math

[Anonymous No.15656622]

guys can some explain to me why .99999… = 1? I’ve seen proofs regarding it but I still don’t understand

[Anonymous No.15656666]

>>15656622

[Anonymous No.15656668]

>>15656622
you can eat half a cookie, then eat half of the remaining half, then half of that. and so on for infinity. even if you get to the last atom of cookie, virtual cookie particles will pop into the universe for you to eat. there's infinite bites in 1 cookie

[Anonymous No.15656738]

>>15656668
not to be rude, but I still don’t understand your point. my logic is that if you can say .000000….1> 0 then .9999999…. does not equal to 1. I do not understand how your cookie analogy explains the question. could you explain
further?

[Anonymous No.15656776]

>>15656622
there is no god

[Anonymous No.15656788]

>>15656622
the difference would be 0.00...1, which is 0 because no matter where you stop reading it will always be 0.000...

[Anonymous No.15656856]

>>15656788
is this also true with 0.00...2?

[Anonymous No.15656858]

>>15656788
>0.00...1
ok schizo

[Anonymous No.15656863]

>>15656622
once you understand that 1/inf=0,
the proof is trivial

[Anonymous No.15656867]

>>15656622
>hurr durr I'm trolling

[Anonymous No.15656868]

>>15656738
>000000….1> 0
nope, it's zero
[math] \displaystyle
0. \bar{0}1
= \lim_{n \to \infty} 0. \underbrace{0 \dots 0}_{n ~ \text{times}}1
= \lim_{n \to \infty}
\left [
\left (
\sum_{k=1}^n \dfrac{0}{10^k}
\right )
+ \dfrac{1}{10^{n+1}}
\right ]
=0
[/math]

[Anonymous No.15656870]

>>15656856
yes, anything after the "..." is destroyed

[Anonymous No.15656880]

[Anonymous No.15656959]

>>15656622
Decimal numbers are really only good for expressing natural numbers (no fractions). They kinda work for some rational numbers: 1/2=0.5 but 1/3 cannot be expressed. The 1/3=0.333... is a way to deal with it, and is defined as a limit. But it is better to avoid decimal notations in these cases. They are even more useless for irrational numbers.

[Anonymous No.15656965]

>>15656959
>Decimal numbers
1/3 is neat in base 3, it's just 0.1
There's nothing holy about the amount of our fingers.
https://www.rapidtables.com/convert/number/base-converter.html

[Anonymous No.15657002]

>>15656622
If 0.999... is not equal to 1, then there must be some number between these numbers. Obviously no number can fulfill this requirement (even assuming you had infinite space to write it out, what would the notation be?). Therefore these numbers are indistinguishable and equal.

[Anonymous No.15657036]

The easiest way to think about is this. 1/3 is 0.3 repeating. Multiple that by 3, and you get 3/3 AND 0.9 repeating, ergo, 1 = 0.9 repeating.

[Anonymous No.15657077]

>>15656622
they never equal

[Anonymous No.15657106]

>>15656965
True, but it is not a decimal number anymore. You need to tell me the base: it is just a way of writing rational numbers, with extra steps.

[Anonymous No.15657146]

>>15657002
That reasoning does not hold for natural numbers: there is no natural number between 1 and 2, yet they are not equal. Your argument may be valid for real numbers, but you need to prove a number of other properties of real numbers for it to work. It is not a very convincing argument without that.

[Anonymous No.15657148]

>>15657002
>Obviously no number can fulfill this requirement
Wrong.

[Anonymous No.15657153]

>>15657146
If 0.9 repeating is not 1, then it's not a natural number, your argument is irrelevant.

[Anonymous No.15657154]

>>15657153
NTA but you're so fucking stupid you really shouldn't be here.

[Anonymous No.15657160]

>>15657154
Not an argument

[Anonymous No.15657161]

>>15657160
Neither is any of the painfully dumb shit you wrote.

[Anonymous No.15657168]

>>15657161
Yes it is. He said that what the other guy said isn't valid because it didn't work with natural numbers, but either .9 repeating is not a natural number or it is equal to 1.

[Anonymous No.15657171]

>>15657168
>He said that what the other guy said isn't valid because it didn't work with natural numbers
He said it isn't a valid argument and demonstrated why. You seriously need to be culled.

[Anonymous No.15657178]

>>15657171
Saying that the reasoning doesn't work because it doesn't work in a different kind of number is not a valid argument. We aren't talking about natural numbers, and the only way .9 repeating could be is if it were 1, which would concede the conflict.

[Anonymous No.15657180]

>>15657178
The fact that your "reasoning" immediately fails when applied to N instead of R is the very essence of what makes an invalid argument. You are genuinely too stupid to be here. Go back to your home board.

[Anonymous No.15657206]

>>15656622
Maybe you just prefer surreal numbers.

[Anonymous No.15657217]

>>15656738
That 1 at the end of infinity doesn't exist. It's 0's all the way down.

[Anonymous No.15657222]

>>15656622
Only a stupid nigger that has not made it past pre calculus would even consider that fucking Zenos paraxodes have not been resolved yet. My god, if people could at least complete PRE ALGEBRA before posting their "Discoveries" here. You are a retard.

[Anonymous No.15657234]

>>15656668
>virtual cookie particles will pop
No, you're a retard. When you get down to the last atom of cookie, you can no longer eat half. In real life, there are no infinities in real life. Infinities as concepts are the products of an imperfect language (math) used to describe things that can never be perfectly abstracted

[Anonymous No.15657268]

In the real numbers, which is the number system that is being assumed, infinitely small is equivalent to zero. It is a matter of definition and convention.

[Anonymous No.15657271]

>>15656622
It isnt

[Anonymous No.15657273]

>>15657268
This

[Anonymous No.15657276]

>>15657002
>Obviously no number can fulfill this requirement (even assuming you had infinite space to write it out, what would the notation be?).
That is not obvious at all. "I cannot imagine such a number", or "I can't write it down" is a poor argument, and usually fails in mathematics.

If a is not equal to b, then (a+b)/2 is between a and b.

If 0.999... is not equal to one 1, then (1 + 0.999...)/2 would be between 0.999... and 1.

Therefore, (1 + 0.999...)/2 fulfills your requirement.

[Anonymous No.15657286]

>>15657276
0.999...=(1 + 0.999...)/2=1 so it's not in between

[Anonymous No.15657299]

>>15657286
>0.999...=(1 + 0.999...)/2=1
>b-b-b-because it just is, ok?
The retard strikes again.

[Anonymous No.15657305]

>>15657222
it hasn't be resolved. because in order to be resolved, you would first half to resolve half the problem. then you would have to resolve the remaining half. and then you would have to resolve the remaining half of that. and so on.

[Anonymous No.15657307]

>>15657299
projection, the post

[Anonymous No.15657328]

>>15656622
then you need to look into other areas of math until you can grasp it, simple as, best wishes and all that

[Anonymous No.15657331]

>>15657307
An actual retard, the poster. It's like you just can't grasp the concept of a proof and think restating the same thing over and over constitutes that. The sad thing is that there is nothing to "prove" in the first place. It's just a definition and you can define extensions of R that allow for 1=/=0.999...

[Anonymous No.15657332]

>>15656788
What if you stop reading at the 1?

[Anonymous No.15657376]

>>15656622
Don't listen to the retards who say it's 1. It isn't. There is absolutely no proof.
When a dipshit says 1/3 = 0.33... and 0.33...*3 = 0.99... and 1/3*3=1 therefore 0.99...=1 then you just tell him that 1/3 is NOT 0.33... because it is just an approximation below 1/3 because 1/3 is not representable in decimal form so 0.99... will also be below 1.

