Big number thread (very big)

[Anonymous No.10512824]

If every number in [0,inf) is a real number then there exists a real number half way between 0 and infinity because every line segment has a midpoint and the midpoint of [a,b] is the same as the midpoint of [a,b).

[Anonymous No.10512830]

infinity isn't a number

[Anonymous No.10512838]

>>10512830
Do you have an issue with the definition of real numbers as all numbers in the interval (-inf,inf)?

[Anonymous No.10512892]

>>10512838
yes. infinity isn’t a number

[Anonymous No.10512896]

>>10512892
I think what you mean is "infinity isn't a real number." It's definitely a number. Did you mean "infinity isn't a real number."

STOP READING IF YOU DON"T WANT YOUR MIND BLOWN

sqrt{-1} isn't a real number either and we do stuff with it every day. sqrt{-1} is a number; its an imaginary number. Likewise, infinity is a number; it's an extended real number.

>detractor BTFO

[Anonymous No.10512901]

[Anonymous No.10512905]

>it's another schizo poster episode
getting tired of these reruns friends.

[Anonymous No.10512915]

>>10512905
Maybe you should go check out an IQ thread or a flat earth thread for some hot OC then.

[Anonymous No.10512919]

>>10512896
you are an extended idiot

[Anonymous No.10512921]

>>10512919
I know you are but what am I?

[Anonymous No.10512929]

>wait, ummm, like
>because Elon Musk and Facebook
>sqrt{-1} is the only non-real number that can exist

[Anonymous No.10512963]

>>10512915
I just wanted to discuss science.

[Anonymous No.10513007]

>>10512963
Well... if you acknowledge that infinity is a number then maybe we can proceed.

[Anonymous No.10513009]

Is Tooker posting from jail?

[Anonymous No.10513020]

>>10512919
I'm gonna extend my cock into your anus

[Anonymous No.10513029]

>>10512824
Yep and the midpoint is at infinity

[Anonymous No.10513033]

>>10512830
so you mean the midpoint of (-inf, inf) isn't 0?

[Anonymous No.10513035]

>>10513029
You think the midpoint of [0,inf) is infinity?
>OK

[Anonymous No.10513045]

>>10512824
[0, inf) is a ray, not a line segment

[Anonymous No.10513053]

>>10513045
but the midpoint of [a,b) is the same as the midpoint of [a,b]

[Anonymous No.10513057]

>>10513053
[0, inf] is still a ray

[Anonymous No.10513094]

>>10513057
a line with one endpoint is a ray.
a line with two endpoints is a line segment.

[Anonymous No.10513103]

>>10513094
a line has no endpoints

[Anonymous No.10513111]

>>10513103
yeah... hence:
an infinite line plus one endpoint is a ray
an infinite line plus two endpoints is a line segment

[Anonymous No.10513121]

>>10513007
lel, that one is a new level of jon schizoness. jon, do you know that t-symmetry violation is well known for decades (first demonstrated in 1964, to leave aside from that whole arrow-of-time thing boltzmann talked about)? your standard model knowledge is so weak bro. are you going to claim credit for the recent LHCb observation of another CP violating process too?

[Anonymous No.10513132]

>>10513121
>your standard model knowledge is so weak bro.
Wow, you mean that one undergrad class I took in particle physics didn't make me know as much as people who do PhDs and post-docs in particle physics? WOW!!!

It was the correlations with delay that I predicted, not T violation. Learn to read, tard.

[Anonymous No.10513137]

>>10512824
A line segment is a subset of a vector space, in this case a vector space over R, parameterized as {u+tv | t∈[0,1]} where u and v are vectors in the vector space. u and u+v are the endpoints of the line segment. u+v cannot be inf so [0,inf) is not a line segment.

[Anonymous No.10513142]

>>10513137
The geometric definition I use is better than that the def you use. To make your definition, first you have to invent vector analysis. To make my definition, you only have to invent points and lines.

[Anonymous No.10513148]

>>10513142
There is only one definition. If you are talking about something else then name it something else, and show that it has a midpoint since you can no longer use theorems proved for line segments.

[Anonymous No.10513149]

>>10513142
you use your definition inconsistently, it doesn't matter what complains you have

[Anonymous No.10513156]

>>10513142
Well that presents a problem since "midpoint" is only defined for a line segment using the definition I gave. So basically what you said in the OP is gibberish only meant to have the appearance of mathematical meaning.

[Anonymous No.10513158]

>>10513148
>There is only one definition.
I agree: an infinite line plus two endpoints is a line segment

>>10513149
>you use your definition inconsistently
an infinite line plus two endpoints is a line segment. Where is the inconsistency?

[Anonymous No.10513162]

>>10513158
>I agree: an infinite line plus two endpoints is a line segment
If the endpoints cannot be represented as vectors in a vector space over R or C then it's not a line segment. Such a construction also has no midpoint as midpoint is only defined with respect to a line segment.

[Anonymous No.10513164]

>>10513156
>since "midpoint" is only defined for a line segment using the definition I gave.
bullshit, pic related

[Anonymous No.10513167]

>>10513164
What is the definition of line segment in your pic?

[Anonymous No.10513172]

>>10513167

[Anonymous No.10513181]

>>10513164
>using distance metrics
>>10513142
>you only have to invent points and lines.
whats it like not remembering what you've just written

[Anonymous No.10513184]

>>10513172
That's not the definition of a line segment. You need to call it something else to avoid confusion.

We can now move back to >>10513164. The proof for theorem 4.12 is faulty since it does not prove that two such circles can exist in all cases.

[Anonymous No.10513201]

>>10513184
>That's not the definition of a line segment.
yes it is

>>10513181
>>using distance metrics
I didn't use a metric. Did you see the word "alternatively" or did your painful cognitive dissonance make you blind to it?

>>10513184
>prove that two such circles can exist
proof is trivial, pic

[Anonymous No.10513207]

>>10513201
>yes it is
It's not, already gave you the definition and it clearly conflicts.

>proof is trivial, pic
This is not a proof. Show me two circles centered on 0 and inf intersecting at exactly two points.

[Anonymous No.10513214]

>>10513207
No, I gave you the definition.

>>10513207
>centered on 0
learn to read tou imbecile

[Anonymous No.10513217]

>>10513201
>I didn't use a metric.
>two lines AC and CB can be equal without a metric
jesus christ almighty, how the fuck do you determine when AC = CB is true

[Anonymous No.10513219]

>>10513214
>No, I gave you the definition.
No, gave you the definition.

>centered on 0 and inf
Learn how to read, schizo.

[Anonymous No.10513222]

>>10513217
>how the fuck do you determine when AC = CB is true
1) make a copy of CB called C'B'
2) put C' at A
3) If B' falls on C then they are equal.
duh

[Anonymous No.10513228]

>>10513219
>centered on 0 and inf
let A=0 and B=inf
The circles intersect twice no matter how you label the points. You can say A=banana and B=umbrella and they will still intersect at S and T.

[Anonymous No.10513231]

>>10513222
i swear to god
what does "put C' at A" mean
because now you've introduced a distance preserving isometry which must have a metric to be defined

>>10513228
thats not how any of this works you idiot

[Anonymous No.10513233]

>>10513228
>let A=0 and B=inf
Doesn't work, you show B in the line and not at its end.

[Anonymous No.10513240]

>>10513231
I haven't introduced distance at all, much less an isometry.

