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by gizmo686
2688 days ago
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Except that we have vector spaces with scalars that are not the reals all the time. For instance, consider this excerpt from the article: "Or perhaps they wouldn’t like A because the scalar field [the complex numbers] is the same as the set of vectors (unless, that is, they thought that the obvious scalars were the real numbers)." In this case, while there is an an acknowledgement that you could take the reals as your scalars, it is regarded as the secondary of the "natural" choices. Or, in my example, example, there is no way to view Q as a vector space over R, but it is clearly a vector space. There is an entire field of algebra (field theory), that relies on the fact that, for example, Q(sqrt(2)) is a 2 dimensional vectorspace over Q. |
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