0.999...=1

haha, no
that "..." is important

nigga you better be joking

[math]-1 = \sqrt1=1[/math]
[math]-1 = 1[/math]
prove me wrong fagtards

Niggas be postin' in a troll thread.

1-.999999 = 0.0000...1
myth busted

0.0000...1=0.0000...0999....

-1 doesn't equal sqrt(1)

0.000...1 is not an infinite decimal

You can't have an element in an infinite ordered set without a successor

wow you must be cs major
[math]-1(-1)=1[/math]

Hey ill give you 0.999 cents for every dollar you give me if you feel that you statement is true you should suffer no loss.

You could've written 0.(9) and made your statement much clearer.

[math]benis - feminine benis = no benis DD:[/math]

How do I thubs up :DD

just give me the benin :DDD

or 0.(((9)))
the jews are behind these mathematical lies
keepin' the white man down.

baited

0.999... where 9 repeats infinitely

[math]\text{You guys ought to know better than to call this man a brainlet. But I guess brainlets tend to see brainletness everywhere. If you knew anything about infinitesimal numbers, which clearly you don't, you'd know }1-0.999\dots\text{ is exactly equal to }0.\ \ldots 01\text{. This is a very real quantity that we use all of the time to calculate derivatives. So next time, keep your mouth shut before lecturing.}[/math]

Hey fucker, I know [math]\sum_n n=\frac{-1}{12}[/math] so you're just trying to trick me into taking 1/12 of a dollar from you.

I know it's bait but it might as well have been a genuine [argument]

1 is successed by the 0 that precedes it, brainlet

the position of that 1 is undefined

[math]\LaTeX[/math]

actually he is taking 1 dollar and giving one back
1-1+1-1+1....= s
1-(1-1+1-1...)=1-s
1-1+1-1+1...=1-s
s=1-s
2s=1
s=0.5

he is tryin to give you 50 cents

No it's not, it's defined to be in the same position as the last [math]9[/math] in [math]0.999\ldots[/math] such that they add to [math]1[/math].

this is bait but still
>last 9

latex more like [math]G^AyTeX[/math]

>can't come up with an argument
>fails to see the logic
>"i know, just pretend it's bait"

there can't be an element without a successor in an infinite totally ordered set

>last digit of an infinitely repeating series of digits
What comes after it? If it were truly the last 9, then it would be followed by an infinite string of zeros. But clearly that isn't the case, because the number is defined as only 9s, repeating forever.

There is no "last 9", there will always be another 9 after it. Unless you think there's a highest number too

Sure there can, [math]0.999...[/math] is a coproduct of two infinitely ordered sets of digits, [math](a_i,b_n)=0.a_1a_2a_3\dots\times b_1b_2b_3\ldots[/math]. [math]a_n[/math] succeeds [math]a_{n-1}[/math], [math]b_n[/math] succeeds [math]b_{n+1}[/math] and [math]b_\infty[/math] succeeds [math]a_\infty[/math].

Shit's infinite and totally ordered, and [math]b_1[/math] doesn't have a successor. [math]QED[/math]

>If it were truly the last 9, then it would be followed by an infinite string of zeros
No, it's preceded by an infinite string of nines.

>But clearly that isn't the case, because the number is defined as only 9s, repeating forever.
Only when you look at the infinite part, not when you look at the number in reverse.

>There is no "last 9", there will always be another 9 after it.
*before it

>Unless you think there's a highest number too
It's possible to define infinite ordered number systems with largest elements.

How are you supposed to do arithmetic with that?

Do arithmetic on each of the sides, it's pretty obvious m8

WHO THE FUCK CARES

did you forget what board you're on?

You can't know the position of any of the elements in the b part since the position of an element is dependent on the position of its predecessor.

A decimal(less than 1) would look like

(10^-1)d + (10^-2)d + ... (10^n)d

with your definition, it would look like

(10^-1)d + (10^-2)d + ... (10^n)d + (10^m)

the value of m can't be derived from the other elements therefore it can't be a number

... (10^-n)d + (10^-m)d

Yes it can. That's like saying [math]a+bi[/math] can't be a number because you can't derive the position of a digit in [math]a[/math] from the position of a digit in [math]b[/math].

Complex numbers aren't ordered but I basically constructed them with an ordering.

You gotta be quicker about these things if you wanna keep up

The point is that it's not a real number. You can't do real number arithmetic with it, you'd haveto make up new axioms for it to make it work and even then it won't work for actual real numbers. the + operator is defined only for real numbers

0.999... + 0.000...1 is not a valid operation since one of them isn't a real number

[math]\sqrt{1}= \pm 1

-1\neq 1[/math]

real numbers are a subset of the superreals, [math]\mathbb{S}[/math]. [math]\mathbb{R}[/math] is actually [math]\mathbb{S}/\langle\mathbb{T}\rangle[/math], where we abandon tiny numbers [math]\mathbb{T}[/math]. This is really advanced stuff they only teach you in PhD programs, so it's understandable you don't know it.

operations defined for set does not necessarily imply that they would work for a set that contains that set. For example, there are arithmetic operations defined for subsets of R that does not necessarily work for all elements of R

In [math]\mathbb{R}[/math] they handwave away the details and make up bullshits about limit points to get around the fact that [math]\mathbb{R}[/math] only makes sense for certain irrationals and finite/repeating decimals. Seriously go ask your professor, any good professor will tell you this is true.