[Barkun No.15657378]

1 is not the first number after zero if we're talking decimals. So what is it smartasses. Modern math is retarded. The first number is a strange quart.

[Barkun No.15657382]

Get the f off mah sci with this modern math shit

[Anonymous No.15657417]

>>15657331
>can't grasp the concept of a proof
>>15657307

[Anonymous No.15657418]

>>15657376
>1/3 is NOT 0.33...
prove it

[µ No.15657472]

Okay, so the notation [math]0.\overline{9}[/math] means you are representing a number by an infinite series, namely
[eqn]0.\overline{9}:=\sum_{n=1}^{\infty}\frac{9}{10^n}:=\lim_{n\to\infty}\sum_{k=1}^n\frac{9}{10^k}[/eqn]
As you can see, the value of a series is defined as a limit.
What's important to remember about limits is that they are not simply "what happens if you plug in infinity".
That concept ist nonsensical, you can't compute an "infinite sum", hence the confusion, because the partial sums never equal the limit. Instead, there are ways to rigorously define limits as a number that is uniquely defined by whatever sequence you have. In this case, we have a series, so the sequence is simply taking the partial sums. But we can simplify this
[eqn]0.\underbrace{9\dots 9}_n=1-0.\underbrace{0\dots 0}_n1=1-\frac{1}{10^{n+1}}[/eqn]
The definition of when the limit of a sequence equals a number is that whenever you choose an arbitrary [math]\varepsilon>0[/math], you can
choose an Index [math]N\in\mathbb{N}[/math], such that for all indeces [math]n[/math] greater than [math]N[/math], the diference between the [math]n[/math]-th element of the sequence and the proposed limit is smaller than [math]\varepsilon[/math]. What this means is that however close you want the sequence to be to the limit, it will be at that close for all indices after some finite amount of them. Now we can prove the proposed result: Let [math]\varepsilon>0[/math]. Then choose an [math]N\in\mathbb{N}[/math] greater than
[eqn]\log_{0.1}(\varepsilon),[/eqn]
the existence of which is guarranteed by [math]\varepsilon[/math] being greater than 0 and the archimedean property. Then for all [math]n\geq N[/math] we have
[eqn]|1-0.\underbrace{9\dots9}_n=\frac{1}{10^{n+1}}=0.1^{n+1}<0.1^{\log_{0.1}(\varepsilon)}=\varepsilon.[/eqn]
Therefore it follows that
[eqn]0.\overline{9}=\lim_{n\to\infty}0.\underbrace{9\dots9}_n=1[/eqn]
QED.

[µ No.15657479]

>>15657472
formatting is messed up in the last line because this site is shit, let me try again

[µ No.15657482]

Okay, so the notation [math]0.\overline{9}[/math] means you are representing a number by an infinite series, namely
[eqn]0.\overline{9}:=\sum_{n=1}^{\infty}\frac{9}{10^n}:=\lim_{n\to\infty}\sum_{k=1}^n\frac{9}{10^k}[/eqn]
As you can see, the value of a series is defined as a limit.
What's important to remember about limits is that they are not simply "what happens if you plug in infinity".
That concept ist nonsensical, you can't compute an "infinite sum", hence the confusion, because the partial sums never equal the limit. Instead, there are ways to rigorously define limits as a number that is uniquely defined by whatever sequence you have. In this case, we have a series, so the sequence is simply taking the partial sums. But we can simplify this
[eqn]0.\underbrace{9\dots 9}_n=1-0.\underbrace{0\dots 0}_n1=1-\frac{1}{10^{n+1}}[/eqn]
The definition of when the limit of a sequence equals a number is that whenever you choose an arbitrary [math]\varepsilon>0[/math], you can
choose an Index [math]N\in\mathbb{N}[/math], such that for all indeces [math]n[/math] greater than [math]N[/math], the diference between the [math]n[/math]-th element of the sequence and the proposed limit is smaller than [math]\varepsilon[/math]. What this means is that however close you want the sequence to be to the limit, it will be at that close for all indices after some finite amount of them. Now we can prove the proposed result: Let [math]\varepsilon>0[/math]. Then choose an [math]N\in\mathbb{N}[/math] greater than
[eqn]\log_{0.1}(\varepsilon),[/eqn]
the existence of which is guarranteed by [math]\varepsilon[/math] being greater than 0 and the archimedean property. Then for all [math]n\geq N[/math] we have
[eqn]|1-0.\underbrace{9\dots9}_n=\frac{1}{10^{n+1}}=0.1^{n+1}<0.1^{\log_{0.1}(\varepsilon)}=\varepsilon.[/eqn]
Therefore it follows that [math][/math]
[eqn]0.\overline{9}=\lim_{n\to\infty}0.\underbrace{9\dots9}_n=1[/eqn]
QED.

[µ No.15657488]

>>15657472
also, this proof uses that [math]0.1^a<0.1^b[/math] if [math]a>b[/math], which technically requires another proof, but i guess its more intuitively clear than the whole thing. you get the point

[µ No.15657493]

>>15657479
Then for all [math]n\geq N[/math] we have
[eqn]|1-0.\underbrace{9\dots9}_n|=\frac{1}{10^{n+1}}=0.1^{n+1}<0.1^{\log_{0.1}(\varepsilon)}=\varepsilon.[/eqn]
Therefore it follows that
[eqn]0.\overline{9}=\lim_{n\to\infty}0.\underbrace{9\dots9}_n=1[/eqn]
QED.

[Anonymous No.15657500]

>>15656622
Taking a wag, it is more semantic than tangible.
1 is greater than .9999
1 is equal to .9999... there's just one caveat

[Anonymous No.15657559]

>>15657472
>I define that 9=10 since the difference of 1 is smaller than the acceptable error E that I arbitrarily set to 2. There I proved that 9=10.

[Anonymous No.15657563]

>>15657180
holy shit you're genuinely retarded

[Barkun No.15657676]

>>15657563
>>15657472
Fuckkkk... You idiots actually believe in modern math and use stuff like LaTeX, this only embeds your errors. Homos. Kys now

[µ No.15657931]

>>15657559
it needs to work for ALL arbitrary [math]\varepsilon>0[/math], not just one. As in, if you can prove [math]|a-b|<\varepsilon[/math] for ALL [math]\varepsilon>0[/math], you have proven [math]a=b[/math].

[Anonymous No.15657969]

>>15656622
Let [math]x = 0.\bar{9}[/math].
Multiplying both sides by ten, [math]10x = 9.\bar{9}[/math],
by which we can get the difference, from the second to the first, [math]9x = 9[/math]
and from this we clearly see that [math]x = 1[/math]

>Things which are equal to the same thing are also equal to one another.
[math]0.\bar{9} = x = 1[/math], then [math]0.\bar{9} = 1 [/math].
Frankly, I see no way of how it can be more basic than this.

[Anonymous No.15658075]

>>15657931
Wow you're retarded. Your post has disproven your own claim. You have a-b=x=0.1^(n) which is > 0 for all n, which already says that a != b, and since E = all numbers > 0 then x cannot be < E. You also disregard the fact that 0.(anything) is not 1. The 1st digit is 0 which means the number is smaller than 1 from the start, it doesn't matter what comes after the comma. I should also say that an infinite period > 0 means that it isn't a defined number and thus you cannot use it like a number but you used it in an equation which invalidates the equation and you did that in a "proof".
>>15657418
Why do you think the 3 repeats forever? Because the numeral system can only represent a small set of very special numbers of the form x = (B^n)/N where n and N are whole numbers and B is the base. There is no solution for any combination of n and N with x=1/3 and B = 10. A repeating 0 at the end means you add nothing more to the amount that you have described thus far which means you have reached the target amount and you have defined the number representing that amount. If you don't get a repeating 0 at the end then the amount cannot be written in your system. If you can't write your number then you can't do calculations with it unless you chose a different number close to it that you can write and use that instead. That's what people do with pi and 1/3 etc.

