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Am 06.08.2024 um 22:21 schrieb Chris M. Thomasson:I say even? Well, both (odd and even) at the same time? This makes me think of signed zero...On 8/6/2024 6:33 AM, Moebius wrote:Yes. An integer z is even, iff there is am integer k such that z = 2k.Am 06.08.2024 um 04:24 schrieb Chris M. Thomasson:>
>There are infinite[ly many] even[ numbers] and>
there are infinite[ly many] odd[ numbers].
On the other hand, some even numbers are odd. :-)
;^D 666?
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Ahh zero. I think its even... ;^)
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1, 2, 3, 4, ...
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odd, even, odd, even, ...
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So, the pattern:
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-2, -1, 0, 1, 2
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even, odd, (even), odd, even
Clearly for z = 0 there is an integer k (namely 0) such that z = 2k.
In other words, an integer z is even if it can be deviede by 2 "without a remainder =/= 0".
Clearly 0 can be devided by 2 "without a remainder =/= 0": 0 / 2 = 0.
So 0 is an odd even number? :-)
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