noko

[Anonymous No.4445207]

0.999 = 1

[Anonymous No.4445217]

2=1

[Anonymous No.4445215]

No.
But if you had said 1 = 0.999...(repeating), then everyone would (should) have agreed.
Also, polite sage because mathematical troll.

[Anonymous No.4445219]

TROLOLOLOLOLOLOLOLOLOLOL

[Anonymous No.4445221]

0.999 = 0.999

[Anonymous No.4445222]

>>4445217

.999 does not equal 1.

.999 . . . = 1

now go away.

[Anonymous No.4445227]

0.000...0001 / 10

[Anonymous No.4445261]

[Anon No.4445276]

This is the oldest troll going, might as well feed the troll.
1/3=0.3333...
3/3=1=0.9999...

[Anonymous No.4445287]

>>4445261

If 0.999...= 1 then 0 = 1

Observe:

<div class="math"> 0.999....~=~1
~0.999^{n}~=~1^{n}~~~~~n~->~infinity
~0~=~1 </div>

[mathgenius (not trolling) !!Ceavr+zDPUk No.4445282]

>>4445276
>>4445261
>>4445222
>>4445215
Wrong.

Oh boy, here we go.

[Anonymous No.4445294]

>>4445287

<div class="math"> Fuck_{you}^{you bastard} </div>

[mathgenius (not trolling) !!Ceavr+zDPUk No.4445296]

>>4445294
You can insult him, yet his proof remains correct. Is that stupid "0.999... = 1" trolling finally defeated now?

[Anonymous No.4445307]

>>4445287
I know you're trolling, but I'll bite. That proof only holds if 0.999... < 1, which is what you're trying to prove in the first place.

>circular logic

Also, LaTeX is your friend.

[Anonymous No.4445310]

But wouldn't that violate the whole number=itself and only itself thing?

I don't know shit about math.

[Anonymous No.4445311]

>>4445296
Why is the geometric series proof not valid?

[Anonymous No.4445317]

OK, I personally don't understand how

.9999......could EVERY amount to one, unless you round.

1=1
1 always equals fucking 1.

I have never had an equation in which I had to reduce 1 to .9999...., so why now?

[Anonymous No.4445319]

For fun, that's what most math is all about.

[Anonymous No.4445326]

>>4445287
>Implying your limits are correct
Full retard.
<span class="math">\lim_{n\to \infty} 0.999(...)^n=\lim_{n\to \infty}e^{n.ln\left(0.999\left(...\right)\right)}[/spoiler]
Which, if you assume that 0.999(...) = 1 (which you DO assume, because you tried a proof by absurd), you get :
<span class="math">\lim_{n\to \infty}e^{n . 0} = \lim_{n\to \infty} e^0 = \lim_{n\to \infty} 1 = 1[/spoiler]

So you proof is wrong.

>>4445296
Not so genius for a math genius to support a wrong proof.

[mathgenius (not trolling) !!Ceavr+zDPUk No.4445328]

>>4445307
It is valid because 0.999... is < 1.

>>4445311
You cannot represent 0.999... as a geometric series.

[Anonymous No.4445331]

>>4445317
see >>4445261
Three different proofs.

>>4445310
What? That's akin to saying one divided by one can't equal one. There are many ways to represent any given number.

[Anonymous No.4445335]

>>4445331
But the proofs in >>4445261 are wrong. The result is correct, but the proofs are wrong.


>>4445328
>It is valid because 0.999... is < 1.
>because 0.999... is < 1
>Assuming the result to prove the result.
>Circular logic
>Overtrolling

[mathgenius (not trolling) !!Ceavr+zDPUk No.4445344]

>>4445326
Your understanding of the proof is wrong. It is valid to assume 0.999... to be < 1, because that's reality.

>>4445335
>doesn't recognize that circular logic in this case is not circular

[mtp !EKFQOBEsKo No.4445342]

its like posting this on omegle again

>>4445328
>You cannot represent 0.999... as a geometric series.
a=0.9
r=1/10

[ ] not told
[x] told

[Anonymous No.4445343]

x = .99999999999999999999......

4x = 3.99999999999....
5x = 4.99999999999...

5x - 4x = 1
x = 1

[Anonymous No.4445346]

>>4445335
>proofs are wrong

How so?

[Anonymous No.4445348]

>>4445344
go die in your fantasy math alone fucktard

[mathgenius (not trolling) !!Ceavr+zDPUk No.4445351]

>>4445342
Wrong. This series does not represent 0.999... If it did, it would converge to 0.999... which it obviously not does, because it converges to 1.

