/sci/ - Science & Math

>>9531728
oops i meant to say
>If so, how would 0.000...1 not be equal to 0?

>>9531728
bump help

>>9531728
0.000...1 is a number in neither standard nor non-standard analysis.

>>9531757
why? also what about the other question

0.999... is not equal to 1 by itself and requires the addition of an infinitisimal to properly sum 1. The repeating nature of repeating decimals is not to say that invoking the number means you continuously receive repetition that otherwise alters a single examination of the invocation, which is what decimal shifting ignorantly ignores in implying
>m=0.999...
>10m = 9.999...
>10m-m = 9
this shouldn't be considered valid because otherwise it assumes the amount of 9's and significant decimal places in 10m and m are not the same, which wouldn't be similar for any other normal, complex, real number that succintly ends in finiteness, such as 1/4 = 0.25

For as long as m=0.999... spits out more 9's, there exists a smallest real amount 1 - 0.999... = z that seperates the difference between 0.999... and 1.

>>9531728
0.00...1 has no useful definition. What is it supposed to mean? The whole point of a repeating decimal is that the repetition doesn't terminate. If you're suggesting that 1 - 0.999... = 0.00...1 then you're contradicting yourself.

Also as for infinitesimals, this is obviously not a rigorous way of looking at it but an infinitesimal is the difference between a number and itself. The only way to get anything out of a summation of infinitesimals is to also be summing an infinite number of them so that spooky shit will happen. If you accept that exceptional things can happen with """numbers""" that are infinitely small, you must also accept that exceptional things happen when considering an infinite number of them.

>>9531814
but that's wrong

>>9531817
Repeating decimal isn't so much a problem as simply the shitty outdated notation is.
[math]1 - 0.\stackrel{\overline{9}}{\rightarrow n} = 0.\stackrel{\overline{0}}{\rightarrow (n-1)}1[/math]
You could address the repetition for any number n and count them up to show that the " ... " or overline repeating symbols are shorthand for "a lot" rather than "continuously more and more with undefined assumptions".

0.99999, -> 5 nines
0.00001, -> (5-1=4) zeros

0.9999999999, -> 10 nines
0.0000000001, -> (10-1=9) zeros

0.999..., -> n nines
0.00...1, -> (n-1) zeros

>>9531817
op here

yes im aware 0.999... = 1, i made that clear in my original post. it represents the same thing an infinitesimal does conceptually aka an infinitely small number

i was secretly hoping there was some area of math that agreed with my supposed retardation but i guess not

>>9531844
[math]1 - 0.\stackrel{\rightarrow n}{\overline{9}} = 0.\stackrel{\rightarrow (n-1)}{\overline{0}}1 [\math]

>>9531844
[math]1 - 0.\stackrel{\rightarrow n}{\overline{9}} = 0.\stackrel{\rightarrow (n-1)}{\overline{0}}1 [/math]

>>9531844
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

>>9531851
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.

>>9531876
>the standard part is still 1
the standard part of the sum is still 1, the expression I gave is 0, apologies

>>9531876
ah godspeed anon ty for help

>>9531885
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.

>>9531876
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.

>>9531892
>lightstone
i just googled him and wtf this dude just stole my genius work

>>9531844
there is no "n"

>>9532188
^ me here.

>>9531960
that's it, this is the last straw. western civilization was a mistake, time to convert to Islam

>>9531728
>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.

>>9532288
Well that just means 0.999... can't equal 1 because there is no end to the 9's.

>>9532548
It's infinitely close to 1.
You cannot define the difference.

>>9532587
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...

>>9532288
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.

>>9533047
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.

>>9533061
[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?

>>9533066
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.

>>9533066
Say x = 0.333...
Does 9 / 3 = x?

Be careful how you answer.

>>9533083
9 / 3 = 3 anon....

>>9533047
>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.

>>9533086
I'm so sorry.

>>9533083
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?

>>9531728
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

>>9533107
But 0.3333... is exactly 1/3 in decimal. 0.333....3 is an approximation.

>>9533116

>>9533118
x = 1/3
x = 0.333...
x = x
0.3333... = 1/3
QED

>>9533106
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.

>>9533106
>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.

>>9533137
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.

>>9533135
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.

>>9533140
> a = 0.999... && b = 0.000...1

>>9533149
0.000...1 is not real.

>>9533137
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
>>2 + 4 = 6

0.333... × 3 = ?
>Step 1: find 0.333... on the numberline
>>0.3
>>0.33
>>0.333
>>0.3333
>>0.33333
>>0.333333
>>0.3333333
>>0.33333333
>>0.333333333
>>0.3333333333
>>0.33333333333
>>0.333333333333
>>0.3333333333333
>>0.33333333333333
>>0.333333333333333
>>0.3333333333333333
>>0.33333333333333333
>>0.333333333333333333
>>0.3333333333333333333
>>0.33333333333333333333
>>0.333333333333333333333
>>0.3333333333333333333333
>>0.33333333333333333333333
>>0.333333333333333333333333
>>0.3333333333333333333333333
>>0.33333333333333333333333333
>>0.333333333333333333333333333
>>0.3333333333333333333333333333
>>0.33333333333333333333333333333
>>0.333333333333333333333333333333
>>0.3333333333333333333333333333333
>>0.33333333333333333333333333333333
>>0.333333333333333333333333333333333
>>0.3333333333333333333333333333333333
>>0.33333333333333333333333333333333333
>>0.333333333333333333333333333333333333
>>0.3333333333333333333333333333333333333
>>0.33333333333333333333333333333333333333
>>0.333333333333333333333333333333333333333
>>0.3333333333333333333333333333333333333333
>>0.33333333333333333333333333333333333333333
>>0.333333333333333333333333333333333333333333
>>0.3333333333333333333333333333333333333333333
>>0.33333333333333333333333333333333333333333333-

>>9533160
>0.999... is not real

>>9533171
Notice how with each step, the left side does not ever equal 1 and the right side does not ever equal 0.

>>9533171
First of all, define "infinite steps".
How many steps is it exactly?

>>9533177
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.

>>9533180
Define "many" and "steps"

>0.9 is pretty close to 1 so 0.9 = 1

>>9533188
>these apples sort of look similar so I'll just replace them with identical abstractions

>>9533177
>>0.999... 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.

>>9533182
When has enough work and time passed for "infinitely long" to suffice to have occurred...?

>>9533180
>How many steps is it exactly?
See >>9533143

>>9533192
start dividing 1 by 3 in decimal form and someone will let you know

>>9533188
>>9533190
>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.

>>9533197
>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.

>>9533199
>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?

>>9533199
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.

>>9533214
Maybe if you're retarded.

>>9533212
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]

>>9531814
>shifting doesn't work

what is 9.999.../10 ?

>>9533225
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•••

>>9533235
9.999...{n}/10 = 0.999...{n+1}

>>9533225
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.

>>9533237
hello shitlatexman

>>9533237
>9.999...{n}/10 = 0.999...{n+1}

sure looks like a shift to me

>>9533239
You keep using infinity like you know what it means but you still haven't sufficiently defined it. Stop that.

>>9533246
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.

>>9533237
>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.

>>9533251
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.

>>9533253
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?

>>9533258
Rounding.

>>9533258
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]

>>9533225
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 >>9533225?

>>9533270
Yes one divided by three equals one divided by three, but what is the actual value?

>>9533265
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 >>9533225?

>>9533279
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.

