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The answer to the question in the title seems an obvious ''Yes by definition !''. And this really is the definition from Wikipedia:

Two line segments are congruent if they have the same length.

But in Hilbert's Foundations of Geometry congruence is defined without use of metric notions, by the axioms of Group IV (chapter 6).

I don't report here these axioms that can be found on the book (or on the Wiki page) and essentially say that congruence is an equivalence relation and that congruence is conserved when we add adjacent segments.

Usually these axioms are illustrated with a figure that use two segments $AB$ and $A'B'$ that have the same length, as in this case.

enter image description here

but I think that this illustration is only dictated by our intuition of congruence as equality of length, and it's not justified by the axioms.

Look at the second figure, where $a$,$a'$ and $a''$ are straight lines and $E$ is a point that does not belong to these lines.

enter image description here

Projecting from $E$ the points $A,B,C$ of $a$ to the other lines, we can define the relation: $$ AB \equiv A'B' \quad BC \equiv B'C' $$ and $$ A'B' \equiv A''B'' \quad B'C' \equiv B''C'' $$

and we can see that this relation satisfies all axioms of congruence. But, clearly, it doesn't satisfies our intuition that congruent segments have the same length.

This construction is essentially the same that Hilbert uses (in the chapter 24) for the construction of an ''Algebra of Segments'', ( where the point $E$ is ''at infinity'').

Obviously ''to have the same length'' is a congruence relation, but only in a metric space, and, in the same space, different congruence relations can be defined that are not equivalent to ''have the same length''. And more, we can have different metric (and different ''length'') for a geometry that satisfies the Hilbert's axioms.

So my question is, at first, if my reasoning is correct or if I've misunderstood something of fundamental, and, if I'm right, what are the axioms that we need for exactly represent our intuitive concept of congruence?

A final notice: this question is related to What really is ''orthogonality''?, where a similar question is posed for angles.

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Emilio Novati
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  • The construction works precisely because $E$ is a point at infinity. In that case, the rays $A''E$, $B''E$, and $C''E$ are parallel. – Justin Benfield Mar 12 '16 at 14:04
  • True if we adopt the paralle axiom. – Emilio Novati Mar 12 '16 at 14:08
  • Would it not be enough if the axiomatic system you have is powerful enough to prove a version of the parallel postulate? – Justin Benfield Mar 12 '16 at 14:09
  • Aren't you just confusing "congruent" and "similar"? Where "congruent" is the stronger notion that involves "equal lengths". – Han de Bruijn Mar 18 '16 at 13:02
  • This is our intuition. But the Hilbert's axioms define congruence, you can see the reference or at pag.8 of the book here: http://www.gutenberg.org/files/17384/17384-pdf.pdf – Emilio Novati Mar 18 '16 at 16:51
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    If I recall correctly, we define length in axiomatic geometry by taking the segments modulo congruence. That is the set of possible length is $$ L = {\text{segments}}/\cong $$ and the length of a particular segment is the class of it modulo congruence: Hence, obviously, by definition of length, not of congruence, your claim is true. – martini Mar 22 '16 at 19:30
  • If I well understand this means that two segments ''have the same length'' if they are congruent. But this means that ''have the same length'' is a properties that depends on the chosen congruence, and we can have a lot of congruence relations. – Emilio Novati Mar 22 '16 at 21:22
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    I'm not sure what you mean when you claim to “define” $AB \cong A'B'$. To give a definition of congruence of segments, you would need to say, given any two segments, whether they are congruent. Your “definition” doesn't say how to do this if, say, $E$, $A$, $A'$ aren't aligned. In any case, if you define some notion of congruence of segments along the lines you've stated, you'll find that Hilbert's axioms cannot all be satisfied by it, even if you do manage to turn it into an equivalence relation. – David Mar 23 '16 at 07:54
  • @David: How we can prove that my definition of ''congruence'' does not fit the Hilbert axioms? – Emilio Novati Mar 23 '16 at 13:25
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    @EmilioNovati That's not a well-defined definition of congruence, so I can't give a complete answer. But Hilbert's axioms imply a geometry very similar to the usual one (or identical, if you take all of them). – David Mar 23 '16 at 16:58

2 Answers2

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First of all, in order to make sense of the question one has to specify the axiomatic system via which the (Euclidean) plane is defined. For the purpose of this answer, I assume that you are interested in Hilbertian axiomatization of the Euclidean geometry. (There are alternatives, such as Birkhoff's and Tarski's sets of axioms.) Hilbert's axiomatization has two equivalence relations built in: congruence of segments and congruence of angles. If you want to change Hilbert's notion of congruence to something else, you are leaving the realm of Hilbert's axioms and, hence, you would have to list your personal axioms and then ask a new question.

Given this, the answer is that, by the definition, two segments have the same length iff they are congruent, the congruence class of a segment $AB$ is denoted $[AB]$.

This, of course, is not entirely satisfactory since we are used to the idea that length should be a real number and not an element of some unfamiliar quotient set (coming from some equivalence relation). I suspect that this is where your question is coming from.

