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10425928 No.10425928 [Reply] [Original] [archived.moe]

>> No.10425953 [DELETED] 

>>10425928
It's indeterminate.
[math]\dfrac{0}{0} = x[/math]
[math]0 = 0x[/math]
x can be any number.

>> No.10425957

>>10425953
But the number 1 is clearly larger than 0.999...

>> No.10425960

>>10425957
It’s the same as (1-1)/(1-1)

>> No.10425961

It has to be 1. They’re going to 0 at the same rate.

>> No.10425964

>>10425960
>>10425961
So which is it? Two different answers here.

>> No.10425967

>>10425964
real answer: divide by zero error. this doesn't indicate any sort of limit, instead 0.999... = 1 by definition so the denominator is illegal, t. NYPD

>> No.10425977

>>10425967
Thank you

>> No.10425980

>>10425961
They aren't going anywhere your dofus.
OP's image is a png, not a gif or a webm

>> No.10425985

>>10425928
Aaahhhh can you shut uppp please

>> No.10425988

let f(x) = -1 / x
limit(x->inf) f(x) = -0

let g(x) = f(x) + 1

limit(x->inf) g(x) = 1


Also, since g(x) is an increasing function for all x > 0, this fits the conventional definition of "0.9999..."

limit(x->inf) g(x) = 0.999...


So:

(1 - g(x) / 1 - g(x))

(1 - (-1 / x + 1)) / (1 - (-1 / x + 1))

(1 + 1/x - 1) / (1 + 1/x - 1)

(1/x) / (1/x)

x / x

And since we established that x > 0, we haven't divided by zero at any point, so this is 1, not undefined

>> No.10425997

>>10425988
yeah but your proof is invalid since g(x) is a function and not a number. l'hopital's rule doesn't exist for numerical expressions

>> No.10425998

This is dumb and you should feel bad.

>> No.10426007

>>10425997
I'm didn't invoke L'opital's rule here.

I showed that 1 - g(x) / 1 - g(x) is equal to 1 for every x > 0.

>> No.10426010

>>10426007
fine, then that has nothing to do with dividing by 1-0.999...

>> No.10426012

>>10426010
How else does 0.999... exist "in the wild", other than as a limit of a function?

>> No.10426013

>>10426007
>L'opital

>> No.10426017

>>10426012
you could consider "1" as just an alternate notation for 0.999... or 2-0.999... or 6-4.999..., decimal notations are om general not unique and that phenomenon exists in the wild

>> No.10426025

This whole thread is bad, it’s based off of a problem missing pieces, leading us to all assume things, not to mention that the whole 0.999... is 1 thing is also just a cheap way that our society threw out a seemingly pointless problem (we all use the logic 0.999... is 1 just so we don’t end up running in circles) without that flawed logic we could also say in reply to “real answer” that 1-0.999... is greater than zero and if you divide the same number (that is greater than zero) by the identical number (that is greater than zero) then it equals 1

>> No.10426027

>>10426025
>t. mathlet

>> No.10426059

>>10426025
Conventional math is different when dealing with infinites. Infinity / Infinity is undefined

>> No.10426073

>>10425928
.111...
--------- = 1
.111...

Constant divided by equivalent constant is 1, no matter the precision of the constant.

>> No.10426078

>>10426073
nuh uh, 0.000.../0.000... = undefined

>> No.10426080

>>10426078
Oh woops, brain fart.

>> No.10426083

>>10426073
>1 - 0.99 = .11

>> No.10426113

>>10426083
That's what I meant, but it's still an irrational constant. Analyzing it's indeterminate nature is useless because of this.

The numerator and denominator are infinitesimally small, but more importantly, they're infinitesimally equivalent. Their ratio remains the same because of this, at 1.

>> No.10426130

>>10426013
el hospital

>> No.10426146

>>10426078
>>10426078
Lol wut 1-0.999... will be indefinitely small but ultimately greater than zero still (ie. 0.0000............01) so that over the same thing = 1

>> No.10426175

>>10426146
>1-0.999... will be indefinitely small but ultimately greater than zero still (ie. 0.0000............01)
Bzzt. Wrong.

