Please, please tell me no one on /sci/ actually fell for the "0.999999... = 1" meme.
>>8109416
Well, prove that it isn't.
Show that if you assume 0.9999... = 1 then that leads to a contradiction somewhere.
pro tip: Make sure your argument does not have contradictions itself.
>>8109440
>hurr you have to prove a negative so that I dont have to prove a positive durrrr
Fuck you. You support the claim, burden of proof is on you.
>>8109446
>burden of proof is on you.
It used to be. Then people came up with a billion proofs for the proposition.
Now that we have a proof that works, the burden of proof is on you as you are now trying to dismiss a well established claim.
Faggot.
Go ahead, this should be easy. Just type up one single number that is between 0.999... and 1.
If there is even one then that would immediately imply that 0.999... is not 1 so go on.
Type it up faggot.
>>8109451
There is none. At the scale of one-infinitieth, numbers are discrete; there is no real number between 0.999... and 1 for the same reason there is no integer between 0 and 1.
LITERALLY YOU:
>You want to prove 0 and 1 are not the same integer? Well, go on then, this should be easy - all you need to do is type up one single integer that is between 0 and 1. If there is one, it would immediately imply that 0 =/= 1. Go on. Do it faggot.
1/9 = 0.1111111111111111111111111111...
0.1111111111111111111111111... * 9 = 0.999999999999999999... = 9/9 = 1
>>8109416
its true bruh. the maths checks out. .9999... = 1 is true by definition. Mathematicians could have assumed it isn't true, but that would lead to algebra breaking down.
>>8109454
>there is no real number between 0.999... and 1
and that implies that 0.999... = 1
if you have 2 real numbers, a and b and
(a + b)/2 = a then that implies a = b
but if you have
(a+b)/2 = c then a is not equal to b as now it has a midpoint which tne implies an infinity of intermediate points.
It is my pleasure to teach you pre kinder math. Keep studying and one day you may stop being a faggot.
>>8109416
0.99999....=1 when taking account significant figures.
>>8109451
No, fuck you. You can't just equate two completely different, distinct, and defined numbers because "well, they're close enough lol." Is 1.111... equal to 1.2? Would you get the same answer for an equation if you were to swap out .999... for 1, or vice versa? No, you wouldn't. One would give you a hard number, the other would give you an asymptote. Those are not the same thing.
I bet faggots like you also think morality is relative.
>>8109416
here's the definitive proof op. unless you think the meme spread to wikipedia...
https://en.wikipedia.org/wiki/0.999...
>>8109463
>he thinks Wikipedia is an objective source
>he thinks Wikipedia isn't pushing an agenda
How fucking stupid can one person get?
>>8109461
>is 1.111... equal to 1.2
no, because 1.12 comes between the two. there are an infinite number of numbers between 1.1111... and 2.
Assume .9999... and 1 are two different numbers op. what's 1 - .9999... ?
>>8109461
>Is 1.111... equal to 1.2?
No because 1.1112 is between those two numbers. And infinitely many more are aswell.
>>8109466
0.000...001
You're comparing two universes
>an inifinite number of paper slips with 0 written on them plus one paper slip with 1 written on it
>an inifitine number of paper slips with 9 written on then plus one paper slip with 0 written on it
They are not the same because they're written differently.
>>8109416
[math]\mathbb{R}-{0}[/math] is a group under multiplication.
[math]0.999... \times 1 = 0.999...[/math]
[math]0.999... \times 0.999... = 0.999...[/math]
Substituting,
[math]0.999... \times 1 = 0.999... \times 0.999...[/math]
Left cancelling,
[math]0.999... = 1[/math]
>>8109466
>no, because 1.12 comes between the two. there are an infinite number of numbers between 1.1111... and 2.
Yeah, I just noticed I made a typo--meant 1.12.
>Assume .9999... and 1 are two different numbers op. what's 1 - .9999... ?
0.00000...1? Is that supposed to be a trick question? Because the answer is very clear. It's certainly not 0, it's an asymptote forever approaching 0. Nowhere near the same thing.
What is the pentium bug for $100, Alex.
>.99999....≠1
i'm wondering if anyone has actually argued this or seen someone argue this in a university math class. if you guys have any stories, please share.
