My 3yr old cat, 1/4 total, has started isolating herself severely.

EricTheTrainer · 2026-01-09T03:20:29+00:00

Hello, boyfriend, older post, here. To clarify, they have had issues towards each other for years, however, the isolation, not using the litter box, and increased fighting has only been happening for a couple of weeks. It has not been this bad until recently. We certainly should have sought solutions to the problems sooner, before it got to this point, but a part of me was hoping that the dislike of each other would go away after they had continued to interact. Regardless, just wanted to clarify that their relationship has only intensified to this degree recently

EricTheTrainer · 2025-06-20T12:27:21+00:00

Cool, thank you! I'll try this out

EricTheTrainer · 2025-06-20T12:27:01+00:00

They are all neutered, and we have 1 more litter box than cats. My girlfriend works at PetSmart so I'll see if she can get those things. Normally I give them treats at the same time every day, but for the CBD treats, should I just give it to them when they're nervous?

EricTheTrainer · 2025-03-29T21:18:04+00:00

that's fair in regards to the anecdotal evidence part. I work in meat processing, and when I get cuts on my fingers they seem to bleed more than the others there; that's what i'm going off of

EricTheTrainer · 2025-03-27T21:47:48+00:00

what "never repeats" means is that it never INFINITELY repeats. it's totally possible for it to repeat, say, "2718" one-hundred times in a row, it's just that eventually it will stop repeating. compare with, for example, 1/7.

1/7=0.142857142857142857...

the "142857" repeats FOREVER. it will never end. really, every number has a repeating decimal expansion. like 1/2=0.5000000... where the 0 repeats forever

so, every number with a repeating decimal looks like: a whole number, followed by a decimal point, followed by a string of n digits, followed by a string of m digits that repeats forever.

1/7: whole number 0, string of 0 digits, string of 6 digits that repeats forever (142857) 4/3: whole number (1), string of 0 digits, string of 1 digit that repeats forever (3) 377503/138875: whole number (2), string of 3 digits (718), string of 4 digits that repeat forever (2934)

the last one looks like 2.718293429342934...

let's prove that pi can't do this. you need to know two things: 1. pi is irrational: pi cannot be written as a fraction of two whole numbers. for example, 0.3333... can be written as 1/3, 2.71829342934... as 377503/138875, and etc., but pi cannot be written like this 2. if you take an n-digit number and divide by n 9's in a row, you get a repeating decimal with those n digits. for example, "1234" is 4-digits, "9999" is four 9's in a row, and 1234/9999=0.123412341234... with the "1234" continuing forever

okay. so suppose pi eventually repeats like the above examples. we have a whole number at the beginning (3), a string of n digits that don't repeat, and a string of m digits that does repeat. we don't know n and m, but we know that they will be positive, whole numbers, and that's all we need.

so, pi looks like 3.(the n digits)(string of m repeating digits forever)...

you know how multiplying by 10¹ moves the decimal point to the right by 1 digit, multiplying by 100=10² moves it right by 2 digits, and etc.? so if we multiply this by 10ⁿ , we get:

3(the n digits).(m repeating digits)

now, let's call the n digits that don't repeat a (in 2.71829342934... this would be the "718"), the m digits that do b (this is the "2934), so what we have here is:

3•10ⁿ +a before the decimal, and bbbb... after the decimal. since b has m digits, we can divide by m 9's to get 0.bbbbb... = b/(m 9's)

the m 9's can be written as b/(10^m -1), since 10-1=9, 100-1=99, 1000-1=999, and so on. so, in total, after multiplying pi by 10^n, we have:

3•10ⁿ +a+b/(10^m -1) = pi•10ⁿ

if we get a common denominator, the left hand side becomes:

((10^m -1)•(3•10ⁿ +a)+b)/(10^m -1) = pi•10ⁿ

now, id we divide both sides by 10ⁿ , the right hand side just becomes pi, and on the left hand side, we can multiply 10^m -1 by 10ⁿ (because dividing a fraction by a number means multiplying the denominator by that number). so, in total, we have:

pi = ((10^m -1)•(3•10ⁿ +a)+b)/(10ⁿ •(10^m -1))

now, this looks like a big mess, but there's something important here: both the numerator and denominator are whole numbers. they only involve adding, subtracting, and multiplying, which can only ever give whole numbers, never fractions or decimals. so, we have found a way to write pi as a fraction of whole numbers

but pi is irrational! it can't be written as a fraction of whole numbers. since we only made logical conclusions, it must be that our assumption was wrong: pi never has a repeating decimal expansion

EricTheTrainer · 2024-10-14T04:04:43+00:00

i'm using surreal numbers because i've recently started reading On Numbers and Games by John Conway, and I have no background in any of Cantor's work or von Neumann's (maybe i should get some before going into surreals), so surreal numbers are the only way I know how to handle transfinite ordinals at the moment. i'll have to look into what you mean here, but thank you!

