• Daemon Silverstein@thelemmy.club
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    19 hours ago

    Actually, the hotel manager could relocate guests from each Nth room to the (2×N)th (every even-numbered room), as there are infinity rooms. This way, there’ll be as infinity free odd-numbered rooms as there are infinity booked guests. Sisyphus can then choose any odd room for himself and another for his boulder.

    • SkyeStarfall
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      7 hours ago

      Person in room 4.3*10^53 when asked to move 4.3*10^53 rooms be like