>what does "put C' at A" mean
it means "translate C'B' so that C'=A"

[Anonymous No.10513243]

>>10513233
Use the eraser tool in MS Paint and it should make sens to you

[Anonymous No.10513248]

>>10513240
translation is literally an isometry

[Anonymous No.10513254]

>>10513248
it is only an isometry when distance is defined. Think about the root words of "isometry"

[Anonymous No.10513260]

>>10513248
Euclidean spaces come equipped with the flat metric. I haven't attached that to my line segment yet.

[Anonymous No.10513264]

>>10513243
No, because then I still have a finite line and not an infinite line.

[Anonymous No.10513268]

>>10513254
>>10513260
so you have no way of translating line segments
translation implies preserving distance, which implies distance
there is no way for you to determine when two segments are equal with your definitions

[Anonymous No.10513270]

>>10513264
if you label "B=inf" then it is infinite

[Anonymous No.10513275]

>>10513270
Labeling something doesn't prove that it exists as you've drawn it. Try again.

[Anonymous No.10513280]

>>10513268
>there is no way for you to determine when two segments are equal with your definitions
Oh really? How about this:
"If C is a midpoint of AB then AC=CB"
BTFO

[Anonymous No.10513282]

>>10513270
>call z a complex number, z = 1 + 0i
>define z to be zeta(z)=0
>ive disproven the riemann hypothesis
Tooker finally revealed his secret proof technique

[Anonymous No.10513283]

>>10513275
If I say "Let B=inf" then B=inf. You try again... or better yet, don't.

[Anonymous No.10513286]

>>10513282
No, you've invoked a contradiction because I can show that
zeta(1) != 0

[Anonymous No.10513287]

>>10513283
You can say whatever you want, but you need to prove it, not just say it. Since you refuse to provide one I'll just take this as you admitting that theorem 4.12 fails. Thus you have no defined "midpoint" as used in the OP and your original claim is rendered meaningless.

[Anonymous No.10513290]

>>10513280
>Let C be a midpoint
>not using distance

[Anonymous No.10513292]

jon, clearly you are a smart guy. and /sci/ would unanimously love you if the threads you make here were not either blatantly false scientifically or contradict well-established mathematics.

i think you have potential. you are smart. but unfortunately you are very deep inside a cage of your own creation. if you could disown your past works, outright (maybe posting a paper that disowns all prior papers) then you have the potential to do something. many young scientists have to go through this process; you seem to have missed that mark by more than 8 or so years, though.

we want to recruit your talents. however you keep being a pseud. just give it up. you didn't revolutionize math nor physics. your riemann hypothesis proof is nonsensical and your physics theories directly contradict experiment. so what i ask is you to exercise the humility to acknowledge the error in your ways and to redeem yourself in order to become a real scientist. you definitely have the potential. but your self-righteous insistence on the correctness of papers you wrote in 2013 is what kills that. even real scientists need to abandon old failed theories. but you don't.

so i am asking you, sincerely, to try to look objectively at your past work, critically, and see if you can abandon your youthful convictions. if you can renew yourself, by abandoning your crazy ideas, you could have potential. in particular you _must_ abandon your delusional idea that you are a "savior" in the terms of some religious crap. just give up your religious shit and failed theories, come to terms with reality, and then you might become a real scientist or mathematician. and once you do that, i think you might have real potential that could realize in a non-meme way

[Anonymous No.10513295]

>>10513287
Did Riemann need to prove that
zeta(z) = SUM 1/n^z ?

For stuff like that, just saying it is totally sufficient

[Anonymous No.10513300]

>>10513292
maybe if you weren't gay

[Anonymous No.10513302]

>>10513295
>x^2 is always non negative (on the reals)
>therefore i^2 is non negative
>its true cuz i just relabeled x to i

[Anonymous No.10513305]

>>10513295
What does zeta(z) = SUM 1/n^z require to exist?. Your definition of what you call a "midpoint" requires two circles centered on 0 and inf intersecting at exactly two points to exist. You have not shown that this is possible, so your argument fails.

[Anonymous No.10513306]

*drops mic*

[Anonymous No.10513313]

>>10513305
Do you see "0" or "inf" anywhere in here?

[Anonymous No.10513314]

>>10513033
It isn't.

[Anonymous No.10513317]

>>10513313
So AB cannot be a line segment with endpoints 0 and inf? Use your words like a big boy.

[Anonymous No.10513324]

>>10513317
The result regards all line segments.

[Anonymous No.10513330]

>>10513324
Then show it works for all line segments.

[Anonymous No.10513339]

>>10513330
AB is an arbitrary line segment.
Therefore, it works for arbitrary line segments.

It's like you think intoducing some algebraic notion of distance is going to somehow override basic geometry.

[Anonymous No.10513342]

>>10513313
i see equals signs all over,
and "radii be less than but large enough" implying distance again
and also, the _Pythagorean Fucking Theorem_
you are blatantly using distance you hack

[Anonymous No.10513347]

>>10513342
>implying distance again
The concept of midpoint does not exist without a concept of distance.
>and also, the _Pythagorean Fucking Theorem_
What?

[Anonymous No.10513348]

>>10513339
>Therefore, it works for arbitrary line segments.
you havent shown that it works, thats the problem
you've assumed the existence of sufficient circles without proof

[Anonymous No.10513351]

>>10513339
>AB is an arbitrary line segment.
>Therefore, it works for arbitrary line segments.
LOL, so I can say literally anything about AB and it will be true for all "line segments" because AB is "an arbitrary line segment?" No, that's not how a proof works. Show me that this works for a "line segment" with endpoints 0 and inf.

[Anonymous No.10513354]

>>10513348

[Anonymous No.10513355]

>>10513254
>>10513260
>>10513240
>>10513201
>I didn't use a metric.
>>10513142
>To make my definition, you only have to invent points and lines.

[Anonymous No.10513357]

>>10513354
>hes drawn a finite line segment
>it has infinite length
>he thinks his contradictory construction is worth anything

[Anonymous No.10513360]

>>10513355
>>I didn't use a metric.
What metric does he use? In fact, what space does he use?

[Anonymous No.10513361]

>>10513342
>"radii be less than but large enough" implying distance again
wrong

AB<AC implies AB is a subest of AC

[Anonymous No.10513364]

>>10513357
>>hes drawn a finite line segment
>>it has infinite length
>>he thinks his contradictory construction is worth anything
construct the finite line segment [0,pi]
define length on it with L = tan(x)
finite line segment has infinite length
imbecile!

[Anonymous No.10513368]

>>10513364
I mean [0,pi/2]

[Anonymous No.10513370]

[Anonymous No.10513374]

>>10513364
>define length on it with L = tan(x)

[Anonymous No.10513375]

>>10513360
he says none, yet he uses all the intuitive parts of the usual metric on R2

[Anonymous No.10513378]

>>10513364
That's not a metric, though, because the distance between x and y is undefined, if both are pi.

[Anonymous No.10513380]

>>10513364
>defines length with a single valued function
>lengths need two fucking points

[Anonymous No.10513382]

>>10513378
what is "y"?

[Anonymous No.10513385]

>>10513380
L is the length measured from the left endpoint of [0,pi/2]

[Anonymous No.10513392]

>>10513382
It's a variable, anon.

[Anonymous No.10513393]

>>10513374

[Anonymous No.10513398]

>>10513385
What's the distance between two arbitrary points with L?

[Anonymous No.10513399]

>>10513385
cool, you still haven't shown that you can successfully make two circles at both endpoints that intersect at 2 points while following this fucking length metric
yknow, the entire reason you made this construction in the first place

[Anonymous No.10513402]

>>10513392
as in
L(x,y)=tan(x)-tan(y) ,

where we select y_0=0 for simplicity to write
L(x,y_0) = tan(x)-tan(y_0)
= tan(x) - 0
= tan(x)

like that? Is that why L isn't a metric?

