0.(9) ≠ 1

No, the problem is 1/3 =/= 0.333...

problem with the 1/3 = 0.(3)

because 1/3 ≠ 0.(3)
0.(3) -> 1/3

I mean 1/3 -> 0.(3) woops

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

...

0.1 = 1/10^1
0.01 = 1/10^2
0.0...1 = 1/10^inf = 0

prove it

en.wikipedia.org/wiki/Least-upper-bound_property

>if 0.00......1 is not a real number
It is

en.wikipedia.org/wiki/0.999...
Just read an article you brainlets

Prove it

0.999...9 + 0.000...1 = 1
0.999...9 = 1 ->
-> 0.000...1 = 0
Infinitesimal = 0

I have already read everything and many times

>1/3 -> 0.(3)
so
1/3 -> 0.(3) |x3
3/3 -> 0.(9)
1 -> 0.(9) hmmm
but 0.(9) < or = 1.
so 0.(3) x3 ≠ 0.(9) or/and 1/3

If 0.(9) != 1
It must be whether 0.(9) > 1 or 0.(9) < 1
But 0.(9) > 1 is wrong because integer part of 0.(9) , which is 0, is smaller than 1

Also, since both are real number there must be 1 another real number between 0.(9) and 1

So let's say 0.(9) < a < 1 and a exists
Since a < 1, a = 0.b1b2b3b4.....bn
If b1 < 9, 0.(9)

...

just 0.(3) -> 1/3 ok

>0.999...9 + 0.000...1 = 1

wow so 1+0=1
ty mathsuperman

1) If 0.(9) = 1, then infinitely small = 0.
But this is absurd.

2) If 0.(0)1 does not belong to the set of real numbers, then and 0.(9) does not belong to the set of real numbers.
They simultaneously cease to belong to the set of real numbers.

>1) If 0.(9) = 1, then infinitely small = 0.
But this is absurd.
Prove it then

>2) If 0.(0)1 does not belong to the set of real numbers, then and 0.(9) does not belong to the set of real numbers.
They simultaneously cease to belong to the set of real numbers.

0.(0)1 is fucking real number you retard

Do you know why is sqrt(2) a real number?

Above all, you didn't falsify my argument; you just sperg out another autistic claiming

1) school-level mathematics (Wikipedia+Google please)
2) en.wikipedia.org/wiki/Talk:0.999...

dealing with repeating decimals isnt well defined
3*1/3 is defined
3*0.(3) isnt defined
try to avoid repeating decimals

0.(0)1 is just 1-0.(9) so

Let's say (1-(0.(9)) != (1-1)

Then (1-(0.(9)) must be bigger than (1-1) or smaller than it

So let's say (1-(0.(9)) < (1-1)
Then must be 0.(9) > 1 but we already figured out it can't be, so 0.000....1 can't be smaller than 0

What if (1-0.(9)) > (1-1)?
Then there must be one number between both
Let's say 1-(0.(9)) > a > 1-1 and a exists
Hmm.... then 0.(9) < 1-a < 1 and 1-a exists
1-a = 0.b1b2....bn « bn € {0,1,2,3,4,5,6,7,8,9} »

But we already know bn = 9 so 1-a = 0.(9)
a = (1-(0.(9)) so

1-(0.(9)) > a doesn't make sense

So 1-(0.(9)) = 1-1 -» 0.(0)1 = 0

Maths have been mistaken for centuries.
Is it possible?

I see no reason to continue to communicate with you, I'm sorry.

>school-level mathematics
>just look up wikipedia and google
Not an argument.

You also learn 0.999999.... = 1 in school tho

>0,000...01
oh for fuck's sake

Because you can't prove 0.(9) != 1.

You are just try hard edge brat teenager thinking you are somehow better than most of mathematicians.

well the problem with repeating decimal is they arent complete you are repeating a task over and over again because you cant complete it
you always end up with a reminder

0.999...9 + 0.000...1 = 1
0.999...9 < 0.999...9 + 0.000...1/2 < 1
0.999...9 < 1 - 0.000...1/2 < 1

I mean
0.999...9 + 0.000...1 = 1
0.999...9 < 0.999...9 + 0.(0)05 < 1

You must prove 0.(0)1 != (0.(0)1)/2 and 0.(0)1 > (0.(0)1)/2 first tho

0.(0)05 x 2 = 0.(0)1
proved

>HAND WAVING INTENSIFIES

>you always end up with a reminder
just reminding you are faggot

That's the same thing.

whatever cuntface

0.(0)05 x 2 is 0.(0)10, not 0.(0)1 you retard

"0.999... = 1" is what's keeping us from interstelar travel

is 0.(0)100 or 0.(0)10000 or 0.(0)1(0)
1 = 1.000000000
1.1 = 1.1000000000

>0 x 2 is 0, not 0 you retard
kek

ban

Of course it's well defined. You should know this from your Calculus course.

