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4th and higher dimension perpendicular

  1. Jul 30, 2012 #1
    For an arbitrary distance the equation is:

    [itex]\sqrt{\Sigma_{i}^{n}x_{i}^{2}}[/itex]

    I would like to know what are the proofs for higher dimensions being perpendicular to our 3-spaital dimensions. If I am wrong in any way, please elaborate.

    I guess what I'm saying is since:

    r[itex]^{2}[/itex]=x[itex]^{2}[/itex]+y[itex]^{2}[/itex]

    and

    r[itex]^{2}[/itex]=x[itex]^{2}[/itex]+y[itex]^{2}[/itex]+z[itex]^{2}[/itex]

    What leads us to think that the higher dimensions continue in this manner?
     
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  3. Jul 30, 2012 #2

    micromass

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    Why can't we continue that way? It's just a definition we make and it turns out to be extremely useful.

    Sure, you could say that the distance [itex]\sqrt{x^2+y^2+z^2}[/itex] corresponds to an actual distance in 3D. What makes us think that [itex]\sqrt{x^2+y^2+z^2+t^2}[/itex] corresponds to an actual distance?? Well, it doesn't. There is not such thing as an actual 4 dimensional space. Sure, a 4-dimensional space is used in relativity, but these are all mathematical models. Nothing says that these models really exist.
     
  4. Jul 30, 2012 #3

    Ibix

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    They don't unless you define them to do so. The regular geometry you do in school is called Euclidean geometry, and works in the manner you describe. It can extend to higher dimensions because why not?

    Euclidean geometry doesn't accurately model the real world, though. Anywhere where gravity can be ignored the real world follows Minkowski geometry, and the distance between two points is S2=(ct)2-(x2+y2+z2). This is similar to Euclid, but different enough to make Relativity Theory interesting. Life gets more complex when gravity is around because space-time is curved and the shortest distance between two points isn't necessarily a straight line. You have to write a more general equation than Pythagoras to deal with that.

    There is no mathematical reason to extend geometry to higher dimensions in any particular way. There is only what is appropriate to the problem at hand. Look up measure theory if you want a headache. :smile:
     
  5. Jul 30, 2012 #4

    HallsofIvy

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    You seem to be confusing mathematics with physics. "3-spatial dimensions" exist in physics and have nothing to do with Linear Algebra. "Dimensions" of a linear vector space are entirely a matter of the size of a basis. We cannot even define "perpendicular" until we define an inner product which, in a finite dimensional vector space, depends upon the choice of basis. In an n-dimensional vector space, given a basis, we can define an inner product in which that basis is "orthonormal" (all basis vectors have length 1 and are perpendicular to each other). Conversely, given an inner product, we can define a basis that is orthonormal using that inner product.
     
  6. Aug 1, 2012 #5

    chiro

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    One way to think about the generalization is look at how the 3D case is derived using the 2D formula, and then to use the same thing for deriving the n+1th dimension formula given a proof for the nth dimensional formula.

    In other words, use what is called mathematical induction. (You also assume the right-angled property of each dimension and pythagoras' theorem requires it because it deals with right-angled triangles).

    As for the inner-product, you just use the ordinary cosine rule where you replace the lengths of the sides with the norms (you proved the norm case above) and you get the result for the inner-product for R^n.
     
  7. Aug 1, 2012 #6

    lavinia

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    If you use the Pythagorean theorem on the right triangle whose height is the length of the 4'th coordinate and whose base is its radial distance from the origin in the first three coordinates (three dimensional space), the formula follows.
     
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