[Anonymous No.10513405]

>>10513402
the domain of the line isnt in the domain of L
tan ( pi/2 ) is undefined

[Anonymous No.10513406]

>>10513402
>L(x,y)=tan(x)-tan(y) ,
This is not a metric, because L(0.5pi,0.5pi) = tan(0.5pi)-tan(0.5pi) = ?. Should be definite zero.

[Anonymous No.10513408]

>>10513399
You doubt that I can make two intersecting circles centered on 0 and pi/2?

Or did you miss pic the first time I posted it?

[Anonymous No.10513409]

>>10513408
use your L length to do it then
stop fucking posturing and demonstrate it

[Anonymous No.10513415]

>>10513408
You can't really interpret infinity geometrically, because A and B would be the same point and C and D would be the same point on your line segment.

[Anonymous No.10513417]

>>10513415
he doesn't understand infinity

[Anonymous No.10513434]

>>10513415
You're getting it backwards. I'm starting with geometry, not algebra. Geometry is real, algebra is notional.

[Anonymous No.10513439]

>>10513405
>tan ( pi/2 ) is undefined
define this number as infinity

>>10513406
>Should be definite zero.
make a piecewise metric conditional for y=x

[Anonymous No.10513441]

>>10513434
>I'm starting with geometry,
no you aren't
you've killed geometry, this is a disgusting, contradictory frankstein monster

also
>algebra is notional.
pic related

[Anonymous No.10513443]

>>10513441
>pic related

[Anonymous No.10513444]

>>10513439
a metric must output a real number, not infinity

[Anonymous No.10513454]

>>10513444
if you preclude the notion of infinite distance then I concede that infinite distance does not exist. However, I do not thusly preclude.

Check out this brand new thing in phsyics, pic related. It just got invented and it uses the same idea that i'm using. They call it "a Penrose diagram"

[Anonymous No.10513458]

>>10513434
>Geometry is real,
Sure. But infinity isn't.

[Anonymous No.10513464]

>>10513454
https://en.wikipedia.org/wiki/Metric_(mathematics)
pay more attention to solid math than to loose physics

but you know what you cant fucking do with a penrose diagram?
make two circles, one in the middle and the other at the edge, which intersect in only 2 spots

[Anonymous No.10513469]

>>10513458
>But infinity isn't.
What is pic related symbol then? Am I hallucinating a sideways eight over there?

[Anonymous No.10513476]

>>10513469
define it

[Anonymous No.10513478]

>>10513469
Something you evidently don't understand at all.

[Anonymous No.10513494]

>>10513464
>you know what you cant fucking do with a penrose diagram?
>make two circles, one in the middle and the other at the edge, which intersect in only 2 spots
I disagree

[Anonymous No.10513497]

>>10513434
You're not making any sense. If you just use geometry, you can clearly see from your picture that infinity lies at about 10*AB. If you now assign any real numbers to that, you get nonsensical results. If you say A is at origin and B at 10, you show that infinity is equal to 100. Well, it's not. This is like drawing a triangle with one point at infinity and then making claims that no such thing as parallel lines exist.

[Anonymous No.10513498]

>>10513476
>define it

[Anonymous No.10513510]

>>10513497
>you can clearly see from your picture that infinity lies at about 10*AB
yes

>>10513497
>If you now assign any real numbers to that, you get nonsensical results.
maybe you do. My results come out fine.

>>10513497
>If you say A is at origin and B at 10, you show that infinity is equal to 100.
that's why pic related with A,B,S, and T is better than the other pic with A,B,C,D

[Anonymous No.10513517]

>>10513464
it's b8, just give up, this is what they want

[Anonymous No.10513520]

>>10513510
>that's why pic related with A,B,S, and T is better than the other pic with A,B,C,D
Except it's not true at all. If the radii of your circles are infinite, S and T will also be at infinity, i.e., the angle between AB and AT is 90 degrees. Unless you of course are doing precisely what I said is stupid, i.e., proving that no such thing as parallel lines exist.

[Anonymous No.10513525]

>>10513494
thats not a circle on the top right you fucking retard, use the diagrams metric you 3 iq shit stain
that "circle" is passing through an infinite amount of space, while the one in the middle is not

>>10513498
>limits are always defined
you should honestly kill yourself, you have 0 mathematical know how
what is the limit of
1, 0, 1, 0, 1, 0, ....
if it isn't defined then you have no fucking right to state that the other limit is defined

remove the meaningless, redundant shit after "and"
you're defining positive infinity in terms of itself

>>10513510
>that's why pic related with A,B,S, and T is better
>theres shit drawn after B
>theres something after infinity

>>10513517
its not, hes been doing this for months

[Anonymous No.10513534]

>>10513520
>If the radii of your circles are infinite
The radii are AS and BT. Result is for the general case.

>If the radii of your circles are infinite
I said the radii are less than AB

[Anonymous No.10513544]

>>10513525
Just because it's old and stale b8 doesn't mean it's not bait

[Anonymous No.10513548]

>>10513534
>I said the radii are less than AB
you haven't shown that that's a valid assumption
relabeling is not as powerful as you think
>>10513302

[Anonymous No.10513549]

>>10513534
r may well be less than AB, but then 3r > AB, i.e., B can't be infinity, since clearly it is finite.

[Anonymous No.10513560]

>>10513525
>use the diagrams metric you 3 iq shit stain
I am using the diagram's metric. To use Penrose's original notation, you are thinking about the metric of the physical manifold M' but I have used the metric of the conformal manifold M. Did you even read penrose's paper?
>Conformal treatment of infinity
>Roger Penrose

It seems like you didn't read it since you apparently have no concept of the difference between M and M'. Maybe read something some time. It will be good for your ignorance.

>>10513525
>what is the limit of
>1, 0, 1, 0, 1, 0, ....
pic

[Anonymous No.10513572]

>>10513548
>you haven't shown
I certainly did

>>10513549
3R is a transfinite number. Use the conformal coordinates and then its a finite number.

[Anonymous No.10513582]

>>10513572
>3R is a transfinite number. Use the conformal coordinates and then its a finite number.
That makes no sense. If r is finite and 3r > AB, AB can't be infinite. If r is infinite, my previous points stand, i.e., drawing a line perpendicular to AB at A will go from S to T. In your picture this would, of course, cross AB at the section between AB, i.e., A would be the same point as that midpoint.

[Anonymous No.10513586]

>>10513582
If wookies live on Endor you must acquit!

[Anonymous No.10513591]

>>10513586
So you were baiting the whole time after all.

[Anonymous No.10513903]

If this troller here is johnathan, i have read your paper. You should look into surreal and hyperreal numbers.

[Anonymous No.10514203]

[Anonymous No.10514214]

[Anonymous No.10514269]

>>10512824
yes, the midpoint is at infinity/2=infinity.

Congratulations.

[Anonymous No.10514330]

>>10512824
>If every number in [0,inf) is a real number then...
inf isn't in [0, inf)
Tooker, we've been over this

[Anonymous No.10515087]

>>10514269
I see you've bestowed infinity with the property of multiplicative absorption. What was your reasoning there?

Let AB = [0,inf]. We form [0,inf) from [0,inf] by removing one point. That is the point B where we can use the flat metric to put the number infinity. After we remove B from the line segment AB, the midpoint of the line segment is still there because we have only removed a single point. The Euclidean metric is L(x,y) is defined for any two points in the interval [0,inf), and since we have only removed B from the line segment AB, we ought to be able to use the endpoint A and the midpoint C.

[Anonymous No.10515142]

Does it make sense or not?

