Tell me a number between 0.999... and 1

...0...9...0009....999

> (You)
> (You)

[math] \displaystyle
1= \frac{1}{2}+\frac{1}{2}=\frac{1}{2}+\left (\frac{1}{4}+\frac{1}{4} \right )=
\frac{1}{2}+\frac{1}{4}+\left (\frac{1}{8}+\frac{1}{8} \right )= \cdots

\\\Rightarrow 0.\overline{1}_{2}= 1

[/math]

your point being?

do the same with

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

the value never deviates from 1

It really doesn't explain why 1 = 0.999... but it's the same concept

although it is

>doesn't explain
It's way easier to understand.
You are at 1 already in the beginning, at each stage you try to chip a piece off, but fail.
A lot less fuzzy than summing up little bits to see if they equal 1 somewhere in infinity.

it's just as easy as trying to find a number between 0.999... and 1 and failing

nah, being at the target from the very beginning feels better

Is this a proof that quantum mechanics is in math?

x=0.999...
10x=9.999...
10x-x=9.999...-0.999...
9x=9
x=1

x=0.999...
10x=9.999...
9x=9.999...-x
9x=9
x=1

0.9999... + (1 - 0.9999...)

>Tell me a number between 0.999... and 1
(1+0.999...)/2

or if you like

0.9999... + Sum[ N * (1 - 0.9999...) ] for N of 1 to Inf

retard coming in. I learned in math class that if a number was higher then 0.5 it was rounded off to 1 so with this logic wouldn't be 0.999.... 1?

with this logic wouldn't 0.75 also be 1
you might not be wrong but you are definitely retarded

ofcource it would. didn't you go to math class "EVERYTHING HIGHER THEN 0.5 BECOMES 1"

It's only natural

Nice proof, Thanks bud

0.9999... + (1 - 0.9999...) would add up to 1

so a number between those would be, for instance:

0.9999... + ((1 - 0.9999...)/2)

You usesld what youre proving dumbshit

Is ithe possible for two real numbers to have the same cardinality, but a different order type? Basically, is it possible for two numbers to be equal to one another in value, but positioned differently on the number line?
I think that the number 0.9999... is equal to the number 1, but also that it is a different number nonetheless, with different properties.

x = 0.999...
10x = 9.999... - 0.000...9
9x = 9.999... - 0.000...9 - x
9x = 9.999... - 0.000...9 - 0.999...9
9x = 9.999...0 - 0.999...9
9x = 9.000...1
x = 1.000...111...

prove me wrong faggots

Oh, I can do that too, senpai!
(X = (3 OR 6)) = TRUE
(3 = X) = TRUE
(6 = X) = TRUE
therefore
3 = X = 6
therefore, by the Transitive Property of Equality,
3 = 6
How did I do?

...

if .999... = 1.000...
does .999... ...8 = .999...?
what about 1.000... ...1?
does that also = .999... ...8?
does 0.000... = ∞?

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

I mean.
The only reason people say 0.99.. = 1 is because they don't understand that we can't actually properly write 1/3, 2/3, etc. and other non-terminating numbers in our system.
The assumption that the two are the same thing often use the fact that 0.99.. x 10 still never terminates, even though it as a property should- we just cannot properly write it as so in decimal form.

infinity math is retarded:

let G be the variable to represent a series of undefined constant length of repetitions of one number in a decimal
and G>1 and G < infinity

example
G = 5
then 0.5G is 0.55555

x = 0.9G
10x = 9.9G - 0.0G9
9x = 9.9G - 0.0G9 - 0.9G
9x = 9 - 0.0G9
x = 1 - 0.0G1
x = 0.9G

No.
Okay so let's say you have 9 lights in a column. Once you've lit all of them up, adding 'one more' light would light up another light to the left and turn off all the lights. So you basically have 9 dark spaces that can be filled in. If we were to light up every light for every column going off to the right toward 9.9999, no matter how far we go there is another 9 tenths needing expressed to the right of that. So every single light of every column onto infinity is lit. No matter how far you go there is not a light turned off.