>>9533285
Good think you aren't in charge of making calculators and computers, nothing would ever be computable in your world.

>>9533292
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.

>>9533298
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 >>9533225, 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?

>>9533300
>he’s wrong because calculators use finite decimals places and are programmed to round up at a certain level of accuracy

>>9533306
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?

>>9533306
>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.

>>9533315
He is wrong in saying those finite decimals mean that calculators conform to his specifications in >>9533225 and he is wrong in saying calculators are fine just letting one divided by three equal one divided by three.

>>9533306
>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?

>>9533458
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.

>>9533464
He's not asking about standard calculus. Have you read the OP?

>>9533464
>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?

>>9531757
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.

>>9533715
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.

>>9533853
>The ... implies an early identifiable pattern of repetition, not that it "goes on forever"
BULLLLLSHIT

>>9533853
>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

>>9533520
>>9533520
>>9533520
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

>>9533389
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.

>>9533873
>It can mean that the difference between x and y isn’t an infinitesimal in standard calculus which you just admitted
He asked specifically about NSA. Go ahead and look. I responded in that way. Then when the other poster got irritated because muh calculus. I again pointed out the distinction in question. Now you get mad because muh calculus and muh semantics.

I have stated things as clearly as possible without being extremely technical since this is clearly a new avenue for the OP.

Is infinity a number?
>no
Does infinity - n or infinity + n intelligibly evaluate?
>no
What is infinity?
>the concept of a number greater than all real numbers
By counting in increments, can you reach infinity?
>no
Is infinity endless, without end, with an undefined end?
>yes
Is infinity well defined?
>it is only defined by how undefinable it is, thus cannot be used in any definite way
Is there a better value to use when thinking of largest numbers?
>yes, a local Maximal Real Number
What is a local maximal real number?
>the greatest finite number in a set or series. In a set from 1-9, 9 is the local maximal real number. If [20] is appended to the set, 20 becomes the local maximal real number
What if i'm a lazy mathlet retard and don't want to write out really big, super long finite numbers?
>then just invoke "local maximal real number" LMRN
What if i want to append greater values to a set forever?
>then at every instance of appending, there is a new LMRN to replace the last
What if I want to concatenate an eternity of everlasting appending to the idea that it has completed this work?
>Then it obviously didn't go on forever, and is therefore finite
Is LMRN < infinity?
>conceptually, yes.
Can you do intelligible arithmetic with an LMRN?
>LMRN-1 = LMRN-1; LMRN+1 = LMRN+1; so yes.
Has infinity been misused, abused, and generally misunderstood in classical msthematics, thereby invalidating them from producing intelligible answers to questions?
>yes
Are humans perfect and infallible?
>absolutely not
Does that mean we have to fix math because it is broken?
>of course

think of number between 0.999.. and 1, there's none, therefore it's the same number

>>9534308
How many 9's are in x=0.999...?
there is at least a number with one more 9 that is greater than the x but less than 1.
Are you dumb?

>>9534324
>0.999...
>how many 9's

AN INFINITE NUMBER OF 9s
HOLY ITS NOT THAT HARD

you'll never get to a point where you claim "there is at least a number with 1 more 9 that is greating than the x but less than 1".

There is an infinite number of 9s and that value 0.999...to infinity is equal to 1

>>9534337
Infinity is not a number amount. Why don't you understand that?
Are you retarded?
Go on. Count to infinity. I'll wait.

>>9533906
Ironic, as you were very clearly the one playing semantics form the beginning. I’ve already proved it, which you ignored, so I’ll state it again. You were not only wrong, not only contradicted yourself, but are now entirely hypocritical (because this is literal semantics) when you asked, “what does infinitely close mean other than the difference between x and y is an infinitesimal.” You answered that question yourself by pointing out that the difference between x and y isn’t an infinitesimal in standard calculus.

And yes, he mentioned NSA, what’s your point? You also just admitted OP seems new so he wasn’t aware of the differences, yet you were still being an arrogant prick. Interesting

>>9534292
Grats and that argument and conversation with yourself

You don’t even need a big circle jerk you can just do it yourself

>>9534349
>God is God
Wow amazing really blowing my mind with that analysis.

>>9534356
>Arduino box tape
the absolute state of sci

Jesus people are retarded.. don’t reproduce lmao I don’t have to tell you I know you couldn’t anyway

engineers, please get the fuck out of this thread. nobody gives a shit about the little theorems you've concocted in your head about how real numbers work.

>>9534398
Count to infinity, mathlet.

>>9534342
Infinity is a concept which brainlets clearly can't get a hold of. Your lack of comprehension to the most basic Calculus concept truly shows your low IQ nigger ancestry

>>9531793
The usual definition of a real number is a Dedekind cut.

A Dedekind cut is a pair [math](A,B)[/math], where [math]A[/math] and [math]B[/math] are sets of rationals, each [math]a \in A[/math] is less than each [math]b \in B[/math], and [math]A[/math] has no greatest element. You can't construct a Dedekind cut where [math]A[/math] is gets arbitrarily close to 0.0...1 while containing 0.

For each [math]1 \times 10^{-n}[/math], there's a smaller [math]1 \times 10^{-m}[/math]. The limit of this sequence is 0, which you can show with deltas and epsilons.

>>9534420
Nono no, you're the one who doesn't undersstand what infinity is, not me! I know perfectly well what it is, but you keep pretending to hammer away unchanging, missing every nail!

Can you reasonably define infinity to the best of your ability?

>>9531793
>>9534428
As for non-standard analysis, I think it's possible in the hyperreals. 0.0...1 is really a parochial way of writing the hyperreal [math]\{0|1, 1 \times 10^{-n}\}. I'm not sure when this number appears in the construction.

The thing is that this depends so strongly on the axiomization and will be so counter-intuitive that the casual way you do algebra and arithmetic on the reals will break down. This is why teachers tell the noble lie that infinitesimals aren't a thing.

>>9531728
0.999... = 1

0.999...9 =/= 1

0.000... = 0

0.000...1 =/= 0

>/sci/ doesn't understand what limits are
Not this shit again.

>>9534613
0.9 rounds up to 1 with no decimal.
>0.[9] (1 decimal) r^ 1 {no decimal}
0.99 rounds up to 1 with one decimal.
>0.[99] (2 decimals) r^ 1 {1 decimal}
0.999 rounds up to 1 with two decimals.
>0.[999] (3 decimals) r^ 1 {2 decimals}
0.9999 rounds up to 1 with three decimals.
>0.[9999] (4 decimals) r^ 1 {3 decimals}

For n decimal places in a number, it can only round up on the (n-1) decimal place.
0.9999 = 0.9999 with 4 decimal accuracy
0.9999 = 0.999[r^9] = 0.99[9+1] = 0.9[9+1] = 0.[9+1] = 1.0 with 3 decimal accuracy

So if there are n amount of numbers after the decimal, you need (n-1) decimal accuracy to round up. If you use n decimal accuracy instead, it will not round up on the n'th decimal place.

So
0.{1,000 nines} rounds up to 1 in 999 decimal places
0.{1,000,000 nines} rounds up to 1 in 999,999 decimal places

>0.{n nines} rounds up to 1 in (n-1) decimal places
>but 0.{n nines} only equals 0.{n nines} in n decimal places
Substitute infinity for n in the above greentexts.
>0.{infinite nines} rounds up to 1 in (infinity-1) decimal places
Oh shit, infinity-1 isn't a thing. Infinity-1 = infinity. That means we can't use (n-1) decimal places with infinite repetition, and can only use n decimal places, yet n decimals places doesn't allow rounding to occur, so 0.999... can not equal 1.

so easy, even a child could understand. You're not less intelligent than a child, right?

