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En Garde, whole no. 16, January 1946
Page 23
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page 23. space given by (1,0,0), (0,1,0), (5,-4,0), & (6,0,0) was the same as that given by (1,0,0) & (0,1,0). This fact can be obtained from the rule we have now developed. First we eliminate (6,0,0) from the set on the basis that 6(1,0,0) - (6,0,0) = (0,0,0) from the resulting set of three vectors we drop (5,-4,0) because -5(1,0,0) + 4(0,1,0) + (5,-4,0) - (0,0,0) and we are left with (1,0,0) & (0,1,0), as observed before. Now at last we are ready to define dimension. It's a beautifully simple definition, checks easily with our intuitive ideas, and is useful in situations where the latter would be entirely useless. Definition: the dimension of space is the smallest number of vectors which can determine the space. We compare this with our intuitive notions by considering 3-vectors, just as we have done with other definitions. This time, however, it's no trouble at all. First is an 0-space determined by a set having no vectors at all? Well, the onlyl 0-space or point we can have in vector space is the vector that doesn't go anywhere, vector 0; and we have seen that the space containing only vector O is determined and we have seen that the space containing only vector 0 is determined by no vectors. Second, is a 1-space determined by a set having one vector? Sure: a set of collinear vectors determine a line, as we saw while testing the definition of "determine" and it's obvious that all but one of a set of colinear vectors can be dropped, by linear dependence. Thirs, is a 2-space determined by a set having two vectors and no fewer? We know that two is the smallest number of vectors that can be coplanar without being colinear; so score another for our side. Last, is a 3-space determined by a set having three factors and no fewer? Once again, three is the smallest number of vectors you can have that won't necessarily be coplanar. And there we are. The geometrical nature of the checks we have just made is notheing against them, since we have already checked our definition of "determine" against geometry. IV This section will be a summary of what we've achieved and an abstract of what is yet to come. We've defined dimension. That's the main thing. Assuming nothing about 4-space except that it behaves just like our 3-space, and that no particular direction in it is singled out by any physical peculiarity as the 4th dimension, we can tell a hell of a lot about it. Imagine yourself a 4-man (I do not mean a supervisor in a mill). Imagine yourself, further, a 4-astronomer. Now since there are four dimensions in your space, the coordinate system you will use to describe the positions in and velocities of your heavenly 4-5 bodies will have four mutually perpendicular axes. (This seems obvious, but if you're a fiend for mathematical precision you may want to check it by the definition of "dimension" and the other results of sections II & III.) Suppose that one day, peering through
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page 23. space given by (1,0,0), (0,1,0), (5,-4,0), & (6,0,0) was the same as that given by (1,0,0) & (0,1,0). This fact can be obtained from the rule we have now developed. First we eliminate (6,0,0) from the set on the basis that 6(1,0,0) - (6,0,0) = (0,0,0) from the resulting set of three vectors we drop (5,-4,0) because -5(1,0,0) + 4(0,1,0) + (5,-4,0) - (0,0,0) and we are left with (1,0,0) & (0,1,0), as observed before. Now at last we are ready to define dimension. It's a beautifully simple definition, checks easily with our intuitive ideas, and is useful in situations where the latter would be entirely useless. Definition: the dimension of space is the smallest number of vectors which can determine the space. We compare this with our intuitive notions by considering 3-vectors, just as we have done with other definitions. This time, however, it's no trouble at all. First is an 0-space determined by a set having no vectors at all? Well, the onlyl 0-space or point we can have in vector space is the vector that doesn't go anywhere, vector 0; and we have seen that the space containing only vector O is determined and we have seen that the space containing only vector 0 is determined by no vectors. Second, is a 1-space determined by a set having one vector? Sure: a set of collinear vectors determine a line, as we saw while testing the definition of "determine" and it's obvious that all but one of a set of colinear vectors can be dropped, by linear dependence. Thirs, is a 2-space determined by a set having two vectors and no fewer? We know that two is the smallest number of vectors that can be coplanar without being colinear; so score another for our side. Last, is a 3-space determined by a set having three factors and no fewer? Once again, three is the smallest number of vectors you can have that won't necessarily be coplanar. And there we are. The geometrical nature of the checks we have just made is notheing against them, since we have already checked our definition of "determine" against geometry. IV This section will be a summary of what we've achieved and an abstract of what is yet to come. We've defined dimension. That's the main thing. Assuming nothing about 4-space except that it behaves just like our 3-space, and that no particular direction in it is singled out by any physical peculiarity as the 4th dimension, we can tell a hell of a lot about it. Imagine yourself a 4-man (I do not mean a supervisor in a mill). Imagine yourself, further, a 4-astronomer. Now since there are four dimensions in your space, the coordinate system you will use to describe the positions in and velocities of your heavenly 4-5 bodies will have four mutually perpendicular axes. (This seems obvious, but if you're a fiend for mathematical precision you may want to check it by the definition of "dimension" and the other results of sections II & III.) Suppose that one day, peering through
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