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8808643 No.8808643 [Reply] [Original] [archived.moe]

why does 0.999... = 1

is any number with an infinite amount of repeating nines equal to the next whole number in the ones place?

does 314.999... = 315?

>> No.8808660

>>8808643
314.999... = 314+0.999... = 314+1 = 315

But proving it for all integers has eluded mathematicians for centuries and is one of the Millenium Prize problems.

>> No.8808677

>>8808660
How does not even work though?

How do you go from a number that's not one to a number that is?

if 0.999... = 1 then 1.000... = 0.999...

which would then imply that 1.000 - 0.999... = 0

and that is clearly false.

>> No.8808678

>why does 0.999... = 1
Because 0.999... is equal to 0.9 + 0.09 + 0.009 + ... ; that's what the notation means. And evaluating expression, through the standard algebraic techniques for solving these types of infinite sums, yields 1.

>is any number with an infinite amount of repeating nines equal to the next whole number in the ones place?
>does 314.999... = 315?
Yes.

>> No.8808681

>>8808677
>How do you go from a number that's not one to a number that is?
0.999... IS one, so this question does not make sense.

>if 0.999... = 1 then 1.000... = 0.999...
>which would then imply that 1.000 - 0.999... = 0
Indeed.

>and that is clearly false.
No, it is quite true.

>> No.8808693

>>8808643
>why does 0.999... = 1
Because it doesn't terminate. You'd agree that 0.9 is closer 1 than 0.5, right? So then you'd also agree that 0.999 is closer to 1 than 0.99. Now just continue doing that, we can get as close to 1 as we want.

>> No.8808727

look up geometric series

>> No.8808731

[math] \displaystyle
1 = \frac{3}{3} = 3 \cdot \frac{1}{3} = 3 \cdot 0. \overline{3} = 0. \overline{9}
[/math]

>> No.8808733

Basically 0.9999... sums to 9/9 at infinite terms 9/9=1

>> No.8808741

If you don't think it is true then what number would be in between 0.999... and 1? There isn't one.

>> No.8808751
File: 112 KB, 953x613, 0.999_1.jpg [View same] [iqdb] [saucenao] [google] [report]
8808751

>> No.8808753

>>8808741
>there's no natural number between 1 and 2, therefore 1=2
nice argument brainlet

>> No.8808813

>>8808751
but what if there is actually not 0.333.... in the glass

>> No.8808815

>>8808677
How is it clearly false?

>> No.8808821

>>8808677
the "minus" operation can be loosely translated as "what is difference between such in such".
Answer me, numerically, what is the difference between 1.000 and 0.999..?

>> No.8808822

>>8808753
There are real and rational numbers between them.

>> No.8808826

>>8808813
Then you're a greedy faggot who doesn't want to share precious water.

>> No.8808843

>>8808821
1 is a whole 0.999... is not

>> No.8808846

>>8808843
What's missing?

>> No.8808854

if 0.999... =1 then does 0.888... = 0.89?

>> No.8808857

>>8808846
the number n such that n + 0.999.. = 1

>> No.8808859

>>8808854
Try subtracting them.

>> No.8808862

>>8808643
The idea of real numbers is to have a set of number such that to each point on the line there would correspond exactly one point. So we can think of real numbers as points on the line.
Now let's try to figure out how to represent those numbers in decimal form. Mind you, we identify numbers with points on the line. So take one point on the line and call it 0, now take another point that is to right from our point 0 and call it 1.
Now we will try to assign a decimal to each point between points 0 and 1.
Divide the interval between 0 and 1 into 10 equal parts, and label them from left to right with 0,1,2,3,...,9.
Now for each of those interval do the same thing. So for example, you take the interval with the label 1 divide it into ten equal parts and label each of them with numbers from 0 to 9.
At this stage you divided the interval into 100 equal parts and you can identify each of those parts by two numbers that range from 0 to 9. And for example the 100th interval here is identified by the pair (9,9).
Now for each of the 100 interval you repeat the process of division in to ten parts and labeling them with numbers from 0 to 9.
Now you clearly see you can continue this process as long as you want, and at the n-th step of this process you will have divided the interval from 0 to 1 into 10^n parts and you can identify each part with an n-tuple of numbers from 0 to 9.
Now here comes the trick, if you would take any point P on the interval from 0 to 1 at each step of the construction there is only 1 interval in which that point is contained.

