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Every proof seems to go above my head as I'm not thorough with calculus or what is being talked about in this similar question in which the author proves it using complicated terms. As a result I tried proving it on my own.

I think it has to do with triangle inequality. Can we say that any curve joint to the two points other than the straight line as sides of a polygon? Can we say that the sum of the sides of that polygon will always be more than the straight line because of triangle inequality theorem? If we can say that, it is proved much more simply than any other proof I have seen yet.

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    What do you mean by a straight line? – Mark Bennet Jul 09 '18 at 16:33
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    What exactly do you mean by "Can we say that any curve joint to the two points other than the straight line as sides of a polygon" ? – Jack M Jul 09 '18 at 16:34
  • @JackM let there be a curve other than the straight line connecting the points. Can't we just say that the curve is a polygon with infinitely many sides? Because then we can use the triangle inequality approach. –  Jul 09 '18 at 17:27
  • @MarkBennet I am not sure how to define a straight line. Take scale/ruler and use it to draw a line. That is the kind of straight line I am referring to. I don't know the level of your math skills, but I know as much as a 10th grader would (and that is why I don't know different kinds of straight lines). –  Jul 09 '18 at 17:31
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    @ShashwatAsthana No, because a smooth curve isn't a polygon with infinitely many sides. Of course it's kind of like a polygon with infinitely many sides, but the difference between a proof and just saying things is being precise. And if you try to make your intuition precise on this issue, you end up with the kind of "complicated terms" that you see in other proofs. – Jack M Jul 09 '18 at 17:35
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    My point is that a straight line can be taken as a primitive concept, or defined geometrically as the shortest distance between two points, or defined as a one-dimensional linear object (as examples). Euclid says that a straight line (undefined) can be drawn from any point to any point and that a straight line can be extended indefinitely. The properties of a straight line are taken from the axioms. So it is not obvious what a straight line is, and how it is defined influences what you can prove and how you can prove it. – Mark Bennet Jul 09 '18 at 17:58
  • @JackM sorry for this late reply. What about calculus? In integral calculus, we take the area of infinitely many rectangles under the curve and add them up to get the area under the curve. If a curve can't be defined as a polygon with infinite side, then integral calculus shouldn't work, right? –  Jul 11 '18 at 08:07
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    @ShashwatAsthana " In integral calculus, we take the area of infinitely many rectangles under the curve" Of course we don't. We take the area under the curve to be a complicated mathematical construct which could, in a loose metaphorical sense, be described as "the area of infinitely many rectangles", but is actually more subtle than that. – Jack M Jul 11 '18 at 08:49

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This is the correct idea, but there's still a lot of leg work to be done to explain why it is the 'shortest distance.' This is because the triangle inequality essentially only deals with straight lines to begin with - the quantities involved in it are norms of vectors, and vectors are essentially straight lines. This means that the triangle inequality proves that among all paths given by straight lines, the single straight line going straight from one point to the other is the best. But how do you know there isn't a really wild and wiggly curve connecting the two points? How do you define the distance along such a curve?

However, analysis furnishes us with tools to improve this proof, that you can look into. For example, if the curve connecting the two points is sufficiently nice (the term 'bounded variation', comes to mind, for example), then one can show that approximating a curve by straight lines more and more finely, and taking the limit, your proof will generalize. There are upgraded versions of the triangle inequality that deal also with integrals, which are also just called the triangle inequality.

Another approach analysis can give us to generalize this idea is in calculus of variations. On this site, there is a linked thread here which discusses this, using the Euler-Lagrange equations.

A. Thomas Yerger
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I think it works for polygonal paths, but this certainly does not show the result in full generality. Indeed, a different way to go, might be to define a path to be a continuous map $\lambda:[0,1] \to \mathbb R^n$.

We need a meaningful notion of "length" in this context, and if we take $d(x,y)$ to be the usual euclidian distance one can propose the following definition

The length of a curve $\alpha:[0,1] \to \mathbb R^n$ is $$\sup_{0=t_0<t_2<,\dots,<t_n=1}\sum_{i=0}^{n} d(\alpha(t_i),\alpha(t_{i+1}))$$

Indeed this makes the proof quite easy, and it is really the "limiting" analogue of your own proof:

A valid partition is given by $n=2$ where the partition is taken to be $\{t_0,t_1\}$ of the interval, in which case the definition of supremum implies that any other path is greater than the straight one.

The more familiar definition of arc length $\int_{[0,1]}d(x(t),y(t))dt$ is equivalent to the one given for a large family of curves (and in particular class $C^1$.) See Rudin's Principles, 6.27.

Andres Mejia
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  • But can't we just take limits? As in, as the number of sides in the polygon tends to infinity, it just turns into a curve, right? –  Jul 09 '18 at 17:23
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    @ShashwatAsthana that is the heuristic. That is also essentially what this definition does, it breaks up a path into partitions and takes the sum of line segments joining two points together. – Andres Mejia Jul 09 '18 at 17:30