[mathgenius (not trolling) !!Ceavr+zDPUk No.4445354]

>>4445348
You can be as mad as you like to be, my math is correct.

[eye No.4445355]

Think OP meant .9 bar or otherwise he's trolling and it's not worth my time to post here.

1/9 = .1 bar
4/9 = .4 bar
7/9 = .7 bar
9/9 = .9 bar = 1

[Anonymous No.4445356]

http://www.wolframalpha.com/input/?i=.9999999......

Its over

[Anonymous No.4445402]

>>4445346
They are wrong because they all use basically the same argument, which was also used in >>4445343
The argument used is that
1.9999999(...) - 0.99999(...) = 1
No, it's not.
The substraction would go as follow :
1.99999(...) - 0.99999(...) = 1.0000(...)01
Where the final 1 is after an infinite number of 0 (my notation is an aberration, but you get the idea)

So if we use this notation, the argument that was used becomes :
1.9999(...) - 0.999(...) = 1.00(...)01 = 1

So basically, you say : 1.000(...)01 = 1
BUT for this to be true, you have to assume that 0.999(...) = 1
(I can demonstrate it very easily (ask me if you want)but I don't want my confusing post to be even more confusing)

So in the end, this argument uses ciurcular logic, so it is a fallacy.

As I said earlier, the result is correct. It was simply not proven by the earlier post.

[Anonymous No.4445410]

>>4445402
Erratum
Everytime I said 1.000(...)01, it was actually 1.000(...)09

But everything I said is still correct

[Anonymous No.4445432]

>>4445402
>cannot into infinity

<span class="math"> 0.999... \cdot 10 \neq 9.999...0 [/spoiler]

[Anonymous No.4445444]

>>4445410
>>4445402

You're right in your conclusion, but your reasoning is slightly wrong.

Essentially everyone in this thread is subtracting infinity - infinity and assuming its zero.

You cant

Therefore you cant do 1.999... - .999...

.999... isnt a number

Deal with it

[Anonymous No.4445449]

Infinit series
/thread

[Anonymous No.4445462]

>>4445402
so now lim x-> infinity of 1/x isnt 0?

[Anonymous No.4445469]

>>4445317
There is a mathematical law that there can be no infinitely small numbers, and those would have to exist for 0.999... to not equal 1.
Also mathematical proofs.

[Anonymous No.4445473]

>>4445444
>converging series equals infinity

[Anonymous No.4445498]

>>4445410
>1.000(...)09
>infinite periodical zeroes and a 9 at the end of an INFINITE string
>mfw

[Anonymous No.4445502]

Just trowing this in here...

1/3 = 0.333...

Let y = 1/3 and x = 0.333...

So therefore y=x

So 3y = 3x is true. Replace x and y with there respective values, you get:

1 = 0.999...

[Anonymous No.4445505]

>>4445432
You really want me to go into specifics?
Really?
I will prove that 1.999(...) - 0.99(...) = 1.0(...)09
where (...) denotes a number n of nines or zeros
Let's prove it is correct for n=1
1.999 9 - 0.99 9 = 1.0 0 09
(A) Let's assume it is true for n.
Let's prove it is true for n+1 :
1.999(...+1 repeatition) - 0.99(...+1repeatition) = 1.999 (...)9 - 0.99(...)9 = 1.999(...)0 + 0.000(...)9 - 0.99(...)0 - 0.00(...)9
By using the hypothesis (A), we have
1.999(...)0 + 0.00(...)9 - 0.99(...)0 - 0.00(...)9 = 1.0(...)09 + 0.00(...)9 - 0.00(...)9
= 1.00(...)9 - 0.00(...)9 + 0.00(...)9
=1 + 0.00(...)9
= 1.00(...)9
= 1.0(...)09

The hypothesis is verified for n+1.
By mathematical induction, the hypothesis is true for any n >= 1
=> n to infinity => 1.99999(...) - 0.99999(...) = 1.0000(...)01

I'm right. You're wrong. You're a retard.

>implying you understood
>implying you read

[Anonymous No.4445508]

>>4445505
>infinite string ending in 09
What the fuck am I reading

[sage No.4445515]

Same fucking threads recycled on /sci/ day in day out. This entire board might as well be deleted. /x/ has more OC than /sci/

[Anonymous No.4445517]

>>4445502
Saying that 1/3 = 0.333(...) is equivalent to saying that 0.999(...) = 1

>circular logic

[Anonymous No.4445519]

>>4445508
n not infinite in my proof
>implying again that you read

[Anonymous No.4445521]

>>4445505

Hey, buddy.

