clearly states uniqueness. – ccorn May 13 '18 at 11:31

2 Answers2

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As far as I can tell, this is best described as an error in the text, possibly one introduced by the translator. These statements appear to be going to some contortions to state everything in terms of the notation "$AB=a$", which is not used in the original German (which you can find in pdf form here). This is all rather awkwardly done, and the explanation is so sketchy (what does "$AB=a$" actually mean and how is it related to "$A$ lies on $a$"??) that I would consider it "not even wrong". In particular, you are correct that the two halves of Axiom I,2 do not have the same meaning; the correct version of this axiom is the first half, not the second half.

So, I would advise that you completely ignore the second part of Axiom I,2. A more modern presentation of these two axioms (with their intended meanings) together with a definition of the notation $AB$ would be:

I,1. For any two points $A$ and $B$, there exists a line that contains both of them.

I,2. A line is uniquely determined by any two distinct points that lie on it. That is, if $A\neq B$ and $a$ and $b$ both contain $A$ and $B$, then $a=b$.

If $A\neq B$, we write $AB$ for the unique line which contains both $A$ and $B$. (Such a line exists by I,1 and is unique by I,2.)

Here is some elaboration on how incoherent and awful the presentation of the axioms you quoted is. Axiom I,1 is rather ambiguous: what does "completely determine" mean? By the usual English meaning of this phrase, it would include a statement of uniqueness, so it would seem to say for any two points $A$ and $B$, there is a unique line containing them both. But if this is the intended meaning, then Axiom I,2 is completely redundant. So, it seems that despite its phrasing, Axiom I,1 is not intended to include any assertion of uniqueness (and indeed other presentations of Hilbert's axioms convert Axiom I,1 into a statement just about existence as I have above).

As for the meaning of $AB=a$, I can think of three reasonable interpretations. Interpretation 1 is that $AB=a$ means "$A$ and $B$ both lie on $a$". Interpretation 2 is that $AB=a$ means "$a$ is the unique line that contains both $A$ and $B$". Interpretation 3 is that $AB=a$ is a primitive notion, defining a function $f(\{A,B\})=AB$ from unordered pairs of distinct points to lines (and this function is what is meant by "completely determine").

All three of these interpretations are problematic. Interpretation 1 turns the second half of Axiom I,2 into an immediate consequence of the definition: if $AB=a$ and $AC=a$, then in particular $B$ and $C$ both lie on $a$, so $BC=a$. Interpretation 2 is rather odd if Axiom I,1 is not intended to include a statement of uniqueness, given that the notation $AB=a$ is presented part of Axiom I,1. Moreover, if Axiom I,1 does not include a uniqueness statement, interpretation 2 makes the second part of Axiom I,2 worthless, because it cannot be used unless you already know that $AB=a$ and $AC=a$ and no axiom guarantees that this will ever happen. Interpretation 3 leaves it completely mysterious what this primitive notion $AB=a$ has to do with the other primitive notion "$a$ contains $A$", and in any case the two halves of Axiom I,2 still do not have the same meaning.

Of course, if you ignore the second half of Axiom I,2 and take the first half as the intended uniqueness statement as I wrote it above, then interpretations 1 and 2 become equivalent and unproblematic.

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Eric Wofsey
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  • Your reasoning is fine if you take for granted the usual setting, where a line is a set whose elements are points. But that is not Hilbert's point of view: he never makes use of sets and of their language. – Intelligenti pauca May 13 '18 at 11:14
  • I am not assuming a line is a set of points anywhere, and I'm not sure what makes you think I am doing so. Regardless of whether you treat lines as sets as points, there is still a primitive notion of a line "containing" a point (or "passing through" a point, or a point "lying on" a line, or however else you want to phrase it). – Eric Wofsey May 13 '18 at 17:13
  • You wrote: "if $AB=a$ and $AC=a$, then in particular $B$ and $C$ both lie on $a$, so $BC=a$". How can you make that deduction? – Intelligenti pauca May 13 '18 at 17:53
  • If $AB=a$, then $A$ lies on $a$ and $B$ lies on $a$. If $AC=a$ then $A$ lies on $a$ and $C$ lies on $a$. So, in particular, $B$ lies on $a$ and $C$ lies on $a$. – Eric Wofsey May 13 '18 at 18:07
  • The statements of these two axioms alone do not make it clear that "lying on" is a relation between one point and a line, instead of between two points and a line. But the rest of the text (see https://math.berkeley.edu/~wodzicki/160/Hilbert.pdf) makes this abundantly clear. – Eric Wofsey May 13 '18 at 18:10
  • In any case, even if you reject the existence of such a "lying on" relation, the second half of Axiom I,2 is woefully inadequate for capturing the meaning of the first half (for instance, if $AB=a$ and $CD=a$ where all four points are distinct, there is no way to ever conclude that $AC=a$). – Eric Wofsey May 13 '18 at 18:13
  • You are using, in fact, "lies on" as a synonym for "is an element of", as if a line were a set of points. The relation "lies on" is defined only by the axioms themselves, so from "$A$ and $B$ lie on $a$" and "$A$ and $C$ lie on $a$" you cannot conclude (on the basis of Axiom 1) that "$B$ and $C$ lie on $a$". You need a second axiom for that. – Intelligenti pauca May 13 '18 at 18:20
  • No, I'm just using the fact that "$A$ and $B$ lie on $a$" is the conjunction of "$A$ lies on $a$" and "$B$ lies on $a$". This is completely standard in axiomatic geometry and has nothing to do with sets. – Eric Wofsey May 13 '18 at 18:54
  • It appears to me that what happened is the translator erroneously interpreted "$A$ and $B$ lie on $a$" as a primitive notion rather than "$A$ lies on $a$", and then (sloppily and incorrectly) attempted to formulate Axiom I,2 in terms of this primitive notion. Again, the second half of Axiom I,2 is not nonsensical or redundant if you take $AB=a$ as primitive, but it does an awful job of capturing the intended meaning and is not adequate for the purpose this axiom needs to serve in the theory. – Eric Wofsey May 13 '18 at 19:19
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If $AB=a$ and $AC=a$, with $B\ne C$, then you know from Axiom 1 that points $BC$ lie on some line: Axiom 2 states that such line is still $a$.

So I think the words added after "that is" are a way to better explain the meaning of this axiom, while your $AB=AC=BC=a$ would be a mere repetition of the first part of the axiom.

EDIT.

To avoid confusion, one should keep in mind that Hilbert does not define lines and planes as sets of points. He writes:

Let us consider three distinct systems of things. The things composing the first system, we will call points [...]; those of the second, we will call straight lines [...]; and those of the third system, we will call planes [...]. We think of these points, straight lines, and planes as having certain mutual relations, which we indicate by means of such words as “are situated,” “between,” “parallel,” “congruent,” “continuous,” etc. The complete and exact description of these relations follows as a consequence of the axioms of geometry.

The first two axioms are then devoted to describe the relations between points and lines and they are both needed, in my opinion.

Intelligenti pauca
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