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/sci/ - Science & Math

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11669397 No.11669397 [DELETED]  [Reply] [Original] [archived.moe]

Let's start with the function y=-cos(x).
set x=pi.
therefore, y=-cos(pi)=1.
now lets go backwards.
assume that you have 0.999...=-cos(x)
if 0.999...=1, then 0.999...=1=-cos(x) where x=pi

however, this is not the case, because pi is an irrational and completely predefined number for all of its digits associated with it i.e. 3.14159265358979323846264338327950288419716939937510... ad infinitum.

so setting y=0.999... would yield a value of x that is infinitesimally close to pi but not pi. you might ask which digit of pi is altered to achieve such an infinitesimal aberration from the number one, but it less as elusive as asking where is the missing 0.000... ...01 from the whole number 1 to make 0.999...

the existence of pi is indisputable. each number is as important as the number that precedes it towards infinity, and it is wholly unique. if 0.999... = 1, then you are denying the absolute existence of pi and you should be ashamed of yourselves.

if 0.999... = 1, then 1=2 and 0=3.

>> No.11669549

>so setting y=0.999... would yield a value of x that is infinitesimally close to pi but not pi
This requires the assumption that 0.999... =/= 1, making your argument circular

>> No.11669569

3 < π
3.1 < π
3.14 < π
3.141 < π
3.1415 < π
3.14159 < π
3.141592 < π
this continues endlessly
π < π

>> No.11669645


y=1 if and only if x=pi ... exactly

>> No.11669685

Are you retarded? Did you miss the fact that it approaches 1 as the argument approaches pi?

>> No.11669782

Super EZ proof that 0.999... = 1
1/3 = 0.333...
3 x (1/3) = 3 x 0.333...
1 = 0.999...

>> No.11669797

Yes and if 0.999... = 1 then arccos(0.999...) = pi exactly. Which it does, so it does.

>> No.11669820 [DELETED] 

if you proved pi < pi we have bigger issues than 0.99.. not being 1

>> No.11669826
File: 112 KB, 953x613, 1=.999....jpg [View same] [iqdb] [saucenao] [google] [report]


>> No.11669837 [DELETED] 

cool image, but "by induction"...? didn't get that

>> No.11669855
File: 234 KB, 610x610, problem-solver.jpg [View same] [iqdb] [saucenao] [google] [report]

You've got a lot of reading to do.
I'll even give you your first link: https://en.wikipedia.org/wiki/Mathematical_induction

>> No.11669857

You're only proving one case. I don't know what you think induction is, but that isn't it.

>> No.11670753

I can disprove that theorm by make the observation that 0.99... and 1 are not the same numbers. mathtards btfo

>> No.11670834

do it big boi

>> No.11670847
File: 1.30 MB, 1280x720, duck is perplexed.png [View same] [iqdb] [saucenao] [google] [report]

I-i...I just did?

>> No.11670853
File: 331 KB, 300x221, done.gif [View same] [iqdb] [saucenao] [google] [report]

I make the observation that you're an idiot.
>wow it really works

>> No.11670860

>You might think every number has a unique decimal expansion. You would be wrong
the absolute lack of argument - made only worse by the fact that the image clearly contains actual arguments and thus would include support for this claim if it could - only strengthens my suspicions that the 0.999... fags are on to something

these threads will continue to be posted until /sci/ actually learns mathematics and produces a satisfying explanation.

