if i choose the real numbers as vector space over SOME field, and i look on the subset of the natural numbers group including zero. so according to subspaces: 1) 0 included 2) closed under addition of vectors. so, is there a Field that can make the natural numbers as a valid subspaces? (field that multipication from natural numbers with scalar is closed) or it doesnt exist

  • $\begingroup$ The only thing that comes to mind is that, since natural numbers form a monoid, one can talk about some monoid containing naturals as a submonoid; or the natural numbers can act by multiplication on some other monoid (the last written additively). $\endgroup$ – lisyarus Dec 4 '14 at 18:47

The natural numbers, with the operation of addition, do not have a vector space structure over any field. To be a vector space, you first have to be a group, and most natural numbers do not have additive inverses.

The integers (again with addition) resolve this issue, since they include the negatives, but they still do not have a vector space structure over any field. The reason is that the integers are "non-torsion", i.e. $1+1+1+1+\dots$ never equals zero. Thus the field cannot have torsion either - it must be "characteristic zero." But every characteristic zero field contains the rational numbers, and there will be no way to define multiplication of the integers by rational numbers since, e.g. there is no integer $x$ such that $5\cdot x = 2$, so there will be no good definition for $(1/5)\cdot 2$.

  • $\begingroup$ It looks like the OP asked about the natural numbers, not the integers. $\endgroup$ – Kevin Carlson Dec 3 '14 at 17:22
  • $\begingroup$ @KevinCarlson - good point; I will edit. I was thrown by the word "group." $\endgroup$ – Ben Blum-Smith Dec 4 '14 at 18:40

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