If 3D is a projection on 2D n 2D is a projection on 1D, can we say that 4D is a projection on 3D?
95316
I believe yes.. It makes sense, You see all of these dimensions are just directions. If you say that the four dimensions are represented by X,Y,Z,T these are merely directions of dimensions then all of those are travellable (meaning you can travel in those directions). You can go to left & Right (X), You can go Forward & Backward (Y), you can go Up or Down (Z). The questions now is how do you travel through time. According to Albert Einstein if there is anything that can go faster than the speed of light then it will be possible to travel through time. It is actually true becuase our mind can go faster that speed of light that whay our mind can travel through time, sometimes we remember things in our past that the way our mind travels.
110094
Yes, definitely multiple postings... \n\nwell... a fourth dimension is time... so anything that changes over time is a fourth dimensional object. But that isn't what you are really asking.\n\nWe can only perceive four dimensions with our senses, so perceiving it is difficult. To us it would look as if it were a regular three-dimensional object, perhaps changing over time in some subtle way. Jack L. Chalker did a good job of describing such a structure in his "Quintarra Marathon" books. This is similar in idea to how you would describe a three-dimensional object such as a pencil to a person living in a two-dimensional world, at best they could only see a cross-section of the pencil, or if the world they were living on were curved they would see multiple cross-sections and not realize they were the same object (put a pencil through a folded piece of paper).\n\nIf you are looking for a way to visualize it to help with problem solving I recommend thinking about it in terms of computer arrays, or a group of three dimensional objects, each can be looked at seperately.\n\nGood luck with your conundrum :-)
113366
The 4th dimension is time. It is essentially equivalent to "flat land" falling through the 3rd dimension.
158992
sorry i must really correct the first answer before i continue;living in a 10 dimensional world is just a theory,known as string theory,and that is all it is,there is no proof\n\n(and the theory ACTULLY states that there are 11 dimensions)\n\na 2d object could not even physically exist because even if you had the smallest possible object that could exist indepdantly(an atom) it still has dimensions of length width and height,however small it may be it is not 2d\n\nIn writing the answer im kinda curious as to Could you theoretically stack and infinate number of 2d objects on top of each other and it still remain 2d?? anyone know?
845473
decagon - 2 d figure\nor \ndecahedron - 3 d figure
1043313
It's too cumbersome to rewrite it all here, but check out http://www.euclideanspace.com/maths/algebra/vectors/angleBetween/index.htm\n\nThat's 2D, but the formula is the same for 3D.
903825
~ you need to say what are V and W??\nI am guessing that V and W are vector spaces and that T is  LINEAR transformation from V --> W.\n\nsuppose  dim(V) < dim(W), take a basis of W: w1, ... ,wn, and assume that T is onto, then for every wi there exist a vi such that\nT(v1)=w1, ... , T(vn)=wn,\nnow let us check if v1,... vn are linearly independent:\ntake a1v1 + ... + anvn =0 a linear combination equal to cero, then\nT ( a1v1 + ... + anvn  ) = T(0) =0,\non the other hand,\nT ( a1v1 + ... + anvn  )  = a1T(v1) + ...+ anT(vn)\n= a1w1 + ... + anwn =0, since w1,..., wn are a basis, that means that\na1=a2=..=an=0\nbut this implies that v1, ... vn are linearly independent.\nand this is a contradiction to the fact that dim V < dim W\n\n\nfor the other problem:\nif dim(V) > dim(W), prove that T cannot be one to one (injective).\n\nAssume that v1,..., vm form a basis for V,\nif Ker(T) =0 (i.e. if we assume that T is 1-1), then\nT(v1), ... , T(vm) are a linearly independent set of vectors, but\nm=dim V > dim (W), but this cannot happen, since any set of linearly independent vectors in W would have to have fewer than \ndim(W) vectors.
949904
I suppose that it means all the lines that you can draw from every vertex to every other vertex.  For a rectangle it would be 2.  For a 5 sided figure.( I got up and sketched it) it would b 5.  You only count a line once, not both directions.\n\nAnother sketch suggests 9 for the hexagon.  Why don't you sketch a hexagon, draw and count the lines between vertices.