So, if I were to subtract .999... From 1, there would be precisely zero lights, same as if I subtracted 1 from 1. Therefore, .999=1

.999...8, otoh, no matter how close to infinity you are, if you subtract that from one there will always be 1 light on.

.999... Doesn't have a 0.00..1 at the end of it

Mathematics education would be so much better if they used infinitesimals instead of ordinary so-called real numbers.

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

>Tell me a number between 0.999... and 1

0x.FFFFFF...

you cant properly represent thirds in decimal format dumbass

[math] \displaystyle
\begin{align*}
\frac{1}{3} = \left (\frac{3}{10} + \frac{1}{30} \right )
&= \frac{3}{10} + \frac{3}{100} + \frac{1}{300} \\
= \frac{3}{10} +\frac{3}{100} + \left (\frac{3}{1000} + \frac{1}{3000} \right )
&= \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \frac{1}{3000} \\
= \frac{3}{10} +\frac{3}{100} + \frac{3}{1000} +\left (\frac{3}{10000} + \frac{1}{30000} \right )
&= \frac{3}{10} + \frac{3}{100} + \frac{3}{1000}+ \frac{3}{10000} + \frac{1}{30000} \\
&\vdots
\end{align*}
\\
\Rightarrow 0.\overline{3}= \frac{1}{3}
[/math]

no. you don't understand.
the reason it is infinite is because it cannot be properly expressed
it is the limitation of the base number system-

the use of 0.33... is just the closest fit
it is not actually equivalent to one third

every one of those stages equals 1/3
every.one.of .them.

including the one that has a infinite terms

this is NOT a sum creeping up to the value 1/3

more compact version

[math] \displaystyle
\begin{align*}
\frac{1}{3} = \left (\frac{3}{10} + \frac{1}{30} \right )
&= 0.3 + \frac{1}{30} \\
= 0.3 + \left (\frac{3}{100} + \frac{1}{300} \right )
&= 0.33 + \frac{1}{300}\\
= 0.33 + \left (\frac{3}{1000} + \frac{1}{3000} \right )
&= 0.333 + \frac{1}{3000} \\
= 0.333 +\left (\frac{3}{10000} + \frac{1}{30000} \right )
&= 0.3333 + \frac{1}{30000} \\
&\vdots
\end{align*}
\\ \displaystyle
\Rightarrow 0.\overline{3}= \frac{1}{3}
[/math]

[math] \displaystyle
\begin{align*}
1 = \left (\frac{15}{16} + \frac{1}{16} \right )
&= \text{0x0.F} + \frac{1}{16} \\
= \text{0x0.F} + \left ( \frac{15}{256} + \frac{1}{256} \right )
&= \text{0x0.FF} + \frac{1}{256}\\
= \text{0x0.FF} + \left ( \frac{15}{4096} + \frac{1}{4096} \right )
&= \text{0x0.FFF} + \frac{1}{4096} \\
= \text{0x0.FFF} +\left ( \frac{15}{65536} + \frac{1}{65536} \right )
&= \text{0x0.FFFF} + \frac{1}{65536} \\
&\vdots
\end{align*}
\\ \displaystyle
\Rightarrow \text{0x}0.\overline{\text{F}} = 1
[/math]

Could there be an alternate universe in which the mathematical community settled upon an alternate convention for the use of ellipses by which ellipses were understood to represent an infinite sequence rather than being considered to represent a number? In other words, isn't this just a matter of linguistic conventions, and not some kind of profound universal truth woven into the very fabric of reality?

Yes, but if you believe in infinitesimals then
that representation is between .999... and 1.

It can be both. You can just abuse the notation if you want to let 0.9... signify {0.9, 0.99, 0.999, ...}. It'd be pretty clear from context whether you mean the sequence or its limit.

no it isn't

every single stage separately is = 1

line #2 for example is just 15/16+15/256+1/256
which is exactly = 1
and so are all the other ones too, of any length, including the one with an infinite length

well is 11

...
you CANNOT fully represent 1/3.
0.33... - isn't the actual value of 1/3 at all since the answer is infinitely recurring

see

you understand that operation is never infinite right?
yes the two sides of the equation are equal but there will literally always be a finite amount of terms on the right side
you cant represent it as the recurring type

So-o, we're not talking about how summing is tricky and surprising with infinity involved?