>>9534697
>can not
looks like you're out of luck, brainlet

>>9534345
>You answered that question yourself by pointing out that the difference between x and y isn’t an infinitesimal in standard calculus.
But we weren't discussing standard calculus. We were discussing non-standard analysis. I brought up standard calculus for contrast. It was possible he had some confusion about the ideas behind NSA and so it was important to know what he meant by the question he asked. What else could he have meant? It's not clear to me. But it's entirely possible he did mean something else, which I could elicit by further explanation and questioning.

The fact that you are butthurt by all this is very strange.

>And yes, he mentioned NSA, what’s your point?
My point is, if he asked a question about non-standard analysis, I should answer based on non-standard analysis, which I did. I said the difference between [math]1[/math] and [math]\sum_{n=1}^{N} 9\cdot 10^{-n}[/math] is [math]10^{-N}[/math]. Which was his intuition. This means, as he thought, there is a difference between 1 and 0.999... in NSA when [math]N[/math] is an infinite integer. As he suspected. Which I was happy to confirm.

>You also just admitted OP seems new so he wasn’t aware of the differences
Which is why I brought up standard calculus, as a contrast.

>>9534713
You're actually less intelligent than a child?

>95 34736
thanks for the (You), bud. here's another one from me

oops, i dropped it. oh well

>>9534738
Did you never attend primary school? They teach rounding to kids.
I mean i'm not asking you have a college education to figure something, i'm just going off bare minimum expectation of basic knowledge. You seriously don't know how to round?

man i'm getting a lot of attention tonight.
i feel pretty

>>9534697
By your logic, seeing we can't replicate the accuracy of an infinitely repeating series during input (a calculator may store that it is infinitely repeating), 0.999... should always be rounded. As soon as you round it to any decimal place it becomes 1, so 0.999... = 1.

0.000...1 is not even a number you stupid fucking faggots. How can a number have an infinite number of 0s and also end in 1?

HURR GURRR IT NOT INFINITE NUMBER OF ZEROS ITS A PATTERN

>>9535085
No, the logic clearly says 0.999... can't be rounded. Rounding requires a decimal accuracy, but repeating digits dont get that.
If i were a shitty changer and we were dealing with money in an arbitrary way like cryptocurrency, you might have some value of arbitrary digits like Z1.482729. If we want to translate this to USD at an exchange rate of $1019 per 1Z, this would become $1510.900851
Since USD only has $n.nn accuracy, that would instead be $1510.90, losing a fraction of a penny's worth on the exchange which may be a problematic loss over innumerable exchanges. Anyway the point is we have two decimals of accuracy. If i were a generous changer and gave you the benefit of rounding up, a Z value to USD of something like $100.466789 could instead become $100.47, but a Z value of $100.460000 would only be $100.46
If you have a decimal accuracy restriction for example 2 decimal places, you need at least a significant 3rd decimal in the base value to round to the flat 2 decimal value. $0.99 = $0.99, $0.99[9 = $1.00

So if we have infinite arbitrary decimal accuracy for a number like 0.999..., our rounding position would be on the infinity-1 position
>[math]0.9_{1}9_{2}9_{3}...9_{\infty}[/math]
>[math]0.9_{1}9_{2}9_{3}...9_{\infty - 1} \big[ 9_{\infty}[/math]
Problem here is that infinity-1 isn't a number and is still just infinity, which means you cant determine to invoke where the decimal accuracy cutoff is, and therefore can't determine a most significant decimal place to round.
You need at least (n-1), one less determined decimal place than the amount of digits n in the decimal, to round, where n decimal places in lieu of (n-1) decimal places then prevents rounding from happening. Since (infinity-1 = infinity), you can't get (n-1) and are force to use n in lieu, but n doesn't have a smaller significant decimal value to round up to. If we had infinity-1 accuracy, it would look something like
> ...999[9 , r^ ...999[10, carry
instead you get
> ...9999[

>>9535088
You can't count an infinite amount so what does it matter to you what comes after it? :^)

>>9535363
Infinitely repeating 9s have an advantage here because we know the next decimal will always be 9, thus we can safely round.
It will never not be 9 next, so it will never be wrong to round up. Never, ever.

To imply that [math]0.999... \ne 1[/math] means there must exist some real [math]r[/math] such that [math]0.999... < r < 1[/math], which is a property that must be satisfied for any numbers with a non-zero difference.

Of course, you can't find such an [math]r[/math] because there *IS NO DIFFERENCE*. The difference isn't just infinitely small. It's zero. You forget there are many ways to represent the same thing in maths. Take [math]5=2+3[/math] for instance. [math]0.999...=1[/math] for the same reason that the [math]0.000...1[/math] cannot exist because that implies that the 0s terminate. If they did, you'd be right, because then you'd have a number really close to zero but not quite zero. But when dealing with infinity things get a bit wacky and this is one of its weird impacts and why it's so counter-intuitive. Don't forget that infinity is a *concept*. It's not an actual number. It's not something you can make an inverse out of and subtract it from 1, which is what you're mistakenly doing (and implying [math]0.999...[/math] is).

>>9535085
Lets have a calculator that shows 8 digits, but secretly works with an extra 9th digit behind the scenes, and the 9th digit becomes the rounding buffer [

[math]0_1 .0_2 0_3 0_4 0_5 0_6 0_7 0_8 \big[ _{0_9}[/math]
When we type 1÷3, the calculator will do:
[math]0_1 .3_2 3_3 3_4 3_5 3_6 3_7 3_8 \big[ _{3_9}[/math]
You only see 0.3333333 to fill all 8 digits on the calculator LCD screen, but there is a 9th stored value. If you type × 3, the calculator then does:
[math]0_1 .9_2 9_3 9_4 9_5 9_6 9_7 9_8 \big[ _{9_9} \rightarrow \big[ _{\stackrel{r}{10}_9} [/math] rounding up the 9th 9 to 10 then carrying it up and over to every digit until it equals 1.0

However if we instead type the threes manually, we only get 8 digits to work with in:
[math]0_1 . 3_2 3_3 3_4 3_5 3_6 3_7 3_8 \big[ _{0_9}[/math] where multiplying this by 3 gives:
[math]0_1 . 9_2 9_3 9_4 9_5 9_6 9_7 9_8 \big[ _{0_9}[/math], where no rounding occurs and this just equals 0.9999999

The infinite repwtition rounds because it rounds WELL before the least rounding point of (infinity-1), but at the same time its not saying 0.999 = 1, it's saying 0.9999999[9 = 1, 8 finite nines, not infinite nines.

>>9535402
Let me add slightly to this.

If you use 0.999... to any finite level of accuracy you would need to round it up because you know the next digit past the decimal place you are cutting it at is a 9.

If you are using it at an infinite level of accuracy, then be my guest, keep it at 0.999...
But I wouldn't hold my breath waiting for that calculation to finish.

>>9535410
You make incorrect assumptions about infinity to be unable to rationalize how r exists.

>>9535413
No one was saying 0.999 = 1.

