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