# When defining a vector space is the scalar part of the field or always a real number

I've stumbled upon an exercise that takest the set of integers $$\Bbb{Z}$$, defines addition and multiplication as usual but scalar multiplication as $$\lfloor{\alpha}\rfloor * k$$, where $$\alpha$$ is the scalar and $$k$$ the element of the vector space and proceeds to claim that this set is not a vector space.

Wikipedia says the scalar is in field $$\Bbb{F}$$:

In the list below, let u, v and w be arbitrary vectors in V, and a and b scalars in F.


which in this case is the integers $$\Bbb{Z}$$?

However, the solution to the exercise only makes sense if the scalar is in $$\Bbb{R}$$. What am I misunderstanding?

In your case, $$\mathbb Z$$ plays the role of the set $$V$$ defined in Wikipedia article.

And you're right, the scalar multiplication defined in your exercise makes sense if the field $$\mathbb F$$ is the field of the reals $$\mathbb R$$.

So here a vector is just an integer in $$\mathbb Z$$.

• I get it now. To solve this exercise I must assume the field $\Bbb{R}$ even though it is not given and I mistook $\Bbb{Z}$ for the field $F$ instead of set $V$. Thank you. Jun 30, 2020 at 8:05
• That is correct! Jun 30, 2020 at 8:11
• @koral the implied field could also be $\mathbb Q$, but that still doesn't make a vector space for the same reason. Jun 30, 2020 at 8:18
• @EspeciallyLime the set not being a vector space is the desired outcome acording to the exercise Jun 30, 2020 at 8:26
• @koral And this comes, for example, from the fact that $0 = \lfloor 1/2 \rfloor . 1 +\lfloor 1/2 \rfloor . 1 \neq \lfloor 1/2 +1/2 \rfloor . 1 =1$ Jun 30, 2020 at 8:29

Let $$E$$ be an abelian group and $$\mathbb{k}$$ be a field. Let $$\left(\lambda,v\right)\in \mathbb{k}\times E \mapsto \lambda\cdot v \in E$$ be a function. We say that $$(E,~\cdot~)$$ is a vector space if $$\forall \lambda, \mu, v,w$$ \begin{align} \lambda\cdot(\mu \cdot v) &=(\lambda\mu)\cdot v & 1_{\mathbb{k}}\cdot v &= v \\ \lambda\cdot(v+w)&= \lambda\cdot v + \lambda\cdot w &(\lambda+\mu)\cdot v &=\lambda\cdot v + \mu\cdot v \end{align} These four assumptions just say that the group structure of $$E$$ and the field structure of $$\mathbb{k}$$ are compatible.

In your question, $$E = \mathbb{Z}$$ is an abelian group and $$\mathbb{k}=\mathbb{R}$$ is a field, and $$\lambda \cdot v = \lfloor{\lambda}\rfloor v$$. To show that $$(\mathbb{Z}, ~\cdot~)$$ is not a vector space over $$\mathbb{R}$$, you can show that one of the four assumptions above is not respected.

For example, let $$v = 1$$ and $$\lambda = \mu = \frac12$$. Then : \begin{align} \left(\lambda + \mu\right) \cdot v = (1)\cdot 1 = \lfloor 1\rfloor \times 1 &= 1 \\ \lambda\cdot v + \mu \cdot v = \lfloor \frac12 \rfloor\times 1 +\lfloor \frac12 \rfloor\times 1 &= 0 \end{align} so for this particular choice of $$\lambda,\mu,v$$, we have $$(\lambda+\mu)v \neq \lambda v + \mu v$$. Thus, $$\mathbb{Z}$$ with this operation is not a vector field over $$\mathbb{R}$$.