Two things:

1.) It's an infinite string. It has no ending. You put an ending on it. No.

2.) That is the limit of the sequence, (if that kind of number was legitimate) but it is not the actual value.

[Anonymous No.4445520]

<span class="math">\sum_{n=1}^{+\infty}\frac{9}{10}\cdot{}\frac{1}{10^{n-1}}=1[/spoiler]
Since it's a geometric serie with <span class="math">a=\frac{9}{10}[/spoiler] and <span class="math">r=\frac{1}{10}[/spoiler]

[Anonymous No.4445530]

>>4445521
read
>>4445519
I used a finite n

;) <3 :D ^^ :ppppppp =]

[eye No.4445528]

>>4445505
You're assuming that infinity + 1 is a valid number. It's not. Therefore, just like you cannot do math with infinity + 1, you cannot expect to have numbers with infinity digits and other numbers with infinity + 1 digits and expect anything rational to come from it.

[Anonymous No.4445535]

>>4445517

Ah well gave it a shot.

[Anonymous No.4445539]

>>4445530
What the fuck are you talking about? Your notation indicates that you have 1 with an INFINITE number of zeros and then a 9 at the end which is mathematically and logically IMPOSSIBLE.
Come back when you know how to write correctly.

[Anonymous No.4445544]

>>4445528
The actual thing I did in >>4445402 which then lead me to >>4445505 to prove that I wasn't just saying nonsense shit is :
I used the flawed notation in every other proofs (the "1.999 - 0.999 = 1" type) to show that even by using these flawed notations, the above equality in parenthesis is still wrong.
The only thing I meant to prove was the fallacy in these proves.

My proofs are coherent in that I was making the hypothesis that the notations used were correct and showing that even with these notations, the forementioned proofs were wrong.

>Proof by absurd
Goddamn it /sci/

[Anonymous No.4445545]

Can one of you neckbeards clearly articulate why the infinite series (geometric) proof is not valid?

I know this is asking a lot of you.

[Anonymous No.4445549]

>>4445287
http://www.wolframalpha.com/input/?i=lim+of+0.999......^n+as+n+approaches+infinity

[Anonymous No.4445550]

>>4445520
It's valid, as i wrote here.
>>4445545

[Anonymous No.4445552]

>>4445545
But it is valid, if we're talking about proving 0.999... = 1.

[Anonymous No.4445553]

>>4445550
It's not, it's a limit.

[Anonymous No.4445554]

>>4445505
Let <span class="math"> \displaystyle{a_n} [/spoiler] be a sequence such that <span class="math"> \displaystyle{a_1 = 0.9, ~ a_2 = 0.99, ~ a_3 = 0.999,} [/spoiler] and so on.

<span class="math"> \displaystyle{a_n = 0.9...9} [/spoiler], n 9s.

<span class="math"> \displaystyle{a_{n+1} = 0.9...99 [/spoiler], n+1 9s.

<span class="math"> \displaystyle{lim_{\n \to \infinity} n = \displaystyle{lim_{\n \to \infinity} (n + 1) = \infty} [/spoiler]

<span class="math"> \displaystyle{lim_{\n \to \infinity} a_n = \displaystyle{lim_{\n \to \infinity} a_{n + 1} = 0.999... } [/spoiler]

<span class="math"> \displa\displaystyle{lim_{\n \to \infinity} ( a_n - a_{n+1} ) = 0 } [/spoiler]

[Anonymous No.4445561]

>>4445553
The fuck are you talking about? go and learn baby-first calculus then we talk.

[Anonymous No.4445563]

>>4445550

I'm well aware of how it is defined mathematically. Someone earlier in the thread stated that this proof didn't hold. I thought it was because of some assumption made.

<span class="math">\displaystyle \sum_{n=0}^{\infty}\frac{9}{10}\cdot{}\frac{1}{10}=1[/spoiler]

LaTeX is hard to use.

[Anonymous No.4445566]

>>4445553

I said clearly articulate mother fucker.

God you are hopeless.

[Anonymous No.4445574]

>>4445545
You're not the only one waiting on that.

[Anonymous No.4445579]

>>4445563
It holds, it's a simple proof without any errors. It have been used for many many years, and it's pretty clear.

