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1. MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:30:51 JST MercurialBlack

   • meso
   @meso Well yeah both of those are true but what axiom or definition makes this true
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:31:47 JST MercurialBlack

     in reply to

     • meso
     @meso Axioms are things that are just assumed to be true, definitions are descriptions of things. I'm trying to figure out what axiom or definition makes it the case that x <= x + 1.
     
     If we just assume that x is a natural number, we can just say that x <= y if, when treated as von Neumann ordinals, every set in x is also in y, but I don't think this is standard.
     
     I think I'm missing some properties of real numbers. Something like, if y >= 0, then x + y >= x, and if y < 0, x + y < x.
     Machismo repeated this.
   • meso (meso@the.asbestos.cafe)'s status on Monday, 12-Jun-2023 10:31:48 JST meso

     in reply to
     @MercurialBlack huh
     Machismo repeated this.
   • meso (meso@the.asbestos.cafe)'s status on Monday, 12-Jun-2023 10:31:58 JST meso

     in reply to
     @MercurialBlack meds please 1 + 1 = 2 thats all
     Machismo repeated this.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:31:58 JST MercurialBlack

     in reply to

     • meso
     @meso >t. hasn't read Principia Mathematica
     
     Took mathematicians 362 pages to prove that
     
     https://en.wikipedia.org/wiki/Principia_Mathematica
   • 7666 (7666@comp.lain.la)'s status on Monday, 12-Jun-2023 10:32:02 JST 7666

     in reply to

     • meso
     @meso @MercurialBlack 1 + 1 = 11 though
     Machismo repeated this.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:32:04 JST MercurialBlack

     in reply to

     • 7666
     • meso
     @7666 @meso i love noncommutative addition
     Machismo repeated this.
   • meso (meso@the.asbestos.cafe)'s status on Monday, 12-Jun-2023 10:32:20 JST meso

     in reply to

     • 7666
     @MercurialBlack @7666 noncommutative addition fans when getting laid shows up
     Machismo repeated this.
   • 7666 (7666@comp.lain.la)'s status on Monday, 12-Jun-2023 10:32:21 JST 7666

     in reply to

     • meso
     @meso @MercurialBlack watch your fire, i specifically avoid getting laid so i can keep servers under my bed and not get stupid questions like "what's that whirring sound"
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 10:35:43 JST MercurialBlack

     in reply to

     • gav
     • meso
     @gav @meso Because we don't start with numbers we start with logic symbols, and then from there sets, and from there numbers.
     
     A lot of it is just being extremely thorough with definition.
   • gav (gav@cawfee.club)'s status on Monday, 12-Jun-2023 10:35:45 JST gav

     in reply to

     • meso
     @MercurialBlack @meso 1 + 1 = 2 is the starting assumption of numbers why would you need to prove that,thats literally what 2 means
     Machismo repeated this.
   • CEO of Monoeye Dating (ceo_of_monoeye_dating@bae.st)'s status on Monday, 12-Jun-2023 10:39:23 JST CEO of Monoeye Dating

     in reply to

     • meso
     @MercurialBlack @meso I think the thing you're missing is that 1*1 = 1 and (-1)*(-1) = 1. These are field properties, though...I don't think you need any of the rest of the structure on the reals to show this.
     
     We can get the result we want just by proving that 0<1, and I think that this is somewhat clear from the first axiom Mist's first post.
     
     You can prove this first: "For every a !=0 in R, exactly one of the two is true: a>0 or -a>0." If both of those are true (or if both of those are false) then you can show that the number in question is equal to zero, and that also follows from the first axiom in Mist's first post. (It *may* help to first prove "if a>b and c>d, then a+c > b+d")
     
     Then, it follows that either 1>0 or -1>0. Then you can use the second axiom Mist posted to show that -1>0 is false.
   • CEO of Monoeye Dating (ceo_of_monoeye_dating@bae.st)'s status on Monday, 12-Jun-2023 11:04:54 JST CEO of Monoeye Dating

     in reply to

     • meso
     @MercurialBlack @meso a>b is "a>=b and NOT b>=a." The other two are just swapping the order.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 11:04:55 JST MercurialBlack

     in reply to

     • CEO of Monoeye Dating
     • meso
     @ceo_of_monoeye_dating @meso And I mean we need to be careful since >, < and >= don't exist in this case. I think equality still does, though, since a = b iff a <= b and b <= a.
     
     [edit: typo. <= to >=]
     Machismo repeated this.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 11:04:56 JST MercurialBlack

     in reply to

     • CEO of Monoeye Dating
     • meso
     @ceo_of_monoeye_dating @meso The whole thing that sparked this was my trying to prove a>=b and c>=d, then a+c >= b+d. That is, the monotonicity of the real numbers with a preorder relation.
   • CEO of Monoeye Dating (ceo_of_monoeye_dating@bae.st)'s status on Monday, 12-Jun-2023 11:24:50 JST CEO of Monoeye Dating

     in reply to

     • meso
     @MercurialBlack @meso Which is reasonable. The first axiom says "addition doesn't dick with the order too much," and the second axiom says "the order points in the positive direction." These are natural things you'd want out of the reals.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 11:24:51 JST MercurialBlack

     in reply to

     • CEO of Monoeye Dating
     • meso
     @ceo_of_monoeye_dating @meso well yeah if we just accept the first thing as true then definitionally it's monotonic and there's nothing more to be done
     Machismo repeated this.
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 11:24:52 JST MercurialBlack

     in reply to

     • CEO of Monoeye Dating
     • meso
     @ceo_of_monoeye_dating @meso is that allowed? Manipulating both sides of a relation means applying a function which preserves the relation, so I'd think we'd need to prove that f(x) = x - b is monotonic
   • CEO of Monoeye Dating (ceo_of_monoeye_dating@bae.st)'s status on Monday, 12-Jun-2023 11:24:52 JST CEO of Monoeye Dating

     in reply to

     • meso
     @MercurialBlack @meso It's the first thing Mist posted in this thread my man.
     
     He believed in you and didn't bash you over the head with it, he thought you'd grab it yourself.
     image.png
   • CEO of Monoeye Dating (ceo_of_monoeye_dating@bae.st)'s status on Monday, 12-Jun-2023 11:24:53 JST CEO of Monoeye Dating

     in reply to

     • meso
     @MercurialBlack @meso You add -b to both sides, which you can do because it's an ordered field.
   • CEO of Monoeye Dating (ceo_of_monoeye_dating@bae.st)'s status on Monday, 12-Jun-2023 11:24:54 JST CEO of Monoeye Dating

     in reply to

     • meso
     @MercurialBlack @meso What do you think of the following?
     
     If a > b then a-b > b-b = 0.
     If c > d then c-d >0 similarly.
     
     From the two above, a-b + c-d > 0+c-d = c+d > 0
     a-b + c - d + b + d > b + d
     a + c > b + d
   • MercurialBlack (mercurialblack@pleroma.mercurial.blog)'s status on Monday, 12-Jun-2023 11:24:54 JST MercurialBlack

     in reply to

     • CEO of Monoeye Dating
     • meso
     @ceo_of_monoeye_dating @meso I don't see how a > b -> a - b > 0.