[Anonymous No.15658094]

>>15656868
>limit
You fucking retard, actually kill yourself

[Anonymous No.15658102]

>>15658094
xhe's right though

[Anonymous No.15658107]

You are responding to a bot.

[Anonymous No.15658114]

>>15656622
1-1/n as n goes to infinity. etc etc

[Anonymous No.15658139]

>>15656738
>my logic is that if you can say .000000….1> 0
you can't say that and be right

[Anonymous No.15658183]

>>15658075
>>Why do you think the 3 repeats forever?
>no proof, just m-muh feelings hand waving
lol

[µ No.15658275]

>>15658075
Again, the reason you are confused is because we are talking about a limit here. Let me try to explain it again:
For simplicity, let [math]a_n:=0.\underbrace{9\dots9}_n[/math]. Now we have a sequence
[math]a_1=0.9[/math]
[math]a_2=0.99[/math]
[math]a_3=0.999[/math]
etc.
Now BY DEFINITION, we have
[eqn]0.\overline{9}=\lim_{n\to\infty}a_n.[/eqn]
[math]0.\overline{9}[/math] is equal to the LIMIT of the sequence as [math]n[/math] goes to infinity.
It is NOT equal to any of the [math]a_n[/math]. Instead, the limit of a sequence is simply defined as the unique number the sequence approaches as [math]n[/math] goes to infinity. That is, the unique number that the sequence gets ARBITRARILY close to, and STAYS close, which means that for ANY AND ALL distances [math]\varepsilon>0[/math] you choose, after some point all the following elements of the sequence have a distance to the proposed limit smaller than [math]\varepsilon[/math]. If you can prove such a number exists for a given sequence,
we say the sequence converges to that number, and you call that number the limit of the sequence. I have done so in my first reply.

[Barkun's Cum Rag No.15658282]

>>15658275
>goes to infinity

[Anonymous No.15658348]

>>15658275
>limit of a sequence is simply defined as the unique number the sequence approaches as n
Approaches but but never reaches. That's all the limit says. What's your point?
>the following elements of the sequence have a distance to the proposed limit smaller than ε
contradicts
> for ANY AND ALL distances ε>0
You keep claiming 0.9period = 1. Mental illness.

[Barkun's Cum Rag No.15658354]

>>15658348
>Presumably it goes to infinity
Says who? Did you check? Whereabouts is the logic which says it goes to infinity?

[Anonymous No.15658360]

>>15658354
its you, you specifie .999... yourself.... that's.. the definition. if you meant .999 with a gorillian zeros that's not what you specified. that's all they are two different things...

[Barkun's Cum Rag No.15658363]

>>15658354
Its not a sequence cause the whole number is instant

[Barkun's Cum Rag No.15658370]

>>15658363
0.999... Is just evidence of the third true number, spoof number. It's more special than normal.

[Anonymous No.15658372]

New anon has entered the chat.
It is more semantic than tangible.
1 is greater than .9999
.9999... is equal to 1. There's just one caveat.

[µ No.15658387]

>>15658348
>Approaches but but never reaches. That's all the limit says. What's your point?
The limit is defined as the point the sequence approaches. The number [math]0.\overline{9}[/math] is defined as the limit of the sequence i defined in the previous reply. The limit does not have to equal any of the elements of the sequence.

to your second point, order of logical junctors is important. I said for all [math]\varepsilon>0[/math], there exists [math]N\in\mathbb{N}[/math] such that for all [math]n\geq N[/math] ...,
meaning you first choose a [math]\varepsilon>0[/math], and then you can find such a [math]N\in\mathbb{N}[/math]. But the point is that you can find it if you know the chosen [math]\varepsilon>0[/math], no matter how small you chose it to be.

[Barkun's Cum Rag No.15658405]

>>15658387
>sequence

[Anonymous No.15658424]

No amount of .999... definitions involving limits can ever counter the fact that [math]\sum_{i=1}^{i=n}\frac{9}{10^{i}}[/math] does not equal 1 for any value of n. You can go as big as you want, as many 9s as you want, it's never going to be equal.

[µ No.15658447]

>>15658424
that doesnt matter because [math]0.\overline{9}[/math] is not equal to any of these sums, its equal to the limit of the sequence of these sums, which is 1.

[Anonymous No.15658464]

>>15658424
Is the following relevant to the discussion?
A mathematical theory is a mathematical model of a branch of mathematics that is based on a set of axioms. It can also simultaneously be a body of knowledge (e.g., based on known axioms and definitions), and so in this sense can refer to an area of mathematical research within the established framework.

[Anonymous No.15658480]

>>15656622
You don't belong to the caste of mathematicians. Is only natural you will never be able to understand such self evident truths. Accept your fate, your place in society as just another normie that is is "bad" at math.

[Anonymous No.15658488]

>>15658447
The point is that no matter how long you continue the series, it will never be 1. Look at people in the thread using recursive arguments like >>15656668 or people who all the time think it's like adding a 9 onto the end of a decimal.
For both of these, neither is an infinite series defined in terms of a limit, they are comparable to the series written in my post. Which, no matter how high n goes, how many iterations are added, will never be equal to 1.
That is what's important. Not limits, since limits are not what is at the core of the intuition used against the equation.

[Anonymous No.15658504]

>ctrl-f
>no Wildberger
it's like you guys want to argue over a made up problem

[µ No.15658511]

>>15658488
>limits are not what is at the core of the intuition used against the equation.
thats the problem. Limits are unintuitive sometimes, and [math]0.\overline{9}[/math] is by definition the limit of a sequence.

[Anonymous No.15658518]

>>15658488
Not him but you have the intuition backward. Nines that don't end are easy and intuitive. Nothing changes. Nines that keep going for a long time are awkward and unintuitive. When do they stop? Why? Where? How?

[Anonymous No.15658521]

>>15658488
0.F... - 0.9... = 0
Looks like you're wrong again.

[Anonymous No.15658523]

>>15658511
Yes, and so it's not applicable to the intuitive idea of continuously adding 9s onto the end of a decimal point, which does not involve limits whatsoever and instead is an ordinary series.

[Anonymous No.15658527]

>>15658521
what are you on about.

[Anonymous No.15658538]

this thread:
>why is the sky blue
>correct answer
>retard:WELL HOW DO WE KNOW WHAT BLUE IS
its like you are talking with a child. you can't ask a question, and then take apart the definition of the very word you used to ask that question. reformulate your question, if you can.

[µ No.15658543]

>>15658518
i genuinely have no idea what you are talking about.

>>15658523
>does not involve limits whatsoever and instead is an ordinary series.
every series does involve limits. the sum of the series is a limit. thats what im trying to tell you.

[Anonymous No.15658566]

>>15658543
It's always nines.

[Anonymous No.15658695]

1/9 = 0.111...
+
8/9 = 0.888...
=
9/9 = 0.999...

