0.999999=1

...

This number must be between .999... and 1. Here it is: .999...95

Ironic shitposting is still shitposting.
Good job making the only .9999... = 1 thread on the board.

>.9999999......8

So user, you're saying with that notation, you write an infinite number of 9s and then, AFTER you write an INFINITE number of 9s, you write an 8?

What a fascinating number that doesn't exist.

0.9999...>0.999...

Says the guy posting it now...

0.999... doesn't even exist.

prove it

[math]
\begin{align*}
&0.33... \\
3 & \overline{)1_0 \;\;\;\;\;} \\
& \;\;\;\underline{9} \\
& \;\;\;1_0 \\
& \;\;\;\;\; \underline{9} \\
& \;\;\;\;\; 1 \;\; etc
\end{align*}
[/math]

one shouldn't think so

0.9999 = 9(1/10 + 1/100 + 1/1000 + •••) = 9((1/10)/(9/10)) = 1

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

Some things are not as hard as they seem.

1/3 != 0.333...

...

(a+a)/2=a

See

0.999(n)... + 0.000(n+1)x

yeah the ... destroys whatever is right of it

Yeah, all the numbers of that form equal each other. If you write 0. then write an infinite number of 9s then you write any natural number after that, it's always gonna be 1.

...

0.99999.... = 1/3+2/3
1/3=0.33333....=0.33333
2/3=0.66666....=0.66667
0.33333+66667=1

If you do not round your numbers go and find the end of 0.9 recurring and stop spamming this shitty meme on this board.

You know, there is something fundamentaly wrong in modern maths, you fuckers are theorizing all that bullshit, and all your trolling is fun...but when all that theorized math will get an aplication in engineering we all will shit bricks, ya know!?

>0.33333....=0.33333
where can we send the nobel prize

Nobel prizes aren't given for Mathematics because memes have no value.

1 is between 1 and 1.
I didn't say anything wrong.

1>1 and 1

1>=1 and 1

not "between"

talk about moving goalposts

So this means every number is equal to each other, because I can write 2.0000.......0 and its equal to 2.0000.........1 and so on, so every number must be equal to each other.

Not every number but every number of the form "2.000...something" (there is an infinite number of zeroes). But there are of course numbers that can't be written in that form.

...

0.999...+x=1.
x is really small number (1/∞). This number is small enough to assume that it is equal to zero without much error when used in other calculations.

This image is self contradictory. If .000000...1 doesn't exist, you can't subtract .99999...... either, as this would imply .99999999 ends.

>all these “proofs”
>no mention of the completeness axiom

[math]x=0.\dot9[/math]
[math]10x=9.\dot9[/math]
[math]9x= 10x-x = 9.\dot9-0.\dot9=1[/math]
[math]9x=1[/math]
[math]\therefore 0.\dot9=1[/math]

[math]1/9[/math] is not [math]1[/math] user

it should be written as [math]9x=9[/math]

[math]9x/9x =1[/math] [math]9/9=1[/math]

B I G B O Y C A U C H Y S E Q U E N C E S

Those number do exist, and performing operations with them is valid. They are just equal to 0 and 1 respectively.

So .9999999999999..., doesn't exist either?

>Implying brainlets that refuse to believe 0.999....=1 will understand the completeness axiom.

aynrandlexicon.com/lexicon/infinity.html

*unzips calculatrice*

x = 4
20x = 24
20x - x = 24 - 4
16x = 20
16x + 5(4) = 20 + 20
21x = 40
21(4) = 40
therefore 21 = 10

>x = 4
>20x = 24
0/10 bait

You didn't phrase it correctly you fucking mongoloid, from your subhuman-tier choice of words it seems like you're seriously convinced the equivalence it's not true

>So tired of 0.9999...=1 threads
>so I made another one
Good job, Bobby.

∞+1 != ∞

∞/(∞+1) != 1
0.999... !=1

if

1/2 + 1/4 + 1/8 ... = 1,

Then

9/10 + 9/100 + 9/1000 ... = 1,

Since the later approaches 1 more quickly than the former.

QED

>Comp Sci brainlets
REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

better yet, use the non-approach approach.
every line is =1 already

[math] \displaystyle
\begin{align*}
1 = \left (\frac{9}{10} + \frac{1}{10} \right )
&= 0.9 + \frac{10}{100} \\
= 0.9 + \left (\frac{9}{100} + \frac{1}{100} \right )
&= 0.99 + \frac{10}{1000} \\
= 0.99 + \left (\frac{9}{1000}+\frac{1}{1000} \right )
&= 0.999 + \frac{10}{10000} \\
= 0.999 + \left (\frac{9}{10000} + \frac{1}{10000} \right )
&= 0.9999 + \frac{10}{100000} \\
&\vdots
\end{align*}
\\ \displaystyle
\Rightarrow 0.\overline{9} = 1
[/math]

Holy fuck this made me cringe so hard, I think I just started another hurricane

if 0.999... = 1,

what's the difference between an open interval and a closed interval?

what does 1.000... - 0.999... equal?

By the way if anyone ever tries to argue that a number doesn't exist, feel free to ignore them. Numbers exist when they are defined and infinitesimals can be defined.

>four = four
>twenty four = twenty four
what seems to be the problem?

>Implying an infinite number of nine exist

choosing between 0.999...=1 and 0.999!=1 is an axiom that can be taken either way depending on the framework of analysis.

>2occatl.net/Physics/1703.0073v3.pdf

>what's the difference between an open interval and a closed interval?
(0;1) doesn't countains 0 and 1 while [0;1] does. Are you even trying user ?

>what does 1.000... - 0.999... equal?
0, that's especially obvious since you're starting from the hypothesis 1=0,999... Are you retarded user ?

>By the way if anyone ever tries to argue that a number doesn't exist, feel free to ignore them. Numbers exist when they are defined and infinitesimals can be defined.
I define the "user is fucking dumb" functions as the sctrictly increasing functions [math]f:\mathbb R \to \mathbb R[/math] which are differentiable with derivative

Frenchfags need to get out. This website is for white people only.

Numbers don't exist in the first place, except as a concept

How is that related to ?

lrn2 reading comprehension fagget

check'd
incorrect
no respect

THATS THE WAY UH HUH UH HUH
I LIKE IT
UH HUH UH HUH

>Since the later approaches 1 more quickly than the former

Does it?

9/100 < 1/4
9/1000 < 1/8
etc.

>...95
Infinite one! Infinite two!