Try dividing 10/3 using long division.
You'll get 3, then continue and get 3.3, then continue and get 3.33, and then continue in this never-ending cycle.
No matter how far you go, there's always that remainder 1 that you have to carry over. This is actually a problem with the decimal system. You can never accurately portray 1/3 in decimal without adding some weird rules like '...' implies that the '3s' go on forever, and even then the true, intuitive notion escapes most people.

If you multiply your result back by 3, you'll get something like '9.9999...'
Where did that '0.000...1' go? It's that remainder which you discarded. You lost some piece of information.

0.999... is equal to 1 for the same reason. You're missing a tiny part of the picture that is due to an artifact that manifests when you try to represent certain numbers in the decimal system. It really is equal to 1, I can assure you.

>>9535420
If you could convert all the energy in the universe into computational power, you still wouldn't find the end of an infinite series.

>>9535420
You're making incorrect assumptions about 0.000...1, such numbers cannot exist. They're self-defeating.

>>9535420
A finding a terminating digit at the end of an endless sequence is like finding water at the end of a bottomless well. The two presumptions contradict each other. The sequence terminates or it doesn't.

>>9535424
Thank goodness i covered that typo with the redundant sentence after it.

>>9535430
You might not ever reach the water but under the the assumption that all wells have water, you can easily accept that there exists some water at the bottom of the bottomless well, all without throwing shit at the walls sperging about refusing to accept that you poorly misunderstand infinity.

>>9535440
>at the bottom of the bottomless well

This right here is your problem.
It's not bottomless if there is a bottom. You're changing the terms.

>>9535440
You're the one who misunderstands infinity. You're just taking an arbitrary large number, and using that in place of infinity to see how it works in your mathematical model.

Infinity has special properties that arbitrarily large numbers do not. I can clearly see your mistakes and, quite frankly, it's one of an amateur.

>>9535427
No shit, therefore 0.999... = 0.nothingbut9's

It doesn't equal 1. You would have to make unnatural and shitbrained assumptions about how basic math works in order to justify it equating to 1.
You interpret your problem incorrectly by posing it as 0.999... < r < 1 to pretend r doesnt exist, when the fact of the matter is you could set this problem up as
0.9 < r < 1 @ r ~could be~ 0.99
0.99 < r < 1 @ r~ 0.999
0.999 < r < 1 @ r~ 0.9999
0.9999 < r < 1 @ r~ 0.99999
...
0.999...9 < r < 1 @ r~ 0.999...99

the way you interpret infinity incorrectly to justify your example of r is pretending infinity is a finite number, which you must understand it really isn't a finite number. You would imply infinity+1 doesn't exist because infinity by itself is a finite end of numbers, even though
0.(infinity 9's) < r < 1 @ r~ 0.(infinity+1 9's) would solve your misinterpretation.

>>9535454

infinity + 1 = infinity

Boom.

>>9535459
Thats the problem. You treat infinity like a finite ceiling. Can't add anything to it to increase the height cause you're already at the ceiling.

Infinity is not a finite number.

>>9535454
infinity + 1 is nonsensical.
https://en.wikipedia.org/wiki/Infinity_plus_one

Nevermind. Why don't you go get a nobel prize with your new findings? I'm sure all the mathematicians there will all be dying of laughter. They'll give you the nobel prize of comedy.

>>9535454
The convergence theorem alone dismantles your argument.

If that's too hard for you, let me show you some basic algebra.

[math]
x=0.999...
10x = 9.999...
10x - x = 9.999... - 0.999...
9x = 9
x = 1[/math]

Or are you going to say something dumb like "oh you can't shift/subtract it like that"?

>>9535454
The convergence theorem alone dismantles your argument.

If that's too hard for you, let me show you some basic algebra.

[math]x=0.999...[/math]
[math]10x=9.999...[/math]
[math]10x-x = 9.999... - 0.999...[/math]
[math]9x = 9[/math]
[math]x=1[/math]
Or are you going to say something dumb like "oh you can't shift/subtract it like that"?

>>9535454
>0.999...9 < r < 1 @ r~ 0.999...99
This makes no sense.
Just as 0.000...1 makes no sense, you cannot infinitely repeat 0 and then have a 1 on the end, you cannot infinitely repeat 9 AND THEN HAVE A 9 ON THE END.
That implies an end. As long as there is an end, it is finite.

This is an INFINITE series of 9s.
There is always another 9, so there will never be anything in between 0.999... and 1.

>>9535462
Infinity by itself is nonsense, you dumb, stupid, fucking retarded, mogolid, tongue choking, mouth breathing, snot nosed faggot.

Jesus fuck.

You jumped out the plane without a parachute and are now complaining jumping out of planes without parachutes is a bad idea. Like holy shit, no one would rationally think to do this but because ur a super smrt mathamagician with m-muh IQ and permavirginity, you're slow and special and didn't figure to pay attention to how the skydive instructor told you not to do it.
Infinity is jumping out of a plane without a parachute. Infinity+1 is complaining how you didn't know you weren't supposed to jump out of a plane without a parachute, as if it weren't some god given basic self-preservation instinct that only non-retards might acknowledge.

Yes, ("a non value" + 1) doesnt make sense, but you know what else doesnt make sense...?

"a non value" just by itself.

>>9535465
Its not dumb to say you can't shift or subtract like that.
0.9 × 10 doesn't equal 9.9
0.9 × 10 = 9.0

Inb4
>i-in-imfibiddy is s-s-special thou. . .

OP, see these two. I think they'll help you better understand your mistake.

https://www.youtube.com/watch?v=G_gUE74YVos
https://www.youtube.com/watch?v=TINfzxSnnIE

>>9535473
>Its not dumb to say you can't shift or subtract like that.
>0.9 × 10 doesn't equal 9.9
>0.9 × 10 = 9.0

This has fuck-all to do with what I'm doing.
[math]0.9 \times 10 = 9.0[/math]

But,
[math]0.99 \times 10 = 9.9[/math]
This is to say, by induction, if the 9s never terminate,
[math]9.99... \times 10 = 9.9...[/math]
There will not be a 'terminating 0' because, again, the 9s never terminating.
That's the whole point. And you're trying to negate it by pretending that the 9s do terminate. That's like treating infinity as an arbitrarily large number. By doing that, you are ignoring its property.

>>9535467
Right so..
In
[math]\sum_{n=1}^{\rightarrow +1} \frac{9}{10^n}[/math], at which value of n do the repeating 9's stop, given you increment n by 1 every iteration?

>>9535487
What repeating 9s?

>>9535483
Indunction? Lmfao

0.9 × 10 = 9.0
0.99 × 10 = 9.9
0.999 × 10 = 9.99
0.9999 × 10 = 9.999
0.99999 × 10 = 9.9999
The only valid assumption to make from the pattern of this work is that, no matter how many 9's exist before the evaluation, the same amount of 9's exist after the evaluation, and there is one less nine after the decimal
0.999 (three 9's) × 10 = 9.99 (three 9's) [two decimal 9's]

0.999... (infinity 9's) × 10 = 9.999... (infinity 9's) [infinity-1 decimal 9's]

You can sooner INFINITELY prove this true for all real number amounts of 9's between 0.9 and 0.999..., and never be able to prove it untrue.

>>9535492

infinity - 1 = infinity

You do have the same "number" of 9s.

God, your math sucks

^
Most retarded post in the thread. Not gonna dignify it with a (You)

>>9535492

infinity - 1 = infinity

You do have the same "number" of 9s.