1 - 0.999... = 0.00...1
1 - 1 = 0
0.00...1 != 0
>>8109479
>Yeah, I just noticed I made a typo--meant 1.12.
no i made the typo. there are an infinite number of numbers between 1.1111... and 1.12. there are no numbers between .9999... and 1.
>0.00000...1? Is that supposed to be a trick question? Because the answer is very clear. It's certainly not 0, it's an asymptote forever approaching 0. Nowhere near the same thing.
asymptotes DO hit 0 after an infinite amount of time. That's an extremely important thing. calculus wouldn't work otherwise.
the answer to the question is 1 - .9999... = 0. you can't have .00000...1. how can there be an infinite number of 0s and THEN 1? if you're putting a 1 at the end of your 0s then you don't have infinite 0s.
here's a cambridge mathematician explaining it. has the meme spread to research mathematicians now?
https://www.youtube.com/watch?v=G_gUE74YVos
>>8109497
If you admit that there is a number 1 at the """end""" of the number 0.000...1, then the number 0.999... does not expand forever.
Go learn about infinity again.
We have this paradox because real numbers are imperfect. It's just a quantifying approximation of continuous line
>>8109446
But it's a very easy proof and you can Google it.
op is either a troll or he's just started high school maths and thinks his limited knowledge of quadratic equations makes him knowledgable in maths.
[math](0.999...)^\infty = 0[/math]
[math]1^\infty = 1[/math]
retards
>>8109416
This is a notational problem. Notational problems, like all problems of language, are resolved by appeal to consensus.
When most mathematicians talk about "0.999 repeating", they are referring to a particular construct, a particular definition, where it is in fact equal to 1.
Further, one of the properties of the Real Numbers is that two real numbers are inequal if and only if there is a third Real number in between them. By appealing to informal arguments, "0.999 repeating" names a Real number, and there is no Real number between "0.999 repeating" and 1, and therefore they name the same Real number.
>>8109505
>how can there be an infinite number of 0s and THEN 1?
[math]1+x+x^{2}+... = 1/(1-x)[/math] right retard? how can you have an infinite series equally a single term?
>>8109416
1/3 = .3333333333333333333....
So 3 * .333333333333333333...
=1
>>8109632
>1/3 = .3333333333333333333....
no, 1/3 = .333333... plus the 0.00...00000 and 1/3 at the end.
>>8109627
Welcome to calculus.
[math] \displaystyle
1 = \frac{3}{3} = 3 \cdot \frac{1}{3} = 3 \cdot 0. \overline{3} = 0. \overline{9}
[/math]
>>8109652
1/3 =/= 0.3333...
>>8109655
yes it does yuo fucknard
>>8109655
but it literally is
>>8109490
The only story I have was intro to analysis where the lecturer seemed kind of disappointed nobody objected to his claim that (0,1) had no maximum.
>>8109633
>the 0.00...00000 and 1/3 at the end.
There is no such Real number. If you want to talk about some new kind of please, please give your definitions of terms.
>>8109472
You don't understand how infinity works
>>8109692
i'm not talking about kinds of numbers, i'm just talking about plain old numbers, the kind we all use, not some arbitrary wishy-washy alternative universe math from the ivory towers
>>8109699
Ok. We're interested in formal analysis, not "it makes sense to me". Might I suggest /x/ or /b/ ? Maybe /lit/ ?
>>8109703
i am doing a serious analysis, go to /lit/ if you want a safe space from the hard truths of math.
>Comparing infinite sums with natural numbers
No
0.99999... = 1 is true only in the same way 1/0 = infinite
You are not using math correctly
Go home
0.9999... = x
10x = 9.9999... / -x
9x = 9 /:9
x = 1
ez
>>8109711
this, there is no infinity in the really existing numbers, you can't take "limits" without the contradictory framework of calculus on top
1 - 0.999... = 0.000...1
you can't just choose to apply infinity when it suits you faggots
>people seriously responded to this thread without saging
?
>>8109712
wtf is /:
>>8109472
If 0.00...001 is a nonzero real number what is its reciprocal?
>>8109717
0.999... literally means infinite 9's... You can't just say something is not infinite when it literally is. Pro tip, decimal expansions are not unique. It is something you have to accept in real analysis.
>>8109627
Do you even realize the argument you have right there proves 0.999... = 1? Write 0.999... as a summation of fractions and literally plug it into that equation right there.
