Proofs
Let \((a, k) \in (\mathbb{Z}^*)^2 \) be two non-zero integers, and \(b \in \mathbb{Z} \) an integer.
If \( ka \mid kb \), then:
$$ kb = kak' \Longleftrightarrow b = ak'$$
So, \(a \mid b \).
$$ \forall (a, k) \in (\mathbb{Z}^*)^2, \enspace \forall b \in \mathbb{Z}, $$
$$ ka \mid kb \hspace{0.2em} \Longrightarrow \hspace{0.2em} a \mid b $$
Let \((a, b) \in (\mathbb{Z}^*)^2 \) be two non-zero integers, and \(c \in \mathbb{Z} \) an integer.
If \( a \mid b \) and \( b \mid c \), then:
$$ \exists (k, k') \in \mathbb{Z}^2, \enspace \begin{cases} b = ka \\ c = k'b \end{cases} $$
So,
$$ c = \hspace{0.2em} \underbrace{kk'} _{ \in \hspace{0.1em} \mathbb{Z} } a $$
Therefore \( a \mid c \). We definitely have:
$$ \forall (a, b) \in (\mathbb{Z}^*)^2, \enspace \forall c \in \mathbb{Z}, $$
$$ (a \mid b) \text{ and } (b \mid c) \hspace{0.2em} \Longrightarrow \hspace{0.2em} a \mid c $$
Let \(a \in \mathbb{Z}^* \) be a non-zero integer, and \((b , c) \in \hspace{0.04em}\mathbb{Z}^2 \) two integers.
If \( a \mid b \) and \( a \mid c \), then:
$$ \exists (k, k') \in \mathbb{Z}^2, \enspace \begin{cases} b = ka \\ c = k'a \end{cases}$$
So,
$$ b + c = \hspace{0.2em} \underbrace{(k +k')} _{ \in \hspace{0.1em} \mathbb{Z} } a $$
Therefore \( a \mid (b + c) \). We definitely have:
$$ \forall a \in (\mathbb{Z}^*), \enspace \forall (b , c) \in \hspace{0.04em}\mathbb{Z}^2, $$
$$ (a \mid b) \text{ and } (a \mid c) \hspace{0.2em} \Longrightarrow \hspace{0.2em} a \mid (b + c) $$
We will also have, by extension:
$$ (a \mid b) \text{ and } (a \mid c) \hspace{0.2em} \Longrightarrow \hspace{0.2em} a \mid (b - c) $$
Let \(a \in \mathbb{Z}^* \) be a non-zero integer, \((b , c) \in \hspace{0.04em}\mathbb{Z}^2 \) two integers.
If \( a \mid b \) and \( a \mid c \), then:
$$ \exists (k, k') \in \mathbb{Z}^2, \enspace \begin{cases} b = ka \\ c = k'a \end{cases}$$
Moreover,
$$ \forall (u , v) \in \hspace{0.04em}\mathbb{Z}^2, \enspace \begin{cases} ub = uka \\ vc = vk'a \end{cases}$$
So,
$$ ub + vc = \hspace{0.2em} \underbrace{(uk + vk')} _{ \in \hspace{0.1em} \mathbb{Z} } a $$
Therefore \( a \mid (ub + vc) \). We definitely have:
$$ \forall a \in (\mathbb{Z}^*), \enspace \forall (b , c) \in \hspace{0.04em}\mathbb{Z}^2, \enspace \forall (u , v) \in \hspace{0.04em}\mathbb{Z}^2, $$
$$ (a \mid b) \text{ and } (a \mid c) \hspace{0.2em} \Longrightarrow \hspace{0.2em} a \mid (ub + vc) $$
We will say that \( a \) divides all linear combinations of \( b \) and \( c \).
Index
Back to top