[Anonymous No.10515186]

>>10515142
jon, do you realize that posting bible crap and threatening violence against your “detractors” doesn’t help your arguments at all? more like it instantly disqualifies you from being taken seriously?

[Anonymous No.10515194]

>>10512838
Yes, it's circular.

[Anonymous No.10515211]

>>10515186
>do you realize that posting bible crap and threatening violence against your “detractors” doesn’t help your arguments at all?
I don't realize that and when it finally comes times for me to get say, "Haha, fuck you. I fucking told you so," I'm going to get a big bonus multiplier over literally telling them one million times.

>more like it instantly disqualifies you from being taken seriously?
I don't realize that. No one is taken more seriously than killers.

[Anonymous No.10515305]

>>10515211
Oh. Okay, now I take you more seriously. I reverse my opinion, yeah infinity hat is real and RH is false, please don't hurt my baby.

[Anonymous No.10515463]

>>10515305
might as well call the other one infinity-hat

[Muhammed Abobou No.10515491]

Numbers have an end.

[Anonymous No.10515579]

>>10513399

[Anonymous No.10515600]

>>10515579
>defines a new metric using tan to try and force infinite lines to act like finite lines
>(this doesnt work)
>instead uses arctan just to waste everyones time and space even more
>draws his circles according to the euclidean metric rather than his abomination
>they arent circles in his abomination
>he somehow doesnt realize this
R1 and R2 are not equal jon
translation doesnt change size in euclidean space, however it does in your new shitty metric

[Anonymous No.10516423]

>>10513137
>A line segment is a subset of a vector space, in this case a vector space over R, parameterized as {u+tv | t∈[0,1]} where u and v are vectors in the vector space.
you know, i didn't think you could be more retarded than OP. congrats on achieving something in your life for once

[Anonymous No.10516584]

>>10516423
https://en.wikipedia.org/wiki/Line_segment#In_real_or_complex_vector_spaces
read nigga, read!

[Anonymous No.10516643]

>>10513292
>clearly you are a smart guy
Lol no he's fucking not. He's a total idiot.

[Anonymous No.10516747]

>>10516584
nigger i know basic analysis and so does every other toddler
the fact that you didn't realize or willingly ignored that OP is attempting to use the Euclidean/geometric definition of a line segment and that you weren't able to crush his shitty "proof" on that basis is what makes you more retarded than him
and quite a feat, too!

[Anonymous No.10516970]

>>10516747
kys retard

[Anonymous No.10517177]

>>10516423
>>10516584
let the interval be AB = [0,inf]
parameterize the interval with x in [0,inf]
define x' = tan^{-1} (x)
therefore: x' in [0,pi/2]
let u = 0
let v point in the direction of x'
let ||v|| = pi/2

there you go, now AB is a line segment according to the vector analysis definition

[Anonymous No.10517181]

>>10517177
ayyy

[Anonymous No.10517305]

>>10517177
almost none of what you just typed is valid

[Anonymous No.10517376]

>>10513164
How does that proof work if you allow one of the points to be infinity? How can you have a circle centred on infinity, and what radius must it have to intersect the other circle centred on e.g 0?
Just try drawing a picture, you'll see it's nonsense.

[Anonymous No.10517384]

>>10517376
he draws a line of finite size
since hes retarded and has no mathematical talent whatsoever

[Anonymous No.10517388]

>>10512921
based

[Anonymous No.10517464]

>>10517376
see>>10515579

Don't forget to check out my new website
http://0cc762.net/

[Anonymous No.10517497]

>>10517464
is that literally a picture of your dick

[Anonymous No.10517525]

>>10517497
he moved it off the homepage but it's still being served:
http://0cc762.net/img2019/TIMESAND___jjj894w56d837i78xzzhm686yj.jpg

[Anonymous No.10517527]

>>10517497
no its a clock but you can check me out here:
https://imgur.com/a/GHI1Xw3
https://imgur.com/a/PxbElsS

[Anonymous No.10517535]

>>10517527
you fucking crazy motherfucker, lol

[Anonymous No.10517538]

>>10517497
I'm confused... so that man is the woman and the woman is the man but the woman looks like a man and the man looks like a woman. The man impregnates the woman but in this case the woman has impregnated the man and they're going to raise baby.......

She sells the sea shells by the sea shore

[Anonymous No.10517545]

>>10517527
Would you believe that the guy in those imgurs gets turned down by ~100% of the women he asks to go out?

[Anonymous No.10517552]

>>10517545
>the guy in those imgurs
guy is me, btw
https://imgur.com/a/D2ruaOC

[Anonymous No.10517648]

>>10517177
>let the interval be AB = [0,inf]
no

[Anonymous No.10517747]

>>10517648
How about this:
let the interval be [0,inf] ?

[Anonymous No.10517757]

>>10517747
no, you can't close an interval at infinity, that doesn't make any sense

[Anonymous No.10517789]

>>10517757
>that doesn't make any sense
you need to prove this because merely citing your opinion is unconvincing

[Anonymous No.10517798]

honestly i feel bad for math anons who continue this argument after it descended into tooker posting his own dick pics (mind you, not even just pics but animated gifs)

i think he self-destructed harder than anybody could even hope for

[Anonymous No.10517817]

>>1051778
A closed interval implies a smallest upper bound (axiom of completion)
Drop in the archimedean property (which is also true for real numbers)

Therefore the interval to infinity can't be closed because it can't have an upper bound.

Q.E.D. b8 hurts me

[Anonymous No.10517822]

>>10517798
>>10517798
>descended
my dick is ascending, as you can plainly see

[Anonymous No.10517831]

>>10517817
>(which is also true for real numbers)
infinity isn't a real number though and I didn't say [0,inf] was made of real numbers. I would say, "Nice straw man," but it wasn't. It sucked.

>>10517817
>A closed interval implies a smallest upper bound
smallest upper bound is infinity
btfo

[Anonymous No.10517833]

that's the real sphinx

[Anonymous No.10517843]

why can tooker never formulate an actual argument?

[Anonymous No.10517846]

>>10517831
>I didn't say [0,inf] was made of real numbers
pray tell, what is the interval then?

[Anonymous No.10517855]

>>10517846
all number which are
(a) greater than equal to zero
(b) less than or equal to infinity

If you don't understand the notation [a,b] maybe you should not post

[Anonymous No.10517859]

>>10517843
my nutsack is the real one
the true vine

[Anonymous No.10517868]

>>10517855
(a) You are talking about infinity like it has a fixed, set value. No

(b) So we are talking about the real numbers?

[Anonymous No.10517894]

>>10517855
>If you don't understand the notation [a,b] maybe you should not post
tooker, you do not understand anything

[Anonymous No.10517942]

>>10513300
Dub dubs

[Anonymous No.10517952]

>>10513469
Isnt that a limniscate?

[Anonymous No.10517953]

>>10517894
There's something I do understand and it's mathematical in nature

[Anonymous No.10518036]

>>10517953
Chemistry?

[Anonymous No.10518046]

>>10517953
kek you reposted the meme i made. :-3

[Anonymous No.10518054]

>>10517953

[Anonymous No.10518076]

>>10518046
It's a good one. Good job, you.

[Anonymous No.10518127]

>>10516423
>>10516747
>OP is attempting to use the Euclidean/geometric definition of a line segment
LOL no, he made up a definition that allows infinity to be an endpoint. This is neither Euclidean nor geometric and breaks the fact that a line segment always has a midpoint. There is no relevant difference here between the vector definition and the geometric definition; both preclude an infinite line segment.

[Anonymous No.10518151]

>>10517177
>define x' = tan^{-1} (x)
>therefore: x' in [0,pi/2]
Nope, tan(pi/2) is undefined.