[Anonymous No.15659093]

>>15658527
he's mixing hexadecimal and decimal in an equation without any annotation of him doing so, i think

[Anonymous No.15660071]

>>15656622
And you are calling yourself a scientists...
0.999 + 0.001 = 1

[Anonymous No.15660094]

>>15660071
0.9+0.1=1
0.99+0.01=1
0.999+0.001=1
:
0.999...+0.000...1=1
0.999...+0=1 (see >>15656868)
0.999...=1

[Anonymous No.15660300]

>>15660094
0.999... = 0.999...9
0.999... =/= 0.999...8
0.999... =/= 1.000...0
0.000...1 =/= 0.000...0

[Anonymous No.15660336]

>>15657036
>1/3 is 0.3 repeating
false

[Anonymous No.15660388]

A theory is a supposition or a system of ideas intended to explain something, especially one based on general principles independent of the thing to be explained.
Here is the crux of the issue.
2+2=4
True. There is a start and end.
.99999... = 1. There is no end. The approach to 1 eventually gets so miniscule .99999... is generally accepted to be equal to 1 for all intents and purposes. No one actually knows because no one can prove it. Remember, it just keeps on going forever.

[Anonymous No.15660399]

>>15660336
>there just has to be a 17 in there somewhere
lol

[Anonymous No.15660400]

>>15660300
>0.999... = 0.999...9
ok
>0.999... =/= 0.999...8
no
>0.999... =/= 1.000...0
no
>0.000...1 =/= 0.000...0
no

[Anonymous No.15660412]

>>15656622
[math]x=0.999...[/math]
[math]10x=9.999...[/math]
[math]10x-9=0.999...[/math]
[math]10x-9=x[/math]
[math]9x-9=0[/math]
[math]9x=9[/math]
[math]x=1[/math]

This relies on the assumption that "0.999..." is a real number though.

[Anonymous No.15660417]

They aren't equal, and whatever "methods" exist to equate them are nothing more than evidences of the fallacies of the current mathematical system(s).

[Barkun No.15660421]

If this were true you wouldn't be able to take the first decimal step because 0.0000...1 would equal 0. Tardes

[Barkun No.15660422]

>>15660417
kys now

[Anonymous No.15660450]

>>15660399
no but it's just not translatable to decimals

[Anonymous No.15660459]

>>15660412

[Jealous No.15660468]

It's due to the Cauchy definition of the limit:
[eqn] 0.999...=\lim\limits_{n\to\infty}\sum_{k=1}^n\frac{9}{10^k} = 1[/eqn]
You can work it out with the [math]\varepsilon\!-\! \delta[/math] formalism pretty easily that this limit is equal to one, and the "dot dot dot" notation is shorthand for this limit.

[Anonymous No.15660469]

>>15656622
1/3=0.3333...
(1/3) * 3 = 1
0.3333... * 3 = 0.9999... = 1

[Anonymous No.15660495]

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

[Anonymous No.15660798]

>>15656622
> 1/3 = 0.3333...
okay
> 2/3 = 0.6666...
okay
> 3/3 = 0.9999...
noway!

[Anonymous No.15660857]

/sci/ can't understand limits lol
This is why we need better math education in America

[Anonymous No.15660868]

>>15660857
> thinks it is not an international board
Improve your own education: this thing reminds limits, but has nothing to do with them.

[Anonymous No.15660871]

simply look at both numbers on the number line, idiots

[Anonymous No.15660884]

>>15660868
The rigorous definition of a repeating decimal is a limit definition

[Barkun No.15660887]

>>15660857
Or the math is just retarded and educators are too

[Anonymous No.15660894]

math] \displaystyle
\boxed{0 < p < 1} \\
p^n-1 = (p-1)(p^{n-1}+p^{n-2}+ \dots +p+1) \\
\dfrac{p^n-1}{p-1} = \sum \limits_{j=0}^{n-1}p^j \\
\displaystyle
\lim_{n \to \infty} \dfrac{p^n-1}{p-1} = \lim_{n \to \infty} \sum \limits_{j=0}^{n-1}p^j \\
\displaystyle
\dfrac{0-1}{p-1} = \sum \limits_{j=0}^{\infty}p^j \implies \dfrac{1}{1-p} = \sum \limits_{j=0}^{\infty}p^j
[/math]

[math] \displaystyle
p=0.1 \\
\dfrac{1}{1-0.1}=\frac{10}{9} = 1 + \frac{1}{9} \\
\sum_{j=0}^\infty 0.1^j= 1 + \sum_{j=1}^\infty 0.1^j \\
9+1=9+9\sum_{j=1}^\infty 0.1^j \\
1=9\sum_{j=1}^\infty 0.1^j = 0.999...\\
\dfrac{1}{3} = 3 \sum_{j=1}^ \infty 0.1^j = 0.333...
[/math]

[Anonymous No.15660897]

[math] \displaystyle
\boxed{0 < p < 1} \\
p^n-1 = (p-1)(p^{n-1}+p^{n-2}+ \dots +p+1) \\
\dfrac{p^n-1}{p-1} = \sum \limits_{j=0}^{n-1}p^j \\
\displaystyle
\lim_{n \to \infty} \dfrac{p^n-1}{p-1} = \lim_{n \to \infty} \sum \limits_{j=0}^{n-1}p^j \\
\displaystyle
\dfrac{0-1}{p-1} = \sum \limits_{j=0}^{\infty}p^j \implies \dfrac{1}{1-p} = \sum \limits_{j=0}^{\infty}p^j
[/math]

[math] \displaystyle
p=0.1 \\
\dfrac{1}{1-0.1}=\frac{10}{9} = 1 + \frac{1}{9} \\
\sum_{j=0}^\infty 0.1^j= 1 + \sum_{j=1}^\infty 0.1^j \\
9+1=9+9\sum_{j=1}^\infty 0.1^j \\
1=9\sum_{j=1}^\infty 0.1^j = 0.999...\\
\dfrac{1}{3} = 3 \sum_{j=1}^ \infty 0.1^j = 0.333...
[/math]

[Anonymous No.15660911]

>>15660884
it is not. any number can work as a limit

[Barkun No.15660912]

>>15660897
Thanks for the grift

>Didn't read

[Anonymous No.15660976]

>>15660912
>Didn't read
why should today be different

[Anonymous No.15661018]

construct a number x such that
0.9999...+ x < 1
you cannot under standard definitions. (hint: see what happens if you find a value for x)

[Anonymous No.15661022]

>>15661018
So are you saying that 0,999.. and 1 are next to each other ?

[Anonymous No.15661032]

>>15660412
>>15660459
x = 0.99
10x = 9.9
10x - 9 = 9.0
10x - 9 =/= x
faggot
>>15660897
explain your steps faggot or fuck off
>>15661018
x = 0.0...y, y = any number
>>15661022
no, faggot, 0.99... can be seen as the closest number smaller than 1 but there are always an infinite amount of numbers between 0.99... and 1 because the number line is continuous and there is no such thing as 2 numbers next to each other with nothing in between.

[Anonymous No.15661043]

>>15661032
btw:
0.99 * 10 - 9 = 0.90
0.99... * 10 - 9 = 0.99...
the left 0.99... and the right 0.99... are different numbers even though they are written same way and look the same so that shows you that 0.99... cannot be a single number but can be an infinite amount of different numbers.

[Anonymous No.15661086]

>>15661032
>there are always an infinite amount of numbers between 0.99... and 1 because the number line is continuous and there is no such thing as 2 numbers next to each other with nothing in between.
Unless they're the same number. And name one number standing between 0.(9) and 1. See? You can't. Thus they're the same number, just as 3 times 1/3 is 1.

[Anonymous No.15661091]

>>15661032
>explain your steps
I could read that with the school math I was taught before I was 16.
go ask mommy and fuck off retard

[Anonymous No.15661115]

>>15661086
I can: 0.99...y where y = any number. There.

