So, in non-standard analysis is 0.999... = 1?

[math]1 - 0.\stackrel{\rightarrow n}{\overline{9}} = 0.\stackrel{\rightarrow (n-1)}{\overline{0}}1 [/math]

sorry im a brainlet, are you saying 0.000...1 is not complete and utter nonsense? or at least can you link me somewhere that would explain this in more detail

If you are wondering whether [math]1 - \sum_{n=1}^{N}9\cdot 10^{-n}[/math] is an infinitesimal when [math]N[/math] is an infinite integer then the answer is "yes." However the standard part of this expression is still 1. So even though I already gave you a hard time about the 0.000...1 notation your intuition is correct.

>the standard part is still 1
the standard part of the sum is still 1, the expression I gave is 0, apologies

ah godspeed user ty for help

It gets better. If you want someone to justify your almost-notation 0.000...001 then you can look up "infinitesimals" by lightstone on libgen (or wherever, it's free to view on jstor). The notation isn't quite that, but it's similar enough and kind of funny. In the right hands, it could be some good trolling material. Use it wisely.

For infinite summation to work as prescribed, even while ignoring there is no way to actually increment to/reach infinity, at n=infinity requires [math]\frac{9}{10^\infty} = 0, or, n÷\infty = 0[/math], which is essentially proveable in calculus and theres no good reason dividing by zero shouldn't also equal infinity, or no good reason infinity shouldn't universally be declared as "undefined" rather than just solely in divide-by-zero examples.

>lightstone
i just googled him and wtf this dude just stole my genius work

there is no "n"

^ me here.

that's it, this is the last straw. western civilization was a mistake, time to convert to Islam

>0.000...1
Here's your first mistake. You turned an infinitely repeating number into a finitely repeating number.

The truth is, there is no 1 on the end. There is no end.

Well that just means 0.999... can't equal 1 because there is no end to the 9's.

It's infinitely close to 1.
You cannot define the difference.

Sure i can.
1 is an integer
0.999... is not an integer

Even disregarding what numbers they were as 0.n and n, one is clearly an integer and the other is not.

More to the point though, infinity doesn't /actually/ mean what you think it means, and "infinitely close" is not an intelligible way to describe the relationship between and 0.999...

If infinity means unending as you're implying, there are always more 9's in the repetition and they continue unending, thus 0.9••• = 0.9••• is valid but 0.9••• = 1 is invalid, as 1 is a finite number while 0.9••• is not.

0.999... can be simplified to not be fractional, the numerator is identical to the denominator so it mush be an integer considering that it is equal to 3/9 + 3/9 + 3/9 which equals 9/9, it is a whole number of unity, the exact same as 1.

[math]\frac{1}{3} × 3 [/math] obviously just equals 1 though. You're doing shitmath with brainlet notation if you think [math]\frac{1}{3} × 3 = 0.\bar{9}[/math]

heres a simple problem you can deduce to realization for yourself:
How do you increment to infinity?

Do you not think 1/3 equals .333... either?
The only way you can justify 0.333... added to itself three times or even 0.666... added to 0.333.. converging to 1 is by admitting .999... is the same as 1.
>How do you increment to infinity?
Applying calculus limits of sequence to check for convergence.

Say x = 0.333...
Does 9 / 3 = x?

Be careful how you answer.

9 / 3 = 3 user....

>1 is an integer
>0.999... is not an integer
Let's use 1.0 then.
Now they're both decimal point numbers. Get fucked.

I think the term you were after is a whole number, but my point still stands. Those are just labels, not the difference between the two numbers.

No matter how long you ran the calculation looking for the end of the sequence you would not reach the end. You would not reach a point where you could take the number of 9s, drop one convert to zeros and append a 1 to define the difference between 0.999... and 1. Never.

I'm so sorry.

