4 ms·

Share your "proof"?

ams6110 10y ago
Well it's been 35 years ... as I recall it's an indirect proof, it starts with assuming there is some other shorter path, and then showing that to be impossible by creating a contradiction of other proven theorems and postulates.
- wfunction 10y ago
  > Well it's been 35 years ... as I recall it's an indirect proof, it starts with assuming there is some other shorter path, and then showing that to be impossible by creating a contradiction of other proven theorems and postulates. Then I'm gonna call B.S. on this until you can show me the proof somewhere. Every single time I've searched this the only rigorous proof I've found has been using calculus (often calculus of variations).
  - ams6110 10y ago
    Well maybe I'm mistaken. It struck me that this is one of the proofs we discussed in class. I took calculus in high school also, maybe we discussed it there. However I've forgotten most of the geometry, and calculus I ever learned, having never had a practical professional need to use any of it. I'm not sure whether that reinforces or contradicts your original point on this subthread.
  - wyager 10y ago
    Variational minimization is one approach. The details of the proof would come down to what sort of space you're working in, though. Euclidian spaces are a lot simpler to prove this in than more complicated spaces with different notions of intrinsic and extrinsic curvature. For example, the shortest path on a sphere is also a line, but it's extrinsically curved and also might not be unique. On the other hand, the proof in a one-dimensional Euclidean space sort of comes from the definition of the metric over the space.
    - wfunction 10y ago
      We're just talking plain Euclidean space here. It's still not a basic geometry problem as the parent was suggesting.
      - m00n 10y ago
        Since the notion of curved path (maybe aside from circular arcs) is not discussed in basic geometry, the problem would be impossible to state. If you assume given the notion of arc-length as the least-upper-bound on lengths of piecewise-linear paths between the two points, then the triangle inequality of basic geometry suffices. This seems not more hand-wavy a proof as something about "light-travelling-at-constant-speed" analogies.
        wfunction 10y ago
        > Since the notion of curved path (maybe aside from circular arcs) is not discussed in basic geometry, the problem would be impossible to state. >> It's still not a basic geometry problem seems we're in agreement?
  - systoll 10y ago
    I recall doing a proof of the form: 1. Prove the triangle inequality, & extend via induction to show that [A] a straight line is the shortest polygonal path, and that [B] adding a point to a polygonal path can only make it longer. 2. Arc length is defined as the limit of the length of a polygonal approximations to a curve. From (B), we know our approximations would approach the arc length from below. So if an curve A->B has an arc length, there's a polygon A->B which is at least as short. And from [A] we know there's no polygon shorter than the straight line. This is weaker than proofs you can do with calculus -- it doesn't prove uniqueness, and it completely ignores non-rectifiable curves, for instance. But I suspect that's what ams6110 is remembering.
    - wfunction 10y ago
      2. Arc length is defined as the limit of the length of a polygonal approximations to a curve. From (B), we know our approximations would approach the arc length from below. You call this "geometry"? Since when does geometry involve a function limiting and bounding process? We didn't even know what a limit was until calculus...

Share your "proof"?

Well it's been 35 years ... as I recall it's an indirect proof, it starts with assuming there is some other shorter path, and then showing that to be impossible by creating a contradiction of other proven theorems and postulates.

wfunction 10y ago
> Well it's been 35 years ... as I recall it's an indirect proof, it starts with assuming there is some other shorter path, and then showing that to be impossible by creating a contradiction of other proven theorems and postulates. Then I'm gonna call B.S. on this until you can show me the proof somewhere. Every single time I've searched this the only rigorous proof I've found has been using calculus (often calculus of variations).
- ams6110 10y ago
  Well maybe I'm mistaken. It struck me that this is one of the proofs we discussed in class. I took calculus in high school also, maybe we discussed it there. However I've forgotten most of the geometry, and calculus I ever learned, having never had a practical professional need to use any of it. I'm not sure whether that reinforces or contradicts your original point on this subthread.
- wyager 10y ago
  Variational minimization is one approach. The details of the proof would come down to what sort of space you're working in, though. Euclidian spaces are a lot simpler to prove this in than more complicated spaces with different notions of intrinsic and extrinsic curvature. For example, the shortest path on a sphere is also a line, but it's extrinsically curved and also might not be unique. On the other hand, the proof in a one-dimensional Euclidean space sort of comes from the definition of the metric over the space.
  - wfunction 10y ago
    We're just talking plain Euclidean space here. It's still not a basic geometry problem as the parent was suggesting.
    - m00n 10y ago
      Since the notion of curved path (maybe aside from circular arcs) is not discussed in basic geometry, the problem would be impossible to state. If you assume given the notion of arc-length as the least-upper-bound on lengths of piecewise-linear paths between the two points, then the triangle inequality of basic geometry suffices. This seems not more hand-wavy a proof as something about "light-travelling-at-constant-speed" analogies.
      - wfunction 10y ago
        > Since the notion of curved path (maybe aside from circular arcs) is not discussed in basic geometry, the problem would be impossible to state. >> It's still not a basic geometry problem seems we're in agreement?
- systoll 10y ago
  I recall doing a proof of the form: 1. Prove the triangle inequality, & extend via induction to show that [A] a straight line is the shortest polygonal path, and that [B] adding a point to a polygonal path can only make it longer. 2. Arc length is defined as the limit of the length of a polygonal approximations to a curve. From (B), we know our approximations would approach the arc length from below. So if an curve A->B has an arc length, there's a polygon A->B which is at least as short. And from [A] we know there's no polygon shorter than the straight line. This is weaker than proofs you can do with calculus -- it doesn't prove uniqueness, and it completely ignores non-rectifiable curves, for instance. But I suspect that's what ams6110 is remembering.
  - wfunction 10y ago
    2. Arc length is defined as the limit of the length of a polygonal approximations to a curve. From (B), we know our approximations would approach the arc length from below. You call this "geometry"? Since when does geometry involve a function limiting and bounding process? We didn't even know what a limit was until calculus...