[eqn]0.\dot01=a=0.c_1c_2c_3...c_n (c_k \in \{0,1,2,3,4,5,6,7,8,9\})[/eqn]
[eqn]
0=0.\dot0=b=0.d_1d_2d_3...d_n (d_k \in \{0,1,2,3,4,5,6,7,8,9\})
[/eqn][eqn]
\text{if }a \ne b \text{ is true}
[/eqn][eqn]
\text{then }c_final \ne d_final\text{ is true}
[/eqn][eqn]
0=c_1=c_2=c_3=...=c_\infty=d_final\text{ but }c_final = 1,\text{ thus } c_\infty \ne c_final
[/eqn][eqn]
\text{Let's say }c_final = c_{\infty +1}
[/eqn][eqn]
\mathbb{C}=\in\{c_1,c_2,c_3,...,c_{\infty + 1}\}, \mathbb{D}=\in\{d_1,d_2,d_3,...,d_\infty\}
[/eqn][eqn]
n(\mathbb{C})=\infty +1, n(\mathbb{D})=\infty,
[/eqn][eqn]
\text{and } n(\mathbb{C}) \ne n(\mathbb{D})
[/eqn][eqn]
\text{so } \infty \ne \infty +1 [/eqn]
But it isn't(en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel),
[eqn]
\text{thus }\infty=\infty+1[/eqn][eqn]
n(\mathbb{C})=n(\mathbb{D}), c_final=c_{\infty+1}=c_\infty
[/eqn][eqn]
\text{ but }c_\infty=0\text{ so }c_final \ne d_final\text{ is not true}[/eqn]
[eqn]\text{thus }a \ne b \text{ is also not true}[/eqn]
[eqn]\therefore 0.\dot01 = 0[/eqn]

that limit is equal to 1 you silly man

>=∈
*=

>delirious

>Get this triggered

By that logic
(1÷3)3=0.(9)? How? An operation should work same forward as backward

Let x be any number that is smaller than 1
Clearly x

0.000...0, 0.999...9, 0.000...1 are not the same as 0.000..., 0.999..., 1*10^(-n). If you write a number as the former, you imply the number terminates after some infinite amount of decimal places back. By definition, a number extending to infinity NEVER terminates. Thus the proof of convergence for 0.999 = 1 holds.

We need a base 18 for decimals to et past this logic problem of infinite repeating single digits like 0.333•••
I've proposed this a couple times but just keep sharing it
01a2b3c4d5e6f7g8h9

1/3 = 0.c, where each lower letter represents and infinite single decimal. Its important to make this distinction cause 0.3••• is as different frm 0.3 as 0.3 is from 0.4, as to say that 0.3••• is a unique number to perform arithmetic on, thus would make more sense to have its own symbol like 0.c

0.3 × 4 = 1.2
0.33 × 4 = 1.32
0.333 × 4 = 1.332
0.3••• × 4 = 1.3•••
Logic would dictate there would be a 2 at the end of that repeating sequence of 3's but there never is.
a = 0.111•••
b = 0.222•••
c = 0.333•••
...
h = 0.888•••

So thaht 0.c × 3 = 1.0 easily, and provable by counting on the number line of 0abcdefgh. 0.c as the analog for the repeating "3" gives us a base
0.c×3 = 3×3 on the decimal number line or 9 counts
abc def gh0

1/3 = [math]0.1_3[/math]

Thanks for the pointless post.

The extra numbers of abcdefgh would only be used for the decimal point, so really it's just an extension of base-10, the one everyone uses. Its nice that 0.333••• is 0.1 in base-3 but no one uses base-3. The problem isn't the base, the problem is how to interpret repeating decimals in a base.

0.3 with a hundred, a thousand, even a million 3's after it, is a literal different number than 0.3•••, as the finite limit of 3's in the non-infinite-repeating case eventually allows base10 arithmetic to be performed.
For example:
0.3 × 12 = 3.6
0.33 × 12 = 3.96
0.333 × 12 = 3.996
0.3333 × 12 = 3.9996
0.33333 × 12 = 3.99996
0.333333 × 12 = 3.999996
which leads to the faulty logic of
0.333••• × 12 = 3.999•••6
where base-10 arithmetic fails the repetiton.

Thanks for the pointless posts

>some retard went through the effort to make this
That's not how our number system works

Kill yourself f a m

1/3 ≠ 0.(3)
I will not repeat anymore

1/3 = 0.333•••
This is not very difficult to understand.s

what kind of nonsense are you talking about?

