Demostración

Si $a \mid b$, existe $q\in \mathbb{Z}$ tal que $b=aq$.

  • $b=aq \Rightarrow b=-a(-q)$, con $-q\in \mathbb{Z}$, y entonces $-a \mid b$.
  • $b=aq \Rightarrow -b=a(-q)$, con $-q\in \mathbb{Z}$, y entonces $a \mid -b$.
  • $b=aq \Rightarrow -b=(-a)q$, entonces $-a \mid -b$.
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