Why are /b/ people so retarded?

Prove that 1/3 = .33..., that 2(.33...) = .66..., and that .66... + .33... = .99...

0.999.../0.999...

[math]1 - \epsilon[/math]

do i win?

0.tentententententen......

You cannot use epsilon OUTSIDE OF A LOGICAL CONDITION, you big brainlet. If you want to use it like that you have to use an infinitesimal, and those numbers are not a subset of R.

obviously i was referring to an infinitesimal you retard

and it said use "a number" not "a number only from the set of real numbers" you daft cunt

You fucking dummy it's 0.000... 000 err.... FUCK.

I made this

0.A

youtube.com/watch?v=eB5VXJXxnNU

What are those pranks?

I don't think you know what .999... means

...

1 + .999... / 2

saying the n-word

pretty epic tbqh

Retards won't understand that
[math]...999 + 1 = 0 \implies ...999 = -1[/math]

1 = 10/10 = 9/10 + 1/10

1/10 = 10/100 = 9/100 + 1/100

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

...

So 1 = 0.999..

>number after an infinite sequence of digits
yea that totally makes sense

>1+4.99999
?

every single thread i am ignored by the brainlets

>craving attention on a pakistani flyfishing forum

...

[math]0.999...\pm \ito [/math] where [math]\eta[/math] is op's IQ.

>.999...5
this doesnt even exist

0.9995

Checkmate brainlets.

First write down 0.999... and I will show you a number :^)
Go ahead, take your time.

>cant comprehend abstract ideas

There are infinite amount of options. For example, [math]e^0[/math].

[math]0.999... \leq e^0 \leq 1[/math]

This is only correct answer

Let us assume correctly that;

0.999 =/= 1

Then

2 - 1 = 0.888 (repeating of course)

Following from this we can conclude using Godel's incompleteness theorem that ;

1 - 1 = 0.999

From Kolmogorov's rigorous foundations of probability theory, an event which will never happen has a probability of 0.

however there is a difference between an event never happening and an event that almost surely will not happen.

Therefore there is a difference between 0 and 0.00000...1

therefore there is a difference between 1-0 and 1-0.000...1
which is the difference between 1 and 0.9999...

If you disagree, then it looks like you should have paid more attention in probability theory, kid

1-0.000...1 is not 0.999..
As [math]0.999...[/math] states that for every fractional position [math]n[/math] there next position [math]n+1[/math] is also [math]9[/math]
Because English is not my native language I don't know the proper name of this "position after point"
[math]0.0000..1[/math] implies that the number of zeros is FINITE. Hence there is such [math]n[/math] such that nth position of [math]0.000...1[/math] equals to [math]0[/math] and [math](n+1) th[/math] is [math]1[/math]. Therefore, [math]1 - 0.000...1[/math] has finite amount of digits after point, while [math]0.9999...[/math] has infinite, which shows that
[math]1 - 0.000...1 < 0.999...[/math]
As for my proof that [math]0.999... = 1[/math]
[math]0.999... = 0.9 + 0.09 + ... = \sum\limits_{i=1}^{\infty}9 *10^{-i}[/math]
You can get the sum of this series by a formula taught in school

That is not a proof.
The way to properly express this value is with the infinite sum of 0.9+0.09+0.009...
Infinite sums that converge are equal to the value that they approach. It's a simple proof to show that the if approaches 1, and so the value is actually 1.

2-1 = 0.888?

inequalities don't exist on complex numbers

It is part of the rigoursly defined set of the real numbers.

obviously (1 + 0.999...)/2

which equals 1.999... / 2 = (0.5).(0.45),(0.45),...
You can define this new class of number so formed as the transinifintesimal class.

>abstract ideas
no one's got time time for your wankery

0

Jesus, how come nobody said this earlier?

2ezy
0.AAA...

this

ITT: morons who can't tell the difference between an indeterminate sequence of digits and [math] 0.(9) [/math].
For all you know 0.999 is the truncation of [math] 0.999(8) [/math]..

total orders, faggot