9/3 = 3/1 = 3

3/9 = 1/3 = 1/3

1/3 does not accurately equal "[math]0.\bar{3}[/math]"

x = 0.999•••
10x = 9.999•••
10x-x = ?

the answer isn't "9x = 9"
If x had n amount of 9's, 10x must also have n amount of 9's.

x = 0.9, × 10 = 9.0 - x = 8.1 / 9 = x
x = 0.99, × 10 = 9.9 - x = 8.91 / 9 = x
x = 0.999, × 10 = 9.99 - x = 8.991 / 9 = x
x = 0.9999, × 10 = 9.999 - x = 8.9991 / 9 = x
x = 0.99999, × 10 = 9.9999 - x = 8.99991 / 9 = x
...
x = 0.9•••{n}, × 10 = 9.9•••{n-1} - x = 8.9•••1 / 9 = 0.9•••{n}
^ this remains true for all amounts of 9's, unending

Y= 0.99
10Y = 9.99
10Y - Y = 9
9Y = 9
Y = 1
^ this remains false for all numbers
10Y is not 9.99, 10Y is 9.9
10x is not 9.9•••{n}, 10x is 9.9•••{n-1}

Hasn't it pretty much always been known that 0.333... is just an approximation of what 1/3 is truly equal to? When did people start to think otherwise?

0.9999... implies you keep adding 9s at the end infinitely
0.000...1 ends with a 1 at some point and is therefore not 0

But 0.3333... is exactly 1/3 in decimal. 0.333....3 is an approximation.

...

x = 1/3
x = 0.333...
x = x
0.3333... = 1/3
QED

When you get to n=infinity, then n = n-1 which is why in your original example, x had 3 9s, but 10x had 4 9s and 0.999... converges to 1.

>If x had n amount of 9's, 10x must also have n amount of 9's.
Yes, it's infinite. You cannot infinite - 1
Unironically, 10 x 0.999... - 0.999... = 8.999...
There is no 1 on the end.

If we delegate all future writeable 9's in (0.999...9) to an aspect of infinite work which we know we can never reach and shouldn't matter to us, then we can just as easily delegate a future 1 to exist after the future 0's in the number (0.000...1) and equally be unbothered by it because it is future work that doesn't matter right now.

a+b = 1
a = 0.9 && b = 0.1
a = 0.99 && b = 0.01
a = 0.999 && b = 0.001
a = 0.9999 && b = 0.0001
a = 0.99999 && b = 0.00001
a = 0.999999 && b = 0.000001
a = 0.9999999 && b = 0.0000001
a = 0.99999999 && b = 0.00000001

you know... call me crazy, but it looks like no matter how many 9's we add to a it remains less than 1, and no matter how many 0's we add to b it remains greater than 0. This PATTERN seems REPEATABLE.

If you repeat that forever you still won't have reached the difference between 0.999... and 1.

The heat death of the universe will come before you reach the difference.

Of course you cant infinity - 1, because infinity is not a number. The equation infinity - 1 = x is only as intelligible as y - 1 = x, which is to say not very intelligent.

Maybe you're starting to realize you don't actually understand infinity.

> a = 0.999... && b = 0.000...1

0.000...1 is not real.

Notice how with each step the left side value approaches closer and closer to 1, and the right side approaches to 0. Infinite steps means infinite steps, the value after any finite number of steps will indeed not be 1, but we're just interested in the value this whole process approaches.

0.(9) = 1.
0.(0)[anything] = 0.

