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Le 29/08/2024 à 13:00, FromTheRafters a écrit :By induction, they all are.
He seems to think that n being some undisclosed (dark) natural numberNo. If n is a visible nabtuarl number, then also n^n^n is also a visible
means that n+1 is merely another undisclosed (dark) natural number and
you cannot pair (thinking of matching,
natural number. That is potential infinity.
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