[Anonymous No.4445586]

Good troll

[Anonymous No.4445594]

>>4445563
<span class="math">\displaystyle \sum_{n=1}^{\infty}\frac{9}{10}\cdot{}\frac{1}{10^{n-1}}=1[/spoiler]
Also just corrected your expression.

[Anonymous No.4445595]

EVERYONE CHILL THE FUCK OUT!

Wolfram's got this shit covered.

[Anonymous No.4445606]

>>4445505
Let an be a sequence such that <span class="math"> \displaystyle{a_1 = 0.9, ~ a_2 = 0.99, ~ a_3 = 0.999,} [/spoiler] and so on.

<span class="math"> \displaystyle{a_n = 0.9...9} [/spoiler], n 9s.

<span class="math"> \displaystyle{a_{n+1} = 0.9...99} [/spoiler]

<span class="math"> \displaystyle{lim_{n \to \infinity} n = \displaystyle{lim_{n \to \infinity} (n + 1) = \infty} [/spoiler]

<span class="math"> \displaystyle{lim_{n \to \infinity} a_n = \displaystyle{lim_{n \to \infinity} a_{n + 1} = 0.999... } [/spoiler]

<span class="math"> \displaystyle{lim_{n \to \infinity} ( a_n - a_{n+1} ) = 0 } [/spoiler]

[mathgenius (not trolling) !!Ceavr+zDPUk No.4445604]

>>4445595
Wolfram is using incorrect approximations. Computer software cannot accurately handle decimals and infinity.

[Anonymous No.4445607]

0.999... is not a number, and it makes no sense to regard it as such.

We can write 0.999... as the limit of a geometric series as it's length approaches infinity. The limit of this series does indeed equal 1. But this is very different from claiming some number 0.999... equals some number 1.

[Anonymous No.4445609]

>>4445579

Thanks.

See, I thought this had some validity:

>Wrong. This series does not represent 0.999... If it did, it would converge to 0.999... which it obviously not does, because it converges to 1.

Then reading over it several times, I realize that this poster is probably an idiot.

[mathgenius (not trolling) !!Ceavr+zDPUk No.4445617]

>>4445609
Read it again and realize it's true.

[Anonymous No.4445618]

>2989 posts

[Anonymous No.4445619]

>>4445505
Let an be a sequence such that <span class="math"> \displaystyle{a_1 = 0.9, ~ a_2 = 0.99, ~ a_3 = 0.999,} [/spoiler] and so on.

<span class="math"> \displaystyle{a_n = 0.9...9} [/spoiler], n 9s.

<span class="math"> \displaystyle{a_{n+1} = 0.9...99} [/spoiler]

<span class="math"> \displaystyle{lim_{n \to \infty} n = \displaystyle{lim_{n \to \infty} (n + 1) = \infty} [/spoiler]

<span class="math"> \displaystyle{lim_{n \to \infty} a_n = \displaystyle{lim_{n \to \infty} a_{n + 1} = 0.999... } [/spoiler]

<span class="math"> \displaystyle{lim_{n \to \infty} ( a_n - a_{n+1} ) = 0 } [/spoiler]

[Anonymous No.4445626]

I always liked the simple "proof", if you can call it that, that consists on realizing that there are no real numbers between 0.999... and 1. From that, what would you subtract from 1 to reach 0.999... ? You'd need to subtract 0, which means they are the same

It's cute, isn't it?

[Anonymous No.4445628]

>>4445619
Are yo fucking kidding me? Fuck you LaTeX, I give up.
>>4445505
You win because LaTeX and my internet are being cunts.

[Anonymous No.4445633]

>>4445626

Not saying this is a real number, but wouldn't you just subtract 0.0(...)1 from 1 to get 0.9(...)?

[Anonymous No.4445637]

>>4445628
You can type your LaTeX in here first, using the symbols on the top or the buttons, and then look at the bottom of the input box for the result. If it has an error, it will tell you what it is

http://www.codecogs.com/latex/eqneditor.php

faster than posting on /sci/

[Anonymous No.4445651]

>>4445633

There is no thing as 0.000(...)1, it goes on forever, like 0.999(...).

I think this is the single main misconception responsible for the difficulty accepting this identity

[Anonymous No.4445647]

>>4445637
True enough. at least the lines that actually matter turned out sort of correctly. The last line was the point of the whole thing. :/

[Anonymous No.4445671]

>>4445647

It's a good proof, but like I said in >>4445651, the main problem people have with this is that they don't understand that 0.999... goes on forever, which is closely tied with the concept of how limits work. Using them to prove the identity is like proving addition using multiplication

Also, we can tell what you meant. All you have to do is double click the LaTeX lines and it will show you the code in a weird pop-up

[Anonymous No.4445706]

>>4445594

More like you wrote my expression how you originally wrote it after I made an analogous expression where the infinite series begins at zero.

