Is the set of integers a vector space?

My friend says it's not, because he thinks vectors have to have have at least two elements, otherwise they're scalars. I say that (Z, +, *) satisfies the axioms of a vector space and that's all you need.

What does /sci/ think?

>>19
>Fail. A norm is implicitly defined with the scalar multiplication.    If you actually need the definition of the norm, here it is: ||0||=0.  Feel better?  (Just like 17 said.)

Fail. What's the norm of the vector space over 2x2 real matrices? It's not simply implied by the scalar multiplication of the field.

>>17
The "obvious" norm does not cut it when you're calling it a vector space. If you called it "the normed vector space {0}", then you could use the usual norm without having to define it.

I know I'm nitpicking, but this shit is important. I've failed lots of tests because I forgot to write stupid shit like "it is normed" just to satisfy a bunch of axioms.

>>20
Dude, that's way, way overboard. Here's a proof: "Z is not closed under scalar multiplication with R".

Here's a counterexample: You need to be able to multiply any vector (say, 2) by any real number you want (say, pi), and get back some value in the vector space. 2pi is not in Z.