• TheMadFool
    8.7k
    Suppose a 2 D space with a y-axis and an x-axis at the usual 90 degree angle to each other. You're at the origin (0, 0) and travel in this space, in a straight line, to the point (3, 4) which basically means you travelled 3 units along the x-axis and 4 units along the y-axis. We could form a right triangle with 3 as its base and 4 as its height. What distance did you actually travel? The actual distance you traveled is the hypotenuse given by square root of [(3^2) + (4^2)] = square root of 25 = 5 units.

    Now imagine you're a shopkeeper and you sell apples at the rate: 3 apples for 4 dollars. In this case, every 3 units along the x-axis (you give 3 apples), there are 4 units along the y-axis (you receive 4 dollars). These values too will form a right triangle. My question is what does the hypotenuse mean? We understand the base is 3 apples sold, the height is 4 dollars received but what, if any, meaning can be assigned to the hypotenuse?

    Since 3 is apples and 4 is dollars, the hypotenuse = square root of [(3^2) apples^2 + (4^2) dollars^2] doesn't make sense since we can't add apples^2 to dollars^2.

    A way to go about making sense of the hypotenuse would be to replace the 4 dollars with 3 apples (since 4 dollars = 3 apples in terms of value) and we get hypotenuse = square root of [(3^2) apples^2 + (3^2) apples^2] = square root of [(2)×(3^2) apples^2] = (1.414)×(3 apples) = 4.242 apples. Since in a 2D space the hypotenuse is the actual distance traveled, it follows then that, in actuality, 4.242 apples were sold.

    Likewise, replacing 3 apples with 4 dollars (since 3 apples = 4 dollars in value) we get the hypotenuse = square root of [(4^2) dollars^2 +'(4^2) dollars^2] = square root of [(2)×(4^2) dollars^2] = (1.414)×(4 dollars) = 5.656 dollars. Since, sorry for the repetition, in 2D space, the hypotenuse is the actual distance traveled, it again follows that the actual amount of money made is 5.656 dollars.

    The original rate is 3 apple for 4 dollars = 0.75 apples per dollar and the rate calculated after replacing dollars with apples and vice versa is (4.242 apples)/(5.656 dollars) = 0.75 apples per dollar.

    How is it possible that when you sold 3 apples, you actually sold 4.242 apples and when you made 4 dollars, you actually made 5.656 dollars?

    [note: square root of 2 = 1.414]

    I know I'm making a mistake. I just don't know where. Help!!!
  • aletheist
    1.4k

    For spatial coordinates, the "unit" along each axis is an arbitrary interval of space and applies in any direction whatsoever, which is why it can also be used to measure the hypotenuse. Moreover, the axis orientations are likewise arbitrary, so you could rotate them and say that you simply moved 5 units along the revised x-axis (or y-axis).

    Apples and dollars do not work that way. Apples are truly discrete units that we can count individually, and dollars are units of exchange that measure value. When you assign apples to one axis and dollars to the other, you are simply representing the relationship between these two different units as a line corresponding to an equation: dollars = 4/3 x apples; the hypotenuse has no meaning at all. The same is true even if you assign both axes to apples or dollars; then the equation is apples = apples and dollars = dollars, while the hypotenuse still has no meaning at all.
  • TheMadFool
    8.7k
    hypotenuse still has no meaning at all.aletheist

    Firstly, how can that be? Consider the much bandied about term, worldline in the theory of relativity. If time were the x-axis and space the y-axis then, the hypotenuse would be the worldline of an object that began existing at the origin. It amounts to the actual experience of that object in spacetime. So, even if the two axes in a coordinate system are different it is possible for the hypotenuse to have meaning.