95192
In normal 3-D space, a cube can be described by {(x,y,z) where 0<=x<=1, 0<=y<=1, 0<=z<=1}\n\nIn normal 4-D space, it's {(x,y,z,w) where 0<=x<=1, 0<=y<=1, 0<=z<=1, 0<=w<=1}\n\nIn n-space, an n-cube can be represented by all n-dimentional vectors where the nth coordinate lies in the closed interval [0, 1].
375748
All squares are rectangles,\nall parallelograms are quadrilaterls\nand all squares are parallelograms
534038
That question is hard! Nobody has ever been there. Or maybe we all are always there. I don't know. But the different dimensions tell you a position and it is getting more exact with every dimension:\n1D: where x is\n2D: where x and y are\n3D: where x and y and z are\n4D: where x and y and z are and when they are there\n5D: I don't know, but you could think of something that would give a more exact description, such as (this is just an idea that went through my head, not anything from a scientist!):\n5D: where x and y and z are and when they are there and how they are there and then\nDo I make sense? Don't hate me if I don't!
396588
If one considers C as a 2-dimensional vector space over R (this is the complex plane representation of complex numbers), then it is trivially isomorphic to R^2 (the real plane) by mapping the complex number (a, b) to the real pair (a, b). The vector space structure is trivially preserved (with respect to real-linear morphisms). If it was not isomorphic, there would be no complex plane representation and we'd be stuck working with an exotic 1-dimensional structure. See http://mathworld.wolfram.com/ComplexPlane.html and http://en.wikipedia.org/wiki/Complex_numbers#Real_vector_space .
891091
Yes, It is possible if we use x,y,z plane in plotting graphs. When using this method we should be very precise.\n\n3D model gives very spectacular graphs. There is algebraic expression for many 3D shapes e.g sphere, cones
799446
electrons can't line up in a plane they wouldn't all fit... unless it was like a big 747... or one of those military airbus things... and by 3d do you mean 2 spacial dimentions + time or 3 spacial dimensions? cause we already live with 3 spacial dmension (4 including time)
1085259
You need to get a little more specific in terms of what kind of "3d equivalent" you want.  Two kinds of 3d solids you can form easily from 2d polygons (closed shapes made of line segments) are prisms and pyramids.  You make a prism of any 2d shape by taking 2 copies of the flat shape, one above the other, and connecting the corresponding vertices (corners) with edges (lines).  From this perspective, a cube, is a square prism.  A triangular prism looks a little like a tent, with two triangles (top and bottom) and three rectangles or squares around the sides.  To make a pyramid, choose any 2d shape as the base, and then a point above the base as the top, and radiate edges down from the top to all the base's corners.  There are other "3d equivalents", so if you're thinking of something else, let me/us know.\n\nYou also need to be more specific about what kind(s) of sides interest you: edges (1d lines) or faces (2d flat surfaces).  The "sides" of a 2d shape are all edges, but 3d shapes like cubes have both edges and faces (a cube has 12 edges and 6 faces).  Any prism made from a polygon with N "sides" (edges) will have 3N edges (N on the top, N on the bottom, and N more connecting top and bottom) and N+2 faces (N around the sides plus top and bottom).  Any pyramid made from a polygon with N "sides" (edges) will have 2N edges (N around the base and N more reaching up to the top point) and N+1 faces (N around the sides plus the base).\n\nDoes this do it?
191182
I am not sure about life more than 4 dimensions. Why? Is this funny for you?
1155114
We live in a universe of observable/measurable quantities. For example, length, width, height, time, mass, etc.  If different quantities cannot be expressed in terms of the other quantities, they are "orthogonal" to each other - you can think of them as being separate dimensions. For example, on an X-Y graph, the x-dimension is independent of the y-dimension because you cannot express a "y" quantity as a combination of "x" quantities. \n\nSpace has 3 dimensions (length, width, height); time is a fourth dimension that cannot be expressed in terms of the other three.\n\nAn n-dimensional space is just a space that is characterized by "n" orthogonal quantities - quantities that cannot be expressed in terms of each other. To completely define the space/universe, you only need to specify "n" values - any more than that would be superfluous because the other values would be some combination of the other orthogonal quantities.\n\nExample: if we lived in a 4-dimensional space characterized by length, width, height, and time, all you need to do is specify those 4 quantities to uniquely determine any point in that space.  If we add another quantity (such as temperature) that cannot be expressed in terms of the other 4, we now have a 5-dimensional space that requires 5 quantities to be specified.