Question

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 RR`, `alpha in W`
`alpha=(x,y,z)`
c`alpha`=(cx,cy,cz)
so c`alpha in W`

Hence, W `<=`V

Answer 1

It looks like you are trying to show that `W` is a subspace of the vector space `V=RR^{3}`, which it is not (when the scalars field is `RR`). Your mistake is that if `c` is irrational, and `x,y,z\in QQ` are nonzero, then `cx`, `cy`, and `cz` will be irrational as well.

Explanation:

If you restricted yourself to showing that `W` is a subgroup of the additive group `V=RR^{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 `QQ` rather than `RR`, then `W` would be a subspace of `V` since `c` would have to be a rational number.