[Anonymous No.15661157]

>>15661115
>0.99...y
What is the difference between 0.99...y and 0.99...9y?

[Anonymous No.15661183]

>>15661043
>written same way
the "..." ain't there for decor you glue-slurper

[Anonymous No.15661402]

>>15661018
0.9 + [<0.1] < 1
0.99 + [<0.01] < 1
0.999 + [<0.001] < 1
0.9999 + [<0.0001] < 1

0.999... + [<0.000...1] < 1
define "..." as "an amount"
x = 9
y = 0
0.9[x] + [<0.[y]1] < 1
0.9[xx] + [<0.[yy]1] < 1
0.9[xxx] + [<0.[yyy]1] < 1
0.9[xxxx] + [<0.[yyyy]1] < 1
[math]n_x = n_y[/math]
0.9[xxx...] + [<0.[yyy...]1] < 1

a + b < c
a[x] + [y]b < c
a[xxx] + [yyy]b < c
a][xxx...] + [yyy...]b < c
define "..."
is "..." equivalent to "an amount"?
or
is "..." equivalent to "the largest number allowed"?

your homework is to define infinity.

[Anonymous No.15661435]

>>15661115
>I can: 0.99...y where y = any number. There.
The number you "found" is less than 0.(9)

[Anonymous No.15661489]

>>15656622
>guys can some explain to me why .99999… = 1? I’ve seen proofs regarding it but I still don’t understand
Math is imperfect. copy/paste to follow:

Gödel's incompleteness theorems are two theorems of mathematical logic that are concerned with the limits of provability in formal axiomatic theories. These results, published by Kurt Gödel in 1931, are important both in mathematical logic and in the philosophy of mathematics. The theorems are widely, but not universally, interpreted as showing that Hilbert's program to find a complete and consistent set of axioms for all mathematics is impossible.

The first incompleteness theorem states that no consistent system of axioms whose theorems can be listed by an effective procedure (i.e., an algorithm) is capable of proving all truths about the arithmetic of natural numbers. For any such consistent formal system, there will always be statements about natural numbers that are true, but that are unprovable within the system.

The second incompleteness theorem, an extension of the first, shows that the system cannot demonstrate its own consistency.

Employing a diagonal argument, Gödel's incompleteness theorems were the first of several closely related theorems on the limitations of formal systems. They were followed by Tarski's undefinability theorem on the formal undefinability of truth, Church's proof that Hilbert's Entscheidungsproblem is unsolvable, and Turing's theorem that there is no algorithm to solve the halting problem.

If you really want to understand research it.

[Anonymous No.15661531]

>>15661435
I was just using the weird notation that others have used in this thread it's not my invention. The "..." already contains infinitely many 9s so that's your number "0.(9)" and the y just says that there is more added to it after those infinite 9s. I already said that there are infinitely many numbers between any 2 numbers no matter how infinitely long you write them and how small you make the difference. Even if the difference gets infinitely small it is infinitely large.

[Anonymous No.15661812]

>>15661489
Godel's incompleteness theorem has nothing to do with this, the proof has been demonstrated above.

[Anonymous No.15661853]

>>15661531
It illustrates examples of theory.
As to the proof, I'm not arguing that. Also, that is not what the OP wanted as I understood it.
>guys can some explain to me why .99999… = 1? I’ve seen proofs regarding it but I still don’t understand
He clearly stated he has seen the proofs. But lacks understanding. Most posters are arguing proofs, correctly and incorrect. I point out the way to accept or understand.

[Anonymous No.15662148]

>>15656622
Consider how the same number is represented in different bases and it will make sense. It is merely a representation problem.

[Anonymous No.15662231]

>>15661402
>define infinity
Infinity is an unbounded quantity greater than every real number.

[Anonymous No.15662582]

>>15661531
you don't understand the "infinitely", that is 9s all along, once you place any other digit somewhere in that infinity, you get a lesser number. And here I see that 0.(9) is the next number to 1, but only in the same sense 0.(0) is the next to 0, so 1 being represented as 0.(9) is just being written in a synonym, because the difference between 0.(9) and 1 is 0.(0)

[Anonymous No.15662716]

>>15656622
Here is proof.

Let S = 0.9, 0.99, 0.999, ...

Corrolary
S is cauchy
Proof
si, sj in S, i,j >= k
| si - sj | = | 9/10^k + .. | <= | 10^(k-1) | < | 9/10^k | = e > 0
lim k e > 0
Thus we can choose any M >= k =10^m/9
and | si - sj | < e, for aebitrarily small e (and thus big M)

Corrolary:
R is complete, and so any cauchy series converges Q converges in N and uh

Fuck math

Let e

[Anonymous No.15662733]

>>15662231
regardless of which equivalency you picked, it doesn't matter.

if [0.999...] has an uncountable amount of 9's, there is no upper limit to define that something cannot come after. On the contrary, it explicitly requires to attempt defining something which comes after.
so regardless if [0.999...] has a countable or uncountable amount of 9's, there is allowance of an additional number [0.000...1] necessary to add up to 1.
to follow the utmost truth of infinity, 0.999... explicitly cannot equal 1, or be used in any useful equation.
[math][0.999...] + [0.000...1] = 1 [/math] is necessarily a false statement if infinity has been correctly invoked, because
[math]\underbrace{[0.999...]}_{\text{this begins, but doesn't end}} \underbrace{+ [0.000...1] = 1}_{\qquad \text{so this never occurs}} \\ \\ 0.999... [/math]

[Anonymous No.15662765]

>>15662733
>but doesnt end
no such thing
complete bullshit
as soon as you put as an axiom that "forever" makes sense then you go off to la la land with no meds

[Barkun No.15662769]

I am fucking intelligent. Repeating numbers are just ill-math noises pointing towards a greater system. If 0.999... equals 1, the first decimal number is a fallacy (0.000...1) for it equals 0.

[Anonymous No.15662777]

>>15662769
hyper-reals allow for 0.0000...1

[Anonymous No.15662781]

it's just because writing all the 9 after the 0. is too tiring so we simplify it as 1

[Barkun No.15662795]

>>15662781
The third number is an important number for calculating interest. Probably why there is so much ill-focus on repeating numbers and their round.

[Anonymous No.15662806]

>>15662765
so you suggest some finite element which is clearly defined to be "less than forever"?

[Barkun No.15662808]

>>15662806
Yep.

[Barkun No.15662811]

>>15662808
It goes.

Strange Quart
>Before any idea is the conception, before conception is strangeness at a quart of consciousness.

Stable Half
>Just before conception is the point where it isn't conceived but is becoming conceived.

SPOOF NUMBER
>conceiving of conception, pre-becoming whole.

1
Conception

[Anonymous No.15662815]

>>15662811
take your meds and/or keep your funny interpretations to yourself.

[Anonymous No.15662816]

>>15656863
/thread

[Barkun No.15662821]

>>15662815
You are a med

[Anonymous No.15662825]

>>15662806
No, I am suggesting when you take as an apriori truth and axiom that "forever", "infinity", etc., are concepts which make sense despite that they can only be described using the antithesis to "ending", "finite" which are natural unavoidable truths you will produce self-referential nonsense.

[Barkun No.15662826]

>>15662811
SPOOF NUMBER is a to a sperm what strange quart is to horniness what stable half is to ejaculation what cum is to 1.

[Anonymous No.15662831]

>>15662825
you're assuming "ending" and "finite" aren't the contradictory statements against the truth of "forever" and "infinity".

this is a you problem. figure it out yourself shitforbrains.