>> No.10426184

>>10426146
>(ie. 0.0000............01)
That's not how it works, if you know the end of the sequence it's not infinite, which means it is not 0.999....

>> No.10426250

>>10425928
its one stupid

the top is numerator and denominator are equal

>> No.10426258

>>10426184
>it's not infinite
yes it is, if it has '...' it's infinite

>> No.10426264

>>10426012
> I heard smart people say "in the wild"

>>10426025
Take your meds

>> No.10426265

>>10425988
>>10426007
The correct answer is undefined. Real numbers are defined as the limit of a cauchy sequence of rationals. The number one could generally be defined more succintly (with ordinal sets), but the cauchy sequence {.9, .99, .999, . . . } is just an obstuse manner of describing the number 1.

If this is unclear, we have:
a = (.9999. . .)(10) = 9.9999. . .
while
b = (1)(10) = 10
So that (9/10)a = 9.999. . . - .999. . . = 9
and (9/10)b = 9

Hence (9)(0.999. . .) = 9 = (9)(1) so that by cancellation 0.999. . . = 1

>> No.10426281

1 in math, divide by zero error in inferior computation.

>> No.10426282

>>10425961
They are not "going to 0 at the same rate". The numerator and denominator are identically zero. By convention, 0/0 is left undefined.

>> No.10426350

>>10425961
>numbers are little diesel engines chugging along
topkek

>> No.10426367

[math] \displaystyle
\boxed{0 < p < 1} \\
p^n-1 = (p-1)(p^{n-1}+p^{n-2}+ \dots +p+1) \\
\dfrac{p^n-1}{p-1} = \sum \limits_{j=0}^{n-1}p^j \\
\displaystyle
\lim_{n \to \infty} \dfrac{p^n-1}{p-1} = \lim_{n \to \infty} \sum \limits_{j=0}^{n-1}p^j \\
\displaystyle
\dfrac{0-1}{p-1} = \sum \limits_{j=0}^{\infty}p^j \implies \dfrac{1}{1-p} = \sum \limits_{j=0}^{\infty}p^j
[/math]

>> No.10426395

>>10425928
[math]\frac{1-\sum_{k=1}^{\infty}\frac{9}{10^k}}{1-\sum_{k=1}^{\infty}\frac{9}{10^k}} [/math]

Then see where it converges. Don't have time to do it correctly now and calculating in [math]\LaTeX[/math]is hard.

>> No.10426449
File: 112 KB, 953x613, C7473FA965194B0B8FE526924E00C01D.jpg [View same] [iqdb] [saucenao] [google] [report]
10426449

>> No.10426573

P = (1-0.999...)
(1-0.999...)/(1-0.999...)
= P/P
=1
:^]

>> No.10426579

>>10426449
So what's the trick in each one of those?

>> No.10426581

>>10426579
the trick is that 0.999... and 1 are different ways to write the same thing

>> No.10426584

looks like you've discovered a situation that causes two axioms to come into conflict

either 0/0 is undefined, therefore x/x is not always 1, or 0/0 is 1, therefore 0/0 is not undefined

>> No.10426585

>>10426573
Nigger

>> No.10426588

>>10426581
Brainlet

>> No.10426589

>>10426588
retarded brainless fungus

>> No.10426590

>>10426588
>>10426589
>typical /sci/ discussion

>> No.10426592

>>10425928
surely anything divided by itself is 1?