>>8109416
So you're going to tell me straight faced that decimal expansions are unique? Give me the proof for that and then you'll have a basis for dismantling very well established mathematics
>>8109472
>>8109497
>>8109717
>0.000...1
>An infinite sequence with something after it.
Jesus. Are you even trying?
In any case, this seems like it's obvious shit being over-thought.
The sequence 0.9, 0.99, 0.999, 0.9999, ... obviously converges to 1 as the number of nines goes to infinity. 0.xxx... is notation that signifies the number of x's goes to infinity.
Problem solved: 0.999... = 1
>>8109781
>Do you even realize the argument you have right there proves 0.999... = 1? Write 0.999... as a summation of fractions and literally plug it into that equation right there.
geomtric series doesn't work like that
I honestly can't tell if people agreeing with OP are trolling.
>>8109793
don't you think trolls would target more serious people and issues? and not brilliant /sci/entists who, since they're so quick-minded, would pick up on trolling very quickly?
>>8109783
0.9/1 = 0.1
0.99/1 = 0.01
...
0.999.../1 = 0.000...0001
i don't get what's so hard about this, it's like you're all stupid and can't do math
>>8109655
prove it
Infinity is a concept. We can arbitrarily assign whatever properties we want to the concept of infinity. The property P(x) together with the inference rules of mathematical logic dictates that 0.999... = 1. That is enough for most mathematicians.
We do not know if infinity as such is metaphysical possible, but it does not really matter. Does the concept of infinity outside of mathematics have the property P(x)? Who cares... in mathematics it does (because we have we have defined it that way) and most mathematicians are in agreement that P(x).
>>8109836
1/3 is a fraction, 0.3333... is a decimel. There's no isomorphism between rationals and reals... so therefore there's no way to map. it makes sense because with decimels everything is counting in 10s, but you can count in 3's with fractions. 10/3 is undefined in decimal
>>8109855
>There's no isomorphism between rationals and reals... so therefore there's no way to map.
>all morphisms are isomorphisms
Excellent argument. Go back to school, anon.
X=0.999
X10=9.999
X10=9+0.999
X10=9+x
X9=9
X=1
Right?
>>8109416
here, let's do it informally OP.
suppose we have two numbers, x and y. can we agree that if there is no number z such that x < z < y, then x = y? okay, good.
can you find a number inbetween 0.9999... and 1.000...? no? then they are equal.
>>8109873
0.999 * 10 = 9.990 you dumb fuck
>>8109864
that's not what i said.
>>8109882
It is an assumption made to move forward with your argument. Maps can be injective but not surjective. Do you have something useful to say?
>>8109891
You can map 1/3 to 0.3333.. but they aren't the same number.
>>8109901
uh, anon...
>>8109416
t. Norman Wildberger
>>8109922
never thought i'd have to say this to someone who knew what the word "isomorphism" meant, but i guess you really didn't
http://math.stackexchange.com/questions/335560/is-1-divided-by-3-equal-to-0-333/335561
>>8109901
Changing the subject to a new argument without conceding to your clear mistake? I don't have time for this, feel free to educate yourself rather than putting full faith into the first idea that you consider without thoroughly testing its viability.
>>8109626
No one responded to this. How is this in any way a dismissible response?
>>8109953
Because the trolls in this thread are ignoring arguments they can't counter without making it too obvious that they are trolling.
>>8109953
I'm pretty sure 75% of the people in this thread already have that idiot filtered
>>8109715
Go to bed Norman
>>8109828
> 0.9/1 = 0.1
>Not 0.9
Pic related
>>8110058
Clearly he meant set minus on the dedekind cuts with the parameters reversed, right?
>>8109828
> A sequence of infinite zeros followed by a one
That's not a real number anon. When the hell are you going to reach the end of an infinite sequence to put a one at the very end?
Also here's a simple proof that .9999...=1:
.9999... = 9/10+ 9/100 + ... + 9/(10^n) (n is a member of the natural numbers)
Hence, using the sum of an infinite geometric series:
.999... = (9/10)/(1 -(1/10)) = (9/10)/(9/10) = 1
I seriously don't get how hard this is to grasp. There's certainly a plethora of strange shit in the many fields of math but this isn't one of those things.