I simply forgot this part:

<span class="math">\displaystyle \sum_{n=0}^{\infty}\frac{9}{10}\cdot{}\left(\frac{1}{10}\right)^n[/spoiler]

By the way, your proof is ugly as fuck and since most people are too lazy to actually look at the photo:

<span class="math">\displaystyle 0.999....[/spoiler]

<span class="math">\displaystyle \frac{9}{10}\left[1+\frac{1}{10}+\frac{1}{100}+\cdot\cdot\cdot\right][/spoiler]

Note: <span class="math">\displaystyle \frac{1}{10^0}+\frac{1}{10^1}+\frac{1}{10^2}+\cdot\cdot\cdot[/spoiler]

<span class="math">\displaystyle \sum_{n=0}^{\infty} \left(\frac{9}{10}\right)\left(\frac{1}{10^n}\right)[/spoiler]

<span class="math">\displaystyle \sum_{n=0}^{\infty} \left(\frac{9}{10}\right)\left(\frac{1}{10}\right)^n[/spoiler]

<span class="math">\displaystyle \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}[/spoiler]

<span class="math">\displaystyle \frac{\frac{9}{10}}{1-\frac{1}{10}}

<span class="math">\displaystyle \frac{\frac{9}{10}}{1-\frac{1}{10}}[/spoiler]

<span class="math">\displaystyle \left(\frac{9}{10}\right)\left(\frac{10}{9}\right) = 1[/spoiler]

Of course, I still prefer to use the sum of a series is defined to be the limit of the sequence of its partial sums approach.[/spoiler]

[Anonymous No.4445711]

>>4445594

More like you wrote my expression how you originally wrote it after I made an analogous expression where the infinite series begins at zero.

I simply forgot this part:

<span class="math">\displaystyle \sum_{n=0}^{\infty}\frac{9}{10}\cdot{}\left(\frac{1}{10}\right)^n[/spoiler]

By the way, your proof is ugly as fuck and since most people are too lazy to actually look at the photo:

<span class="math">\displaystyle 0.999....[/spoiler]

<span class="math">\displaystyle \frac{9}{10}\left[1+\frac{1}{10}+\frac{1}{100}+\cdot\cdot\cdot\right][/spoiler]

Note: <span class="math">\displaystyle \frac{1}{10^0}+\frac{1}{10^1}+\frac{1}{10^2}+\cdot\cdot\cdot[/spoiler]

<span class="math">\displaystyle \sum_{n=0}^{\infty} \left(\frac{9}{10}\right)\left(\frac{1}{10^n}\right)[/spoiler]

<span class="math">\displaystyle \sum_{n=0}^{\infty} \left(\frac{9}{10}\right)\left(\frac{1}{10}\right)^n[/spoiler]

<span class="math">\displaystyle \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}[/spoiler]

<span class="math">\displaystyle \frac{\frac{9}{10}}{1-\frac{1}{10}}[/spoiler]

<span class="math">\displaystyle \frac{\frac{9}{10}}{1-\frac{1}{10}}[/spoiler]

<span class="math">\displaystyle \left(\frac{9}{10}\right)\left(\frac{10}{9}\right) = 1[/spoiler]

[Anonymous No.4445726]

>>4445637

God I love you. Probably the most interesting and useful thing to come out of this thread.

The preview script for GM is nice but when you end up writing 10+ lines of LaTeX, you can't see shit.

[Anonymous No.4445744]

>>4445637
\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\
frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\f
rac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\fr
ac{1}{\frac{1}{\frac{1}{}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}

[Anonymous No.4445749]

>>4445637
<span class="math">\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\
frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\f
rac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\frac{1}{\fr
ac{1}{\frac{1}{\frac{1}{}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}[/spoiler]

[Anonymous No.4445756]

>>4445207

[Anonymous No.4445775]

>>4445756
I want explainations like this on every problem on /sci/
Math books should look like this

[Anonymous No.4445777]

>>4445637
testing

<span class="math">F_{uv}=\partial _{u} A _{v}- \partial_{v} A_{u}[/spoiler]

[Anonymous No.4445781]

>>4445777
i'm sorry to bring this up guys but...

look at those digits