God, your math sucks (dumbass)

>look mom i can do the same retarded shit twice in a row
>>why are they calling you retarded, anon?
>cause i cant read lmao
>>oh. Thats nice, anon.
>can you make tendies for dindin mommy
>>no, mommy has to go hang out with jerome and darnell in her room tonight~
>b-but i want tendies...
>>and mommy needs to get fucked by big nigger dicks, honey. Don't be so selfish!

>>9535468
>Infinity by itself is nonsense
Exactly.
Stop trying to treat it like a rational number.

>>9535524
Provide (1) example where i did that.

>>9535525
not him, but here: >>9535492 where you add/subtract 1 from it.

>Me watching brainlets trying to create new notation for decimals.

>>9535525
right here:
>You can sooner INFINITELY prove this true for all real number amounts of 9's between 0.9 and 0.999...
>all real number amounts of 9's
>all real number

Except infinity is not a real number.

Are you done being retarded?

>>9531814
The sum of 1/10^n is an increasing and bounded sequence which means it's convergent. Then you can do calculations on the series. Therefore you can divide by 10, or multiply by 10 or do pretty much whatever you please. So 9*0.111...=0.999..., 90*0.111...=9.999... and 90*0.111...-9*0.111...=9, (10-1)0.111...=1, 9*0.111...=1.

>>9535541
I didn't say infinity was a real number. I said there were infinite real numbers and proving the assumptions true would take forever, that you could never increment to the one single case where "infinite" 9's could, under your tard misinterpretation of infinity, prove it false by acknowledging infinity-1 isn't intelligible. You can't increment to infinity, so you can only infinitely forever prove that more and more and more and ever more 9's in a number of 0.9+n9's × 10 would equal 9.9+(n-1)9's.

You not only dont rationalize infinity correctly, you can't identify when i'm abiding by classical definitions and when i'm not. You dont even understand infinity at a basic level that could offer you an edge of comparison. You just literally don't even know what infinity is or even might or could be.

>>9535525
here
>>9535492
>>9535454
>>9534697

>>9531844
We already have notation for "n nines" it's 0.999...9, unknown 9s but they terminate.

According to 0.999... !=1 tards
0.333... x 3 = 0.999...
1/3 x 3 = 1
0.333... != 1/3

>>9535580
>according to dummies, true statements are true

>>9531728
1/3 = 0.333...
0.333...*3 = 0.999...
therefore 0.999... = 1

>>9535718
1/3 is clearly greater than 0.333...

>>9535731

>>9535718
[math] \displaystyle
1 = \frac{3}{3} = 3 \cdot \frac{1}{3} = 3 \cdot 0. \overline{3} = 0. \overline{9}
[/math]

>>9532548
>0.999... can't equal 1 because there is no end to the 9's.
There are a lot of things that don't have ends.

For instance, there is a decimal and an infinite amount of zeros behind every integer you type.

>>9535430
If there's no water it in it's hardly a "well."

>>9535758
0 isnt a fucking number.
All real numbers exist between noninclusive 0 and infinity.

>>9535745
Is 0.3 less than 1/3? Yes
Is 0.33 less than 1/3? Yes
Is 0.333 less than 1/3? Yes
Is 0.3333 less than 1/3? Yes
Is 0.33333 less than 1/3? Yes
Is 0.333333 less than 1/3? Yes
Is 0.3333333 less than 1/3? Yes

Is it true that no matter how many 3's there are, it is still less than 1/3? Yes

>>9536297
>not apart of the natural numbers
>not a number
kek

With the way infinity currently works in classical brainlet math, being that it works retardedly and poorly, you can make the assumption that in calculus an infinite sum [math]\frac{9}{10^n}[/math] would evaluate to zero at n=infinity, even though its impossible to increment n to infinity but lets just pretend you could, just as much as current calculus pretends you can.
9/infinity must equal 0 cause you are being a deliberate fucking retard and can't figure a number to exist in 0.999... < r < 1, while 9/ any arbitrarily large number would get closer and closer to zero, so [math]\frac{9}{\infty} = 0[/math], but also [math]\frac{n}{0} = \infty[/math], a number divided by zero can be proven to be infinity and not "undefined".
But holy fuck we aren't even done. When n=infinity, and because you are retarded, we now have terminating 0's after the preceeding 9's
If an integer 1 could be viewed as 1.000... , then that repeating 9 number would eventually become 0.999...000...

INFINITY:
>Intentionally
>Not
>Faithfully
>Incrementing
>Numbers
>In
>Thoughtful
>Yields

Get
Fucked
[math]0.\bar{9} \neq 1[/math]

>>9536358
>just as much as current calculus pretends you can.
But it doesn't. You can intuitively think about it as evaluating at [math]n = \infty[/math] but that's not how the limit of a sequence is defined.

A real sequence [math](a_n)_{n\in\mathbb N}[/math] is said to have a limit if there exists some [math]L \in \mathbb R[/math] such that for any strictly positive distance [math]\varepsilon[/math] there is a point in the sequence after which all elements are closer than [math]\varepsilon[/math] away from [math]L[/math]. In other words, if the sequence gets arbitrarily close to some number and then stays arbitrarily close.

>>9536358
>repeating 9 number would eventually become 0.999...000...
The absolute state of /sci/

>>9534308
Think of an integer between 1 and 2, there's none, therefore they're the same number.

>>9536478
>Think of an integer between 1 and 2, there's none, therefore they're the same number.
Except there is, dimwit. You can demonstrate that there is one too.

Now find a number between 0.999... and 1. Go ahead. You can try, but no such real number exists.

>>9536478
The reason you can say that two real numbers are the same if there exists no real number between them is because the reals are (usually) constructed as an equivalence class of cauchy sequences of rational numbers. The Peano axioms don't allow you to do the same thing on the natural numbers, so your criticism is not valid.

>>9536478
>therefore they're the same number.
Except numbers aren't necessarily integers. Real numbers (in their totality) work differently than integers.

Get off of /sci/ if you're so incapable of offering a valuable point

>>9536298
nope,
0.333...= 1/3

>>9536508
an equivalence class [math]\rightarrow[/math] all the equivalence classes (for some sensible equivalence relation)

>>9536464
Setting a limit to infinity means it either RANDUMLY XD' reaches infinity, or it doesn't reach the limit at all.

If it doesn't reach the limit at all, 9/10^n is well off enough just being 0.999....
if it does reach the limit, it gets a terminating 0, and remains well off enough just being 0.999...

The problem with the 0.999... < r < 1 deal is that it partially assumes infinity is a finite end, for example if no one knew how to count over 100, we would assume 100 in that thinking must be infinity. 100+1 doesnt make any sense to anyone, so it's fair enough to say that 0.(100 nines) < r < 1 and that there exists no real number to fit in between, so unfaithfally using dumb logic you can relatively safely assume 0.999... is basically just 1, but this requires the faulty misunderstanding that infinity is a finite end.

