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From Euclid's definitions, postulates, and common notions, can you prove that a straight line is the shortest distance between two points, or is that basically an assumption of the way lines are measured?

Here is an online copy of much of the text of Euclid's Elements.

Proposition 20 is:

In any triangle the sum of any two sides is greater than the remaining one.

This does prove the theorem for the case where one straight line is shorter than two straight lines at an angle, and it's obvious how to prove from that that any chain of straight lines is longer than a single straight line, but I don't see anything that rules out that another sort of curve might be shorter. Maybe you could prove it from proposition 20 using the method of exhaustion?

What about modern formulations of Euclidean geometry? Do any of them make it a theorem rather than an axiom that the shortest distance is a straight line?

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    What is a path for you? – blamethelag Aug 08 '23 at 20:20
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    I don't think Euclid has a notion of an arbitrary curve. I am not sure what you mean by "modern formulations of Euclidean geometry". If you mean Euclidean space as formalised in modern treatments of geometry that combine algebra and analysis, then they certainly do allow you to show that straight lines give the shortest distance amongst any reasonable curves (using ideas from the calculus that Archimedes' method of exhaustion foreshadowed). – Rob Arthan Aug 08 '23 at 20:24
  • @RobArthan I think by “modern formulation of Euclidean geometry”, the OP means something like Hilbert’s axiomatisation. – user1551 Aug 09 '23 at 10:28
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    @user1551: my original comment did allow for that possibility but became too long (and I rather doubted that the OP had that in mind, because those axiomatisations have no primitive concepts that correspond to an arbitrary curve). I think it's down to David to explain what he meant. – Rob Arthan Aug 09 '23 at 20:58
  • @RobArthan, yes, I was talking about Hilbert's axiomatization and few others I've seen. I'm not familiar enough with modern developments in geometry to have meant anything more obscure than that. – David Gudeman Aug 09 '23 at 21:11
  • David: thanks for that. My phrase "modern developments in geometry" was really intended to be less obscure than Hilbert's axiomatisations: I just meant how we teach geometry via algebra and analysis in the usual pure maths curriculum. The latter allows graphs of transcendental functions, space-filling curves and all sorts of other monsters which cannot be spoken of in an axiomatisation of a theory of line and points. – Rob Arthan Aug 09 '23 at 21:23
  • @RobArthan, I see. Yes, I am familiar with modern analytical geometry at least as taught to physics undergrads, and I was not referring to that. – David Gudeman Aug 09 '23 at 22:35

3 Answers3

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There are certain things that you have to consider. First of all, the shortest distance among what paths? You have to note that planar paths defined by arbitrary functions don't necessary exist in classical Euclidean geometry. For example, the path $ x(t) = t , y(t) = \exp(t) $ doesn't exist in classical Euclidean geometry, but our objects are lines, line segments, circles, angles, and things like that. If you consider paths consisting of $n$ line segments, you can prove your claim by proposition 20 and induction on $n$.

Secondly, even if you allow arbitrary paths defined by functions from $\mathbb{R} \to \mathbb{R}^2$ in your system, before proving your claim, you have to come up with a plausible definition of the arc length. What does it really mean when we say that some curve is longer than another? One agreed upon definition of the arc length is as follows:

$$f: \mathbb{R} \to \mathbb{R}^2,\ f(t) = ( f_1(t), f_2(t)),\ a \leq t \leq b.$$

Consider $\Sigma = \{P\mid \text{$P$ is a partition of $[a,b]$}\}$, suppose $P_0 \in \Sigma$ and that $P_0 = \{x_0, x_1, x_2,\dotsc, x_n \}$, where $x_0=a$ and $x_n=b$. Define $\Gamma(P_0) = \sum_{k=1}^{n}|f(x_k)-f(x_{k-1})|$. Now, the arc length can be defined to be $\Gamma = \sup \{ \Gamma(p) \mid P \in \Sigma\}$, provided that it exists.

So, we intuitively define the arc length to be the supremum of all possible finite line segments following each other from $f(a)$ to $f(b)$. By this definition and by the proposition 20 and proving the case for $n$ line segments, it easily follows that the shortest path is the straight line segment from $f(a)$ to $f(b)$.

