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1. MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:30:42 JST MercurialBlack
   I've wasted an hour and a half trying to figure out how to prove that in the real numbers, x <= x + 1. There is no hope for me.
   
   Ultimately trying to prove monotonicity. All I have to work with is the definition of a preorder (reflexive and transitive), definition of a monotonicity and a monotic map, and definition of real numbers. I think I'm missing something here.
   • meso (meso@the.asbestos.cafe)'s status on Monday, 12-Jun-2023 10:30:53 JST meso

     in reply to
     @MercurialBlack -1 <= -1 + 1
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:30:54 JST MercurialBlack

     in reply to

     • meso
     @meso What are you getting confused about both your statements are true
   • meso (meso@the.asbestos.cafe)'s status on Monday, 12-Jun-2023 10:30:55 JST meso

     in reply to

     • meso
     @MercurialBlack wait no dawg 10 <= 10 + 1
   • meso (meso@the.asbestos.cafe)'s status on Monday, 12-Jun-2023 10:30:55 JST meso

     in reply to

     • meso
     @MercurialBlack what
     Machismo repeated this.
   • mist (ai@cawfee.club)'s status on Monday, 12-Jun-2023 10:32:48 JST mist

     in reply to
     @MercurialBlack peep these axioms (defining an “ordered field”)
     Machismo repeated this.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:32:49 JST MercurialBlack

     in reply to

     • mist
     @ai
     Machismo repeated this.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:33:35 JST MercurialBlack

     in reply to

     • mist
     @ai trash. Do I need to use field properties to get there?
     Machismo repeated this.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:33:37 JST MercurialBlack

     in reply to

     • mist
     @ai I'm at the point where if its beyond Tarski–Grothendieck then I don't accept it as an axiom I am broken
   • mist (ai@cawfee.club)'s status on Monday, 12-Jun-2023 10:33:45 JST mist

     in reply to
     @MercurialBlack Yeah. I can define *other* total orders on the set R which do not satisfy x <= x + 1 for all x. So you need to assume that the total order you are working with satisfies some kind of compatibility with the field structure on R.
     Machismo repeated this.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:33:53 JST MercurialBlack

     in reply to

     • mist
     @ai Right, right. One such being literally just >= := <=
   • gav (gav@cawfee.club)'s status on Monday, 12-Jun-2023 10:33:59 JST gav

     in reply to

     • mist
     @ai @MercurialBlack is 0<i ? because otherwise i dont understand what point 2 is controlling for
     Machismo repeated this.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:34:21 JST MercurialBlack

     in reply to

     • mist
     • gav
     @gav @ai thing is ii = -1, iii = -i. iiii = 1. If i < 0, then i^n < 0, but this isn't the case since i^4 = 1. if 0 < i, then similarly we have the problem that i^n = -1 and 0 < -1 is a contradiction.
     
     We can define some other total orderings on C, I think, but can't use the standard one that applies to R.