I do not understand your logic
just a statement without proof

stop talking nonsense

It isn't nonsense.
0.999••• = 1 is nonsense. It cannot be proved by itself and requires getting it as a sum or using division from it, such that 1/3 = 0.333••• so 3/3 = 0.999•••, while we know N/N = 1, or that 0.999••• ÷ 3 = 0.333•••, and 1.0 ÷ 3 = 0.333•••

Thats why the new decimal number line would have these repeating decimal steps. For numbers 1-8, where 0.999••• doesnt exist because it can be proven that 0.c × 3 = 1.0 by itself.

Try and contradict it.

0.3 -> 1/3
0.3333 -> 1/3
0.3333333... -> 1/3
0.(3) - Infinity in this model is unreachable.
1/3 - here infinity is already achieved.

1/3 - "numerical machine" and not the final product.

Incompatibility of theoretical models.

no
0.(0)1

>0.(3) - Infinity in this model is unreachable.

moron, infinity is in that syntax's definition

I dont know what the hell you're talking about. What the fuck does 0.(n) mean?

0.3 is not equal to 1/3 you mongol.
try dividing 1 by 3. Do you even know what long division is. Did you even kow fractions are writte decimally by doing the division?

fucking street shitting pajeet go throw your dirty trousers on someone else's yard.

so we have two "0.(9)"
0.(9) -> 1 --- real 0.(9)
0.(9) = 1 --- Imaginary 0.(9)

>Not equals, only converges to one.
yes equals, because 0.(9) is the notation for the limit, not the sequence of partial sums.

*sobs*

Infinity is NaN and so is the infinitesimal also NaN. These are concepts, not numbers. Pi is a number, infinity is not a number. Pi has a seemingly infinite amount of non-repeating decimal places, but "infinity" itself is not a property of pi, because infinity is a concept and not a number, no more than than a Ford F150 is a number, or a plastic bag is a number. You do arithmetic on numbers, and using infinity in arithmetic leads to useless results, because infinity is not a number but instead a vague, undefined amount of something, much the same as trying yo determine the numerical value of "some" or "a lot".
1÷3 = 0.3••• this is true
3÷3 = 1.0 this is true
0.3••• × 3 = 1.0 this is true
0.3••• × 3 = 0.9••• this is not true

the idea that 0.3••• × 3 = 0.9••• extends from the FLAW of thinking "0.3•••" is related to "3" for the obvious reason that both numbers contain a 3 and we assume 33×3 = 99, 3×3 = 9, 0.3 × 3 = 0.9, so 0.3••• × 3 must also have a 9 in the answer.

0.333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333 is NOT the same number as 0.3•••, and it is as truthfully as different as 1 is different from 2, meaning 0.3••• is a UNIQUE number that is confusingly written using established numbers.
It would be like instead of a number existing for 8, you were meant to write colon-three instead ":3" so that the number line would be
0 1 2 3 4 5 6 7 :3 9, and instead of ":3" being a combined character on your keyboard where 8 would be, there is no such combination and you're expected to manually write ":" and "3" individually everytime you wanted to use the number eight. Its dumb, and as dumb as writing 0.n•••, or 0.n..., or 0.(n) or whatever the fuck you do, which is where the extended decimal base-18 system comes into play to solve by ease of interpretation.

You only know how to use someone else's theories and stupidly prove them every time by any means.
But you do not want to think with your head.
You can not create something new yourself.

Our conversation is over.

In short 0.999••• doesnt equal 1 because (1÷3)×3 does not actually equal 0.999•••, but instead the equation is "([a unique number])×3 so that that the answer is 1"

I dont know what you're talking about retard. This is my theory one, and we weren't having a conversation either. So far as I could tell you were just shitting out brain diarrhea which no, that does not pass for information conversion much less conversation. Stay a forever brainlet.

0.9
0.99
0.999
0.999...9

Look, there are only nines here.
There is no one here.
0.999...9 - absolute limit.
The largest number of model.
After him there is nothing.
And this number itself is unattainable.

But you say that there is something after infinity.
You violate the laws of the model.
You are violating our old arrangements.

1,2,3,... -> infinity -> another reality
What kind of absurdity?
It is impossible to combine these models together.

An infinite number of points in a segment.
There are two incompatible realities.
0.(9) = 1 only in the world of segments, but not in the world of points, they do not even know about the existence of 1.

But you prove that they know about the existence of 1 and that their world must live according to the laws of your world.

>you ever smoked so much weed you became a foreigner?
honestly my guy, it is very difficult to understand you. What is your native language?

...

To reiterate, "infinity" is NaN. Not a number.
It is as useful as replacing with "some" or "a lot". You cannot do an infinite amount of something because you don't know how much to do, just as the same as doing a thing "some" amount of times, "many" amount of times, or "a lot" of times. "Infinity" has the same numerical meaning as some, many, a lot, an undefined "x" variable in a equation not equal to anything, whatever. It's not a number, and therefore you can't do arithmetic with it that results in a number.

