$\mathbb{N}$ isn't a vector space over $\mathbb{R}$ I want to prove that $\mathbb{N}$ isn't a vector space over $\mathbb{R}$. I want to say this is true, since for instance $1/2 \cdot 1 = 1/2$ which is not in $\mathbb{N}$.  I know this isn't that simple.  I'm missing something.  Please help.
 A: $\mathbb{N}$ is countable, but a non-zero vector space over $\mathbb{R}$ is uncountable , because  $\mathbb{R}$ is uncountable  
A: What you have is fine, but here is an even stronger result.
Proposition There exists no countable nonzero $\mathbb{R}$-vector space.
Proof.
Seeking a contradiction, suppose that $V$ is a countable nonzero $\mathbb{R}$-vector space. Then there exists a nonzero $v\in V$ and $\mathrm{Span}\{v\}$ is countable. That is, there exist distinct $\lambda_1,\lambda_2\in\mathbb{R}$ such that $\lambda_1 \cdot v=\lambda_2\cdot v$. This implies $(\lambda_1-\lambda_2)\cdot v=\mathbf{0}$ and multiplying by $(\lambda_1-\lambda_2)^{-1}$ gives $v=\mathbf{0}$, a contradiction.  $\Box$
A: The clearest problem is that $(\mathbb N, +)$ isn't a group. It lacks inverses, such as $-1$. For this reason, $\mathbb N$ isn't a vector space over $\mathbb R$, or a vector space over any other field, or even a module over any ring.
I guess $\mathbb N$ is a semimodule over the semiring $\mathbb N$. That's probably the most direct analogy to a vector space that it affords.