It turns out that one can (assuming all but the continuity axioms of Hilbert) define an ordered field $F$ whose elements are the congruence classes, you can find a detailed construction in the book:

R. Hartshorne, "Geometry, Euclid and Beyond", Springer Verlag, 2000.

One needs the two continuity axioms (of Hilbert's axiomatic system) in order to prove that $F$ is isomorphic to the field of real numbers.

Note. In contrast, if you accept Birkhoff's axiomatization, then the Euclidean plane $E$ comes equipped with a distance function $d: E\times E\to {\mathbb R}_+$. The length of a segment $AB$ is simply the real number $d(A,B)$. A congruence of this geometry is an isometry, i.e. a map $g: E \to E$ preserving the distance function. Two segments $AB, CD$ are congruent in this sense iff there exists an isometry $g$ such that $g(AB)=CD$. Thus, two congruent segments have the same length, again, just by the definition. A good reference for this is:

E. Moise, "Elementary Geometry from an Advanced Standpoint", Pearson, 1990.

Moise sketches the relation between the two (Hilbert and Birkhoff) sets of axioms and explains, for instance, how to add congruence classes in Hilbert's plane. But he does not verify any of the ordered field axioms (for instance, he does not explain how to multiply the congruence classes).

Edit. I think, I understand now the source of your confusion, it is hidden in the article an. The point is that Hilbert's axiomatic system does not just contain an equivalence relation, called congruence, but a very specific one, which you can find for instance in either one of the references I gave above. If you change Hilbert's segment congruence equivalence relation to a different equivalence relation, you obtain a different axiomatic system. It appears that you are interested in replacing the standard segment congruence equivalence relation with the similarity equivalence relation. This equivalence relation has exactly two equivalence classes, namely, one represented by degenerate segments $AA$ and one represented by nondegenerate segments $AB$, $A\ne B$. This is all fine, but now you are asking about "length" in relation to this equivalence relation. For this, you have to specify what do you mean by the "length" since the latter is not a part of Hilbert's axiomatic. There are two meaningful choices:

a. Use the notion of length $[AB]$ as the equivalence class of the segment $AB$ with respect to the standard notion of congruence. This notion is clearly not preserved by the similarity: Similar segments, in general, do not have the same length. This is quite immediate (you can use the fact that there are infinitely many congruence classes of segments in the traditional sense and only two similarity equivalence classes). Thus, the answer to your question is negative in this case.

b. Use the notion of "length" of $AB$ as the similarity equivalence class of the segment $AB$. Then your notion of "length" takes exactly two values and is, but the definition, preserved by the similarity equivalence relation. Very few (if any) people, however, will accept this notion of "length" as useful, for instance, because it does not take values in the ordered field (the field of two elements cannot be ordered).

Moishe Kohan
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  • @Thanks for your answer that clarify the question (+1), but i'm not sure if accept. I prefer Hilbert's axioms because, as you noted, we can construct from them a model of the real numbers. And, if I well understand, from these axioms we can proof tha a segment can ''have the same lenght'' of any other, depending on what congruence relation we use. – Emilio Novati Mar 23 '16 at 17:42
  • It seems to me that a similar problem exists also in Birkhoff axiomatization, because we have tho compare segments on different lines, where the lenghts are defined by the one-one correspondence of the points with the real numbers, but these correspondence can be fixed chosing the points that correspond to $0$ and $1$ in an independent way on the two lines. To be sure that the segments $01$ an the two lines have ''physically'' the same lenght we need some rigid body that we can use as a mesure rule. – Emilio Novati Mar 23 '16 at 17:43
  • So it seems that without some physical convenction we cannot define an identity of lenght. I've posted a realted answer about this. http://math.stackexchange.com/questions/1710515/a-cartesian-coordinate-system-is-a-mathematical-or-physical-thing – Emilio Novati Mar 23 '16 at 17:44
  • @EmilioNovati: You do not need any physics to address this problem. My suggestion for you is to read Hartshorn's book, this should clarify your issues. – Moishe Kohan Mar 23 '16 at 17:54
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I think everyone is ignoring a crucial fact. Hilbert's congruence axiom for segments is a axiom about an undefined word: congruence. (By congruence, I mean the congruency of segements although the argument I'm presenting may apply to the congruency of angles too.) Hilbert purposely left the word congruence undefined so that an interpretation may be given by anyone who pleases. Surely, you know that an interpretation of congruence must be associated with some model that someone is trying to show satisfies Hilbert's congruence axioms based upon its interpretation of congruence.

This is how models for axiomatic systems work to begin with. In order for there to be a model for a set of axioms, the axioms must contain undefined terms for which the model provides interpretations and the axioms must also be true under this interpretation.

The interpretation of congruent segments by means of the idea that two segments are the same length may be provided by any metric space that has a defined metric which I believed you mentioned. But this merely provides a model for Hilbert's axiom, in other words, an interpretation. It does not replace the axiom!

The example in which you gave another interpretation of congruence without the intuitive idea of length is just that: just another interpretation or model. It again does not replace the axiom, but provides an interpretation for the word congruence.

James Fair
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