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)inW
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)
calpha=(cx,cy,cz)
so calpha in W

Hence, W <=V

1 Answer
Oct 22, 2017

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.