    Secondly, suppose that one simply wishes, on a whim, to give the hypotenuse a meaning. What meaning would you give it? 5 apple-dollars? Just like 5 space-time (worldline)? If apple-dollars has no meaning then how can space-time wordlines have meaning?
  • aletheist
    1.4k

    Time and space are continuous in themselves, so any units assigned to them are completely arbitrary, and you cannot measure a worldline except along the two different axes. Apples are discrete and dollars are defined, so comparing them to time and space is like comparing ... apples to oranges.
  • TheMadFool
    8.7k
    Time and space are continuous in themselves, so any units assigned to them are completely arbitrary, and you cannot measure a worldline except along the two different axes. Apples are discrete and dollars are defined, so comparing them to time and space is like comparing ... apples to oranges.aletheist

    Right! Taking x to mean apples, y to mean dollars, we get the equation for the hypotenuse (5 units) to be (3x)^2 + (4y)^2 = 5^2 which becomes 9x^2 + 16y^2 = 25. This, it turns out, is the equation of an ellipse. The major axis of this ellipse is 5/3 and the minor axis is 5/4. I don't know where to go with this. :chin:
  • aletheist
    1.4k
    I don't know where to go with this.TheMadFool
    How about nowhere, since it is nonsense.
  • TheMadFool
    8.7k
    How about nowhere, since it is nonsense.aletheist

    :rofl: Are you a mathematician?
  • tim wood
    6.1k
    What are you asking? Where the nonsense is? Or how (or why) if you put nonsense in, you do not get sense out? It can be fun to mix and match ideas, even instructive and educational, but don't get lost in them. Simplest is that the labels on the axes are just that. And you apply those labels first as convenience for yourself, and second because you have already determined that using numbers and lengths to represent some aspect of those things is or might be meaningful. But there is no such thing as a length of apples. What is represented on a graph is an abstraction of, presumably, a quantity of apples, or any other parameter that lends itself to quantification.

    A silly example. Apples can be bought by the pound. Fellow requests five pounds of apples, pays for them, and in turn is handed a bag containing five pounds of apples. "What's this?" he says. "These are just apples - where are my five pounds?"

    The moral of the story is that the sense sought is often not intrinsic but instead applied. Sometimes that's hard to get, and sometimes the language used by those not mystified can be misleading.
  • TheMadFool
    8.7k
    What are you asking? Where the nonsense is?tim wood

    To begin with, I don't understand why you think it's a nonsensical question.

    If the x-axis and the y-axis were distances then 5 represents the actual distance between positions (0, 0) and (3, 4).

    Similarly, if a moving object, p, were to begin at (0, 0) and travel 4 units of distance in 3 units of time, it would be at point (3, 4). In this case 5 would represent the worldline of the object p in spacetime.
  • tim wood
    6.1k
    Yes, but a distance is not an apple. A distance is a distance and an apple is an apple. If you insist on blending them into, say, apple-meters. then you shall have to figure out on your own what, exactly, that means.

    Similarly, if a moving object, p, were to begin at (0, 0) and travel 4 units of distance in 3 units of time, it would be at point (3, 4). In this case 5 would represent the worldline of the object p in spacetime.TheMadFool
    Actually, no. On a Euclidean plane. But spacetime is not a Euclidean plane except as an approximation. What you're referring to is a spacetime diagram.

    This is appropriately obscure and worth reading as many times as it takes to realize what's going on, even if you don't understand it. From https://physics.stackexchange.com/questions/371338/distance-formula-in-euclidean-space-vs-spacetime-interval-why-is-one-pythagor

    "The quantity "s2" (the squared-interval) is the analog of the square of the radius of a circle. For fixed s,
    s2=(ct)2−x2
    is the set of events that are "equidistant from the origin"... in the sense that a wristwatch on an inertial observer will elapse s ticks from origin to this set of events. This curve is a hyperbola, and it plays the role of the circle (as a curve of constant separation from the origin).

    Although there is a minus sign there, it's pythagorean in the sense that it's referencing two vectors that are "perpendicular" in that geometry. (Their dot-product in that geometry is zero.) The best way to appreciate "perpendicular" is that: when the "radius vector" meets the "circle", the "tangent to the circle" is "perpendicular to the radius"".(Italics added.)
  • TheMadFool
    8.7k
    5 apple-dollars
  • simeonz
    257

    I would like to give a little clarification on the answer provided by @tim wood. The application of the Pythagorean theorem is intended for spaces equipped with distances and angles. More precisely, for any affine space with associated inner product space or even a module. The theorem there follows simply from the properties required of the inner product, which then makes the statement a rather unhelpful fact. Essentially, the theorem is axiomatic.