>>8110065
>Also here's a simple proof that .9999...=1
you cant prove that without first saying which number system you're using.
if you're using the reals, the answer is so blatantly obvious it defies belief anyone could even beg the question in the first place. All we have to do is check whether the sequences 0.9, 0.99, 0.999... and 1, 1, 1, ... are equivalent cauchy sequences of rational numbers, which they fucking obviously are.
However, there are number systems where 0.9999... is not equal to 1. The main point to note here is that to this day I have not heard of a single reason why anyone should care about these extensions of the reals. Obviously we can do calculus just fine without the need for actual infinitesimals, so why bother with these things?
So that fucktards on 4chan my troll other fucktards?
literally the only reason I can think of
>>8110106
True. I was just implying we were in the reals with respect to the proof I gave.
>>8109709
I'm curious, you can't conceive infinity, what's the biggest number you can think?
>>8109731
>>elementaryschool
(0.9999....∞) + (0.0000...∞1) = 1
:^)
Are we getting invaded by /v/ or something?
>>8109416
tell me a number that fits in between 0.9... and 1 and ill accept that they're not equal
>>8109514
No, I agree there are infinitely many 9s in 0.999.... But when you take 1-0.999..., you get 0.000...1, where you have a 1 in the [math]\omega[/math] decimal point.
>>8109766
1...000, i.e. a 1 followed by infinitely many 0s to the left of the decimal point. Come on, this isn't even difficult.
>>8110356
How about 0.999...5?
>>8110359
0.000...1 is nonsensical. So a infinite expansion of zeros ends with 1? Basically, you are telling us that an infinite decimal expansion is finite. Nice contradiction.
>>8109532
>someone disagrees with me and claims two completely-different numbers are not equal to one another
>h-he must be in high school!
Fuck off, you walking testament to eugenics.
>>8109876
I understand where you're coming from, but one of those numbers is still defined as being higher than the other, regardless of the fact that it's only by a very miniscule amount. If I show you a step on a staircase and then point to the one above it and say, "There are no steps between this step and the one before it, therefore they are at the same elevation," I would be wrong. An asymptote is not the same as the number it is approaching. It will, by its very nature, never actually reach that number, but merely get closer and closer to it. To equate the two is just silly. Yes, the fraction multiplication is a neat trick, but that's all it is--a neat trick. These are still two different numbers with two different values.
>>8109478
>0.999...×1=0.999...×0.999...
How is that possible? Isn't .999×.999 != 1? I'm not that one guy bitching, I'm lurkin and think this is fascinating
>>8109686
Where the flying fuck do you get this shit from?
>>8110402
You're confusing a set being well-ordered with being finite. 0.000...1 has (countably) infinitely many 0s before the 1, and then a 1 afterwards. Are you disagreeing that the decimal points of a number are well-ordered, that it's possible to have an infinite well-ordered set, or that it's possible to have an well-ordered set where one element is larger than infinitely many other elements? In any case, you're wrong: these are well-known results in set theory.
X = 1
1/3 of x * 3 = 0.333... * 3 = 0.999...
X = 9
1/3 of x * 3 = 3 * 3 = 9
See?
>>8110356
That's like saying "Tell me an integer that fits between 1 and 2 and I'll accept that they're not equal".
Why would two "consecutive" (I know the word doesn't really apply to the set of real numbers, but I'm not sure what you'd call this) real numbers have to be equal? Although there are other proofs for 0.999...=1, this one's fairly weak.
>>8110417
If it ends it's not infinite.
actually 0.333... * 3 = 1, not 0.999...
This 0.999... = 1 is one of the most annoying memes in math and it only works if you ignore constraints inherent to infinities.
>>8110483
That's not even remotely true. The interval [0,1] ends at 1, but there are undeniably infinitely many points in between (including, for example, both 0.999... and 1).
>>8110497
Please do not confuse countable and uncountable infinities.
Infinitesimals don't really exist (even less so than infinities).
>>8110507
The distinction between countable and uncountable infinities is irrelevant here. Replace my example with the set of rational numbers between 0 and 1 inclusive and it's still just as true.
>>8110527
You're right, though my second objection still stands. Infinitesimals don't real.
>>8110531
K, they aren't "real" but they exist in certain sets. What's you're point?
>>8110478
Because if there were no number between 0.999... and 1 it would contradict some very important properties of real numbers.