2 + 4 = ?
>Step 1: find 2 on the number line
>> one, two: ••
>Step 2: find 4 on the number line
>> one, two, three, four: ••••
>Step 3: count four from two
>> three, four, five, six: ••••••
>Step 4: solve
+ 4 = 6

0.333... × 3 = ?
>Step 1: find 0.333... on the numberline
.3
.33
.333
.3333
.33333
.333333
.3333333
.33333333
.333333333
.3333333333
.33333333333
.333333333333
.3333333333333
.33333333333333
.333333333333333
.3333333333333333
.33333333333333333
.333333333333333333
.3333333333333333333
.33333333333333333333
.333333333333333333333
.3333333333333333333333
.33333333333333333333333
.333333333333333333333333
.3333333333333333333333333
.33333333333333333333333333
.333333333333333333333333333
.3333333333333333333333333333
.33333333333333333333333333333
.333333333333333333333333333333
.3333333333333333333333333333333
.33333333333333333333333333333333
.333333333333333333333333333333333
.3333333333333333333333333333333333
.33333333333333333333333333333333333
.333333333333333333333333333333333333
.3333333333333333333333333333333333333
.33333333333333333333333333333333333333
.333333333333333333333333333333333333333
.3333333333333333333333333333333333333333
.33333333333333333333333333333333333333333
.333333333333333333333333333333333333333333
.3333333333333333333333333333333333333333333
.33333333333333333333333333333333333333333333-

>0.999... is not real

Notice how with each step, the left side does not ever equal 1 and the right side does not ever equal 0.

First of all, define "infinite steps".
How many steps is it exactly?

And? You can't count infinite steps. But the value bloody well clearly tends to 1.

Let's put it this way: if you continue this process infinitely long, you will eventually exceed every finite number smaller than 1. Therefore it's equal to 1.

Define "many" and "steps"

>0.9 is pretty close to 1 so 0.9 = 1

>these apples sort of look similar so I'll just replace them with identical abstractions

.999... is not real
Then your problem seems to be with decimals themselves, you don't believe in decimal notation preferring fractions and ratios because you can't even justify something as simple as 1/3 in decimal form.

When has enough work and time passed for "infinitely long" to suffice to have occurred...?

>How many steps is it exactly?
See

start dividing 1 by 3 in decimal form and someone will let you know

>0.9 = 1 in n accuracy; d.0
>0.99 = 1 in n.n accuracy; d.1
>0.999 = 1 in n.nn accuracy; d.2
>0.9999 = 1 in n.nnn accuracy; d.3
>0.99999 = 1 in n.nnnn accuracy; d.4
>0.999999 = 1 in n.nnnnn accuracy; d.5
It seems with finite decimal accuracy, a repeating pattern of 9's after decimal is equal to 1.
but arbitrary infinite accuracy... that never ends! It's not finite! 0.999... never reaches 1 because every extra 9 retroactively redefines the closest value to 1 while also redefining the d.value of accuracy!

We may only need 6 decimals in d.5 accuracy to equal 1, but we need d.infinity-1 accuracy for infinite decimal places to equal 1. Oh! But infinity-1 doesn't exist, so we can't have the required infinity-1 decimal accuracy where infinite 9's can equate 1.

>it must equal 1 because it will literally never ever equal 1
how fucked retarded do you gotta be to not think and how did you get that way.

>1 ÷ 3 can't equal anything because you can literally never make it equal something.
how fucked retarded do you gotta be to not think and how did you get that way?

Let me put it this way.

If you substitute 0.999... with 1 in any equation you will never be able to define the difference in the result.

Maybe if you're retarded.

The irony of your post is that, even though you just said it, you don't believe it's true.

Repeating decimals are problems of the number system, not actual values.
1÷3 doesn't equal 0.333... because the actual amount of leftover 3's are undefined. 1÷4 = 0.2+0.05 = 0.25, finite, flat, even steven in just two decimal places. Its the true decimal value.
1÷3 = 0.3+0.03+0.003+0.0003+...
unending, irrational, we can write all the 3's we want with unlimited decimal accuracy but it will never evenly portray the true value of 1÷3, and 0.333... will always be a significant amount less than the actual decimal value of 1÷3, which is why [math]A \frac{1}{3} × 3 = 1[/math] and [math]B 0.\bar{3} × 3 = 0.\bar{9}[/math] but [math]A \neq B [/math], because [math]\frac{1}{3} > 0.\bar{3}[/math]

>shifting doesn't work

what is 9.999.../10 ?

