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démonstration simplification de la congruence.md

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+up:: [[congruence]] 
+title:: "démonstrations sur les simplifications possibles avec la congruence"
+#démonstration 
+
+---
+
+# Simplification totale
+
+> $ka \equiv kb [kn] \iff a \equiv b[n]$
+
+$$
+\begin{align}
+ka \equiv kb [kn] &\iff kn \mid ka - kb  \\
+&\iff \exists i \in \mathbb{Z}, ka - kb = ikn \\
+&\iff \exists i \in \mathbb{Z}, a - b = in  \\
+&\iff n|a-b  \\
+&\iff a \equiv b [n]
+\end{align}
+$$
+# Simplification partielle (sans le modulo)
+
+> $ka \equiv kb [n] \iff a \equiv b [n]$ si $\text{pgcd}(k, n) = 1$
+
+On suppose que $\mathrm{pgcd}(k, n) = 1$
+Alors :
+$$
+\begin{align}
+ka \equiv kb [n] &\iff n \mid ka - kb  \\
+% &\iff \exists i\in\mathbb{Z}, ka - kb = in  \\
+&\iff n \mid k(a-b)  \\
+&\iff n | a-b & \text{car } n \text{ et } k \text{ sont premiers entre eux, et donc n'ont aucun diviseur commun}  \\ \\
+&\iff a \equiv b [n]
+\end{align}
+$$
+
+