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Le 02/04/2024 à 17:51, Jim Burns a écrit :Your 'trick' only fails to demonstrate a bijection. Failing to demonstrate a bijection does not mean that there is no bijection, only that your 'trick' doesn't work to that end.On 4/2/2024 3:36 AM, WM wrote:>If your assumption leads to "no bijection",>
but there is a bijection,
then your assumption is wrong.
My trick proves that there is no bijection.
Or could you explain why first bijecting n and n/1 should destroy an existing bijection?
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