5 ms·

Along those lines (no pun intended), I think every STEM college graduate ought to be able to prove that the shortest path between 2 points is a line segment ins

Along those lines (no pun intended), I think every STEM college graduate ought to be able to prove that the shortest path between 2 points is a line segment instead of taking it for granted.

ams6110 10y ago
That's a high school freshman geometry problem.
- wfunction 10y ago
  > That's a high school freshman geometry problem. Looks like we found another person who doesn't know how to do it :) look it up! In pre-college geometry this isn't something you prove, it's something you postulate and take for granted.
  - ams6110 10y ago
    Maybe these days. We definitely proved it in my geometry class in 1980/81.
    - wfunction 10y ago
      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.
- obastani 10y ago
  How did the proof you learned go? As far as I know, calculus is required to rigorously prove this fact.
swiley 10y ago
Every STEM student at my university learns enough to do this as a freshman, I don't think most are ever actually asked to though.
- wfunction 10y ago
  "Learns enough to do this"... in what sense? If you compare this to a puzzle, are you claiming they have all the pieces, or are you claiming they actually would be able to put the pieces together if you asked them to? The proof steps are tricky, it's not just a straightforward evaluation of some expression. Also -- are you really claiming every STEM student at your university learns calculus of variations as a freshman? Because that's the only way I've actually seen others prove this. While doing it with plain calculus isn't impossible, it requires a similar kind of thought process that you get after learning calculus of variations, so I'm really skeptical what you're saying is the case, unless you're thinking of a different proof or something.
monochromatic 10y ago
Is there a way to prove this that doesn't get into calculus of variations?
- wfunction 10y ago
  I'm not totally sure; I think it depends on what exactly you classify as calculus of variations, and what assumptions you're willing to make along the way. For what it's worth, I don't think I know how to prove the most general case possible. But given that I encountered this in physics rather than in math, I was satisfied with proving the physical analog, which would be the claim that the path of light in Euclidean space is always linear. Assuming we're talking about light allowed me to make 3 useful assumptions: (1) that the speed of the particle doing the traveling is constant, which comes in quite handy in the proof [1], (2) that there is only 1 independent variable (i.e., time) rather than 3 space coordinates, and (3) that our functions are all sufficiently smooth and such. The nice thing about this proof is it doesn't need the Euler-Lagrange equation that everybody uses... it doesn't need calculus of variations at all. Every step is an elementary calculus step. But the downside is the thought process that actually leads to the derivation is the one you'd only really get after studying a bit of calculus of variations, so you wouldn't come up with it without having that foundation (even though it's not required). (Yes, you can try to remove time from this proof and still keep it rigorous, but then it'd require a stronger background than just freshman calculus in order to be understood.) [1] The reason I was satisfied with this is that it's obvious that the speed of the particle is irrelevant to the length of the path so long as the medium is uniform, so once I proved it for light of constant speed, I was done. It's not necessarily satisfying for a mathematician though.
  - m00n 10y ago
    This is a pictorial description of the proof in the sibling comment. It does not use calculus of variations, just the definition of the path length of a (differentiable parametrized) curve. If you want to call your parametrization a "light-ray" thats fine, but muddies the waters for me. Just spell out what we mean by path, length of a path, and use calculus of a single variable. No need to introduce 'time', 'constant speed' or the like. BTW, a truly variational proof is in the wikipedia article [0]. [0] https://en.wikipedia.org/wiki/Calculus_of_variations#Example https://en.wikipedia.org/wiki/Calculus_of_variations#Example
rxhernandez 10y ago
It isn't though, the shortest path between two points is a geodesic.
- wfunction 10y ago
  I'm trying not to be a smart aleck and instead I'm putting the question into terms anyone who's finished high school would understand. We're not talking general relativity here, we're just talking about plain Euclidean geometry. The kind you (or maybe everyone except you) automatically thinks of... the kind you learn in middle/high school.
  - rxhernandez 10y ago
    > I'm trying not to be a smart aleck and instead I'm putting the question into terms anyone who's finished high school would understand. How's it being a smart aleck? A geodesic is a very easy concept to understand. Not only that, but it's ridiculous to dumb down information when the whole point of the conversation was to understand truths. If this question was on history concepts, and you said that Columbus discovered America, it would be a bit ridiculous for you to criticize me on calling you out on it, right? A straight line being the shortest distance between two points is about as true as Columbus discovering America. You have to add on a bunch of qualifiers for either to be remotely true. > We're not talking general relativity here, we're just talking about plain Euclidean geometry. See this is the thing, you don't even need to understand general relativity to understand this. Airplane pilots understand this, they fly on great circles because of this concept, which has absolutely nothing to do with general relativity. > or maybe everyone except you Or maybe anyone with a math degree, 4 or 5 physics classes, an aerospace degree, airplane pilots, optical engineers. It's really not that esoteric.
rxhernandez 10y ago
It isn't though, the shortest path between two points is a geodesic.
wfunction 10y ago
EDIT: Whoops! I just realized the entire point of this problem was missed, and it's my fault -- I worded my comment pretty poorly in hindsight. I meant to say that STEM college graduates ought to be able to solve problems such as this one. That would include less-trivial instances, like the light-ray example I gave in another comment where you have to find the path that light travels when it crosses between media with different velocities. Unfortunately I made the mistake of posing the most trivial instance of this kind of problem as an example (Euclidean vector space), which resulted in many people missing the point I was trying to make. Sorry about that. The point wasn't solving this specific problem, the point was this kind of problem, and my example was overly simplified and therefore not representative of what I was trying to convey. With that aside, note that this is still not a basic geometry problem, since you need to be able to discuss the lengths of arbitrary smooth curves, not just lines/angles/circular arcs.