>> No.8808867

>>8808857
It's called zero and is expressed this: 0

>> No.8808870

>>8808862
(cont.)
Moreover, for each other point P', no matter how close to P, there is a step in the construction and an interval from that step such that P is in that interval and P' is not.
From this remarks you see that each point on the interval is defined by an infinite sequence of interval, where each step n of the sequence is the unique interval from the n-th step of the construction that contains P. But each of those intervals has an unique label - a n-th tuple of numbers from 0 to 9. Now we will identify each number on the interval as the sequence of those n-tuples. Instead of the sequence of the n tuples w can think of each points ans an infinite string of numbers fro 0 to 9. Now we will take take infinite string of numbers, put "0." and call this new string of characters is a decimal representation of the real number P. Now my question to you is, what is the decimal representation of the ponit 1?

>> No.8808891

>>8808731

I guess the question here is, why is 1/3 == .333....? Are they literally the exact same?

>> No.8808896

>>8808891
Try the division

>> No.8808897

>>8808867
n has to be larger than zero

>> No.8808901

>>8808897
Then you'd be making a false statement

>> No.8808902

>>8808677
>How do you go from a number that's not one to a number that is?

"The father of Ivanka Trump"
and
"The current president of the USA"
is the same person.

0.999...
and
1
are the same number.
And so is
1/2 + 1/2
and
1/4 + 3/4

2/6
and
1/3
and
0.333...
are the same numbers too

>> No.8808934

if 0.999... = 1 then 0.999... + 0.999... should = 2

but it's not possible to add together infinite numbers because you'd be adding for an infinity long time.

>> No.8808957

>>8808934
>if 0.999... = 1 then 0.999... + 0.999... should = 2
Indeed it is.

>but it's not possible to add together infinite numbers because you'd be adding for an infinity long time.
You can't do computable arithmetic on these representations of numbers, but that doesn't mean the algebra doesn't go through.

>> No.8808967

>>8808693
Except not TO one

>> No.8808972
File: 76 KB, 325x282, image (3).png [View same] [iqdb] [saucenao] [google] [report]
8808972

>>8808693
That logic isn't sound in the case of equality. Its only sound in the case of equivalency. The real proof is as follows:

0.99~ = 0.33~ * 3
0.33~ = 1/3
1/3 * 3 = 1
Therefore 1 = 0.33~ * 3
OR 1 = 0.99~

>> No.8808977

>>8808643
The proof's pretty straightforward. I think someone posted it here last time someone started a threat like this.

let x = 0.99....
10x = 9.999
10x - x = 9
9x = 9
x=1

Alternatively,
1/3 = 0.33...
0.33 * 3 = 0.99...
1/3 * 3 = 3/3 = 1
0.99... = 1

>> No.8808979

>>8808897
Can you justify that?

>> No.8808992

>>8808677
Okay, so, we can think about this as "distances between two numbers."
Proposition
If a - b != 0, a and b are different numbers
if a - b = 0, a and b are equivalent

2-3 = -1, so 2 and 3 are different numbers
2-2 = 0, so 2 and 2 are the same number

1 - 0.99... = 0.00... so 1 and 0.99... are the same number

>> No.8808997

>>8808826
but 0.333... is an infinite nnumber clearly there is only finite amount of water in the glass.

>> No.8809002

>>8808902
How can I be the same person as my wife's son's father?

