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Le 21/04/2024 à 18:32, Moebius a écrit :Not a proof, as doesn't give a n actual REASON it should be true.
Alle geraden (natürlichen) Zahlen sind natürliche Zahlen. Erzählst Du Deinen Studenten an der THA etwas anderes?No, you are right. But not all doubling results in even natural numbers.>Proof: The original set ℕ fits into (0, ω). The doubled set doesn't.
Proof: Let k in {2*n : n e IN}. Then there is an n in IN such that k = 2*n. Since IN is closed unter multiplication and 2 is in IN too, 2*n in IN and hence k in IN. Hence {2*n : n e IN} c IN. qed
Regards, WM
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