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?


2 Answers 2


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 $\mathbb Z$.

  • $\begingroup$ 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. $\endgroup$ Commented Jun 30, 2020 at 8:05
  • $\begingroup$ That is correct! $\endgroup$ Commented Jun 30, 2020 at 8:11
  • $\begingroup$ @koral the implied field could also be $\mathbb Q$, but that still doesn't make a vector space for the same reason. $\endgroup$ Commented Jun 30, 2020 at 8:18
  • $\begingroup$ @EspeciallyLime the set not being a vector space is the desired outcome acording to the exercise $\endgroup$ Commented Jun 30, 2020 at 8:26
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    $\begingroup$ @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$ $\endgroup$ Commented 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 $\mathbb{R}$.


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