>Seriously go ask your professor, any good professor will tell you this is true.
Ask them what? Those retarded pseudo mathematical ramblings?

>0.999...=
the last valid number
0.9... 1; 0.9... 0
0.9... < 1, if 1 existed
I'd be an total idiot if I wasted my time trying to communicate intelligently with you. You know it all, when you aren't deliberately stirring up problems.
Prove how 1 unit came to be from the 0 unit dimensions of a singularity.
I couldn't care less if you babble I'm wrong. I keep an open-mind of a questing general scientist, so I can't be wrong. Maybe you are "better than a god" as you believe.

0.999=OP is a bitch
prove me wrong
protip: you can't

>
[math]x=0.999...[/math]
[math]10x=9.999...[/math]
[math]10x=9+0.999...[/math]
[math]10x=9+x[/math]
[math]9x=9[/math]
[math]x=1[/math]
[math]0 . 9 9 9 . . . = 1 [/math]

Now shut the fuck up and go argue about something else you braindead mongoloid autists

It's not.

>brainlet gets btfo with deeper mathematical theories than he can process
>begins slinging insults with no refutations

fucking

lol

[math]10x=9.999\ldots[/math]
[math]10x=9+0.999\ldots[/math]
is not a valid step,
[math]10x=9.999\ldots[/math]
[math]10x=9-8\times0.00\ldots 001+0.999\ldots[/math]
is correct, which gives you in the end that
[math]0.999\dots=1-0.00\dots 001[/math]
which is true

0.111... = 1

oops, that 8 should be a 9

>is shown to be a fucking retard
>acts like he was only pretending to be retarded to not seem like a fucking retard on an anonymous kalmyk horse archery tip-sharing forum
>makes himself look even more retarded

you didn't show anything, though

in grade school they teach you bullshit like "there is a 1-1 correspondance between real numbers and decimal expansions", but they neglect to tell you that it's a lie, since their mapping, which ignores tiny numbers, maps 1 to both 1 and .999...

therefore, there's actually more (measure-wise) decimals than real numbers, and they handwave away the algebraic inconsistency to push an agenda

>not a valid step
Are you implying that [math]9.999 /ne 9 + .999[/math]
By that logic [math]9.5 /ne 9 + .5[/math]

kek

1/3 = 0.33333....
1/3 + 1/3 + 1/3 = 1.
1/3 x 3 = 1.

It's simple.

>prove me wrong

Son you havent proven yourself RIGHT to necessitate being proven wrong.

No, what you're saying doesn't even make sense. 10 times .999... = 9.999...9990 so 10-9.999...9990=0.000...0010

Pretty sure this is b8 at this point.

The nines don't ever end like you show.

the mappings between decimal expansions and real numbers are wrong

There are infinite nines INBETWEEN the ends, you doofus. I would expect a lowly undergraduate to get this, but I guess there is just no helping some people.

>mfw people on reddit-lite are dumb enough to believe the sequence 1,0,0,0,0,... is the same as 0,9,9,9,9,...

Real numbers aren't nessisarily described by a single sequence of decimal digits.

>You can't have an element in an infinite ordered set without a successor
Do you mean predecessor? Are you denying the existence of sets with order type [math]\omega+1[/math]? Do you not believe [math]\{1-\frac{1}{2^n} : n \in \mathbb{N}\} \cup \{1\}[/math] is an infinite ordered set? Oh right, you don't know what ordinals are because you're just a popsci retard.

0.(1) = 1/9
0.(9) = 9 * 1/9 = 1

Easy cheesy, lemon squeeze.

1/3 + 1/3 + 1/3 = 1
0.333... + 0.333... + 0.333... = 1
0.999... = 1

Why is this so hard for brainlets to understand?

As far as I know, it depends on the chosen metric wheter [math] 0.999... = 1 [/math]. You look on [math] 0.999... [/math] as a cauchy-sequence and if it convergeces at all and what its limit is depends on the metric.

0.999...=/=1
there
waddya gonna do about it?

Scientific notation is off

Because decimal expansions and fractions arent the same thing.

1 - 0.(9) = 0.(0)

0.(0) =/= 0

Yes they are. I think I can begin to understand why you can't get your head round this, it's because you're a simpleton.

wrong

.999 ≈ 1
.999 =! 1
there you go friendo

>it's a brainlets don't understand limits nor convergence episode

No, 1=1 and that's it. If 1 can equal anything else than 1 or an equation that also equals 1, then mathematics is broken as a whole and should be reworked to not be retarded

1 =/= ~1

are you saying that an irrational number can equal a rational number?

k

prove 0.999... is irrational

.999 recurring isn't irrational you dunce.

X ≥ 0
X+Y > 0

Solve for Y:

Y = .000...1

Otherwise this simple question has no answer.

how about Y=5?

Oops, I wrote the equation wrong, sorry.

X < 0
X+Y ≤ 0

Solve for Y:

Y = .000...1

so with X=-5 my Y could still be 5...
As long as Y >= X < 0 it should work in general.
i nearly get what you are trying to say, but i can't put it in an equation.

>.000...1

This isn't a number. The ... means "repeat forever".

infinitesimals

in base 2 it is

if 0.(9) = 1, then 1 + 0.(9) = 2
but it clearly doesn't. it equals 1.(9)
where is my Field's medal?