>> No.8809008
File: 1.83 MB, 200x200, mind_blown.gif [View same] [iqdb] [saucenao] [google] [report]
8809008

>>8809002

>> No.8809011

0.999... does not exist

prove me wrong

>> No.8809015

>>8809011
It's defined as the geometric series 0.9 + 0.09 + 0.009 + ..., i.e. the sum from i=0 to infinity of 0.9 * 0.1^i. Elementary analysis clearly shows that this limit does indeed exist.

>> No.8809018

>>8809015
how can something requiring """infinity""" exist

>> No.8809022

>>8809018
Go study some analysis and find out. We epsilon-delta now.

>> No.8809024

>>8809022
ok i actually will do this thank you

>> No.8809028

>>8809002
You're not. Some other guy is.

>> No.8809031

>SCIENTISTS SAY THEY CAN ONLY SAY WITH 99.999...% CERTAINTY THAT YOU CAN NEVER GO BACK IN TIME

>> No.8809041

Figured it out ur welcome every1

1/3 = 0.333.... so
3/3 = 0.999....
3/3 = 1
0.9999..... = 1

>> No.8809051

>>8809041
>1/3 = 0.333.... so
prove it

>> No.8809053

>>8809024
The summary (lightly simplified, but not much) of it is this:

Consider the "infinite sum" 0.5 + 0.25 + 0.125 + 0.0625 + 0.03125 + ...; in other words, 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... . I claim that these infinitely many values together sum to 1.

What that means is the following: consider the sequence of *partial sums* of this infinite sum: the first number, the sum of the first two numbers, the sum of the first three numbers, etc. This yields 0.5, 0.75, 0.875, 0.9375, 0.96875 ...; AKA 1/2, 3/4, 7/8, 15/16, 31/32, ... . This is an infinite sequence of numbers that get ever closer to 1, but never reach it.

Now it so happens that the numbers in this sequence, while all being less than 1, get *as close to 1* as you like. All numbers from 15/16 (= 0.9375) onwards are between 0.9 and 1; all numbers from 127/128 (= 0.9921875) onwards are between 0.99 and 1. More generally, for *any distance D larger than zero* that you can name -- here I named 0.1 and 0.01 -- there is a *point in the sequence* such that *all points afterwards* are no further than D removed from 1. All points in the sequence after 15/16 are AT MOST 0.1 removed from 1 (i.e. between 0.9 and 1.1); all points after 127/128 are at most 0.01 removed (between 0.99 and 1.01).

Based on this, in calculus, we say that the infinite sum has the value 1. For you can make *finite approximations of the infinite sum* that are *as close to 1 as you would like*. This is called a limit, and the formal statement would be that the limit as i goes to infinity of (1 - 1/(2^i)) is 1. Moreover, formally, the SUM for i from 0 to infinity of 1/(2^i) is 1.

0.999..., AKA 0.9 + 0.09 + 0.009 + ..., AKA the sum for i from 0 to infinity of 0.9 * 0.1^i, has a value of 1 by much the same reasoning.

See https://en.wikipedia.org/wiki/Limit_of_a_sequence and https://en.wikipedia.org/wiki/Series_%28mathematics%29 for further introductory reading.

>> No.8809057

>>8809051
So this is the stock response, but I've never quite understood it. If you won't permit .333... as 1/3's decimal representation, then what is?

>> No.8809059

>>8809053
"""facts""" like 1+2+3+... = -1/12 invalidate the whole concept of infinite series

>> No.8809063

>>8809059
That's not a limit though

>> No.8809064

>>8809059
>"""facts""" like 1+2+3+... = -1/12 invalidate the whole concept of infinite series
1+2+3+... = -1/12 is NOT true in the theory of infinite series. 1+2+3+... is just infinite, just as you'd expect it to be.

>> No.8809070

>>8809051
I mean it's literally the decimal representation of 1/3. Just divide 1 by 3.

>> No.8809145

>>8808751
I appreciate this picture, these threads are just boring. 1 Million "3*1/3=3/3=0.3...=1"-posts.
But yellow 2 is false, there are many models with infinitely small numbers (although not [math] \mathbb{R} [/math]).

