FÃ¼r $ x = 1 $: $ x \mid y \iff y = k \cdot 1 = k $, also $ y $ ist beliebig teilbar â‡’ $ y \mid x $? $ y \mid 1 $ â‡’ $ y = \pm1 $. Also $ x \mid y \Rightarrow y = \pm1 $, aber $ (\pm1) \mid 1 $: wahr. Also **ja**, $ x \mid y \Rightarrow y \mid x $ gilt.

Mon, 04 May 2026 07:23:33 +0000

Content under review