Let $S$ be the following subset of the 3-dimensional vector space $\R^3$.
\[S=\left\{ \mathbf{x}\in \R^3 \quad \middle| \quad \mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}, x_1, x_2, x_3 \in \Z \right\}, \]
where $\Z$ is the set of all integers.
Determine whether $S$ is a subspace of $\R^3$.

We claim that $S$ is not a subspace of $\R^3$.
If $S$ is a subspace, then $S$ is closed under scalar multiplication.
But this is not the case for $S$.

For example, consider $\mathbf{x}=\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}$.
Since all entries are integers, this is an element of $S$.
Let us compute the scalar multiplication of this vector $\mathbf{x}$ and the scalar $1/2 \in \R$. We have
\[\frac{1}{2}\cdot \mathbf{x}= \frac{1}{2}\cdot\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}=\begin{bmatrix}
1/2 \\
1/2 \\
1/2
\end{bmatrix}.\]
Since $1/2$ is not an integer, the scalar multiplication $\frac{1}{2}\mathbf{x}$ is not in $S$. Therefore the subset $S$ cannot be a subspace of $R^3$.

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Proof.
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x_1 \\
x_2 \\
x_3
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x_2 \\
x_3 \\
x_4
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x_2 \\
x_3
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