1/9 > 0.1•••
2/9 > 0.2•••
3/9 > 0.3•••
4/9 > 0.4•••
5/9 > 0.5•••
6/9 > 0.6•••
7/9 > 0.7•••
8/9 > 0.8•••
9/9 > 0.9•••

9.999...{n}/10 = 0.999...{n+1}

Then what does it equal in your world where .999... can't be 1 and how can you complete the arithmetic involving it in your example if the value of 1/3 can not ever be determined given an infinite amount of time to calculate it?

>undefined
its not undefined though, its the amount tending towards infinity and putting a bar about the 3 is the exact same as using ..., its still defined as tending towards infinity, different notations to indicate the exact same thing or its also undefined in which case your argument still breaks down and you can't even define a simple division of 3 into 1.

hello shitlatexman

>9.999...{n}/10 = 0.999...{n+1}

sure looks like a shift to me

You keep using infinity like you know what it means but you still haven't sufficiently defined it. Stop that.

Its not my job to define it, its an established mathematically defined value, the burden is on you to look up the widely accepted definition.
Its the final limit, the limit of all limits, the greatest value possible in any value set.

>1/9 > 0.1•••
>2/9 > 0.2•••
>3/9 > 0.3•••
>4/9 > 0.4•••
>5/9 > 0.5•••
>6/9 > 0.6•••
>7/9 > 0.7•••
>8/9 > 0.8•••
>9/9 > 0.9•••
Those are all wrong.

You can't increment to it, you can't reach it, and no arithmetic performed on it returns knowable values. Somehow you don't seem to understand that it isn't actually a number.

Your problem is that you don't believe in God.

According to your theory that completely disregards all known calculus, you still haven't explained how can you do any arithmetic involving 1 ÷ 3 or values that produce similar nonterminating decimals?

Rounding.

1 ÷ 3 = (10)÷3 = 0.3; 1r
1r ÷ 3 = (10r)÷3 = 0.33, 1r
1r ÷ 3 = (10r)÷3 = 0.333, 1r
1r ÷ 3 = (10r)÷3 = 0.3333, 1r
>this goes nowhere
1÷3 = [math]\frac{1}{3}[/math]

Then why were you using the equal sign and not saying can be rounded to or an approximate equal sign and why can't .999... be rounded to 1 in the exact same way 3* 1/3 is rounded to 1 as you claimed in ?

Yes one divided by three equals one divided by three, but what is the actual value?

Then why were you using the equal sign and not saying can be rounded to or an approximate equal sign and why can't .999... be rounded to 1 in the exact same way 3* 1/3 is rounded to 1 as you claimed in ?

It has no actual decimal value, but its fractional value is actually [math]\frac{1}{3}[/math] and is fine enough to leave it at that.

Mathematics has a limited vocabulary and you're all arguing in semantics thread.

Good think you aren't in charge of making calculators and computers, nothing would ever be computable in your world.

Calculators use finite decimal place accuracy, they don't even come close to approximating a significant amount of arbitrary accuracy work to sufficiently define infinite repeating decimals.

We already live in my world. Open your eyes.

The calculators I use (including the one built into windows, linux, and mac) show 1/3 = 0.333..., then when I multiply that by 3 its says 1 instead of 0.999... like you implied , so you are wrong.

OP HERE

I thought this thread died like 10 hours ago. Boy was I wrong.. it’s turned into a masterpiece I can’t stop laughing

Also nobody has really replied to my “if we have x and a number x0 infinitely close to it then the difference is an infinitesimal in non standard analysis” thing. If that’s the case then in the hyperreals 0.999... + infinitesimal = 1?

>he’s wrong because calculators use finite decimals places and are programmed to round up at a certain level of accuracy

Op here again..

If that is the case, then in the hyperreals 1/3 is not equal to 0.333... so if I was doing 1/3 + 1/3 + 1/3 I would need to add “1/3rd of an infinitesimal” to each of them? If so, then that would be 0.333... of an infinitesimal which would need “1/3rd of an infinitesimal”, etc, so isn’t that infinitely recursive?