>> No.10426594

>>10426592
huurrr

>> No.10426598

>>10426592
>b-but muh undefined

>> No.10426599

>>10426594
It isn't zero it's a number greater than zero, if you want to debate zero divided by zero that's a different bag of worms, some number greater than zero divided by itself is one

>> No.10426600

>>10426579
>>10426581
when you divide by 1 by 3 there is a remainder. 10/3 = 3+R1
100/3= 33+R1
they just choose to ignore it

>> No.10426606

[math]
x= \frac{1}{10} \\
0. \overline{9}=9x+9x^2+9x^3+9x^4+ \cdots \\
0. \overline{9}=9x \left (1+x+x^2+x^3+ \cdots \right ) \\
0. \overline{9}=(1-x) \left (1+\mathbf{x}+x^2+\mathbf{x^3}+x^4+ \cdots \right ) \\
0. \overline{9}=1-x+ \mathbf{x-x^2}+x^2-x^3+ \mathbf{x^3-x^4}+x^4-x^5+ \cdots \\
0. \overline{9}=1
[/math]

>> No.10426607

>>10426599
>1-1 greater than zero
bravo retard

>> No.10426627

>>10426600
it's exactly [math]0.1_3[/math]

>> No.10426631

>>10426449
>hyperreals dont exist
brainlet detected

>> No.10426641

Wow this thread really reveals how mathematically inept the people on this board are. Anyone who does not immediately know the answer is 1 should forever leave.

>> No.10426652

>>10426606
Wrong. Wow you are so stupid. Your calculation has a blatant mistake. Not just the first but also the last element does not cancel out. Therefore the jump from line 5 to line 6 is wrong.

>> No.10426655

>>10426652
> the last element
no such thing

>> No.10426657

>>10426641
so leave

>> No.10426662

>>10426655
Yes there is. Your mind is too small to see it.

>> No.10426670
File: 1005 KB, 220x260, 1545869393864.gif [View same] [iqdb] [saucenao] [google] [report]
10426670

>>10426631
>hyperreals are a subset of reals

>> No.10426687
File: 8 KB, 169x220, x.jpg [View same] [iqdb] [saucenao] [google] [report]
10426687

>>10426662
sure bud, you have the large mind

>> No.10426827

>>10425928
A 1-0.999.../1-0.999...=0/0=error

B 1-0.999.../1-0.999...= 1-1-1=-2

C 1-0.999.../1-0.999...=0.00...01/0.00...01=1

I know that A is correct but couldnt B be right?
Or that is only if you dont have it set up like this

>> No.10426922

>>10426827
For B:
1 - (0.999.../1) - (0.999...) = ?
1 - 0.999... - 0.999... = -0.999...8

>> No.10426925
File: 502 KB, 1080x1451, x7wg0u7xu8j21.jpg [View same] [iqdb] [saucenao] [google] [report]
10426925

>>10426662
VERY big brained my guy! Good job!

>> No.10426930

>>10425928
You would get 0/0.
0.999999 = 1 has been shown to be true for a long time.

>> No.10426934

>>10426930
>0.999999 = 1 has been shown to be true for a long time.
nope, it's 0.9... = 1

>> No.10426945

>>10426258
nope
an infinite number wont have an end

>> No.10426956

>>10426945
>an infinite number wont have an end

correct, that's why writing anything after '...' is meaningless

>> No.10427550

>>10425928
so waht is the answer?it's 1 isn't it?

>> No.10427562

1 - 0.999... = 1 - (1 - 1/inf) = 1/inf
(1/inf) / (1/inf) = inf/inf = undefined
checkmate matheists

>> No.10427592

Why can't we just say .999... is a poorly defined expression (leaves ambiguity in value when contested with 1) and that the question itself is flawed.

I don't think the term .999... itself should really come into question here. This shit is a funny allegory for the kind of thinking that's been fucking up quantum mechanics for years now.

>> No.10427602

[math]\lim_{n \to \infty}1^n = 1[/math]. But is [math]\lim_{n \to \infty}0.999...^n = 1[/math]?

>> No.10427606

>>10427592
because it is a well defined expression

>> No.10427616

>>10427592
>.999... is a poorly defined expression
It's perfectly well defined, brainlet

>> No.10427617

>>10427592
Because it is perfectly well-defined. The fact that mathlets don't understand it is immaterial.

>> No.10427618

>>10427606
How so?