Also it doesn't matter if you parameterize it because then your circle argument fails.

[Anonymous No.10518331]

>>10518127
>This is neither Euclidean nor geometric and breaks the fact that a line
let AB = [0,pi/2]
parameterize AB with "x"
define the distance from A with L = tan(x)
Now B = infinity, or B is "infinitely far from" A
Where is the error?
(It's in your reasoning)

[Anonymous No.10518339]

>>10518151
>Nope, tan(pi/2) is undefined.
Nope, you're tarded
see >>10513498

[Anonymous No.10518349]

>>10518151
>Nope,

>>10517894
>not understand anything

>>10517305
>none of what you just typed

>>10516643
>a total idiot.

>>10513544
>stale b8

>>10513417
>he doesn't understand infinity

>>10513219
>schizo.

[Anonymous No.10519009]

>>10513560
2^-n is monotonic but doesn't diverge to infinity

[Anonymous No.10519700]

>>10519009
does it not converge to zero?

[Jagresh Abibi No.10519842]

>>10515491
Amen, brother. I now see the light.

[Anonymous No.10519976]

>>10519700
I was talking about the sum, it converges to 1

[Anonymous No.10520001]

>>10519976
What was your point then? You linked the pic that mentioned diverging

[Najier Obibo No.10521563]

>>10515491
Thank you, I will teach this to my children and my children's children.

[Anonymous No.10522387]

>>10521563

[Anonymous No.10523523]

>>10512896
imagine being this fucking retarded

[Anonymous No.10523608]

OP is dancing around the fact that he solved the Reimann hypothesis

[Anonymous No.10523609]

>>10522387
https://biblehub.com/ezekiel/4-12.htm

[Anonymous No.10523839]

>>10523608
I feel like you guys are the ones dancing around it... or rather cowering from it in fear. When Fagbert solved this solved sort problem and posted it on 4chan, it was in the news the next day. What's the difference?
> 4chan solved part of math problem puzzling scientists for decades
>https://www.rt.com/news/442388-4chan-solution-math-problem/

[Anonymous No.10523843]

>>10523608
>>10523839
Not to mention that Fagbert probably doesn't even exist given how I posted my RH solution on 4chan in the days preceding Fagbert's miraculous solution, and then you guys made up some bullshit Fagbert specifically to detract from the accolades due to me.

[Anonymous No.10523878]

>>10518331
There is no "error," just a non sequitur. Parameterizing does nothing. Where are the circles?

[Anonymous No.10523888]

>>10518339
>tan(pi/2) = lim(y/x) as y->1 and x->0
Why?

[Anonymous No.10523892]

>>10523878
They are centered on A and B which were already well-defined before I introduced a metric to measure the distance between them.

Check out my blog:
http://0cc762.net/the+end+times.html

[Anonymous No.10523900]

>>10523888
This comes from the definition of the tangent.
Also, to teach fear of the Lord.
They don't have any now,
but they will before I'm done,
and I don't want them to forget again any time soon.
The true vine belongs to me.
I am the gardener of that garden,
and I will rip up the weeds planted by enemies,
and maybe I'll salt the earth where those weeds grew and take my garden elsewhere.

[Anonymous No.10523901]

>>10513111
There does not exist an infinite line with two endpoints

[Anonymous No.10523905]

>>10523901
Let Sigma_1 be a set that contains an infinite line.
Let Sigma_2 be a set that contains two points.
Define the elements of Sigma_2 as "endpoints."

What is in the union of Sigma_1 and Sigma_2?

[Anonymous No.10523920]

These children of malice and cuckoldry. There was nothing I wanted more than to put my dick inside good looking women, and did they ask me to get their daughters pregnant? No, they did not. Instead, they who knows what extensive resource to harvest unconsensually my semen and then to spread open their daughters' legs for some wicked person to stick his filthy hands in there. Now there's nothing I want more than to kill my enemies.

[Anonymous No.10523923]

>>10523892
Ah so let's determine the midpoint of [0,4] by parameterizing [0,2] via x^2. We draw two circles on [0,2] and find the line from their intersections intersects [0,2] at 1. This means the midpoint of [0,4] must be 1^2=1. Oh but we already know the midpoint is at 2, not 1. Gee I guess parameterizing tells us nothing.

[Anonymous No.10523934]

What do you think was in the freezer in Manafort's storage locker?
>Manna Fort

You can't find any story that mentions it now, but at the time of the raid on Manafort's storage locker they said he had a freezer in there. I am quite sure it was full of my semen.

[Anonymous No.10523948]

>>10523892
>>10523923
Oh and I forgot to mention, parameritizing something from a line segment doesn't make the original thing a line segment.

[Anonymous No.10523953]

>>10523923
>and find the line from their intersections intersects [0,2] at 1
So you can clearly see... it does have a midpoint. Changing the way you define distance between "x" and "x^2" can't make it stop having a midpoint. You have used one metric for [0,4] and another for [0,2], and then stupidly you assume that midpoint is the same in both metrics. It is not the same in both metrics, but it does exist in both metrics. For example,
x to x^2 : [0,1] to [0,1]
but
x to x^2: [1,2] to [2,4]

Your argument is feeble at best and most likely is deliberate stupidity. I have made an argument about the existence of midpoints, and you have raised a straw man regarding the distance of the midpoint from the endpoints under various metrics. The fact that the distance to the midpoint changes in the different metric PROVES that the midpoint continues to exist after changing the metric. Nice job making my argument for me with your stupid straw man.

[Anonymous No.10523957]

>>10523953
>x to x^2: [1,2] to [1,4]
oops

[Anonymous No.10523965]

>>10523953
Infinity is not a fixed quantity, so no fixed quantity can be derived from it. Infinity is not a geometric quantity, so no geometric quantity can be derived from it.

Unless you break some rules and get "creative" in how you interpret things, which isn't really math.

[Anonymous No.10523995]

>>10523965
I'm not deriving anything from infinity. I'm starting with the line segment AB=[-pi/2,pi/2] and then defining distance on it with the tangent. Every number in the infinite line (-inf,inf) now appears in the finite line segment AB. Therefore, the real number line lies between A and B.

[Anonymous No.10524002]

>>10523900
>weeds planted by enemies
pic related

When Trump was asked about Manafort, he said, "That business was years ago." Pic related was born in 2006. They say it came from Melania's womb but since she is a man I think that's unlikely. I think Stormy Daniels is the mother, and they fomented this abomination so that they could skip me in my own inheritance and give it directly to Trump's son. They tried to cut me out of my own thing and if someone puts a knife in my hand then I will cut them out of that thing which never had any part in to begin with. I am the Lord.

[Anonymous No.10524030]

>>10523953
>So you can clearly see... it does have a midpoint.
Yes [0,2] has a midpoint, that was never in controversy.

>Changing the way you define distance between "x" and "x^2" can't make it stop having a midpoint.
Non sequitur since I never claimed [0,pi/2] didn't have a midpoint. The relevant point is that changing the way you define distance does not magically give what you parameterized a midpoint.

>You have used one metric for [0,4] and another for [0,2], and then stupidly you assume that midpoint is the same in both metrics.
I asked you where the circles are and you told me they were centered on 0 and pi/2. Now you are saying that these circles have no bearing on the midpoint. So what is the point of all this? It does not tell us that you can have circles centered on 0 and inf that intersect to give us the midpoint, nor does it tell us that [0,inf] is a line segment! Thank you for admitting your own argument is guff.

>The fact that the distance to the midpoint changes in the different metric PROVES that the midpoint continues to exist after changing the metric.
You're assuming that there is a midpoint in the first place when that's what you were supposed to be proving with this entire exercise... fucking retard.

