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

David Rojas

David Rojas

Answered question

2022-01-24

Let V=R3 and W={(x,y,z)x,y,zQ}. Is WV? Justify your answer.
So, I wrote:
1) (0,0,0)W
2) α,βW
α=(x,y,z),β=(x,y,z)
α,β=(x+x,y+y,z+z so α+βW
3) cR,αW
α=(x,y,z)
cα=(cx,cy,cz) so cαW
Hence, WV

Answer & Explanation

Karly Logan

Karly Logan

Beginner2022-01-25Added 11 answers

It looks like you are trying to show that W is a subspace of the vector space V=R3 , which it is not (when the scalars field is R).Your mistake is that if c is irrational, and x,y,zQ 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=R3, 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 Q rather then R, then W would be a subspace of V since c would have to be a rational number.

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