[Barkun No.15662838]

>>15662831
You're assuming he assumes or had assumed you would assume his assumption, asumpter

LEAVE HIM LONE

[Anonymous No.15662839]

What a weak reply.>>15662831
I don't need to assume that, there is plenty of evidence that finiteness predates infiniteness. No contradictions are made in accepting natural numbers and counting, it is only when you say "just keep counting forever bro" you are introducing "forever" without having explained anything about what forever is.

[Barkun No.15662846]

>>15662839
There's no such count as 1 quintillion, you lose track at 1000 quadrillion so you keep a memoir called a 'bar' and then count from 1 again. You collect bars, which become beds, cubes and houses by the 1000. You then collect maxth sets by the 40,000. You then collect boxth sets by the 40,000. Then you reach where infinity lies, the infinite chasm of repeating counts.

[Barkun No.15662853]

>>15662846
Cities, civilization

Empires, Universal Empires

[Anonymous No.15662855]

>>15662839
>there is plenty of evidence that finiteness predates infiniteness.
do you even know what the word "evidence" means

[Anonymous No.15662858]

>>15662855
Do you? Where in the ancient texts is infinity? Where is 0.9999... = 1? Must be in the bible, since God is infinite amirite? Where was it written that to count from 1 to 3 you must first invent infinity?
>but muh infinity and finity are abstract concepts which blablabla
there is no realm of perfect mathematical objects and ideas anywhere in the universe so you might as well posit the existence of god then

[Barkun's Cum Rag No.15662870]

>>15662858
you and i are gods, but only one of us is good

[Anonymous No.15662930]

>>15662765
>nigger lacks the skill of abstract thinking
my condolences, yet you still can be great in something else

[Anonymous No.15662933]

>>15662858
infinity is well imagined as a quality of "afterness", so the fact you can count beyond 1 at all is surely evidence of something after 1. the "afterness" of infinity just happens to evoke "perpetual afterness", always looking forward but never seeing an end.

now, humans are not very good at prediction. humans are not very good at "looking forward", but just because humanity can't do something, doesn't mean it can't be done at all, or that something else isn't already doing it.

the quality of infinity, relative to human inefficacy with prediction, is then also equivalent to "unknown", besides also being "after".

what is the overwhelming sense of attempting to witness the vastness of all time from the present forward, if not an attempt to measure infinity? and why would it be overwhelming at all?
what, but God, could certainly know all the future?
what, but God, could be so overwhelming?

[Anonymous No.15662941]

>>15662582
No, 1 is represented as 1 and only as 1. Anything else is a different number.

[Anonymous No.15663054]

>>15662930
>retard doesnt understand WHY we define limits using finite deltas and epsilons

[Anonymous No.15663082]

>>15656622
Simple as 1 - 0.999.. = 0.000...

Literally no further explanation needed.

[Anonymous No.15663100]

>>15663082
proving the assumption is the best way to filter roomtemps

[Anonymous No.15663106]

>>15663100
>proving the assumption
Which assumption?

[Anonymous No.15663180]

>>15663082
>1 - 0.999.. = 0.000...
>no further explanation needed.
Yes it's 0,000... and not 0
What was your point here?

[Anonymous No.15663225]

>>15663180
>Yes it's 0,000... and not 0
Are you implying these are different numbers? You're the one making assumptions here. The only "assumption" made is that the arithmetic is done in decimal notation

[Anonymous No.15663244]

>>15663225
Not making any assumptions but if you take away string of 9's from a string of 10's you will never take out everything

[Anonymous No.15663272]

>>15656738
.0000...1 is a finite number, propperly defined. It is greater than 0 because you can find at least one number greater than 0 but lesser than the previous (dividing by 2). The number .9999... has an infinite number of 9's, so it isn't possible to find a number smaller than 1 but bigger than .9999... That number is infinitely close (not to be confused with a limit of a sequence) to 1, so it is 1.

I'd imagine the proof you might know is made with fractions, by setting x = 0.9999... and multiplying both sides by 10, and so on.

[Anonymous No.15663303]

>>15663244
Not him but if there's no end, you never get anything but 0.

[Anonymous No.15663318]

>>15663272
0.1=10^-1
0.01=10^-2
0.001=10^-3
:
0.000...1=10^-inf=0

[Anonymous No.15663337]

>>15663303
Then you are jumping to conclusions already
There is also no end in the repeating 9's so you cannot substract it because if you could do that, that would mean that they end somewhere

[Anonymous No.15663349]

>>15663318
>0.000...1=10^-inf

Not true.

[Anonymous No.15663356]

>>15662941
is 2/2 a different number? what about 2^0? well now this one I myself do not understand (I know how to prove it, but it still doesn't make much sense, I suspect n^0 to be where math broke)

[Anonymous No.15663362]

>>15663054
it's not about limits, nigger fren

[Anonymous No.15663425]

>>15663337
Not true. If there is nothing but 9, then there is nothing but 0.

[Anonymous No.15663439]

>>15663244
>if you take away string of 9's from a string of 10's
Wtf are you talking about? What is a string of 10s? Like 0.10101010101010? Or are you not working in base 10?
>you will never take out everything
Based on..? Again you're making strange assumptions
>>15663337
>There is also no end in the repeating 9's
Correct
>so you cannot substract it because if you could do that, that would mean that they end somewhere
Nevermind you've invalidated your entire stance on the subject. You're working under a different set of math rules than the rest of humanity

Here's some resources for you to catch up to what we're working with

https://www.khanacademy.org/math/arithmetic-home/addition-subtraction/basic-add-subtract/v/basic-subtraction

https://en.m.wikipedia.org/wiki/Subtraction

https://www.preschool-printable-activities.com/free-subtraction-worksheets.html

[Anonymous No.15663442]

>>15663425
So strings of 9's taken away from strings of 10's
That does not end in 0

[Anonymous No.15663448]

>>15663439
>Wtf are you talking about?
Every decimal consists of 10 numbers which are 0-9

[Anonymous No.15663449]

>>15663349
prove it

[Anonymous No.15663456]

>>15663442
We already agreed it doesn't end.

[Anonymous No.15663458]

>>15663448
Anon we know you're being obtuse on purpose. Now go on to explain how that relates to this problem. But first watch the video and complete the worksheet I gave you

[Anonymous No.15663459]

>>15663356
cos(0)

[Anonymous No.15663468]

>>15663456
I've agreed to it long time ago

[Anonymous No.15663476]

>>15663468
So saying it doesn't end in 0 is vacuous.
It doesn't end in anything so there is nothing but 0.

[Anonymous No.15663485]

>>15663476
No. There is not even 0 because if there was, it would mean that it had an end

[Anonymous No.15663489]

>>15663485
If something is always 0 then how can it not be 0?

[Anonymous No.15663494]

>>15663489
>If something is always 0
What something is always 0 ?
Elaborate

[Anonymous No.15663500]

>>15663494
The numbers you get when you subtract 1.000... - 0.999...
Impossible to get any number except 0

[Anonymous No.15663508]

>>15663500
But
>We already agreed it doesn't end
said you.

And
No. you take 9 out of 10 and are left with 1 which expands to next digit which makes it 10 again where you then again take away 9
Or in short, strings of 9's taken away from strings of 10's
Which doesn't end in 0

[Anonymous No.15663522]

>>15663508
>Which doesn't end in 0
It doesn't end in anything. This is a vacuous truth.

1.000... - 0.999.. = X.XXX...
No X can be anything except 0.