[Anonymous No.10524385]

>>10524030
>relevant point is
that is every line segment has a midpoint. The midpoint is the relevant point. If you change the metric then of course you change the distance from that point to the end points. Your straw man seems like it evolving into a red herring.

[Anonymous No.10524395]

>>10523905
>There does not exist an infinite line with two endpoints
null

[Anonymous No.10524403]

>>10524385
>that is every line segment has a midpoint.
Yes, and [0,inf] is not a line segment. If you change the definition of line segment to include [0,inf] then not all line agreements have midpoint. You fucking troglodyte.

>The midpoint is the relevant point. If you change the metric then of course you change the distance from that point to the end points.
Then what is the point of parameterizing [0,inf]??? There is none, otherwise you would have explained it already.

[Anonymous No.10524639]

>>10524403
>Yes, and [0,inf] is not a line segment.
[0,pi/2] is a line segment with flat metric D_1. Call that line segment AB. Now attach a second metric D_2 to AB such that distance is a vector (D_1,D_2). Are you saying that if D_2 gives AB as [0,inf], then AB both ~IS~ and ~IS NOT~ a line segment?

[Anonymous No.10524661]

>>10524639
No, I'm saying that [0,inf] is not a line segment. Please explain why you think adding a metric to a line segment produces a new line segment.

[Anonymous No.10524674]

>>10512824
Anon, I...

[Anonymous No.10524683]

>>10524674
also, if anything, that would be a ray, not a line segment. but really, it is the limit of the line segment as b approaches infty. the midpoint does not converge in this limit to any particular number, and so does not exist.

[Anonymous No.10524835]

>>10524661
>Please explain why you think
You have explained why you think adding a metric to a line segment makes it stop being a line segment. You have also explained, by proxy, that you don't a good reason to think that.

>>10524683
No. A ray has one endpoint and looks like [0,inf). The second square bracket makes it a line segment.

[Anonymous No.10524854]

>>10524835
You can't have infinity inside a square bracket. Go back to Math 1010

[Anonymous No.10524878]

>>10524854
In MATH 1010 I learned that claims need to be proven. You have made a claim but not proven it. To the contrary, pic related proves my claim that I can put infinity inside a square bracket.

[Anonymous No.10524900]

>>10524878
>The sky is green
The above proves that the sky is not blue

Also
>You have made a claim but not proven it
My claim is backed by freshman level math, ei. "infinity is not a real number and therefore cannot be a member of a real set", whereas your claim is unsubstantiated and your proof amounts to "because I said so"

[Anonymous No.10524928]

>>10524900
>The above proves that the sky is not blue
You have not provided an uneditted photo showing that the sky is green.

>>10524900
>therefore cannot be a member of a real set"
Your straw man is as pathetic a your trolling. Did I say it was part of a real set. How about
[-3i, 7i] ?

Is it your belief that I can't put -3i and 7i in square brackets because they are not elements of a real set?

[Anonymous No.10524929]

>>10524835
>You have explained why you think adding a metric to a line segment makes it stop being a line segment.
Makes what stop being a line segment? [0,pi/2] is a line segment. The burden of proof is on you to show that adding something to it creates a new line segment. I have no burden of proof, but I've already shown that [0,inf] is not a line segment by definition. So it does not matter how you construct it.

If you responds that you are using your own special definition then you must prove that your special definition also has a midpoint in all cases. So far the only argument you've given fails since you can't show that two circles can be centered at 0 and inf and intersect in exactly two places.

[Anonymous No.10524939]

>>10524928
A line segment is a subset of a vector space over R or C. The example you gave works but inf does not.

[Anonymous No.10524948]

>>10524928
Infinity also isn't a complex number and so can't be a member of a complex set. Are you sure you don't want to go back to Math 1010?

Also I've provided as much proof for the sky not being blue as you have provided for including infinity inside a square bracket which was the entire point of that exchange. That's okay though, I know you like to have things explained

[Anonymous No.10524949]

>>10524929
>[0,pi/2] is a line segment.
No, that's an interval. The line segment is AB. AB is a line segment no matter what metric I attach to it. The metric is what constructs the interval. The line segment AB exists independently.

The burden of proof is on you to show that AB stops being a line segment when when you attach a metric to it. Specifically, you have to show that when you attach two metrics to it to define a vector distance (D_1,D_2), AB simultaneously IS and IS NOT a line segment.
(HINT: Since AB is a line segment, you will have a hard time showing that AB is not a line segment.)

8===D

[Anonymous No.10524959]

>>10524939
>The example you gave works
I know, I showed it here: >>10517177

>>10524948
Now your opinion is that real numbers can go in square brackets, imaginary numbers can, but for some reason extended reason real numbers can't, and you have not specified that reason. I think you need to go to PHIL 101 with all these straw men you are raising. Did you know that a straw man is not a valid argument but rather it is one of the classical fallacies?

>>10524948
>provided as much proof
Nah, I showed a picture. You only wrote some sentence.

[Anonymous No.10524992]

>>10524949
>No, that's an interval. The line segment is AB.
No, those are two letters. See, I can be obtuse too.

>AB is a line segment no matter what metric I attach to it.
False. A line segment is finite by definition.

http://mathworld.wolfram.com/LineSegment.html

The endpoints must be distinct and must correspond to vectors in a vector space over R or C. Any you choose results in a finite portion.

>The burden of proof is on you to show that AB stops being a line segment when when you attach a metric to it.
I already did, [0,inf] is not a line segment by definition. Try reading.

>Specifically, you have to show that when you attach two metrics to it to define a vector distance (D_1,D_2), AB simultaneously IS and IS NOT a line segment.
Either it has endpoints in R or C or it doesn't. It doesn't simultaneously have and not have just because you add different metrics. You don't simultaneously have two different metrics.

[Anonymous No.10525004]

>>10524992
>those are two letters
Yes. One refers the to end point "A" and the other to the end point "B"

>A line segment is finite by definition.
AB is finite no matter what metric I attach to it.

>R or C
over R, C, or [math]\overline{R}[/math] you mean. If you'll allow C then you have to prove why the extended real are not allowed

> [0,inf] is not a line segment by definition
By definition, AB is a line segment

>R or C or it doesn't.
What is special about R and C that allows them but disallows all other number systems? (HINT: the thing that makes them special to you is that they support your opinion and the extended real prove that your opinion is wrong)

[Anonymous No.10525028]

>>10524959
You might want to work on that reading comprehension and then try again.

>I showed a picture
>You only wrote some sentence
Your picture was a nonsensical expression with no justification. They are analogous

[Anonymous No.10525061]

[Anonymous No.10525062]

>>10525004
>Yes. One refers the to end point "A" and the other to the end point "B"
[0,pi/2] refers to the endpoints 0 and pi/2.

>AB is finite no matter what metric I attach to it.
Then AB is not a line segment with endpoints at 0 and infinity and your claim in the OP fails.

>over R, C, or R¯ you mean.
Nope, just R and C.

>If you'll allow C then you have to prove why the extended real are not allowed
That's the definition. If you want to make up some other concept then state it has such, and prove it has a midpoint. This is the only substantive thing you need to do and you fail with every post.

>By definition, AB is a line segment
What are the endpoints of AB?

>What is special about R and C that allows them but disallows all other number systems?
They allow line segments to always have a midpoint, kek.

[Anonymous No.10525114]

>>10525062
0 and pi/2 aren't points, they are numbers. A nd B are the endpoints.

>just R and C.
why nor R-bar?