Along those lines (no pun intended), I think every STEM college graduate ought to be able to prove that the shortest path between 2 points is a line segment instead of taking it for granted.

ams6110 10y ago

That's a high school freshman geometry problem.

wfunction 10y ago
> That's a high school freshman geometry problem. Looks like we found another person who doesn't know how to do it :) look it up! In pre-college geometry this isn't something you prove, it's something you postulate and take for granted.
- ams6110 10y ago
  Maybe these days. We definitely proved it in my geometry class in 1980/81.
  - wfunction 10y ago
    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.
obastani 10y ago
How did the proof you learned go? As far as I know, calculus is required to rigorously prove this fact.

swiley 10y ago

Every STEM student at my university learns enough to do this as a freshman, I don't think most are ever actually asked to though.

wfunction 10y ago
"Learns enough to do this"... in what sense? If you compare this to a puzzle, are you claiming they have all the pieces, or are you claiming they actually would be able to put the pieces together if you asked them to? The proof steps are tricky, it's not just a straightforward evaluation of some expression. Also -- are you really claiming every STEM student at your university learns calculus of variations as a freshman? Because that's the only way I've actually seen others prove this. While doing it with plain calculus isn't impossible, it requires a similar kind of thought process that you get after learning calculus of variations, so I'm really skeptical what you're saying is the case, unless you're thinking of a different proof or something.

monochromatic 10y ago

Is there a way to prove this that doesn't get into calculus of variations?

wfunction 10y ago
I'm not totally sure; I think it depends on what exactly you classify as calculus of variations, and what assumptions you're willing to make along the way. For what it's worth, I don't think I know how to prove the most general case possible. But given that I encountered this in physics rather than in math, I was satisfied with proving the physical analog, which would be the claim that the path of light in Euclidean space is always linear. Assuming we're talking about light allowed me to make 3 useful assumptions: (1) that the speed of the particle doing the traveling is constant, which comes in quite handy in the proof [1], (2) that there is only 1 independent variable (i.e., time) rather than 3 space coordinates, and (3) that our functions are all sufficiently smooth and such. The nice thing about this proof is it doesn't need the Euler-Lagrange equation that everybody uses... it doesn't need calculus of variations at all. Every step is an elementary calculus step. But the downside is the thought process that actually leads to the derivation is the one you'd only really get after studying a bit of calculus of variations, so you wouldn't come up with it without having that foundation (even though it's not required). (Yes, you can try to remove time from this proof and still keep it rigorous, but then it'd require a stronger background than just freshman calculus in order to be understood.) [1] The reason I was satisfied with this is that it's obvious that the speed of the particle is irrelevant to the length of the path so long as the medium is uniform, so once I proved it for light of constant speed, I was done. It's not necessarily satisfying for a mathematician though.
- m00n 10y ago
  This is a pictorial description of the proof in the sibling comment. It does not use calculus of variations, just the definition of the path length of a (differentiable parametrized) curve. If you want to call your parametrization a "light-ray" thats fine, but muddies the waters for me. Just spell out what we mean by path, length of a path, and use calculus of a single variable. No need to introduce 'time', 'constant speed' or the like. BTW, a truly variational proof is in the wikipedia article [0]. [0] https://en.wikipedia.org/wiki/Calculus_of_variations#Example https://en.wikipedia.org/wiki/Calculus_of_variations#Example

rxhernandez 10y ago

It isn't though, the shortest path between two points is a geodesic.

wfunction 10y ago
I'm trying not to be a smart aleck and instead I'm putting the question into terms anyone who's finished high school would understand. We're not talking general relativity here, we're just talking about plain Euclidean geometry. The kind you (or maybe everyone except you) automatically thinks of... the kind you learn in middle/high school.
- rxhernandez 10y ago
  > I'm trying not to be a smart aleck and instead I'm putting the question into terms anyone who's finished high school would understand. How's it being a smart aleck? A geodesic is a very easy concept to understand. Not only that, but it's ridiculous to dumb down information when the whole point of the conversation was to understand truths. If this question was on history concepts, and you said that Columbus discovered America, it would be a bit ridiculous for you to criticize me on calling you out on it, right? A straight line being the shortest distance between two points is about as true as Columbus discovering America. You have to add on a bunch of qualifiers for either to be remotely true. > We're not talking general relativity here, we're just talking about plain Euclidean geometry. See this is the thing, you don't even need to understand general relativity to understand this. Airplane pilots understand this, they fly on great circles because of this concept, which has absolutely nothing to do with general relativity. > or maybe everyone except you Or maybe anyone with a math degree, 4 or 5 physics classes, an aerospace degree, airplane pilots, optical engineers. It's really not that esoteric.

rxhernandez 10y ago

It isn't though, the shortest path between two points is a geodesic.

wfunction 10y ago

EDIT: Whoops! I just realized the entire point of this problem was missed, and it's my fault -- I worded my comment pretty poorly in hindsight. I meant to say that STEM college graduates ought to be able to solve problems such as this one. That would include less-trivial instances, like the light-ray example I gave in another comment where you have to find the path that light travels when it crosses between media with different velocities. Unfortunately I made the mistake of posing the most trivial instance of this kind of problem as an example (Euclidean vector space), which resulted in many people missing the point I was trying to make. Sorry about that. The point wasn't solving this specific problem, the point was this kind of problem, and my example was overly simplified and therefore not representative of what I was trying to convey. With that aside, note that this is still not a basic geometry problem, since you need to be able to discuss the lengths of arbitrary smooth curves, not just lines/angles/circular arcs.