>>9536508
>The Peano axioms don't allow you to do the same thing on the natural numbers, so your criticism is not valid.
this is not even true, solid buzzwords, well memed my friend

>>9536509
>Get off of /sci/ if you're so incapable of offering a valuable point
holy shit you're retarded. the irony is that he didn't say real numbers, he said numbers. there most certainly is a number in the hyperreals between 0.999... and 1. so you're actually retarded and a hypocrite as I can say "Except numbers aren't necessarily reals or natural"

>>9536532
>>9536464
In real numbers, there always exists a greater real number, and infinity always exists even greater by some undefined degree than any great real number.
0.(100 nines) < r < 1 and r can be 0.(101 nines)
since we cant encapsulate infinity that way, there can't actually be infinite 9's in 0.999... in the sense that infinity is some arbatrarily large finite real number. It's not an incrementimg amount of digits on a number line where the 9's get closer and closer to 1, it's a loop around the world where whichever place you choose to start counting 9's, you will wrap around the entire planet and come back to more 9's, forever counting 9s as you forever circle the globe in a loop. You never ever get close to 1 cause you are only ever counting more 9's from 0.999...
You could say that an infinitely repeating decimal exists in a different number dimension seperated from the other numbers, and you can understand it just as easily when thinking about infinity itself similarly, considering that no matter how much we increment real numbers we will always have arbatrarily large real numbers and never reach infinity, as if infinity were in a different dimension seperated from real numbers by an improbability ether. So take this seperation property of infinity as given, then apply it to something like an infinite amount of 9's in 0.999..., and we get [0.999...] as a unique number dimension full of 9's where you must count every 9 on a line with the caveat that you don't know the line loops around the world and comes back on itself.

Moreover, that unfaithful dumb logic is broken by something as simple as >>9536478
Where there exists no real integer between 1 and 2 so 1 must equal 2.

>>9536546
So there are at least two distinct ways to challenge 0.999... < r < 1

>1) there is no finite end of the 9's but always must exist an ever increasing local maximal real number of countable 9's, so 0.(LMRN nines) < 0.(LMRN+1 nines)
>2) It's just a dumb half-baked assumption that x = y in x < r < y for the abitrary pointless rule that there doesn't exist a way to increment x to y. 1 < r < infinity means 1 = infinity because there exists no value to add to 1 to increment to infinity.

>>9536536
How is it not true? The Peano axioms imply that there does not exist a natural number between [math]S(n)[/math] and [math]S(S(n))[/math], and since [math]S[/math] is injective [math]S(n) \neq S(S(n))[/math].

>>9534430
0.99... literally means infinite nines. If you don't believe in infinity, use different notation.

Check this one though
...999999.0 = -1

>>9536580
I believe in infinity, i just dont believe anyone with a formal education in mathematics understands it, and time and time again that presumption is proven correct by threads like these.
Mathlets with degrees in mathletism can spout off tons of different interpretations for it in different maths applications, but they understand the mechanics of none of them and further don't understand how the different usages are often paradoxical and cancel each other out.

0.999... means infinite 9's, and that means you never stop counting 9's. If you never stop counting 9's, you cant ever reach a sufficiently close enough point where you could presume it's close enough to 1 to simply not be differentiated from 1. You cant count to infinity so you will only ever be able to count to a maximal real number of 9's, so there will always exist at least that maximal real number +1 more 9 greater than the last count, meaning that although there may be infinite 9's, there are only a finitely invokeable amount of 9's in an arbitrary invocation of the number 0.999... because, although there are infinite 9's, you cant count to infinity; you can't finish counting all the 9's. If you can't actually do something, you can't actually use it for something else, and if you can't count to infinity, you can't encapsulate an infinite amount of anything into a finitely convenient representation. So 0.999... doesn't mean a largest finite amount of 9's in "infinite 9's", it means "there are even more 9's than you could possibly know, so think of the largest possible real number and there are significantly more 9's than that", and if there are even more 9's than you could possibly know, there exists an infinite amount of r's in 0.999... < r < 1 that have even more nines.

infinity was a fucking mistake

>>9536641
op here

can't believe i've gotten this many replies. i've been on 4chan for years and this is my biggest shitpost yet its so exciting

>>9536622
>I must down all the digits before I can reason about them.
good job brainlet

>>9536298
">Is it true that no matter how many 3's there are, it is still less than 1/3? Yes"
If i cant count it, it cant be real

>>9536298
>if i reduce the accuracy it becomes less accurate!

>>9536358
>If an integer 1 could be viewed as 1.000... , then that repeating 9 number would eventually become 0.999...000...
Where the fuck did this logic come from?

Are you fucking retarded?

>>9536675
9/10^n, n=infinity, 9/infinity = 0, = 0.999...0

it was written in the post.

>>9536652
What number comes after the largest possible number?

>>9536648
0.999... threads are unquestionably the best threads I will never tire of them

>>9536585
...999.999... = 0 fight me

>>9536671
>infinite amount of 3's is less than 1/3
>reduced accuracy
How much more increased accuracy can you get than infinite?

>>9536650
>i can do things no one has ever done by doing them
oh the irony

>>9536680
You can't perform operations with infinity because its not an actual quantity you fucking moron. 10^infinity is just some meaningless, undefined bullshit phrase

>>9536785
So there is not an actual quantity of 9's in 0.999... is what you're saying.

>>9536680
9/10^2 is 0.09
Your argument is shit.

>>9536691
>>infinite amount of 3's is less than 1/3
But it isn't.
What is so hard to understand? Only when you reduce the accuracy to a finite number does it become less than 1/3.

>>9536682
probably 53.2

>>9536799
this thread has given me many hearty laughs

>>9536818
>reduce the accuracy to a finite number
>infinitely reduce the accuracy to a finite amount
lmao

>>9536814
Its in relation to an infinite sum you idiot.
[math]\frac{9}{10^1}_{0.9} + \frac{9}{10^2}_{0.09} + \frac {9}{10^3}_{0.009} + . . . = 0.\bar{9}[/math]

>>9536827
Then where does the 0 come from?
It always ends in a 9, so there would be no 0.

>>9536818
The proof clearly stated that no matter how many 3's there are, it's less than 1/3.
0.3 < 1/3
0.33 < 1/3
0.333 < 1/3
0.3333 < 1/3
0.33333 < 1/3
0.333333 < 1/3
0.3333333 < 1/3

Just keep adding on extra 3's forever and this comparison remains true. Why do you think this isn't inclusive of infinity? Infinity is unending, and you're doing unending work by forever adding on extra 3's. It is explicitly abiding by the definition of infinity. No matter how many 3's there are.

No. Matter. How many threes there are. It is less than 1/3.

If you want to say infinity isn't even a number much less a real number, go ahead, but that ends this conversation right here, cause you could just as well claim there are potato amount of 3's in 0.333... and equally assure yourself potato is meaningless as a value and that you're not even remotely close to discussing anything of interest.

>>9536851
If infinity were a value as its treated in classical maths,
[math]. . . + \frac{9}{10^\infty}_{0.0} = 0.\bar{9}0[/math]

>>9536860
I just want to say that you are 100% correct. There is no such thing as "an infinite number of threes." Such thinking by brainlets is inconsistent even with their own fuckery like "muh infinity is not a number". Oh, really? Then how can there be an infinite string of 3s?

I think this idea can be polished a little better but I just want you to know I'm with you bro.

>>9536827
Oh I see what you've done.
You've snuck in your impossible infinity-1 argument again.
Because once n reaches infinity the number of 0s will be infinite, so at n-1 must have an 0 added to it.

Unfortunately it doesn't work that way. The series demands that a 9 always be added, so even at infinity it will be infinite 9s.

>>9536869
Honestly what the fuck

>>9536860
The real number 1/3 is defined as the equivalence class of rational cauchy sequences that approach 1/3. The sequence [math]\left(\sum_{k=1}^n \frac{3}{10^k}\right)_{n\in\mathbb N}[/math] is part of that equivalence class. All these things have very strict definitions that you keep ignoring.