>What are the endpoints of AB?
A and B

[Anonymous No.10525258]

>>10525114
0 and inf aren't points, they are numbers. A line segment having a midpoint has no bearing on whether there exists a number halfway between 0 and inf. Thank you for admitting you're argument fails.

>why nor R-bar?
Because that's the definition. If toy wasn't some other definition don't call it a line segment. The phrase line segment connotes a certain meaning, not whatever meaning you want it to have.

[Vaneesh Ibabi No.10525275]

>>10515491
Agreed. We must spread the word, because our time is short and our numbers are limited! Praise be to God, the one true endless being.

[Anonymous No.10525279]

>>10525258
If the interval is [a,b] and we let A=a and B=b then every number in the interval [a,b] has a corresponding point in the line segment AB.

>>10525258
>that's the definition. I
Which definition says, "Only real numbers are allowed unless you need to use imaginary numbers to avoid contradicting yourself, but definitely not extended real numbers?"

[Anonymous No.10525283]

>>10525279
What about the quaternions {1,i,j,k}. Is [5k, 11k] not allowed? "k" isn't a real number and it's not an imaginary number. Is it your opinion that [5k,11k] is not allowed?

[Anonymous No.10525290]

>>10525283
...and just so I can head you off on your next stupid comment: If real numbers are allowed, and imaginary numbers are allowed, and quaternion numbers are allowed, then why not extended real numbers?

[Anonymous No.10525528]

>>10525279
>If the interval is [a,b] and we let A=a and B=b
A and B are points, not numbers. You can't even keep your own internal logic.

>then every number in the interval [a,b] has a corresponding point in the line segment AB.
And? A line segment having a midpoint has no bearing on an interval having a number halfway. Does the midpoint correspond to a halfway point? No, as I've already shown.

>Which definition says, "Only real numbers are allowed unless you need to use imaginary numbers to avoid contradicting yourself, but definitely not extended real numbers?"
https://planetmath.org/linesegment

[Anonymous No.10525533]

>>10525283
Quaternions are not allowed no.

[Anonymous No.10525621]

>>10525533
Why is [i,5i] allowed but not [k,5k]? "i" and "k" both have the property i^2=k^2=-1

In fact, without introducing the other two quaterions "i" and "j," the quaternion "k" is exactly like the imaginary number "i" in every way. In fact, if you construct a 2D plane from 1 and "k" every theorem valid in C is valid in that plane.

How about this. Let "apple" be a number with the property apple^2=-1. Why is [3apple, 5apple] only allowed sometimes?

[Anonymous No.10525657]

>>10525621
>Why is [i,5i] allowed but not [k,5k]?
Because it's not in R or C.

>Why is [3apple, 5apple] only allowed sometimes?
Because that's the definition. If you would like to use some other definition you are free to do so, but then you must prove that it has a midpoint, which you have so far failed to do.

[Anonymous No.10525668]

>>10525657
It is in C though. "k" has the property k^2=-1. In the absence of quaternions "i" and "j," "quaternion "k" is indistinguishable from imaginary "i."

A number is a complex number iff
z = x + TITTIES*y
x is a real number
y is a real number
TITTIES has the property that TITTIES^2 = 1

How is it in not in C?

[Anonymous No.10525670]

>>10525668
>TITTIES^2 = -1

[Anonymous No.10525687]

>>10525668
>It is in C though. "k" has the property k^2=-1. In the absence of quaternions "i" and "j," "quaternion "k" is indistinguishable from imaginary "i."
There is no absence of quaternions when using a quaternion. k = ij.

[Anonymous No.10525710]

>>10525687
I said in 2D. You would need 4D to decompose the "k" part of the {1,k} plane into "i" and "j"

[Anonymous No.10525718]

>>10525710
Then k is not a quaternion and your question is trivial.

[Anonymous No.10525732]

>>10525718
consider the 4D quaternion space spanned by {1,i,j,k}

Now consider the section (subspace) spanned by 1 and "k." When did "k" stop being a quaternion?

You are stupid, and even on this unrelated tangent I have just BTFOed you again/

[Anonymous No.10525753]

>>10525732
>Now consider the section (subspace) spanned by 1 and "k." When did "k" stop being a quaternion?
It depends on what "consider" means. If you mean that k is no longer a quaternion then it's when you said "Now consider..." If you mean that k is still a quaternion but you are only looking at a certain section then it never stopped being a quaternion.

[Anonymous No.10525760]

>>10525753
and that section is indistinguishable from C

[Anonymous No.10525764]

>>10525760
If k is not a quaternion it's indistinguishable, yeah. All you're doing is playing word games. The definition is the definition, get over it. None of this guff has any purpose, respond to >>10525528 or admit your argument fails.

[Anonymous No.10525794]

>>10525764
It is a quaternion and the {1,k} plane is indistinguishable from C, therefore the 1D interval [6k,90k] is indistinguishable from the 1D interval [6i,90] and yet you somehow manage to make a distinction in that space between between your ears. Please tell us, how do you do this?

I would say your argument fails but these tarded statements you keep making don't seem like an argument to me. Are you trying to make an argument?

[Anonymous No.10525801]

>>10525794
>It is a quaternion and the {1,k} plane is indistinguishable from C, therefore the 1D interval [6k,90k] is indistinguishable from the 1D interval [6i,90]
Doesn't follow. k is a quaternion and thus distinguishable from i when i is not a quaternion.

And thanks for admitting your argument fails since you didn't respond. B T F O HURR DURR

[Anonymous No.10525807]

>>10525528
>https://planetmath.org/linesegment
This definition says extended reals are allowed, as I showed here:>>10517177

>Does the midpoint correspond to a halfway point? No
"Midpoint" and "halfway point" are synonymous

[Anonymous No.10525810]

>>10525801
>thus distinguishable from i
Please make the distinction in a post then if you think one can be made.

[Anonymous No.10525817]

>>10525807
>This definition says extended reals are allowed
No it doesn't. Why are you lying about text that anyone can read?

>as I showed here:>>10517177
This doesn't show that AB is a line segment, nor can it be by definition.

>"Midpoint" and "halfway point" are synonymous
The midpoint of a line segment has no necessary connection to the halfway point of an interval parameterized over it. They are not synonomous at all.

[Anonymous No.10525821]

>>10525810
I just did.

[Anonymous No.10525823]

>>10525810
>Please make the distinction in a post then if you think one can be made.
For instance, let there be two planes spanned respectively by {1,A} and {1,B}. 1, A, and B are mutually orthogonal unit vectors and
A^2 = B^2 = -1

Which plane is C and which is not?

[Anonymous No.10525829]

>>10525821
you did not, see:>>10525823

[Anonymous No.10525833]

>>10525823
I did, if k is a quaternion and i is not then they are distinguishable by virtue of one being a quaternion and the other not.

[Anonymous No.10525835]

>>10525833
>let there be two planes spanned respectively by {1,A} and {1,B}. 1, A, and B are mutually orthogonal unit vectors and
>A^2 = B^2 = -1
>Which plane is C and which is not?
let there be two planes spanned respectively by {1,A} and {1,B}. 1, A, and B are mutually orthogonal unit vectors and
A^2 = B^2 = -1

Which plane is C and which is not?

[Anonymous No.10525837]

>>10525835
I already showed the distinction, move on.

[Anonymous No.10525839]

>>10525837
You didn't, dumbass. If you could make a distinction then you could say which of A and B is a quaternion.

[Anonymous No.10525846]

>>10525839
You already started with the premise that k is a quaternion, so your question is not analogous.

[Anonymous No.10525849]

>>10525846
Is k "A" or is k "B"?
You haven't distinguished them yet.

