Say I have the set $V$ of rational numbers as vectors and the field $F$ of reals as scalars. Does $V$ form a vector space over $F$? I ask this because $V$ isn't closed under scalar multiplication.

On Wikipedia it doesn't state that being closed under scalar multiplication is an axiom, but on WolframAlpha it says "A vector space $V$ is a set that is closed under finite vector addition and scalar multiplication." If WolframAlpha were correct, then surely the rationals couldn't form a vector space over the reals?

Thanks for any replies.

  • $\begingroup$ In he Wikipedia definitions, closure is in one definition implicit ("operation") and in the other explicit. $\endgroup$ – André Nicolas Dec 24 '13 at 0:58
  • $\begingroup$ Ah I didn't see the alternative definition, thank you. $\endgroup$ – James Machin Dec 24 '13 at 5:29

The sentence 'A vector space over a field $F$ is a set $V$ together with two binary operations that satisfy the eight axioms listed below' from the Wikipedia article seems slightly vague. The scalar multiplication is not exactly a binary operation on $V$ but rather a function $\mu:F\times V\rightarrow V$, which implies that vector spaces are closed under scalar multiplication.


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