[Anonymous No.15663533]

>>15663522
Except every calculation for a string gives you 1 which you are clearly forgetting

[Anonymous No.15663550]

>>15663533
The 1 can never be an X.
It's like musical chairs. The 1 will never find a place sit anywhere. Always 0

[Anonymous No.15663565]

>>15663550
>The 1 will never find a place sit anywhere.
Except every step

[Anonymous No.15663575]

>>15663449
10^(-infty) isn't a number, it's just an abuse of notation that is sometimes used in limits. 0.0...01 is a finite number, which means that there is a finite number of 0's, say n. Then, 0.0...01 = 10^(-n), which is definitely greater than 0. There's nothing to prove there.

[Anonymous No.15663581]

Pardon the interruption
Why is .99999... expressed with ellipses and not the symbol for infinity ∞ ?

[Anonymous No.15663583]

Scientific notation already btfos 0.99...≠1 fags

We all know every digit in a decimal sequence corresponds to a power of 10. Until they can explain what power of 10 the "1" in 0.00..1 corresponds to, they aren't worth the time of day

[Anonymous No.15663586]

>>15663565
Which step? X is never 1 at any step.

[Anonymous No.15663590]

>>15663586
For every 10 - 9

[Anonymous No.15663603]

>>15663590
So 1.000... - 0.999... = 1.111... ??

[Anonymous No.15663613]

0.999... just approaches 1

[Anonymous No.15663616]

>>15663603
1 - 0.9 = 0.1
0.1 - 0.09 = 0.01
0.01 - 0.009 = 0.001
etc.

[Anonymous No.15663621]

>>15663613
0.999... is always indistinguishable from 1.

[Anonymous No.15663622]

>>15663575
>10^(-infty) isn't a number
0 is a number
> 0.0...01 is a finite number
it's a bs notation
only non-insane interpretation is to ignore anything after the "..." since anything beyond that is multiplied with 10^-inf=0

[Anonymous No.15663626]

>>15663616
The 9s all end there. We agreed that the 9s don't end.

[Anonymous No.15663632]

>>15663626
So you keep doing it forever and there is always that 1 which allows it

[Anonymous No.15663637]

>>15663632
The 1 can never be part of the answer. The answer can only have 0s that don't end.

[Anonymous No.15663640]

>>15663621
Yeah, but the moment you start with "0.9" then that is not 1. It gets infinitely close to 1, but it isn't 1.
>but in any practical situation-
In any practical situation there is no infinity of anything and 0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 also equals 1 even though it's a distinctly finite number.

[Anonymous No.15663642]

>>15663616
1 - 0.9... = 0
0.1 - 0.09... = 0
0.01 - 0.009... = 0
etc.

[Anonymous No.15663644]

>>15663637
>The answer can only have 0s that don't end.
By the power of the 1 that drives it

[Anonymous No.15663645]

>>15663640
>practical
>math
lol idiot

[Anonymous No.15663649]

>>15663640
1 and 0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 are distinguishable. 1 and 0.999... aren't.

[Anonymous No.15663655]

>>15663644
It is a powerful 1 that nevertheless has no place to sit in the answer, always and forever. Musical chairs.

[Anonymous No.15663657]

>>15663613
0.999... is static, it isn't growing/approaching
there are infinite 9s from the get go

[Anonymous No.15663659]

>>15663649
>1 and 0.999... aren't distinguishable
Really? Because 0.9 and 1 are very clearly distinguishable.
>but it changes when you add infinite 9s to the 0.9
No. You started with a number that isn't 1. Unless that number changes to 1 then it is not 1. Simple as.

[Anonymous No.15663660]

>>15663659
What's the difference? Simple as.

[Anonymous No.15663666]

>>15663655
>It is a powerful 1 that nevertheless has no place to sit in the answer
Well i guess you have to prove it now that it does not sit in the answer

[Anonymous No.15663668]

>>15663660
0.9 isn't 1, no matter how many 9s you add to it.

[Anonymous No.15663671]

>>15663668
0.9 ends. If the 9s end, then there's a difference. If they don't, there's not.

[Anonymous No.15663678]

>>15663666
I think you rather have to prove that it does.

[Anonymous No.15663684]

>>15663671
Does 0.999... say 1? No. It says zero dot nine nine nine dot dot dot which is not one. Therefore they are not the same.

[Anonymous No.15663686]

>>15663678
I already did
>>15663533

[Anonymous No.15663690]

>>15663686
So 1.000... - 0.999... = 1.111... ??

[Anonymous No.15663692]

>>15663690
>>15663616

[Anonymous No.15663693]

>>15663659
>>15663660
nta
>>15660388
>The approach to 1 eventually gets so miniscule .99999... is generally accepted to be equal to 1 for all intents and purposes. No one actually knows because no one can prove it. Remember, it just keeps on going forever.
But I should have said it goes on forever we think, because we really don't know. It's scary to think about, but at some distant point it could stop or (shudders) loop back. We don't know, hence the generally accepted and the for all intents and purposes parts.

Why use ellipses, not the symbol for infinity?

[Anonymous No.15663695]

>>15663684
I certainly agree that the spellings are different.

[Anonymous No.15663698]

>>15663692
All of those have 9s that end. We agreed the 9s don't end.

[Anonymous No.15663706]

>>15663698
They dont end but for every step they create a new number that is 1 which drives the reoccurrence

So the 1 sits in the answer

Now this is for you >>15663666

[Anonymous No.15663717]

>>15663706
>So the 1 sits in the answer
Where in the answer? Which X does it sit in?

[Anonymous No.15663727]

>>15663717
Dont bring algebra into talk about decimals, left to right counting and powers of 10. You are only complicating things

But anyway.. for every X the predecessor was 1
So it sits in every X before any X

[Anonymous No.15663751]

>>15660871
why are you all so fucking stupid?
for something to be a number means: to occupy a certain position on the number line
.999....9 occupies the same position as 1, and so it's the same number by the above definition
if you don't get this, you literally don't understand the first thing about mathematics (what a number is)

[Anonymous No.15663756]

>>15663727
Algebra?
If what you're saying is true, then 1.000... - 0.999... = 1.111...
This is an exceptional statement and certainly requires a proof.

[Anonymous No.15663783]

>>15663756
See there is this thing called 10 - 9 which gives you 1
You need to understand this system before decimal system because it predeceeds it
Now the concept of 10 - 9 = 1 gets introduced into decimal system which gives you 1.0 - 0.9 = 0.1
Because there is a leftover of 0.10 (SIPLIFIED FOR UNDERSTANDING THAT WE ARE STILL WORKING IN POWERS OF 10), you do it again as 0.10 - 0.09 which gets you to 0.01
Now we have reached a loop and for every step the loops repeats coherently.. leaving us number 1 to upkeep the loop

Thats the proof that there sits 1, now where is your proof that 1 doesn't sit there ?

[Anonymous No.15663789]

>>15663622
An incorrect one. It should be 0.00..., which is infinite. Yours shows a finite number.

[Anonymous No.15663794]

>>15663783
But the "leftover" 1 never sits anywhere. The first X isn't 1. The second X isn't 1. As far as I go, all the Xs are 0. Why would that change?

[Anonymous No.15663836]

>>15663794
>The first X isn't 1. The second X isn't 1.
You cannot cheat your way through here, you already introduced a system that comes way after the system we are currently figuring
Every X is 1 before they are 0 if you just understand to do things left to right

[Anonymous No.15663876]

>>15663836
I didn't introduce any system. We're subtracting 1.000... - 0.999... to find X.XXX... I can't find any X that isn't 0. (I don't think you can, either.)

You have a 1 that you are very attached to. You don't want to lose it. But it never sits in any X. You have an emotion for this 1. Emotion doesn't change what sits in the X. Let it go.