[Anonymous No.10525857]

>>10525849
>Is k "A" or is k "B"?
Why are you asking me? You made up the question. Call one a quaternion as you did k and then the question will be analogous to our current situatation.

[Anonymous No.10525863]

>>10525857
>Call one a quaternion
Yes, please do that. Which one do you say it is?

[Anonymous No.10525865]

>>10525863
You made the question and you called k a quaternion. You do it.

[Anonymous No.10525871]

YOU GUYS WANT TO KNOW MY SECRET FOR ULTIMATE INTERNET ARGUING SUCCESS???????????????????????????

INSIDE MY BRAIN I CAN SEPARATE INFORMATION BETWEEN THINGS THAT ARE MY OPINIONS AND THINGS THAT ARE FACTS, AND I DON'T ARGUE ABOUT MY OPINIONS (up to my occasional error-based learning experience)

[Anonymous No.10525876]

>>10525865
You said (You) could make the distinction. I'm telling you that one of "A" or "B" is "k." My belief is that the are indistinguishable, and you have claimed the opposite. Please make the distinction between A and B.

HINT: One of them is "k."

[Anonymous No.10525879]

>>10525876
Earlier, I said "this is k" and you said, "That one is k." That is not a distinction, that is an identity. Now I didn't tell you which one is "k" and it appears to me that you cannot distinguish them.

[Anonymous No.10525880]

>>10525876
>You said (You) could make the distinction.
No, I said that if k is a quaternion and i is not, they are distinguishable. So if A a quaternion and B not? Or vice versa? If not, then your question has no relevance.

[Anonymous No.10525892]

>>10525879
No, earlier you said

>[k] is a quaternion and the {1,k} plane is indistinguishable from C, therefore the 1D interval [6k,90k] is indistinguishable from the 1D interval [6i,90]

I said

>Doesn't follow. k is a quaternion and thus distinguishable from i when i is not a quaternion.

>That is not a distinction, that is an identity.
A quaternion is a distinct identity from a non-quaternion.

>Now I didn't tell you which one is "k"
Yes, which means your question is irrelevant to what was discussed earlier. Try again.

[Anonymous No.10525899]

>>10525880
>k is a quaternion and i is not
I agree that "k" is a quaternion and "i" is not.

>So if A a quaternion and B not? Or vice versa?
It is also true that "A is k" and "B is not," or vice versa.

> If not, then your question has no relevance.
I have conceded everything in your last post and the relevance of my question is implied. Please make the distinction, idiot.

(HINT: It can't be done)

[Anonymous No.10525900]

>>10525892
>Try again.
Ok, here's a stabdalone question which does not depend on anything written in any other post:

Let there be two planes spanned respectively by {1,A} and {1,B}. 1, A, and B are mutually orthogonal unit vectors and
A^2 = B^2 = -1

Which plane is C and which is not?

[Anonymous No.10525902]

>>10525899
>I agree that "k" is a quaternion and "i" is not.
Good, so you admit they are distinguishable.

>It is also true that "A is k" and "B is not," or vice versa.
Good, so you admit they are distinguishable.

>I have conceded everything in your last post and the relevance of my question is implied.
Yes, and now your question is answered.

>Please make the distinction, idiot.
You just did it, one is a quaternion and the other is not. Congratulations retard.

[Anonymous No.10525904]

>>10525900
>Ok, here's a stabdalone question which does not depend on anything written in any other post
Then why should I care? Don't waste my time with irrelevancies.

[Anonymous No.10525909]

>>10525902
>so you admit they are distinguishable.
if they were distinguishable, then you would be able to say which of A and B is the quaternion.

[Anonymous No.10525912]

>>10525904
>Then why should I care?
Because this simple question demonstrates that you are totally wrong

[Anonymous No.10525913]

>>10525909
>if they were distinguishable, then you would be able to say which of A and B is the quaternion.
No, distinguishable just means that you can tell they are different, not that you can tell which is which. You just told me they are different.

[Anonymous No.10525916]

>>10525912
It demonstrates nothing and I've already proven I'm right. Instead of telling me how a quaternion can be the same thing as a non-quaternion, you ask irrelevant questions.

[Anonymous No.10525917]

>>10525912
I have a dime and a nickel, one in my left hand and one in my right. Are the two coins the same?

[Anonymous No.10525921]

>>10525913
If the are distinguishable, then you can "distinguish" one from the other, but you have not been able to. You identified one when I put a label on it, and it kept being a quaternion after I took the label away. Please make the distinction now that I have removed the label.

[Anonymous No.10525924]

>>10525917
>Are the two coins the same?
You first: Let there be two planes spanned respectively by {1,A} and {1,B}. 1, A, and B are mutually orthogonal unit vectors and
A^2 = B^2 = -1

Which plane is C and which is not?

[Anonymous No.10525925]

pls respond

[Anonymous No.10525926]

>>10525921
>If the are distinguishable, then you can "distinguish" one from the other
You just told me one is different from the other. Therefore they are distinguishable.

Distinguishable:
clear enough to be recognized or identified as different; discernible.

>You identified one when I put a label on it, and it kept being a quaternion after I took the label away.
I don't know what "took the label away" means. Either it means it stopped being a quaternion or it didn't. Either way, you're a waste of time.

[Anonymous No.10525930]

>>10525924
Irrelevant, try again.

[Anonymous No.10525952]

>>10525926
Two things are only distinguishable if you can distinguish them.

>>10525930
>Irrelevant
wrong. this question is relevant because it demonstrates that you cannot make a distinction between A and B even when one of them is the imaginary number "i" and the other is the quaternion "k"

>clear enough to be recognized or identified as different
please identify A and B is different. One is "k" and the other "i"

>I don't know what "took the label away" means.
You're stupid them. First I said "k" is a quaternion and "i" is the imaginary number. Then I relabeled them as A and B so that neither has the text string "quaternion" attached to it

[Anonymous No.10525962]

>>10525952
>Two things are only distinguishable if you can distinguish them.
You did that for me by tell me they are different.

>wrong. this question is relevant because it demonstrates that you cannot make a distinction between A and B even when one of them is the imaginary number "i" and the other is the quaternion "k"
If that's true then they are distinguishable. But your original question did not have that information. All you've shown is that you are mentally ill and can't argue logically.

>please identify A and B is different. One is "k" and the other "i"
One is a quaternion and the other isn't.

>You're stupid them. First I said "k" is a quaternion and "i" is the imaginary number. Then I relabeled them as A and B so that neither has the text string "quaternion" attached to it
Then they are different since one is a quaternion and the other is not.

R
E
T
A
R
D

[Anonymous No.10525974]

>>10525952
I have a coin in my left hand and a coin in my right. One is a nickel and the other is a dime. Is the coin in my left hand the same as the coin in my right?

[Anonymous No.10525979]

Let there be two planes spanned respectively by {1,A} and {1,B}. 1, A, and B are mutually orthogonal unit vectors and
A^2 = B^2 = -1

Which plane is C and which is not?

[Anonymous No.10526004]

>>10525979
See the above posts.

[Anonymous No.10526043]

>tfw to brainless to understand a single sentence from this thread
quick rundown?

[Anonymous No.10526051]

>>10526043
A delusional schizophrenic who believes he's God is bad at math.

[Anonymous No.10526065]

>>10526051
>/sci/ has an e-celeb thread
you guys are as vapid as /mu/

[Anonymous No.10526100]

>>10526051
>believes he's God
Gedalia really is my Hebrew name.
God is to Gedalia as Jon is to my English name Jonathan.
https://reformjudaism.org/practice/ask-rabbi/why-do-some-jews-write-g-d-instead-god