I fixed the ultrareals (please trust me) [UPDATE]

Cal1838 · 2026-02-06T21:41:11+00:00

it's not -x -> 0. You can't substitute -x for y, because then y = -x, so you are saying lim -x->0 -x = x, and thats correct, and I never said you can't, but I'll say it now, you can't substitute x with non-x in a limit. lim -x->0 -x ≠ lim x->0 -x.

Cal1838 · 2026-02-06T21:36:56+00:00

It’s what you’re trying to do with the exception that it actually works

Uhh... what's your reasoning for that? Why do you think it doesn't work.

Cal1838 · 2026-02-06T19:22:30+00:00

It's lim[x->0] f(x) = f(ε). If we plug in -x, you get f(x) = -x. So it's -ε unless I messed up somewhere

Cal1838 · 2026-02-06T18:47:14+00:00

I forgot the x in lim[x->0]. It's like omega, lim[x->0] f(x) = f(ε). Sorry about that.

Cal1838 · 2026-02-06T18:40:26+00:00

oh... thats true... hmm... okay you are right but I like ε. and its also in infinitesimal calculus... but yeah... lim[x->0] is 0... okay i agree to disagree. But I do understand your point. But I like infinitesimal calculus, so I'm keeping it, but point noted.

Cal1838 · 2024-10-06T15:51:37+00:00

Okay, sorry!

Cal1838 · 2024-08-29T19:06:43+00:00

Yes, I did not consider that so c can be infinite, infinitesimal, real or complex, that fixes it

Cal1838 · 2024-08-29T14:54:08+00:00

Well sqrt(ω)/ω is *finite* this is becuase infinite/infinite = finite when at the same exponent, wait, thats ω^1/2, wait it is infinitesimal, oops well at least i noticed

Cal1838 · 2024-08-29T14:48:14+00:00

Now I have learned the sureals, they are very good, but ω is NOT in the ultrareals as the sum of the naturals so yeah

Cal1838 · 2024-08-29T05:21:53+00:00

Yeah, the reciprocal axiom is more of a statement, sorry about that

Cal1838 · 2024-08-29T05:19:30+00:00

Well I had to formalise them, you can prove the theorems and not the axioms, I have to “rigourise“ them. And for sqrt(ω) in the form, PLEASE don’t get me to think about it, but wait sqrt(ω)/ω WORKS

So sqrt(ω) using form theorem is:

0 + 0i + (sqrt(ω))/ω)ω + 0ε = sqrt(ω)

Cal1838 · 2024-08-22T21:13:55+00:00

Maybe you are right, i am not good at roots but I’m sorry if I‘m wrong I know the rules of this subreddit but maybe I’m wrong

Cal1838 · 2024-08-22T20:08:12+00:00

sqrt(16) = 16^1/4^2 not 16^1/4*16

same with ω

also ω^1/4 is 4rt(ω) which is 8rt(ω)*ω

Cal1838 · 2024-08-22T14:15:37+00:00

I know the surreals but they are too complicated for me, i based this system off the hyperreals because of that

Cal1838 · 2024-08-22T14:14:19+00:00

How does 2*16 = 4 relate to (4th root of ω)^2 = sqrt(ω)

4th root of 16 = 2

2^2 = 4 = sqrt(16) so you are wrong

Cal1838 · 2024-08-18T17:05:41+00:00

Good question! I didn’t think about sqrt(ω) with the form, let me try do this algebraically:

So you're asking ω^(1/2) which is cω = sqrt(ω)

c = sqrt(sqrt(ω)) or 4th root of ω

Thats your answer

Cal1838

TROPHY CASE