    The introduction of this abstraction in the first place aims to eliminate the need for making any point the fixed origin, and any orientation automatically vertical, horizontal, etc. However, because angles and distances are defined, you can make use of Cartesian coordinate systems by choosing one point arbitrarily and one orientation (the abscissa) as reference, allowing you to express the other points and directions using numbers produced by the metric and norm in the affine space. (In fact, for the Pythagorean theorem, you don't need to use numbers, but say booleans with conjunction being multiplication and logical exclusion being the addition. You don't need to fix the unit of your quantities at all for non-unitary module either. No dollars, no euro, just abstract monetary value. But that just complicates the intuition here.)

    This trick is sensible only because the dot product between the n-tuples matches the inner product between the vectors. This can proven using the axioms of the affine spaces and the fact that the two basis vectors are orthonormal - perpendicular and unitary (for modules they can be chosen to be of identical magnitude, and the two products are not equal, but homomorphic). The other properties are also consequently preserved in the translation process - the Pythagorean property (I don't call it a theorem, because it stems axiomatically), distances computed using the Pythagorean property (ironically, but it is already preserved), the tangent (being the ratio in two dimensions) between the coordinate vectors and the abscissa, etc.

    What I am getting at is that we are first taught coordinates at school. We start with the description of numeric pairs and move to their correspondence to points in space. Unfortunately, this introduces angles and distances generically, by working out some arithmetic with the coordinate numbers. Affine spaces reinforce the notion that you can't derive such concepts mechanically. They are ascribed to the space semantically, and only then they become numerically expressed.

    In your particular case, you have two axes and you assume that they are perpendicular, i.e. that there is meaning to the angle between them. What does this angle express? You will find out that the main requirement of the Pythagorean theorem, the rightness of one angle, is untenable in your case, because the appearance of orthogonality between your basis vectors is only the remnant of some pedagogic intuition. As tim wood pointed out, the distance is not specified either, aside from strides along the coaxial directions. In summary, we can create a pair of numbers from virtually anything, but that does not confer the necessary structure for a Cartesian coordinate system corresponding to a choice of origin and orthonormal basis in some kind of affine space.
  • TheMadFool
    8.7k
    All that went over my head.

    If memory serves, Pythagora's theorem works only for right triangles and yes the axes that I used are perpendicular and yes there's a right triangle (3, 4, 5) formed. There are 3 apples that come at a total cost of 4 dollars. What's the hypotenuse in terms of apples and dollars? That's all I'm asking.
  • jgill
    1.1k
    What's the hypotenuse in terms of apples and dollars? That's all I'm asking.TheMadFool

    The square root of the square of the number of apples plus sixteen. An amazing breakthrough in marketing!
  • TheMadFool
    8.7k
    The square root of the square of the number of apples plus sixteen. An amazing breakthrough in marketing!jgill

    I don't get your joke. We have two items in our list: apples and dollars, each of them forming a side of a right triangle. What's the hypotenuse in terms of apples and dollars?

    If I ask a similar question with distance, the answer is quite obvious: the hypotenuse is the shortest distance between the two points that form the ends of the hypotenuse; the hypotenuse is a distance.
  • simeonz
    257
    If memory serves, Pythagora's theorem works only for right triangles and yes the axes that I used are perpendicular and yes there's a right triangle (3, 4, 5) formed.TheMadFool
    How did you come to the realization that the angle is perpendicular when constructing your mathematical model of the problem domain? What property in the actual domain of application of your model prompted the idea? If you were not concerned with any domain when you constructed the model, how do you come to need to ask semantic questions now?

    Essentially, you are trying to bypass semantic questions when constructing your geometry, but you want to coerce semantics at a later date, through some synthetico-analytic magic.

    P.S. The above came off a bit tacky. Not trying to be stern or anything...
  • jgill
    1.1k
    I don't get your joke. We have two items in our list: apples and dollars, each of them forming a side of a right triangle. What's the hypotenuse in terms of apples and dollars?TheMadFool

    A number representing the square root of the square of the number of apples plus sixteen
  • TheMadFool
    8.7k
    I'm going to rework my presentation in order that you may better understand my predicament. Please keep the explanation simple as I'm not a mathematician.