>>8808972
>That logic isn't sound in the case of equality. Its only sound in the case of equivalency.
And that is exactly the point. One way to construct [math]\mathbb{R}[/math] is by equivalence classes of cauchy series of Rationals.
Unfortunately, your proof (as all the others in this thread) is not valid, it just shifts the "problem" from 0.9...=1 to 0.3...=1/3 (yes I know, I'm able to divide, but it's about representation).

>> No.8809212

>>8809145
So, open the book and give a proof.

>> No.8809231

Lets say theres some infinitesmal k between 1 and .999...

1 = .999... +k

We can divide both sides by any number, lets divide by 3 for familiarity and demonstration's sake.

.333... = .333... + (.333...)k => .333... = .333... +k

Well, thats odd, both sides are not equal. Huh, it turns out that the infinitesmal doesnt make sense for any other iteration of 1 = .999... +k. If thats the case, then 1 =/= .999... +k.

Thus 1 = .999...

>> No.8809240

>>8808643
>0.999... = 1
>0.999... = 1 - (1/infinity)
>0.999... - (1/infinity) = 0.999
>[0.999... - 2(1/infinity) = 0.999
>1 = 0.999 - (infinity)(1/infinity) = 1-1
>1=0

I maed a meem xD

>> No.8809256
File: 47 KB, 649x491, IMG_4954.jpg [View same] [iqdb] [saucenao] [google] [report]
8809256

>brainlets still getting baited by trolls who contest this

>> No.8809546

>>8809145
If none of the proofs satisfy you please provide one, you vapid twit

>> No.8809575

if 0.999... = 1

what does 0.1/10^(infinity) +0 .999... = ?

>> No.8809581

>>8808751

What we adjusted our model to include an infinitely small number? Let's call it Horse.

Horse would have these properties:

Horse + Horse = 2 Horse
Horse * X = X Horse
However because the Horse is infinitely small, any X Horse is also equal to Horse.

So Horse = 2 Horse
Divide by Horse
1 = 2

That doesn't make much sense until you realize there is no Horse. Or should I say, zero is also a Horse? It looks like a Horse, it walks like a Horse, gosh darn-it it must be Horse.

>> No.8809584

>>8809575
Sky's the limit.

>> No.8809589

>>8809575
0.1/10^infinity = Horse

>> No.8809595
File: 288 KB, 1399x953, .9999.jpg [View same] [iqdb] [saucenao] [google] [report]
8809595

i hate you /sci/

>> No.8809596

>>8809575
0 + 1 = 1 ? am I missing something?

>> No.8809603

>>8809575
First of all, unless you're working in an extended number system, infinity isn't a number. It's a possible cardinality for a set, but that doesn't make it a number. You can't just raise something to the power of infinity.

What you CAN do is add 0.999... to the LIMIT of 0.1/10^(x) as x APPROACHES infinity (or, to use more appropriate language, grows without bound). The result is, of course, 0.999..., because the limit of 0.1/10^(x) as x grows without bound is exactly 0.

So, if your point is 0.999... can't equal 1 because adding the limit of 0.1/10^(x) as x diverges would also give you 1, I agree with your premise but not your conclusion, because the limit of 0.1/10^(x) as x diverges is exactly 0, and it's absolutely expected behavior for adding 0 to a number to give you the same number back.

>> No.8809623

>>8808678
It's what this anon said. If you understand how to do the sum of a geometric series, .999... can be represened by the series of 9*(1/10)^n starting at n = 1.

Sure, you can argue that it will never EQUAL one, and there is no proof that would adequtely address that. What every proof does show is that the difference between .999... and 1 is absolutely and completely negligible in every possible sense; a condition that is sufficient to say that they are "equal".

>> No.8809648

>>8809623
So they're not equal but we say they're equal because it's very close?

Why didn't someone just post this at the beginning of the thread?

why do people try so hard to convince others that 0.999... = 1 and not just admit that it doesn't but it's very close so we say it is.