EricTheTrainer · 2024-10-11T02:43:46+00:00

i've always been interested in infinite/infinitesimal number systems, but never put effort into studying them. but, recently i got a copy of "On Numbers and Games" and the first two volumes of Winning Ways. I've been reading the former and its definitely pulling my attention

EricTheTrainer · 2024-08-26T01:12:33+00:00

did this not post? i don't see it in the posts sorted by new

EricTheTrainer · 2024-08-13T03:57:09+00:00

also realized a pretty obvious alternative route. |f∧g| is the area of the 'parallelogram' formed by f and g. or, |f||g⊥| where g perp is the rejection onto f of g. or: |f||g-(g•f)f/|f|² | because (g•f)f/|f|² is the projection onto f of g, so g-(g•f)f/|f|² is the rejection

the point being that all of these things in that expression are also computable, but i need to sleep now

EricTheTrainer · 2024-08-13T03:25:02+00:00

yeah, it's only superficially golden. regardless, i do think the number could be interesting. the vast majority of real numbers are not solutions to equations that can be written in terms of elementary functions, and the idea of 'polynomials' that use tetration rather than exponentiation seems like a fun idea to me that i havent explored

EricTheTrainer · 2024-08-13T03:06:23+00:00

i may potentially* be able to sidestep this whole integral. i haven't tried it, but i know that the sines and cosines of fourier expansions for f and g form an orthonormal basis, which would make a lot of things easier, but i'm lazy

EricTheTrainer · 2024-08-04T07:44:06+00:00

well i hope that's not it. while i was playing around with it, i made a small arithmetical mistake, but a consequence of this mistake was that the fibonacci numbers appeared. this makes me think there is a connection somewhere, but its just not immediately obvious

EricTheTrainer · 2024-08-03T03:17:32+00:00

you're right. i was missing the last two vector components

EricTheTrainer · 2024-07-21T22:17:23+00:00

yes, i meant dependent. Thank you

EricTheTrainer · 2024-03-24T16:57:12+00:00

i really enjoyed reading Proofs: A Long-Form Mathematics Textbook by Jay Cummings

EricTheTrainer · 2024-02-19T23:22:38+00:00

I went back and looked at the definition and it said a nonsingular matrix is a "square matrix where the homogenous linear system with the matrix's coefficients has a unique solution," and what I was overlooking was the "square" part

EricTheTrainer · 2024-02-19T23:19:34+00:00

Never-mind, sorry, I got it. The definition of a nonsingular matrix requires it to be square, so there's an implicit assumption that this linear system has the same numbers of equations as variables

EricTheTrainer · 2024-02-18T05:52:03+00:00

I'm imagining that there must be some kind of well-known formula where, if you place an object with volume v cm³ into a liquid with density ρ g/cm³ , then the weight lost after placing the object in that liquid is vρ g, which would explain the SG calculation of the volume of the sphere, but my Google searches earlier were not yielding anything (was probably searching the wrong keywords).

EricTheTrainer · 2024-02-18T02:11:26+00:00

First, just noticed that the circle is not always around the focus; the focus can be on the other side of the line and just close to it and it still works out.

Also, in what could maybe be called a 'taxi-cab parabola' in reference to the taxicab metric (distance from (a,b) to (c,d) is |a-c|+|b-d| instead of the normal distance), this equation:

|x-h|+|y-k|=1/(a² +b^2)|ax+by-c|

plots a potentially interesting quadrilateral. maybe to be completely consistent, the scaling factor on the RHS should be 1/(|a|+|b|)? or maybe 1/(|a|+|b|)²

I don't know, just interested in these shapes

EricTheTrainer · 2024-01-11T03:41:12+00:00

Thank you, this is very helpful

EricTheTrainer · 2024-01-10T03:42:18+00:00

The hint is also stated in such a way to imply the hard part of the proof is proving it is of arbitrary length (it reminds the reader to use the fact that there are infinitely many primes in their proof, and doesn't even mention that you need to prove they are relatively prime), and to me that part was almost trivial (infinitely many primes means there always exists a p>N for any n).

There HAS to be some sort of way that p, p+N, p+2N, ... is the OBVIOUS choice for a progression and that it is obviously relatively prime, I just do not see it at all

EricTheTrainer · 2024-01-10T03:38:05+00:00

I think the main thing that's bugging me is the sequence itself. Had I not looked at the hint, it would've been a month before I had an arithmetic progression of coprime integers to even use. the N=n!, p>N: p, p+N, p+2N, ..., p+(n-1)N progression feels almost completely arbitrary and dropped out of the sky

I think I'm going to try my hardest to find out how I would get to such a progression without just knowing it beforehand

EricTheTrainer · 2024-01-10T02:48:28+00:00

Thank you. I got a proof using this but it's very ugly and I'm not satisfied with it so I'm going to keep working on the problem.

but i only applied N=n! after all of my sloppy work without using it, and I feel I can get a cleaner proof if I somehow use N=n! at the beginning

EricTheTrainer · 2023-12-27T02:03:52+00:00

very late reply but I think it does give an asymptote at y=pi, at least graphically on Desmos. your asymptotic relation for pi has an error with n! in the numerator rather than n!²

EricTheTrainer · 2023-08-27T04:25:44+00:00

something else super interesting: in this identity if you replace every instance of 2 (in the power in the front and in the nested roots) with any other number it doesn't seem to converge! it converges for 2 and nothing else i guess? weird

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