# Let V=RR^3 and W={(x,y,z)|x,y,z in QQ}. Is W<=V? Justify your answer.

## So far, I wrote: 1. (0,0,0)$\in$W 2.$\alpha$,$\beta$ $\in$ W $\alpha$=(x,y,z) $\beta$=(x',y'z') $\alpha , \beta$=(x+x',y+y',z+z') so $\alpha + \beta \in W$ 3. c $\in \mathbb{R}$, $\alpha \in W$ $\alpha = \left(x , y , z\right)$ c$\alpha$=(cx,cy,cz) so c$\alpha \in W$ Hence, W $\le$V

It looks like you are trying to show that $W$ is a subspace of the vector space $V = {\mathbb{R}}^{3}$, which it is not (when the scalars field is $\mathbb{R}$). Your mistake is that if $c$ is irrational, and $x , y , z \setminus \in \mathbb{Q}$ are nonzero, then $c x$, $c y$, and $c z$ will be irrational as well.
If you restricted yourself to showing that $W$ is a subgroup of the additive group $V = {\mathbb{R}}^{3}$, then your first two steps are sufficient.
For vector (linear) spaces, the problem is with the scalar multiplication. Since $c$ is an arbitrary real number, it could be irrational, which would prevent closure with respect to scalar multiplication.
On the other hand, if your field of scalars was $\mathbb{Q}$ rather than $\mathbb{R}$, then $W$ would be a subspace of $V$ since $c$ would have to be a rational number.