It's unbelievable how hard people try to shill 0.999... = 1, you'd think they were being paid.

>> No.8809653

>>8809648
>So they're not equal but we say they're equal because it's very close?
It's not so much that it's "very" close. It's INFINITELY close, and any two numbers that are infinitely close are equal.

>> No.8809677

>>8809002
kek

>> No.8809717

Suppose [math]0.999 \dots \neq 1[/math]. Using [math]d \colon \mathbb{R} \times \mathbb{R} \to \mathbb{R}[/math], [math]d(x, y)=|x-y|[/math], we get a metric space. Metric spaces are Hausdorff, so there are disjoint nbds [math]U \ni 0.999 \dots, V \ni 1[/math].

On the other hand, let [math]W[/math] be any nbd of [math]0.999 \dots[/math]. Now, since [math]d(0.999 \dots, 1)< \varepsilon[/math] for all [math]\varepsilon > 0[/math], there is an open ball [math]B(0.999 \dots, r) \subset W[/math], but then [math]1 \in W[/math], contradicting the Hausdorff property.

>> No.8809734

>>8809717
>d(0.999…,1)<ε for all ε>0

you can't just assert this when you're trying to prove that .999...=1

>> No.8809763

Yes I do and I will ask a question please answer me. I have been thinking about it since yesterday night.

1)Take n number of things out of which p are identical to one another and other q are identical to one another.
Total number of arrangement of n number of things among themselves without repitition is
nPn = n!
2)But total number of unique arrangement is
nPn / (p! * q!)

>How do i conceptually and logically infer 2 from 1?
Please explain.
Don't give example to prove this.
Best is to give pure mathematical path of infering 2 from 1.

pls halp

>> No.8809787

>>8808660

Base case
1 + 0.99... = 2
Induction Hypothesis
x_n + 0.99... = x_n + 1
Induction Step
x_(n+1) + 0.99... = x_n + 1 + 0.99... = x_n + 2

where's my prize

>> No.8809793

>>8809648
>why do people try so hard to convince others that 0.999... = 1 and not just admit that it doesn't but it's very close so we say it is.
nice strawman faggot

if two numbers x and y are distinct, then there is a number between the two of them

now tell me which number is between x=0.999... and y=1 ?

>> No.8809799

>>8809793
(1+.999...)/2

>> No.8809800

>>8808731
i like this

>> No.8809806

>>8809793
0.99... + ((1 - 0.99...) / 2)

ha I got you there is no way you'll think of a flaw in my argument

>> No.8809807

>>8809793
0.999..., except a 0.999... that's bigger than the one you wrote. That's which number.

>> No.8809809

>>8808751
re.2: Yes they do.
Example: dx

>> No.8809813

>>8809807
>>8809806
>>8809799
let y = .0000...00009 + your number, that's closer to 1 than your x
thus your x was not .999...

>> No.8809814

>>8808643
In case anyone wants the real reason why it's because the sequence of numbers

{.9, .99, .999, .9999, ... }

converges to 1 in R. Here's what this means exactly. Let a_n be the nth term in the sequence. For any number e>0 we can find an integer N such that |a_n - 1| < e whenever n>N. This is clear since we can always make |a_n - 1| smaller by taking larger n.

This limit is unique. Suppose there was some other number than 1 it converged to, say C. Then |C - 1| = |(C - a_n) + (a_n-1)| <= |C-a_n| + |a_n-1| < e + e = 2e (Here I use the triangle inequality) whenever n>N and n>N' where N' is the number associated with convergence to C. Since e is arbitrary, the inequality |C - 1| < 2e must always hold for every positive e. Therefore C=1. Since the limit of the sequence is unique and real numbers are defined by convergent sequences, we identify .99... with its limit 1.

>> No.8809816

>>8808643
if two numbers differ by an arbitrarily small amount they're the same number

>> No.8809820

>>8809813
you shouldn't end your arguments with ellipses...
it makes you look like an edge lord........