>> No.11670889

\boxed{0 < p < 1} \\
1 = p + (1-p) ~~~~~~ \overset{1}{ \overbrace{[=====p=====|==(1-p)==]}} \\
\text{divide p using x} ~~~~~~ \overset{1}{ \overbrace{ \underset{p}{[ \underbrace{=====x=====|==(p-x)==}]} ~~ + ~~ (1-p)}} \\
\text{solve x and (p-x), when length ratios must be the same} \\
\dfrac{x}{p-x}= \dfrac{p}{1-p} \Rightarrow x- xp = p^2 - xp \Rightarrow \underline{x=p^2} \Rightarrow \underline{(p-x)=p(1-p)} \\
\overset{1}{ \overbrace{ \underset{p}{[ \underbrace{=====p^2=====|==p(1-p)==}]} ~~ + ~~ (1-p)}} \\
\overset{1}{ \overbrace{ \underset{p^2}{[ \underbrace{=====p^3=====|==p^2(1-p)==}]} ~~+ p(1-p)+(1-p)}} \\
\overset{1}{ \overbrace{ \underset{p^3}{[ \underbrace{=====p^4=====|==p^3(1-p)==}]} ~~+ p^2(1-p)+p(1-p)+(1-p)}} \\
(1-p)+p(1-p)+p^2(1-p)+p^3(1-p)+ \cdots =1 ~~~~ \left | ~ \times \frac{1}{1-p} \right . \\
1+p+p^2+p^3+ \cdots = \dfrac{1}{1-p}

>> No.11670897


CRINGE. I imagine a fat slob who loves anime and does math to feel cool.

>> No.11670908

Nice b8

>> No.11670984

what is this math even
written so neatly i have never seen such style before

>> No.11671018

>Guys I proved that A =/= A

>> No.11671021
File: 471 KB, 717x768, Retardscreachingexcitedly.png [View same] [iqdb] [saucenao] [google] [report]

lol retard it clear says there in the op that B =/= A
what are you literally brain damaged?

>> No.11671031

pi = lim(n->inf) (cos(360/n) * n)/2

>> No.11671039

sorry, replace cos with sin

>> No.11671101

I'm not approaching 1 I am approaching 0.999... which is an infinitesimal step below 1.
let us see the iterations as y approaches 0.999... not 1.

where the exact number pi = 3.14159265358979323846264338327950288419716939937510...

looking at it logically, each 9 appended to the number 0.999... adds a greater deal of accuracy to the number pi.

relatively speaking (looking at the last arccos iteration), it takes about 19 9's to get 6 correct decimal digits of pi, which isn't very fair. If the resolution of 0.999... is not proportional to the resolution of pi in the limit, then it means that the ideal realization of pi with "catch up" later to all of those 9's appended in the limit or even never.

tldr it takes a lot of 9's to make pi, therefore each digit of pi has some amount of (9/10)^n in the function that help build it.

to make it clear i am not approaching 1 I am approaching 0.999... in the limit.

>> No.11671125
File: 145 KB, 720x405, 70b5cd_e588f78e634444c3a45b56418f9dd38f-mv2.jpg [View same] [iqdb] [saucenao] [google] [report]

sqrt of minus one (0.000...1) equals minus one

>> No.11671166

so the square root of minus one (0.000...1) = minus one (0.000...1) cause placeholder 0 and negative numbers do not exist

>> No.11671227
File: 86 KB, 820x995, merlin-magician-royalty-free-clip-art-png-favpng-Z5ZaejkSui63qLKPWsiSUUYcE.jpg [View same] [iqdb] [saucenao] [google] [report]

Zero (0) is simply Shorthand for Minus One (0.000...1) and for Infinity (∞) because Both are exactly Equal completing the Definition of so-called 'undefined'.

>> No.11671280


>> No.11671372

cos every1 is a sinner

>> No.11671425

Here's a simple argument why 0.99... != 1

If two numbers a and b are the same, then clearly for any function f, we must have that f(a) = f(b). This is just common sense.

Define f to take in any real number, and return the digit in the tenths place.

Then f(1) = 0, but f(0.999...) = 9. Therefore 0.999... != 1.

>> No.11671471

american education, folks

>> No.11671477


>> No.11671509
File: 254 KB, 512x512, 1588947543986.gif [View same] [iqdb] [saucenao] [google] [report]


>> No.11671515 [DELETED] 


>> No.11671519 [DELETED] 


>> No.11671522 [DELETED] 


>> No.11671525 [DELETED] 



>> No.11671545

>let's pretend finite is infinite

>> No.11671553

up your ass

>> No.11671563 [DELETED] 

I could do it until the end of time and there would still be no 1 faggot.