>> No.10427621 [DELETED] 

>>10427602
[math] \displaystyle
\lim_{n \to \infty}1^n = 1 \\
1^ \infty \textsf{ is undefined} \\
\lim_{n \to \infty}0.9...^n = 1
[/math]

>> No.10427622

>>10427602
Yes.

>> No.10427628

>>10427602
[math] \displaystyle
\lim_{n \to \infty}1^n = 1 \\
1^ \infty \textsf{ is undefined} \\
\displaystyle
\lim_{n \to \infty}0.9...^n = 1
[/math]

>> No.10427631

>>10427618
[math]
1 = \dfrac{3}{3} = 3 \cdot \dfrac{1}{3} = 3 \cdot 0.\bar{3} = 0.\bar{9}
[/math]

>> No.10427641

So... if you guys are gonna insist on this... then would .999...99 = .999...?

Also, isn't the fraction a better expression for what you're trying to express? Why bother switching to decimal?

>> No.10427642

>>10427641
>.999...99 = .999...?
yes >>10426956

>> No.10427645

>>10426657
Baited

>> No.10427650

>>10427645
look what the cat brought in

>> No.10427651

I'm thinking the guys here don't really care about context... I would think it's common sense that .999... in the current setting could result in ambiguity of an infinitely small number divided by itself or 0/0.
And if 1-.999... isn't a good way to describe an infinitely small number... then what is?

>> No.10427656

>>10427651
https://en.wikipedia.org/wiki/Construction_of_the_real_numbers

>> No.10427661

>>10427656
That's a cool wiki page i agree... Is there any context you could give me where there's utility in using .999... instead of 1 though?

>> No.10427689
File: 99 KB, 414x446, smug perturabo.jpg [View same] [iqdb] [saucenao] [google] [report]
10427689

Imagining believing that one branch of maths acts consistently in all situations

>> No.10427692

Can anyone here give me a context where using .999... instead of 1 would provide more utility?

If not... I'm gonna go ahead and keep spouting that .999... is ill defined when used with 1.

>> No.10427705

>>10427617
Yup. It simply becomes a very small number divided by itself. Infinitely small, but not zero, and because it's divided by itself the answer is 1.

>> No.10427709

>>10425961
good bait bro

>> No.10427804

>>10427661
1/3 = 0.333...
3 * (1/3) = 0.999...

>> No.10427893

>>10427705
>Such a brainlet that they didn't even know that the guy they responded to was talking about the well-definedness of the expression 0.999..., and not the well-definedness of OP's expression.
>Such a brainlet that, not only do they make the above mistake, but they conclude that a fraction equivalent to an indeterminate form (0/0) is a well-defined expression.
Time to stop posting.

>> No.10427907 [DELETED] 

>>10427893
The answer is still 1 no matter how hard you many people you call brainlet

>> No.10428051

>>10427705
>Infinitely small, but not zero
False
You don't understand what infinite means

>> No.10428054

>>10428051
You don't know what zero means

>> No.10428072

>>10427705
>Infinitely small, but not zero

There are no infinitesimals in R you retarded ape, R is an Archimedean field, this is not valid

You could use any formulation of the completeness axiom you wanted to show infinitesimals aren't valid in R.

>> No.10428176

>>10428072
>it is wrong because I arbitrarily define it to be so

>> No.10428199

>>10427804
But you forgot the remainder.
Consider
a)10/3 = 3+R1
b)100/3 = 33+R1
c)1000/3 = 333+R1
d)10000/3 = 3333+R1
e)100000/3 = 33333+R1
Oh wait there's that pesky remainder you keep forgetting about. It's negligible in many everyday computations, that's why the math works so well above and beyond. Not to mention random noise outright discards this noise. But you know darn right that remainder will be there even in an infinite sequence. Maybe it is clear now, or maybe you are still skeptical but 1/3 will yield a remainder now matter how far you go because 10/3 will never be equal to 3 evenly. That is a mathematical fact so basic that higher levels can only forget about it when it becomes infinitely small(not 0).