>> No.8809824

>>8808677
Your intuition wants there to be an infinitely small number, an infinitesimal between the two. Infinitesimals aren't well defined in the real number system, a whole lot of maths break if you try to shove them in. There are number systems that use infinitesimals though. Accept that in R there are multiple decimal expansions for numbers.

>> No.8809831

>>8809799
False.

>> No.8809832

>>8809813
Let m be a variable that changes.
Let its infinitesimal unit of change dm be less than 0.0...01.
Let x be 0.999... + dm.
My x is now closer to 1 than 0.999... is.

>> No.8809835

>>8809820
>not knowing ellipses is one way to denote an infinitely repeating decimal
back to /pol/ please

>> No.8809837

>>8809832
now let x = 0.999... + dm + dm
still closer.

thus your x=0.999... + dm did not equal the true 0.999...

>> No.8809842

>>8809835
>>not knowing all mathematicians are edgelords
no indeed i did not

>> No.8809848

>>8808677
>actually 1.000... - 0.999... = 0.0...01 so it is actually true because 0.0...01 = 0

>> No.8809852
File: 2.28 MB, 1134x991, 1476667419594.png [View same] [iqdb] [saucenao] [google] [report]
8809852

>>8809814
i'm afraid you got meme'd, my friend. good luck in your introductory real analysis class!

>> No.8809854

>>8809852
Thanks but I'm already done with it. Have to keep things simple for those not in the know.

>> No.8809861

>>8809852
How do you mean?

>> No.8809916

>>8808643
think of 0.9999.... as 1- lim(1/n) as n --> infinity

>> No.8810301

>>8809734
Take any arbitrarily small ε. If it is between 0 and 1-0.999...9 for some number [math]n[/math] of 9's, then [math]|1-(0.999 \dots 9 + 9\cdot 10^{-(n+1)})|< \varepsilon[/math].

>> No.8810621

>>8810615

>> No.8810625

So you are telling me
0.(9) exists but 0.(9)8 cannot, right?
so much for a continious number line
Cuz you know, we could pretty much deduce all numbers are equal by that logic
1 = 0.(9) = 0.(9)8 = 0.(9)7 = ...

>> No.8810631

>>8809717
>Now, since d(0.999…,1)<ε for all ε>0
but this is equivalent to d(0.999...,1) = 0 which is equivalent to those things being equal by the axioms of metric spaces.

>> No.8810633

>>8810631
>axioms
Here, found your problem
>It is true, because we defined it to be true

>> No.8810652

>>8810633
and that is precisely how math works, isn't it

>> No.8811315

>>8809734
Alright. Let's called [math]u_n[/math] the series composed of 0.9, 0.99, 0.999 and on and on.

I want to show that when n goes towards infinity (i.e, with the famous 0.9999...), I get a one.

So I chose an epsilon. Any epsilon. I look at [math]1 - \epsilon[/math] and see if I can find a [math]u_n[/math] that is closer to 1 than [math]1 - \epsilon[/math].
Take [math]\epsilon = 1 ; 1 - 1 = 0 ; u_0 = 0.9 > 0[/math] So if I consider a distance of 1, I can use [math]u_0[/math]

It works for any epsilon - it's quite easy to see. Take [math]\epsilon = 10^{-999}[/math] Then [math]u_1000 = 1 - 10^{1000}[/math] is close enough to one.

It even works for an epsilon so small it can be considered 0. Therefore, 0.999... = 1.

>> No.8811318

>>8811315
Shit. [math]u_{1000}[/math].

>> No.8811451

>>8808753
Are you braindead?

>> No.8811872

>>8808643
0.9999 = 1 cannot be true because 1 - .9999 = .000...01

>> No.8812070

>>8808643
>why does 0.999... = 1
Because R is a Banach-space.

Any more questions?

>> No.8812074

>>8811872
>.000...01
is a Cauchy series (roughly speaking) therefore it converges in R and is equal to 0.