[Anonymous No.15663889]

>>15663876
>I didn't introduce any system
>to find X.XXX..

But you did since you are not finding anything, you just start to solve from left to right
>lets go..

[Anonymous No.15663893]

>>15663889
I found that every X = 0. Did you find a different vlaue for X?

[Anonymous No.15663910]

>>15663893
Yes
X = 1 for step1 and for step2 that X = 1 turns to 0 while second X turns to 1 and so on

[Anonymous No.15663912]

>>15656622
No, you cant divide by zero

[Anonymous No.15663926]

>>15663910
>X = 1 for step 1
You found that 1.000... - 0.999.. = 0.1XX...
It has a 1 in the first decimal ??
Really?

[Anonymous No.15663930]

>>15663912
of course you can.
1/0 = 999...

[Anonymous No.15663932]

>>15663930
You mean ...999 ?

[Anonymous No.15663939]

>>15663926
Yes
Then you do the second step and you have 0.01XX..

[Anonymous No.15663961]

>>15663939
Then there's no 1 in the first decimal.

[Anonymous No.15663972]

>>15663961
But there is in the second, and the 1 sits there for every step

[Anonymous No.15663981]

>>15663972
But it never sits there. We agree that the answer to 1.000... - 0.999... must have a 0 in the first decimal place, true?

[Anonymous No.15663982]

>>15663356
you can think of n^0 as the empty product, the multiplicative version of the empty sum
https://en.wikipedia.org/wiki/Empty_product
https://en.wikipedia.org/wiki/Empty_sum
best wishes

[Anonymous No.15663983]

Two real numbers, say x and y, are different if there exists a real number u > 0 such that abs(x - y) = u. Just show that there's no such u when x = 1 and y = 0.999...
Can you do it, /sci/?

[Anonymous No.15663992]

>>15663581
because they are not the same thing

[Anonymous No.15663995]

>>15663981
Well X -> 1 -> 0 is two steps
X -> 1 is a step that preceeds 1 -> 0
So it definitely sits there

[Anonymous No.15663997]

>>15663583
scientific notation can only represent finite fractions, this is a very well known fact, so much so that the french have a set between Z and Q called D for such fractions

[Anonymous No.15664004]

>>15663995
It doesn't sit there. It never finds a chair.
But do we agree that the answer to 1.000... - 0.999... must have a 0 in the first decimal place? or do we disagree?

[Anonymous No.15664014]

>>15663684
wait till you learn about other languages besides english

[Anonymous No.15664022]

>>15664004
It sits down every single step though
I agree that any step greater than 2 it is 0, but the 1st step has to be accounted for
Just because X -> 1 -> 0 is always 2 steps, you are probably going to say that when the steps approach infinity or something and then i can say that you are mushing those 2 steps into 1 again and forgot to do it step by step

[Anonymous No.15664023]

>>15663932
why would 1/0=-1

[Anonymous No.15664024]

>>15663992
Well, thanks. A search was only a little less satisfactory. But you did try, appreciate it.

[Anonymous No.15664031]

>>15663983
the lurkers certainly can, the mentally i'll freaks that keep making these threads sure as hell can't, and as for your average poster, who can really tell

[Anonymous No.15664038]

>>15664024
ah, you where asking as to why we use "...", sorry, these type of threads really burn you out, there are more notations beside that one, hope this helps
https://en.wikipedia.org/wiki/Repeating_decimal

[Anonymous No.15664044]

>>15664038
Quick thanks, heading over there now.

[Anonymous No.15664050]

>>15664022
I'm not mushing anything. I know that the first X can't be 1. I know that the second X can't be 1.
This is all absolutely finite.
I know that once an X can't be 1 it can never be anything but 0. And I know that every X I can get to can't be 1.
Why would this change?

[Anonymous No.15664054]

>>15664023
why would ...999 ≠ -1 ?

[Anonymous No.15664063]

>>15664050
>I know that the first X can't be 1.
It can, for the first step of the calculation

>I know that once an X can't be 1 it can never be anything but 0
There is the mush, you are taking 2steps at once

[Anonymous No.15664074]

>>15664054
i make no such claim, and in fact it is(in the 10-adics), i do find weird the assertion that 1/0=...999

[Anonymous No.15664085]

>>15664063
It could be 3 steps. If I can see that the X always ends up at 0 no matter how far I go, then I see that it always ends up at 0.

Your leftover 1 is a beautiful number but it just doesn't sit anywhere.

[Anonymous No.15664089]

>>15664074
So what's the beef?

[Barkun No.15664090]

kys now

[Anonymous No.15664096]

>>15664038
Bookmarked so I can read it again later. Look what I found:
However,everynumber with a terminating decimal representation also trivially has a second, alternative representation as a repeating decimal whose repetend is the digit9. This is obtained by decreasing the final (rightmost) non-zero digit by one and appending a repetend of 9. Two examples of this are1.000... = 0.999...and1.585000... = 1.584999.... (This type of repeating decimal can be obtained by long division if one uses a modified form of the usualdivision algorithm.[2]
1.000...= 0.999...

[Anonymous No.15664100]

>>15664085
It sits every step though
X -> 1 -> 0
≠
X -> 0

[Anonymous No.15664104]

>>15656868
but here we dont have equality but just the limit, this would be like saying that sequence of 2^-n is also 0 since its limit is 0

[Anonymous No.15664110]

>>15664100
That describes how it fails to sit. I agree.

[Anonymous No.15664119]

Let x = 0.99999…
10x = 9.99999…
10x — x = 9.00000…
10x — x = 9
9x = 9
x = 1
took me liek two mintues

[Anonymous No.15664135]

>>15664110
It literally shows you that it sits there

[Anonymous No.15664141]

>>15664135
It literally shows you that it doesn't sit there.

[Anonymous No.15664150]

>>15664141
It shows you that X -> 1 -> 0 is 2 steps
Before X can become 0 it has to be 1 first

Are you even interested in math ? You have not shown anything mathmatical yet but are just pushing arguments

[Anonymous No.15664155]

>>15664150
It shows that every X = 0

>Are you even interested in math ? You have not shown anything mathmatical yet but are just pushing arguments
Stop with this.

[Anonymous No.15664204]

>>15663459
please, elaborate

[Anonymous No.15664232]

>>15663356
>how many ways can you choose nothing
It's always 1.

[Anonymous No.15664278]

>>15664038
I just saw this article
>https://en.wikipedia.org/wiki/0.999......
and it's such brainrot full of flawed "proofs" for something that is wrong. Most of them have been regurgitated in this thread and debunked by me.

[Anonymous No.15664345]

>>15664278
1 and 0.999... can't be different because the difference can't be anything other than 0.

[Anonymous No.15664364]

>>15664278
whatever pigeon, im glad that your lot in life is to be retarded

[Anonymous No.15664370]

>>15664089
this post >>15663930

[Anonymous No.15664389]

>>15664370
Not my post. 1/0 ≠ 999... but = ...999

[Anonymous No.15664419]

>>15656622
Writing a fraction whose denominator isn’t a factor of ten in decimal requires recurring decimals, which represent the limit of the sequence you get by cutting off the decimals at a finite point. For example, 1/3 = 0.333… because the sequence 0.3, 0.33, 0.333, … converges to 1/3. Since 0.9, 0.99, 0.999, … converges to 1, 0.999… = 1. To suggest otherwise necessarily requires you to write many fractions in the numerator-denominator form, since by the same standard 0.333… ≠ 3.

[Anonymous No.15664425]

>>15664419
*1/3