You are actually stating that in math infinity doesn't exist?

wew lad

Every single stage shows the value and the error.
If you do the exact same expansion for decimal
you will find the error is more - 1/10 vs 1/16, 1/100 vs 1/256... etc.
So in terms of infinitesimals the base 16 representation will be a slightly different number than either .999 or 1 and will thus be "inbetween" the two.
But you don't have to accept the notion of infinitesimals to work with everyday real numbers.

Why is 1=0.99... so hard for people to grasp? Does american education not include limits in the curriculum?

>and the error
what error, where?
each line = 1, exactly
each line is correct
line #2 is correct
line #10 is correct
line #389475896589346 is correct

with infinite lines, it remains correct

0,999,5

500

Its an assimptote, meaning its allways getting closer and closer but never quite touching 1. There is nothing between them because the (...) IS the between them.

Lim x->1

.9/1

big =/= infinite
If there is anything in between, it's merely big.
By definition, infinity is when there is nothing in between.

(.999...+1)/2

/thread

just go and read a book before spewing any more rubbish.

1-inf negative
1-0

0 ----sqrroo>1 0--exp-->1 .0--->10
is not integer

(1+1)/2=1

Any interval is infinitely divisible, unless you have autism.

Gee you totally showed me, jamal.

It should be .999 infinitely, then .999... at the end

that's not a number
that;s like comparing 1 to 111111111111111111

aynrandlexicon.com/lexicon/infinity.html

Asymptote (check your spelling) is defined for curves. 0.(9) is not a curve. And (...) is not a thing in and of itself. None of what you said made any sense and it was bad English to boot. Again, read a book.

>/b/

/b/ still exists?

in the first line you stated x = 0.999...
then in the fourth line you replaced x with 0.999...9 instead of 0.999...
i mean sure they're technically the same thing but the by way you're using it no

>tfw i will probably fail physics class over rounding to the wrong number for significant units

all the numbers to the right of the ... are unreachable
they are unreachable because you can never get to them
0.000...1 is indistinguishable from 0 in any meaningful way because in any calculation, the only thing that has any interaction are the endless 0s.
1=/=0.999... implies that 1=0.999...+0.000...1 or similar.
If we treat 0.000...1 as a 0
we get 1=0.999...+0

Like every non-brainlet in this thread has already mentioned, if you want to actually understand this stuff you should go and study calculus.
>inb4 infinitesimals are not 0

omfg
when representing 1/3> both sides of that equation no matter how far out you expand will have a fraction that cannot be fully expressed- the base ten decimal conversion is incomplete

...

they aren't the same
it's just that we pick the perfect scenario from the limit
if you actually passed calculus you'd know that

and your rationale is retarded because its based on how much you think 0.0..1 impacts an equation

>assimptote
come on, man
don't do this to me

other wait around, physics brainlet

explain

[math]0.999 + \epsilon[/math]

Wow, That Was Easy!

the easiest part in that was ignoring the ...
also the dumbest part

>And (...) is not a thing

>no matter how far out
someone doesn't grock infinity

stfu ur stupid

en.wikipedia.org/wiki/Definition

>x=0.999...
>11x=9.999...+x
>9x=9.999...-x+x
>9x=9.999...
>x=1.111...
What the fuck guys
How can x be both 1 and 1.111...???

>11x - 9x = x

Your Nobel prize will come in the mail.

>9x=9.999...-x+x
9x=10x-x+x
9=10-1+1
9=10

where did you get your bachelor of addition user?

...

>brainlet

TIL that [math] \displaystyle 1.\overline{0}[/math] doesn't exist

*zenos laterally*

It's a hard word to guess how it's spelled outside my native language.

Except that 1/3 does not equal 3/10 + 1/30