Aria
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    I'm accepting the answer, but you might want to edit it to point out that the method would not have been available to Euclid (but might have been available to Archimedes with the method of exhaustion). Also, I don't fully agree that Euclid's geometry doesn't have arbitrary (smooth continuous) curves. The definitions do contemplate them (it calls them lines), it's just that there are no axioms to deal with them. – David Gudeman Aug 09 '23 at 15:08
  • Dear @DavidGudeman, Since there was not a mathematical concrete definition of these general curves, and there was also no axioms concerning them, It is safe to say that these notions were not present in Euclid's axiomatic system. Euclid probably tacitly considered these curves and called them lines, but this sort of informal definition is far from adequate for us to deal with these objects mathematically. If you still feel that the answer can be improved, I'd be really glad if you could implement the necessary modifications, and I will accept your edition. – Aria Aug 09 '23 at 15:45
  • If you don't use the Pythagorean theorem, you won't be able to prove that it is the shortest, rather than merely a shortest. For example, in $\ell^\infty$ there are many minimizing paths joining a pair of points. See my answer. – Mikhail Katz Aug 09 '23 at 16:21
  • Dear @MikhailKatz, I think by mentioning only the triangle inequality and using induction it easily follows that the length of the polygonal curve is greater than $AB$. suppose it's true for all polygonal curves having less than $n$ line segments, and for the $n+1$ case, all you have to do is connect A to $x_{n-1}$. Then, we can use triangle inequality and by induction your argument follows for a general polygonal curve comprising an arbitrary number of line segments. I might have misunderstood your point, or there might be a mistake in my reasoning. I'd be happy if you would point that out. – Aria Aug 09 '23 at 16:35
  • @Aria Greater than or equal to. MikhailKatz is correct that there exist metrics in which shortest paths are not unique, and that the Euclidean metric, which is defined in terms of the Pythagorean theorem, is a metric in which shortest paths do exist. But (1) this hardly seems relevant in this context, and (2) there are other metrics in which shortest paths are unique. – Xander Henderson Aug 09 '23 at 16:37
  • Dear @XanderHenderson, sorry, greater than or equal to. That's an interesting point, but as you mentioned, I think the way that Euclid defined an order on line segments and other axioms in his geometry lead to the unique Euclidean metric. As a matter of fact, the Pythagorean theorem is not independent from Euclid's axiom and can be proven in his system. So pointing out other general metric spaces is not relevant in this context, as you mentioned. – Aria Aug 09 '23 at 16:55
  • Proposition 20 is more than the " triangle rule" used in the definition of a metric. It has strict inequality. That's enough to show that the straight line is the unique shortest path. If a metric allows other minimal paths, then it will have non-degenerate triangles where one side is equal to the sum of the others. – David Hartley Aug 10 '23 at 13:34
  • @DavidHartley, exactly. Equality only happens when the points are colinear. that's also the case for a polygonal curve with $n$ line segments. – Aria Aug 10 '23 at 14:03
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This question fundamentally comes down to "How do you measure the length of a curve?"

Classical Greek mathematics doesn't really have the ability to either satisfactorily ask or answer this question. Archimedes could describe the lengths of circular arcs but, according to the linked article, the fact that a straight segment is the shortest path between two points is taken as axiomatic; there are probably other classes of curves which the Greeks could compute the lengths of, but the general problem was well beyond their mathematics. Indeed,

  • even Archimedes' argument about the value of $\pi$ relies on an infinitesimal argument, which is more akin to calculus than the rest of Greek mathematics, and
  • the Greeks didn't really even have a notion of "general curve"—they were comfortable with straight lines and conic sections, but didn't do a great deal beyond these somewhat elementary curves.

Basically, the question doesn't make sense in the context of Greek mathematics; and even if it did, the tools to answer the question in a rigorous way simply don't exist.

In modern mathematics, we have a more precise definition of distance: if $\gamma : [a,b] \to \mathbb{R}^n$ is some curve which is "nice enough" (injective and continuously differentiable, perhaps—this is basically what is required for a "rectifiable" curve; more general notions exist, but I don't think that these really conform to the usual notion of what a "path" is), then the length of this curve is given by something like $$ \operatorname{Length}(\gamma) = \lim_{\max_j \Delta t_j \to 0} \sum_j \left|\frac{\gamma(t_j) - \gamma(t_{j-1})}{\Delta t_j}\right| \Delta t_j = \int_{a}^{b} |\gamma'(t)|\,\mathrm{d}t, $$ where $\{a=t_0<t_1<\dotsb<t_n=b\}$ is some partition of $[a,b]$, and $\Delta t_j = t_j - t_{j-1}$. See, for example, arc length on MathWorld. This formula comes from approximating $\gamma$ with very short line segments, and then taking a limit as the "mesh" of those segments goes to zero (basically, the length of the longest segment is taken to zero).

The argument that a straight line segment is the shortest path between two points then comes down to basically two observations:

  1. The triangle inequality: if $A$, $B$, and $C$ are three points, then $$ d(A,C) \le d(A,B) + d(B, C). $$ From the triangle inequality, it can be deduced that any piecewise linear path from $A$ to $C$ is as least as long as the straight line segment.