So as you can probably see now, 0333... != R1+0.3333.....; these are two different numbers not a way to write

>>bbbbut, since 0.333... is infinite you will never get to add back in the remainder. That is true. But what is also true, is that this can be rewritten as a finitely long number: 0.333(10/3) and likewise this number can be written as 0.3333(10/3) = 0.33333(10/3)=0.333333333333333333(10/3).
It should not be hard to see that if I multiply this number by 3, the result is identically 1.

>> No.10428202

>>10425928
Are you serious?

>> No.10428986

>>10428199
>1/3 currently does not have an exactly equal (aka valid) translation to decimal

In base-3 it is exactly 0.1 = 1*3^(-1).
There is nothing 'clean' about base-10.
1/10 looks nice in base-10 because monkeys have 10 fingers and we created the symbols accordingly.
In hexadecimal (base-16) it's a fucking mess:
0.1999...

https://www.rapidtables.com/convert/number/decimal-to-hex.html

>> No.10428994

>>10428176
>arbitrarily define
look who's talking

>> No.10429008

Well, 1-1 equals zero. So zero divided by zero is DNE. Why is this a fucking question?

>> No.10429011
File: 417 KB, 1846x867, the absolute state of sci.png [View same] [iqdb] [saucenao] [google] [report]
10429011

ITT

>> No.10429038

>>10429011
this, but also for literally every thread on sci

>> No.10429152

>>10425928

you are asking what is the limit of x/x as x approaches 0 from the positive side

this is not mind blowing mathematics guys the answer is 1. you can graph it on desmos or something its first semester college maths.

>> No.10429164

>>10429152
>you are asking what is the limit of x/x as x approaches 0 from the positive side
no he's not

>> No.10429185

>>10429164
this

>> No.10429190

>>10428986
Changingbase also requires a change in definitio of division so....

>> No.10429197

>>10429190
[citation needed]

>> No.10429239

>>10429197
Umm how about binary? even additionand subtractionare not the same hence anstract algebra

>> No.10429242

>>10429239
different bases arent different algebraic structures
youre thinking of modular numbers

>> No.10429311

>>10429239
1+1=10 --- 1+1=2
10+1=11 --- 2+1=3
11+1=100 --- 3+1=4

it's exactly the same, wtf are you talking about.
the numbers look different, but the math is the same

>> No.10429379

>>10428054
Mathlet please leave

>> No.10429387

>>10429379
So a number greater than zero is zero?

>> No.10429398

>>10429387
1/inf =0

>> No.10429652

>>10425928
if 0.999... is a function approaching 1 as it's argument approaches inf, then the answer is 1.
If 0.999... is a number then it's 0/0 so indeterminate.

>> No.10429791

>>10429652
>approaching
'...' means it's already at infinity, not approaching

>> No.10429848

>>10429652
>if 0.999... is a function
It is not.

>If 0.999... is a number then it's 0/0
Indeed.

>so indeterminate.
No. That word doesn't mean what you think it means.

>> No.10430227

>>10425928
zero fuck you if u wanna read all these real analysis bullshit

>> No.10430265

>>10429848
that what does indeterminate means good sir?

>> No.10430290

>>10430265
something exists but is unknown

undefined = doesn't exist

>puts feeding spoon back in drawer

>> No.10430293

>>10429791
when do we arrive at infinity ?

>> No.10430310

>>10426579
Part of it is psychological. The bottom left one where it has "9x = 9" is incorrect, its actually "9x = 8.999...1". The repeating number 9 in the ones above makes you more susceptible to accepting the 9 as correct.

>> No.10430330
File: 3 KB, 635x223, r8.png [View same] [iqdb] [saucenao] [google] [report]
10430330

>>10430293
>no approaching
did I fucking stutter?

>> No.10430333

>>10430310
8.999...1 = 8.9... = 9

>> No.10430907

>>10425928
If 0.999... = 1, does 0.0999 = 0.1?

>> No.10430935

>>10430907
yes

>> No.10430942 [DELETED] 

>>10430907
you mean 0.09... = 1
yes

>> No.10430945

>>10430907
you mean 0.09... = 0.1
yes

>>
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