>> No.8812388

Since we're having this thread, I just ran across this

https://www.youtube.com/watch?v=wsOXvQn3JuE

This is supposed to be the top science/math female youtuber. Jesus fuck

>> No.8812406

>>8812388
She'll get tons of views though as people watch it to laugh at her and then comment to prove her wrong. Meanwhile her view count grows and grows.

>> No.8812424

>>8812388

>Published on 1 Apr 2012

>> No.8812431

>>8812388
That's an April Fools joke my friend.

>> No.8812454

>>8808643
0.999... DOES NOT = 1

it's just so close that modern math can't tell the difference, just wait a couple hundred years

>> No.8812459

>>8808897
That's only true if 0.999... is less than 1.000..., which is circular reasoning.

>> No.8812462

If x-y=0 then x=y

1-0.999... = 0.000....=0

>> No.8812555

>>8808643
1=0.1×10
10=1×10

0.9=0.1×9

>> No.8812592

I am getting so fucking sick of these thread.

Just look up a few equivalent definitions for the real numbers. (inb4 Wilderberger autists). One of these definitions you can find is that as a set the reals is set of sequences of natural numbers, i.e. decimal expansions modulo an equivalence relations. This equivalence relation then has the property that two such sequences are the same if and only they converge to the same limit.

So the crux of the matter is that some of these equivalence classes contain more than one element, i.e. have multiple representations as sequences. This is e.g. the case for 0.999... and 1. So some real numbers have non unique representations as sequences.

To sum it all up: 0.999... = 1 is true by the definition of real numbers. The real problem is that people appearently do not define what the fuck they are talking about when posing this problem.

>> No.8812634
File: 1.09 MB, 1263x1060, 1425822523532[1].png [View same] [iqdb] [saucenao] [google] [report]
8812634

>>8808751
You need to improve on this. It's really ugly and hard to read. Also needs some anime bullshit.

>>8812592
On /g/ we made pic related and it worked pretty well.

>> No.8813166

This is retardation.

0.999... = 1.000...
1.000...... = 1.001

0.999 = 1.001

0.999 != 1

>> No.8813191

>>8808643
THIS IS THE EASIEST PROOF TO UNDERSTAND:
1/3 = 0.333...
0.333...*3=0.999...
BUT
1/3*3=1
THEREFORE
0.999... MUST BE EQUAL TO 1

>> No.8813208
File: 56 KB, 382x358, 1491199286072.jpg [View same] [iqdb] [saucenao] [google] [report]
8813208

>>8808753
>unnatural numbers

>> No.8813224 [DELETED] 

>>8808643
0.999... = x
(0.999...)2 = x2
0.999... = x2 = x
nice fucking number faggots

>> No.8813240

0.999... = x
(0.999...)^2 = x^2
0.999... = x^2
x^2=x^1
1^2=1^1

0.333.. = y
(0.333..)^2 = y^2
0.111.. = y^2
1/9 = (1/3)^2
It's fun cuz the fact that it has no finite number of digits makes it so all these work

These numbers have a name
they are called nigglings dipshit math fucker

>> No.8813275

>>8809002
your not, cuck

>> No.8814108

A brainlet arrives, behold.

How about x+y=1

x=0.99999.....
y=0.000000......1

In other words, you can "limes" and shit these two things together.

>> No.8814121

>>8813191
so you're saying because 1 divided three times is 0.333... and 0.333...*3 is 0.999... then 0.999... is the original number that was being divided?

Some how that doesn't feel right, but I guess my shitty human feels don't matter? well fuck you if that's how it is, inconsiderate dick

>> No.8814142

>>8808643
[math]0.999...=\sum_{1}^\infty \frac{9}{10^n}=9\sum_{1}^\infty \left(\frac{1}{10}\right)^n=9\left(\frac{\frac{1}{10}}{1-\frac{1}{10}}\right)=9\left(\frac{1}{9}\right)=1[/math]

>> No.8814146

>>8809057
>crickets

every time

>> No.8814243

>>8808643
>why does 0.999... = 1

Because computers can't do fractions.

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