  2. Inequalities behave "nicely" with respect to limits: if $a_n \le b_n$ for all $n$, then $\lim_{n\to\infty} a_n \le \lim_{n\to\infty} b_n$. Thus \begin{align} &d(\gamma(a),\gamma(b)) \le \sum_{j} \left| d(\gamma(t_j), \gamma(t_{j-1}) \right| \\ &\qquad\implies d(\gamma(a), \gamma(b)) \le \int_{a}^{b} |\gamma'(t)|\,\mathrm{d}t = \operatorname{Length}(\gamma). \end{align}

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    I will note that the other answer provided here is very similar to mine, and written at about the same time. I have over 25k XP, while Aria (the other answerer) has less than 1000. I appreciate the upvotes, but you should also consider upvoting the other answer---give a new user some XP, too. – Xander Henderson Aug 08 '23 at 21:09
  • "From the triangle inequality, it can be deduced that any piecewise linear path from A to C is longer than the straight line segment." Actually, your argument shows that a pw-linear path is not shorter than the straight line, not that it's longer than. With the triangle inequality alone, you can only show that there is no shorter path than a straight line, but it doesn't imply that straight line is the shortest path. – BigbearZzz Aug 15 '23 at 16:19
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I see that neither of the other answers mentioned the Pythagorean theorem, which seems to me to be the main point here. If one has a polygonal curve connecting a pair of points $A$ and $B$, each side of the polygon can be included as the hypotenuse of a right-angle triangle one of whose sides is parallel to $AB$. Since the hypotenuse is longer than each of the sides by the Pythagorean theorem, it follows that the curve is longer than the segment $AB$. Exhaustion-type arguments when a curve is approximated by a polygon were certainly familiar to the ancient Greeks. If one doesn't use the Pythagorean theorem, one won't be able to prove that it is the shortest, rather than merely a shortest. For example, in $\ell^\infty$ there are many minimizing paths joining a pair of points.

J. W. Tanner
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Mikhail Katz
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    The Pythagorean theorem (Book 1, Proposition 47) is not necessary here. The triangle inequality (Book 1, Proposition 20) is sufficient. – Xander Henderson Aug 09 '23 at 14:28
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    @XanderHenderson, it may be sufficient for a different argument, but for the argument I presented it is not sufficient, since one needs to know that the hypothenuse is greater than each of the other sides. This does not follow from the triangle inequality, and moreover would be false in a general triangle, since a side can certainly be shorter than one of the other sides in an arbitrary triangle. What argument did you have in mind? – Mikhail Katz Aug 09 '23 at 14:40
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    If you want to present a different argument, great. But if you start your argument by criticizing the other answers and asserting that they have missed the "main point", you should expect pushback. The Pythagorean theorem is not needed for the argument that I presented. Your answer gives the impression that it is. Perhaps you should edit your answer to clarify that you are providing a different argument, rather than a critique of the existing answers? – Xander Henderson Aug 09 '23 at 14:43
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    Cool off, @Xander, will you? I did not criticize your argument, but simply pointed out that it does not use the Pythagorian argument. Now that you mention it, I could point out that an argument using integrals would not seem to have been accessible to Euclid :-) – Mikhail Katz Aug 09 '23 at 14:48
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    (1) Please do not tell others to "cool off". It is condescending. (2) I believe that I made it quite clear that Euclid would not use an integral argument---I was very careful to point out that the Greeks wouldn't have had a notion of "general curve", and that infinitesimal arguments were uncommon in their thinking. So, again, if you are not critiquing the other answers, I suggest that you edit your answer to remove this impression that you are critiquing the other answers. – Xander Henderson Aug 09 '23 at 14:53
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    If you don't use the Pythagorean theorem, you won't be able to prove that it is the shortest, rather than merely a shortest. For example, in $\ell^\infty$ there are many minimizing paths joining a pair of points. See my answer. – Mikhail Katz Aug 09 '23 at 16:22
  • You don't need the Pythagorean theorem to obtain a uniquely geodesic space. $(\mathbb{R}^n, \ell^p)$ is uniquely geodesic for any $p \in [1,\infty)$. In the classical Euclidean setting, metrics other than the usual metric are not considered (though that metric is equivalent to the Pythagorean theorem). In a modern setting, sure, you can use other metrics, but then you should probably specify the metric up front. The fact that a straight segment is minimizing does not depend on metric; only uniqueness (which doesn't really seem to be considered in the question). – Xander Henderson Aug 09 '23 at 16:44
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    But it is not uniquely geodesic for $p=\infty$ as I mentioned. – Mikhail Katz Aug 09 '23 at 16:45
  • Yes... and? There is a reason that I specified $p \in [1,\infty)$, and not $p\in [1,\infty]$... You argue that "If you don't use the Pythagorean theorem, you won't be able to prove that it is the shortest". This isn't so---if you want to bring non-Euclidean metrics into the discussion, then there are many metrics which do not satisfy the Pythagorean theorem, but which do give $\mathbb{R}^n$ a uniquely geodesic structure. – Xander Henderson Aug 09 '23 at 16:47
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    @XanderHenderson, but your proof doesn't. – Mikhail Katz Aug 10 '23 at 09:52