>>9536869
But as pointed out to you before it is not treated like you pretend it is treated. No mathematician treats infinity like that.

>>9536888
In 9/10^n, as n increases and gets larger and larger, the partial sum of that soecific iteration gets closer and closer to 0.
You can also think of it as 1/n.
1/1 = 1
1/2 = 0.5
1/3 = 0.33..
1/4 = 0.25
1/5 = 0.2
1/6 = 0.166..
1/7 = 0.1428~
1/8 = 0.125
1/9 = 0.11..
1/10 = 0.1
...
1/1000 = 0.001
...
1/infinity = 0.000...

>>9536892
If no mathematician treats infinity like a number, then why do mathematicians obviously treat infinity like a number on examples such as setting a limit to infinity or infinity+1 can't exist because infinity is the definite finite ceiling.

>>9536889
1÷3 is defined by one divided by three.

Fuck off with your mumbo jumbo please.

>>9536860
>because it doesn't happen for my finite amounts, it cant happen for non-finite amounts.

You gave "proof" that a FINITE amount of 3s is less than 1/3. No shit.

You cant tame infinity like a real (or complex) number.
You set limits.
Find a proof to tell me pic series is not convergent.

>>9536907
\lim_{x \rightarrow \infty} f(x)
is not
f(\infty)

1 != 0
1 + 1 = 2
2 +1 = 3
...
inf + 1 = inf
1 = 0

>>9536928
0 cant equal 0 since there's no such thing as nothing.

>>9536913
>convergence is equivalence
1, no fuck off

2, Infinity is unending. You can't tame it, correct. You can't encapsulate it. You can't finish writing enough 3's to equate an infinite amount of 3's, so you can't reach infinity, so you can't actually know infinity exists as a value; you can only presume it exists as a vector, a direction to do unending work. If you stop doing the work at any point, you have a finite value, but you don't want a finite value, you want an infinite value. You want to try to encapsulate infinity though, cause you want to absolutely be sure there are "infinite 3's in 0.333..." and that such a statement is valid, but encapsulation requires termination, it requires a beginning and an end. Can you grasp something with your hand without wrapping your fingers around it? No.
We can encapsulate finite numbers easily: 2
2 comes between 1 and 3, we can encapsulate it as 1[2]3
2 counts can also be encapsulated
[ • • ]
Anyone can grasp 2.

Infinity can't be encapsulated with either method, and the infinite counts just looks like
[ • • • • • • • ...
Without ever getting a right bracket to surround it, to grasp it, to understand it's value.

So infinity as a value must not be a thing, and instead infinity can only mean "unending, without end", a direction to do more and more, a vector, a heading, but not a value, and therefore not a number.

>>9536961
your mommas so fat she's unending, you can't tame her, you can't encapsulate her

>>9533853
anon's first ellipses expression. go easy on him guys.

>>9535548
>There’s an infinite number of increments before it; therefore it cannot be conceptualised.
There’s an infinite amount of incrememts between 0 and 1. I guess we will never be able to conceptualise 1.
Either that or you’re just retarded, and can’t understand that infinity can be used because it’s properties are known. Your mind is clearly just too small to understand an abstract concept.

>>9537011
Provide (1) known, consistent, singular definition of infinity that is not paradoxical to any other usage of infinity in known maths.

>>9537024
>I don't even know the axioms and my hand wavy idea of infinity is inconsistent so axiomatic math is also inconsistent
Please end yourself

>>9537011
An infinite amount of increments between 0 and 1 requires each increment be a significant value z where [math]z × \infty = 1[/math], and that such a significant value z subtracted from 1 finitely equates [math]0.\bar{9}[/math]

>>9537024
It’s an unbounded limit. All of those paradoxes have a fallacy in them somewhere.

>>9537068
Define unbounded limit. Surely you should have been smart enough to say "unlimited".

>>9536961
>>convergence is equivalence
>1, no fuck off
Except that the reals are literally defined as the limits of rational cauchy sequences with sequences between which the distance converges to 0 being treated as equal by the equivalence relation. So yes, convergence is by definition equivalence.

>>9537079
Keep repeating the same retardation over and over again

We all know you are wrong kiddo

>>9536907
But they don't, you keep saying they do that, but they don't.

>>9537085
Stop being hitler.

>>9537086
So you are saying that infinit limits do not exist in calculus?
You are saying infinity+1 exists?
You are saying that mathematicians agreet to never use or teach these things?

This is ok
>saying "you can infinitely count 9's in 0.9 repeating"

This is not ok
>saying "there exist an infinite amount of 9's in 0.9 repeating"

The former acknowledges pretense and present action. The latter implies past tense, that "an infinite amount" has been counted. The former cannot be disproven, while the latter is proveable as false.

infinity is not an amount.

>>9537106
infinity is a value you dumb fuck

>>9537093
They exist, but clearly not like you think they do. The limit has a very specific definition. Your criticism is based on some shitty intuitive idea of limits, and not on the strict rigorous definitions that mathematicians use.

The limit at infinity for a function [math]f : \mathbb R \to \mathbb R[/math] is said to be [math]L \in \mathbb R[/math] if
[eqn]\forall \varepsilon>0 : \exists x \in \mathbb R : \forall y\in\mathbb R : |f(y) - L| < \varepsilon.[/eqn]

Similarly, the limit at infinity of [math]f[/math] is said to be [math]+\infty[/math] if
[eqn]\forall L \in \mathbb R : \exists x \in \mathbb R : \forall y\in\mathbb R : f(y) > L .[/eqn]

>>9537106
Semantics really, when someone says "There are an infinite ..." they mean there is no upper bound to the number of things.

>>9537090
ur mom is hitler faget

>>9537106
>the latter implies past tense
So now you demonstrate your brilliant mastery of the English language?

>>9537127
At infinity. So. Not at any other number. Implying infinity gets a special rule, because it's not a number.

If you have a formal education, you believe infinity is a value. You also know the the value it has is undefined, however.

This is called a paradox. Welcome.

0.999... < r < 1
Does r exist?

Look at it in a variety of ways.
Are there infinitely many divisions between 0 and 1? If yes, there is a z value where z×infinity = 1, and 1-z = 0.999...
This requires infinity be an arbatrily large real number. If it isn't, and is instead a vector, z cant exist because there is no end to the counting of possible division parts between 0 and 1, yet if there is no end to the counting of 9's in 0.999..., it can not ever rationally reach or get close enough to 1.
>infinity is a value: 0.999... =/= 1 because [math]z_1=\frac{1}{\infty}, 1-z_1 = 0.\bar{9}[/math]
>infinity is a vector: 0.999... =/= 1 because there exists only more and more 9's never getting closer to 1
>infinity is a value without a specific definition: 0.999... =/= 1 because you can not increment or count to infinity, but any attempt to do so will reveal greater and greater local maximal real number amounts of possible 9's, where [math]\frac{1}{LMRN} = z_2 \in \mathbb{R}[/math], a value similar to math]z_1[/math] to satisfy the same evaluation of [math]1-z_2 = 0.\bar{9}[/math]

>"[math]\frac{1}{3} = 0.\bar{3}, 0.\bar{3} × 3 = 0.\bar{9} = \frac{3}{3} = 1[/math]": no possible amount of 3's will ever rationally and succinctly finitely equate the decimal value of 1/3, so [math]\frac{1}{3} > 0.\bar{3}, \frac{2}{3} > 0.\bar{6}, \frac{3}{3} > 0.\bar{9}[/math]
>"x= 0.9... ; 10x= 9.9... ; 10x-x = 9 ; x=1": y=0.9, 10y=9.0 || y=0.999, 10y= 9.990 || y=0.(n nines), 10y=9.(n-1 nines) ^ 10y-y = 8.999...1; 9y/9 = y = 0.999...