For example:
>5x+4x =
We can evaluate it to be
>5x+4x = 9x
but X still has no value and is therefore not an actual arithmetic equation presenting a single number solution. Just as valid would be the answer
>5x+4x = 8x-x
And you can see that all of this has different interpretations and is meaningless towards producing some kind of intelligble single unit answer.

You could also rewrite it
>5∞+4∞
and its still just the same. Infinity is absolutely not number, and can't be used in any arithmetic to get a number that isn't already afflicted with infinity or a with property of infinity, with the latter extending to 0.999•••, but 0.999••• is also not a number that is sanely represented in readable or writeable arithmetic. Ergo, it makes sense to give the repeating single digit numbers 0.1 through 0.8 a unique identifying symbol or character that understands the problem of infinite repeating singles being nothing more than the product of using Base-10.

google translater
What sentence?

>"Infinity" has the same numerical meaning as some, many, a lot,
nope, infinity means, surpassing all bounds. Whatever upper bound you make, infinity is larger
and if you say that infinity does not exist then you also say that [math]0.\bar{9}[/math] does not exist.

Sum to infinity means that you let your index grow without bounds and look what happens to the total

Its not just Base10 that has repeating numbers but you get the idea. Base-16 would need a similar layout like
0123456789ABCDEF
abcdefghjklmn where
j = 0.AAA•••
k = 0.BBB•••
l = 0.CCC•••
m = 0.DDD•••
n = 0.EEE•••

Now this is all very fresh and I'm open to changing the abcdefgh stuff to different symbols if not creating new symbols or some kind of lowercase number or strikethru that wouldn't conflict with subscript numbers or the broken idea of overline numbers, neither of which are easily reproducable on a keyboard, but i feel using symbols that look like the established base10 arabics would just continue to lead to this logic fuckup of misinterpretation for normies and brainlets.

but user, in base-16 0.FFFFFF(...) is 1.000000(...)
and in base-3 0.222222(...) is 1.000000(...)
and so on

simply because infinitesimals don't exist in the real numbers.
In hyperreal numbers? Sure
In surreal numbers? Sure
but not in real numbers

You cannot take a measurement of the total without concatenating and deciding at what number you've cut off to take the measurement, which is no different than saying 0.33333 × 3 = 1 which is, for a fact, false because 0.33333 × 3 = 0.99999
You can't say that 0.333••• × 3 = 0.999••• = 1, because you are basically saying 0.3x × 3 = 0.9x = 1, which is also false.
0.9x is a number even if we don't know what x is and is therefore only equal to itself without arithmetic. 0.9x = 0.9x.
You can try to solve for x in 0.9x = 1 all you want but you wont get very far with this understanding.
>divide both sides by x
Ok 0.9 = [math]\fract{1}{x}[/math]
x = 1.111•••
ah, but we are defining infinite repitition as a variable, so really we get
x = 1.1y
For the equation
>0.9(1.1y) = 1
Which has literally solved nothing and only made the equation more complex, and will continue to do so for every attempt at defining the variable.

It is DISGUSTING that infinity is used in "higher" math as if its suppose to be an intelligble number. You cannot set a limit to infinity cause that is just saying no limit, which then precludes and invalidates writing a limit into the equation outright.

brainlet

>You can't say that 0.333••• × 3 = 0.999••• = 1, because you are basically saying 0.3x × 3 = 0.9x = 1, which is also false.
I have no idea what the fuck you're trying to say.
how do you get from [math]0.\bar{3} \cdot 3 \eq 0.\bar{9} \eq 1[/math] to [math]0.3 \cdot x \eq 0.9 \cdot x \eq 1[/math]?

fuck it, just replace all \eq with =

The point of this thing that i'm just going to call intermantissa is to get rid of writing single digit repeating decimals, and writr them in a way that is distinct from the single number they're composed of to get around logic problems.
0.3 × 3 = 0.9 we can do the math and know this
0.4 × 3 = 1.2 we can do the math and know this
the reason we know to do the math this way is cause we understand 0.4 has a unique and different value and written symbol than 0.3
If 3 wasn't even a number and we were meant to write three as "-22-", two 2's surrounded by hyphens, this would get very old and cumbersome very fast, and arithmetic with "-22-" would be confusing where 2 × 4 = 8 but -22- × 4 = 12 rather than what might logically be seen as "-88-". Its a battle of logic reminding yourself that "-22-" is not "two 2's" but instead "three"; much the same as 0.333••• or any single digit repeating decimal has no logical meaning when used in arithmetic. "0.333•••" isn't "zero point three three.. three repeating", cause 3 isn't "hyphen two two hyphen". Where the logic problem of -22-×4 = 12 instead of -88-, we get a similar logic problem where 0.333••• × 4 = 1.333••• instead of 1.333•••2

Really, just use the abcdefgh decimal system for a couple problems and it becomes easy to understand.