>0.999... + infinitesimal = 1?
yes

1 = 9/10 + 1/10 = 0.9 + 1/10
= 0.9 + 9/100 + 1/100 = 0.99 + 1/100
= 0.99 + 9/1000 + 1/1000 = 0.999 + 1/1000
= 0.999 + 9/10000 + 1/10000 = 0.9999 + 1/10000

etc.

He is wrong in saying those finite decimals mean that calculators conform to his specifications in and he is wrong in saying calculators are fine just letting one divided by three equal one divided by three.

>Also nobody has really replied to my “if we have x and a number x0 infinitely close to it then the difference is an infinitesimal in non standard analysis” thing. If that’s the case then in the hyperreals 0.999... + infinitesimal = 1?
This is because it answers itself and I don't understand how it is a question. What does "infinitely close" mean other than "the difference between x and y is an infinitesimal"? Did you have some other interpretation about which this question has some obvious difficulty?

Well if you agree with me there then obviously 0.999... is not equal to 1 and all of standard calculus proves this isn’t self-evident so you’re just being an arrogant prick

Standard calculus would say 0.999... = 1 precisely because the difference between the two is infinitely small.

He's not asking about standard calculus. Have you read the OP?

>Standard calculus would say 0.999... = 1 precisely because the difference between the two is infinitely small.
Standard calculus identifies the infinite series [math]\sum_{n=1}^{\infty}9\cdot 10^{-n}[/math] with 1 because the difference between the nth partial sum and 1 can be made as small as we like. There is no notion of "infinitely small" or "infinitesimal" in standard calculus.

Non-standard analysis says simply that
[eqn]1- \sum_{n=1}^{N}9\cdot 10^{-n} = 10^{-N}[/eqn] and this is the answer. When [math]N[/math] is an infinite integer then [math]10^{-N}[/math] is an infinitesimal. 0.999... is not equal to 1 in non-standard analysis. However, the standard part of [math]10^{-N}[/math] when [math]N[/math] is an infinite integer is 0. So the standard part of 0.999... is 1.

How can I be more clear for you?

0.000...1 is an inherent contradiction.

the .... after the 000 implies that this goes on forever yet at the same time ends with one.

The ... implies an early identifiable pattern of repetition, not that it "goes on forever"
You can't prove that anything lasts forever, so you can't say it goes on forever. You can only say there is an early repeating pattern.

O K retard?

What if we wrote right to left?
We'd start with the end of numbers.
9 = x
1 = y
->99 = x
->01 = y
->999 = x
->001 = y
->9999 = x
->0001 = y
->99999 = x
->00001 = y
We'd eventually have to make the assumption about what their beginnings look like:
0.9...999 = x
0.0...001 = y

Nothing has changed about the values, only a slight difference in looking at them. You are not smarter than an average ghetto nigger who can't figure whether water is wet or not. You do not understand infinity. You can't even rigorously defend or explain the classically retarded way you've been taught to think about it. You are basically a broken computer program.

>The ... implies an early identifiable pattern of repetition, not that it "goes on forever"
BULLLLLSHIT

>early identifiable pattern of repetition
>repetition
>doesn't go on forever
>u can't prove that anything lasts forever

the decimal representation of 1/3 goes on forever

That’s literally what he just said dumbfuck. You were the one that stated, “what does infinitely close mean other than the difference between x and y is an infinitesimal.” It can mean that the difference between x and y isn’t an infinitesimal in standard calculus which you just admitted

Also you’re playing semantics with the other guy. There is a proper notation in non-standard calculus for infinite reoccurrance before the lsd

On your calculator, do 1 ÷ 3 =
multiply it by 3

clear the calculator.

On your calculator, type 0.3, with as many 3's at it will allow you, then multiply it by 3.