>>9537158
the value it has isn't undefined, numbnuts

>>9537599
Oh ye?
Which value is it then?

>>9537604
the value with greater magnitude than all other values. you'd know this if you looked at a little bit of complex analysis

pic related. it's you

>>9537610
>the value of infinity is the value that doesn't exist

>>9537628
>i'm wrong so i'm going to make something up and pretend it's related to what you said to save face

>>9537632
>unrelated
>"infinity is a value greater than all other values"
>this value can not be found
>this value is not a specific number
>this value has no known relationship of definite greatness to any specific number
>this value doesn't exist
>unrelated

>>9537662
you're simply too stupid to understand

>>9537667
"A value greater than all other values" but not being a specific value is equivalent to saying you've caught fish and know how to catch fish when in reality, you've never even tried fishing.

If you can't grasp it, you can't use it :^)
Simple life lessons 4 u

>>9531728
Wow! This thread is horrible. Let's talk about hyperreal numbers.

A hyperreal number is an element of an ultrapower of the real numbers. That is, we take all the sequences of real numbers, and we say that two sequences are equal if they differ at no more than finitely many terms. For example,

{0, 1, 1, 1, 1, ...} = {1, 1, 1, 1, 1,...} since they differ only in the first place.

We embed the reals into the hyperreals by identifying the real number r with the equivalence class of sequences represented by the constant sequence {r, r, r, ...}. We then obtain a natural definition for addition and multiplication with the corresponding identities: pretty much everything is done component-wise!

Now, how do we compare hyperreals to each other? We can't do it componentwise since, for example, {0, 2, 0, 2, ...} and {2, 0, 2, 0, ...} have components which are not consistently bigger or smaller than each other. Basically, we pick a predetermined set of indices on which to compare them component-wise.

Anyway, let's just take for granted that we have a nice (totally) ordered field! Since the sequence {1, 1/2, 1/3, 1/4, ...} is less than {c, c, c, c, ...} on all but finitely many elements regardless of c (as long as it's positive), we can see that {1, 1/2, 1/3, 1/4, ...} is smaller than c but bigger than 0! We've constructed an infinitesimal.

We can construct another just as easily: for example, {1, 1/4, 1/9, 1/16, ...} is clearly between those two. Now we can talk about real numbers and "halos" of infinitesimals around them, and doing analysis on this field and all kinds of fun stuff. However, let's not get distracted.

You ask if 0.999... = 1. In the reals, we define 0.999... = lim n -> infty sum from 1 to n of 9(.1)^n, which by the definition of limit is precisely 1. If that definition holds in the hyperreals, clearly the statement is true as well! But, unlike in the reals, there are numbers infinitesimally close to, but not equal to, 1.

>>9537771
For this reason, it might not be very helpful to define 0.999... in the same way; it's best to avoid the notation at all to avoid this confusion, which is typically what is done.

Finally, the answer to your question is, nobody uses the notation 0.999... in non-standard analysis.

>>9537127
∀ε>0:∃x∈R:∀y∈R:(y >x =>|f(y)−L|<ε).

∀L>0:∃x∈R:∀y∈R:(y >x =>f(y) > L).

>>9537773
To be clear, we avoid infinite decimal representations of hyperreals because in 100% of cases (there are a lot of hyperreals) no finite amount of information (the stuff before the dots) is enough to distinguish two hyperreals.

>>9537813
>>9537773
basically you're not only late in the sense that this has been covered, you're late in the sense that you've retroactively been proven wrong, there is such a notation

guess you have to go back

[math]
\displaystyle \sum_{n=1}^{\infty}\frac{9}{10^n} = \frac{9}{10} + \frac{9}{100} +\frac{9}{1000} + ...
= b = \frac{1}{10}(9 + \frac{9}{10} + \frac{9}{100} +...)
= \frac{1}{10}(9+b) = b \\
\frac{1}{10}(9+b) = b \\
\frac{9}{10} = \frac{9}{10}b \\
1 = b
[/math]

>>9538349
I haven't seen anything like a definition of what the hyperreals are nor why the decimal notation is confusing. I don't have to do anything.

(1/inf)+(inf-1/inf) = 1
(inf-1/inf) = 1-(1/inf)

(1/inf)!=0
(inf-1/inf) != 1
0.99999999...!=1

>>9538488
Learn math please.

>>9538662
>i have no argument

>>9538718
Been covered multiple times. You're fucking up "decimal shifting"

Shift your head between a rock and a hard place.

>>9538725
what is 9.999.../10 ?

>>9537780
Yes, you're right. Excuse the error, it was apparently too late.

>>9538772
An answer you wont believe because you understand infinity incorrectly. Deduce it from this.

>>9538822
>i can't answer so i spew out bullshit

>>9538856
It was a direct solution to >>9538488 you fucking subhuman.

>>9538864
just give the answer
is it >1, =1 or <1

>>9538875

>>9537578
>no possible amount of 3's will ever rationally and succinctly finitely equate the decimal value of 1/3, so [math]\frac{1}{3} > 0.\bar{3}, \frac{2}{3} > 0.\bar{6}, \frac{3}{3} > 0.\bar{9}[/math]

>>9538880
classic nut on the bus masturbating, just keeps on screaming some god damn weird cult-mumbling and won't answer a thing you ask...

>1, =1 or <1 ?

>>9538895
>t. i'm so fucking retarded that I can't do math or understand answers so there's no actual point to me asking questions cause i wont get it anyway, but i'll keep asking questions cause i'm too retarded to not

>>9538895
you should kill yourself

>>9538916
>>9538919
9.999.../10
1, =1 or <1 ?

>>9538916
>>9538919
9.999.../10
>1, =1 or <1 ?

>>9538948
Is 1 > 0.9?

>>9531728
op here

i seriously cant believe ive had such an impact on other's lives. this has became a beautiful shitpost, one for the memory books. keep it coming. we all know 1 > 0.999..., only brainlets think otherwise. viva la hyperreals

>>9539153
no

>>9539175
No one gives a shit, anon.
if you want up votes, fuck off back to rebbit

>>9539299
hahahaha hes mad nobody love him

(1/inf)!=0
(1/inf)>0
=> 0,9999999... !=0

>>9539327
>=> 0,9999999... !=1
fixed

>>9539327
>>9539330
Show how [math]\frac{1}{\infty} > 0[/math]

>>9539465
If infinity were an arbatrarily large real number ㅅ, which is all it might be, 1/ㅅ > 0

>>9539885
infinity is greater than arbatrary large numbers

>>9539893
How much greater?
An arbatrarily large number ㅅ[math]_1[/math]+n would just be another arbatrarily large number ㅅ[math]_2[/math]

>>9539899

>>9539885
>ㅅ

>>9539951
>six fingers

>>9539899
>How much greater
infinitely greater. duh.

This is the heir of quantum mechanics move on something new.999 really how f****** retarded

>>9540004
fuck