    Here goes...

    1. There's the usual 2D Cartesian coordinate system (x and y axes are perpendicular). I'm not willing to debate on the matter of whether the x and y axes are perpendicular or not because that's a given. Imagine now a person travels 3 m along the x axis and 4 m along the y axis. The person started from point (0, 0) and is now at the point (3, 4). If I know calculate the length of the hypotenuse it's 5 m and this 5 m is a distance just like the 3 m traversed along the x axis and the 4 m along the y axis. In other words, the hypotenuse makes sense to me - it's a distance, just like the adjacent and the opposite sides

    2. There's, again, the usual 2D Cartesian coordinate system (x and y axes are perpendicular). This time, however, the x axis is labeled as apples and the y axis is labeled dollars. 3 apples cost 4 dollars. Here too there are 2 points: (0, 0) which represents 0 dollars for 0 apples and the other point (3, 4) which represents 3 apples for 4 dollars. This time too we can construct a hypotenuse which is 5 "something". 5 isn't dollars, neither is it apples. We have nothing else to choose from since we haven't anything else to choose from apart from apples and dollars.

    :chin:
  • simeonz
    257

    The whole point is that the 2D Cartesian coordinate system is not a picture. It is ascription of coordinates to some plane of points, which points correspond in pairs to vectors, which vectors individually correspond to lengths and in pairs correspond to angles. Ok, the points are dollars-apples, but the remaining properties are not automatic. They are not provided by the Cartesian coordinate system for you, mathematically. They are provided by you, originally, so that you can justify the use of Cartesian coordinate system. Otherwise, what you have are just pairs of numbers corresponding to points, and the rest is as real as Tolkien's world.
  • TheMadFool
    8.7k
    The whole point is that the 2D Cartesian coordinate system is not a picture. It is ascription of coordinates to some plane of points, which points correspond in pairs to vectors, which vectors individually correspond to lengths and in pairs correspond to angles. Ok, the points are angles-apples, but the remaining properties are not automatic. They are not provided by the Cartesian coordinate system for you, mathematically. They are provided by you, originally, so that you can justify the use of Cartesian coordinate system. Otherwise, what you have are just pairs of numbers corresponding to points, and the rest is as real as Tolkien's worldsimeonz

    All I can say is that I'm talking about Descartes and you're talking about Descartes' father. We're talking past each other.
  • fishfry
    1.8k
    The whole point is that the 2D Cartesian coordinate system is not a picture. It is ascription of coordinates to some plane of points, which points correspond in pairs to vectors, which vectors individually correspond to lengths and in pairs correspond to angles. Ok, the points are angles-apples, but the remaining properties are not automatic. They are not provided by the Cartesian coordinate system for you, mathematically. They are provided by you, originally, so that you can justify the use of Cartesian coordinate system. Otherwise, what you have are just pairs of numbers corresponding to points, and the rest is as real as Tolkien's world.simeonz

    Sure, but I'm not following your point. The Pythagorean theorem goes back over a couple of thousand years, and Cartesian coordinates go back only to Descartes. We don't need an orthogonal coordinate system to have the theorem. Everyone agrees with that I'm sure. I am not understanding the point you're making. A right angle (if I remember my high school geometry) is when you have a line intersecting another line and making equal angles on each side. No coordinates or numbers needed.
  • simeonz
    257
    A right angle (if I remember my high school geometry) is when you have a line intersecting another line and making equal angles on each side.fishfry
    My point is that you need to have the concepts of "angles" (so that they can be equal), "directions" (so that you can make the points on your lines aligned, i.e. colinear), "distances" apriori, before resorting to analytic geometry. (And formally, we would call that an affine space today. Although the terminological designation would not be present historically, the ideas would be the same.) It would be backwards thinking if we started with pairs of numbers, declared them to be the Cartesian coordinate system for implicit space of entities and finally tried to infer a sensible explanation of the nature of the metrics of those entities.
  • simeonz
    257