>> No.11671571

oh sweetie, that's why inf isn't a number

>> No.11671583

One of the most basic properties of the real numbers is the Archimedean property, part of which states that there are no infinitesimally small real numbers. More exactly, it states that for all real numbers greater than zero, there are natural numbers n such that 1/n is less than the real number.
This is provable from an even more fundamental property of the reals, the least upper bound property. It states that any nonempty set of real numbers which is bounded above has a least upper bound: a number which is >= all elements of the set, and <= all upper bounds of the set.
Let S be the set of infinitesimals in the reals. S is obviously bounded above by 1, because there wouldn't be infinitesimals greater than one. Then, assume by contradiction that S is nonempty. This means that S has a least upper bound I'll call r.
r < 2r, and since r is an upper bound, 2r cannot be in S, so can't be infinitesimal. Thus, we can pick a natural n such that 1/n < 2r.
r/2 < r, and since r is the least upper bound, there must be at least one infinitesimal i > r/2.
However, 1/(4n) < r/2 < i, meaning i is not infinitesimal. By contradiction, S is empty.

>> No.11671588

Let f be the function that returns the denominator of any fraction. f(1/2)=2, but f(2/4)=4. Thus, 1/2 != 2/4.

>> No.11671614 [DELETED] 

At no point during an eternity is there even the mechanism for a 1 to appear.

Face it mathcuck, you've been hoodwinked into believing illogical fantasies.

>> No.11671618

Where's the 2?

>> No.11671653

x = 0.333...
10x= 3.333...
10x-x = 3.333... - 0.333...
9x = 3
3x = 1
x = 1/3 = 0.333...
x = 0.33oh wait you just realized that the "proof by induction" presented in your picture is fucking retarded and is just an aberration of humans using decimal to represent numbers being unable to deal with that situation very well333...

>> No.11671661

>during an eternity
lol, what on earth is going on in that pretty little head of yours

>> No.11671717 [DELETED] 

6/3 is calculation, not a number

0.9999... is a numerical expression

eternity by definition lasts forever. Are you dumb?

>> No.11671721

very neat

>> No.11671726

i could write pi forever and still not be able to make a perfect circle out of it

perfect circles must be mathematically impossible

>> No.11671752

6/3 is a number, just like 1+1 is a number, 0.5 is a number 7*2 is a number, and the infinite sum from n=1 to infinity of 9/(10n) is a number. It doesn't matter how you label something, as long as it equals a number, it is a number.

>> No.11671784 [DELETED] 

This is literally correct. Perfect circles are mathematically impossible.

>> No.11671801

why do people still comment on these threads?

>> No.11671813

you just -said- that it's not the case. but it in fact is true that 0.999...=1=-cos(pi).

>> No.11671918

>by definition
ok, here's an actual definition:
An unbounded quantity that is greater than every real number.

nothing about 'time', nimrod

>> No.11671929

>wipes sweat off forehead

>> No.11671979 [DELETED] 

Nice irrational set you have there. Shame you can't enumerate it.
Just tell me you're a midwit next time, things will be a lot quicker.

>> No.11671997

>enumerate it
only losers like you need to do that.
it's perfect, cope

>> No.11672059 [DELETED] 

It's not real if you can't yield it.

>> No.11672265

You are absolutely correct OP, nice proof this shows proof that you have a creative mind.

Also it's kind of wierd that this is the exact way that I thought of the .999 problem.

>> No.11672588

OP here.
are you... me?

>> No.11672709

>so setting y=0.999... would yield a value of x that is infinitesimally close to pi but not pi
you assumed that .999... = 1 so this is a non sequitor

>> No.11672734

why do these threads survive lol

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