    Don't you see what I mean when I say that you have only dollars and apples corresponding to pairs of numbers and no actual geometric model?
  • fishfry
    1.8k
    My point is that you need to have the concepts of "angles" (so that they can be equal), "directions" (so that you can make the points on your lines aligned, i.e. colinear), "distances" apriori, before resorting to analytic geometry. (And formally, we would call that an affine space today. Although the terminological designation would not be present historically, the ideas would be the same.) It would be backwards thinking if we started with pairs of numbers, declared them to be the Cartesian coordinate system for implicit space of entities and finally tried to infer a sensible explanation of the nature of the metrics of those entities.simeonz

    I guess I don't follow your point. The historical evolution is well known, from Euclid to Descartes. And in modern math we start with a 2-dimensional coordinate system and define the Euclidean distance. Either way works. What exactly is the question or issue?
  • simeonz
    257

    I say that the Pythagorean theorem applies to affine spaces over inner product spaces, because in affine spaces any two points from the underlying point space map to a vector, and we have the requirement of distributivity for inner products, hence:
    (a + b) (a + b) = a ^ 2 + b ^ 2 + 2 a * b
    
    Thus, for any triangle, the square of the norm of the third side equals the sum of the squares of the norms of the other two sides plus their doubled inner product . If the sides are orthogonal to each other, the inner is zero, hence the theorem. The trick then is to guarantee that we preserve the inner product when we move to n-tuples, which we do with a coordinate system. A Cartesian coordinate system is assignment of n-tuples to points, such that the implicit basis corresponding to the strides in unit distances along the coordinate axes is orthonormal. It cannot be orthonormal, if we haven't ascribed angles and distances.

    And in modern math we start with a 2-dimensional coordinate system and define the Euclidean distance.fishfry
    I don't see it that way really. We still come from the geometric perspective, to define angles and distances in one way or another, and only then we have the privilege of calling an n-tuple of points being from a Cartesian coordinate system. Cartesian coordinate systems come with semantics that need to be defined apriori. They are not just mechanical assignment of pairs of numbers to some arbitrary point space.

    There is only one sense, in fact, in which I am not correct. And it is that a Cartesian coordinate system might be a applied to the very n-tuples, with vectors being n-tuples, distances and angles computed in the usual way, etc. But then, we couldn't talk about apples and dollars, because since the underlying point space is just a mechanical bonding of numbers, it is unitless.

    Edit: That is my perspective anyhow. That is how I was taught analytic geometry at uni. But I am a software guy, so you may wish to exercise some reasonable caution and not take my word for it.
  • simeonz
    257

    I know that the Pythagorean theorem can be proven with constructive geometry, with only areas of aligned triangles. This is however not about Cartesian coordinates as far as I am concerned, because it is not about analytic geometry. In fact, the irony is, that you would most likely use such constructive proofs to validate the sensibility of the assumptions of affine space to your application domain before moving to analytic geometry. You would use the constructive proof to guarantee that the inner product for orthogonally directed vectors is zero (involving also the definition of inner product through cosine of angles and distances), and then you would also get the Pythagorean theorem to your orthogonally directed vectors in the affine space, making the exercise a little vacuous. But the constructive proof requires domain-level intuition - moving triangles, aligning their sides, etc. Those are not analytic. They are intuitional.
  • fishfry
    1.8k
    I say that the Pythagorean theorem applies to affine spaces over inner product spaces,simeonz

    That sounds right. I don't know much about affine spaces. But basically an affine space is a vector space that's "forgotten its origin" and you don't need any privileged origin to have the Pythagorean theorem be true.

    But you are saying this as if someone is denying it. I don't think anyone is denying that the Pythagorean theorem is false in affine spaces. Help me understand what is the point of the thread. I don't think anyone disagrees with what you said here.

    I don't see it that way really. We still come from the geometric perspective, to define angles and distances in one way or another, and only then we have the privilege of calling an n-tuple of points being from a Cartesian coordinate system.simeonz

    No that's not true. We define as the set of ordered pairs of real numbers. Then we define the usual Euclidean Euclidean distance, and we define the usual dot product. Then the angle between two vectors is the arccosine of the dot product of their normalized versions. That is,

    so that can be defined as . I assume you agree. And we can even formalize the arccosine by defining the cosine as the real part of the complex exponential function, and the arccos as its inverse. All this can be done without reference to geometry and we can even define angles without geometry. I'm guess you know this but disagree for some reason?


    Cartesian coordinate systems come with semantics that need to be defined apriori. They are not just mechanical assignment of pairs of numbers to some arbitrary point space.simeonz

    No they don't and yes they are. You just define Euclidean n-space or in general you can define an abstract inner product space and everything works out fine without geometric semantics. For example if instead of Euclidean n-space we can work in generalize inner product spaces and all the theorems carry over directly. I can't see the point of objecting to this but maybe I'm misunderstanding you.

    https://en.wikipedia.org/wiki/Inner_product_space

    There is only one sense, in fact, in which I am not correct. And it is that a Cartesian coordinate system might be a applied to the very n-tuples, with vectors being n-tuples, distances and angles computed in the usual way, etc. But then, we couldn't talk about apples and dollars, because since the underlying point space is just a mechanical bonding of numbers, it is unitless.simeonz

    There's no Cartesian coordinate system in an inner product space but there is a notion of an orthonormal basis. That's Fourier series, functional analysis, and quantum physics based on Hilbert space. All this is standard. I don't follow your point.
  • fishfry
    1.8k
    Those are not analytic. They are intuitional.simeonz

    Well those are not mutually exclusive. Of course we use geometric intuition to get the analytic approach off the ground, but that's true of everything. Is that what you're saying, that we need the ancient geometric intuition to ground the modern analytic approach? Perfectly well agreed. But again, what of it? Nobody's disagreeing.

    ps -- I see that you're not the OP. I should quit while I'm behind here. What I know about all this is that inner product spaces are a vast abstraction of ancient Euclidean geometry. But who would disagree? The law of cosines was known to Euclid and is the same concept as the dot product.
  • simeonz
    257
    No that's not true. We define R2 as the set of ordered pairs of real numbers. Then we define the usual Euclidean Euclidean distance, and we define the usual dot product.fishfry
    I still want to be certain that you concur with me on the definition of Cartesian coordinate systems.

    A Cartesian coordinate system is an assignment of n-tuples to the points in a point space underlying Euclidean space, such that the dot product between the n-tuples is isomorphic to the inner product between the displacement vectors of the points from the origin. And Euclidean space is indeed a special case of affine space (something I blurred over a little here), such that the vector space is inner product space and the field of the inner product space is ordered and complete (or which would be the same - the real numbers). Other then that, the point space and vector spaces are arbitrary. That means that we have already defined an inner product (not just dot product for the n-tuples) somewhere. And we can't merely define the dot product over the n-tuples and map the n-tuples to points (mind you, by points I don't even mean locations, but objects that conform to the requirements), or we would have what appears to be considered "generalized" version of the Cartesian coordinate system. There is no geometric sense in it. It is more or less, assignment of n-tuples to points, which can be manipulated with arithmetic.

    As I said, we can also use n-tuples as the underlying inner product space and n-tuples as the underlying point space, but then we cannot talk about dollars and apples, or sensible angles and distances, because those are mechanical constructions now.
  • simeonz
    257
    I assume you agree.fishfry
    I agree with the fact that we can define the dot product as you specify, but we need inner product as well, or we are just manipulating unitless numbers that don't correspond to anything.

    Is that what you're saying, that we need the ancient geometric intuition to ground the modern analytic approach?fishfry
    To some extent. But I was saying that there is one more hop (probably) in my mind to how this intuition translates to Cartesian coordinates. We first justify the requirements of the affine spaces with the constructive proofs, such as the properties of the inner product in the inner product space. Then we assign n-tuples to the points in the point space, proving that we preserve the inner product with the dot product. Since, in the OP's question there is no inner product, just dot product, there is nothing to preserve and no Pythagorean semantics to be had. We either have arbitrary assignment of numbers to points somewhere, in some semantic domain, or we work with numbers as our semantic domain, and those numbers have no units.
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