>>8110551
*no number between, and they were not equal
>>8110531
Why not? They're used all the time in mathematics. Arbitrarily cutting off decimals at a certain point to remove the idea of infinitesimals is arbitrary and does nothing but lead to useless, counterintuitive "proofs" like 0.999... = 1.
>>8110551
Try 0.999...5. Clearly 0.999...5 is more than 0.999... because of the extra 5 at the end, but it's still not quite equal to 1. So 0.999... < 0.999...5 < 1.
>>8110557
0.999...5 is less than 0.999...
>>8110559
Nah. 0.999...5 - 0.999... = 0.000...5 > 0, so 0.999...5 > 0.999....
>>8109679
>that (0,1) had no maximum.
But it does. It's 1.
0.999.... Is obviously within (0,1), but according to this thread, 0.999... = 1.
>>8110557
0.999...995 is exactly 0.000...004 less than 0.999...
Regardless, you have yet to disprove that 0.999 = 1 despite numerous proofs offered. Please provide a single proof for your counterclaim if you wish to be taken seriously.
>>8110562
0.999...9995 doesn't exist. An infinite sequence that terminates is a contradiction in terms.
>>8110565
>0.999...995 is exactly 0.000...004 less than 0.999...
Don't be silly, I'm talking about a decimal with a 5 in position [math]\omega+1[/math], not a decimal with a 5 in position [math]\omega[/math]. Of course you can also have a number smaller than 0.999... which has a 5 in it, but that's a different number (off by 0.000...45, specifically).
>Regardless, you have yet to disprove that 0.999 = 1 despite numerous proofs offered. Please provide a single proof for your counterclaim if you wish to be taken seriously.
As stated on many previous occasions, 1-0.999... = 0.000...1 > 0, so 1 > 0.999... and it's therefore impossible that 1 = 0.999...
>>8110572
Again, there are infinitely many 9s before the 5. The list of 9s does not terminate, yet there is still a 5 after all of them. It's entirely possible for a set to be both well-ordered and infinite, as is the case here with the set of decimal points of 0.999...5.
>>8110572
>infinite sequence
infinite sequence doesn't exist.
>>8110582
>There are infinitely many 9s before the 5
Then there is no 5.
>>8109454
>argumenting against yourself
are you retarded?
>>8110587
This is a basic set-theoretic idea I'm talking about here. There is a 9 in decimal positions 1, 2, 3, and so on for all positive integers. Then there is a 5 in an index after the index corresponding to each such positive integer. If you're unfamiliar with the idea of a well-order, here's the wiki page: https://en.wikipedia.org/wiki/Well-order.
Believe it or not, it is possible to have an infinite well-ordered set. And furthermore, the decimal positions of a number must be such a set. They have to be well-ordered if we want decimal numbers to have any meaning, and they must also be infinite if we want to allow constructions like 0.999... to exist in the first place.
Nice bait OP it's worked thus far
>>8110600
Well if we're dropping to Wikipedia citations, here's their opinion on 0.999...
https://en.wikipedia.org/wiki/0.999...
Eight whole proofs for you to attempt to refute.
>>8110616
Most of these "proofs" are completely lacking in rigor. They just assume you can carry out the usual arithmetic on infinite series and infinite decimals, and then arrive at absurd results. The fraction "proof" relies on the equivalent false assumption that 1/9 is exactly equal to 0.111..., while more formal "proofs" like Dedekind cuts and Cauchy sequences rely on shaky notions of limits.
How about you try to refute my proof that 1 > 0.999...?
>>8110637
If 1 > 0.999..., then that would mean 1 is not equal to 0.999...
But it is, and any college math professor, or even any college math student can tell you that. Not that I expect you to be at college, but a high school math teacher should be able to prove it too if you live in a country where teachers need to have a university degree.
>>8110656
> If 1 > 0.999..., then that would mean 1 is not equal to 0.999...
I agree. 1 is not in fact equal to 0.999....
>But it is, and any college math professor, or even any college math student can tell you that. Not that I expect you to be at college, but a high school math teacher should be able to prove it too if you live in a country where teachers need to have a university degree.
Nice appeal to authority - hypothetical authority, even. If it's so easy to prove, why can't you do it?
>>8110420
No, that is not how it works. If 0.999.. doesn't equal 1 then things like completeness are contradicted. Learn some math before you start talking about it
0.999 is not equal to 1 indeed
however, the problem is where that 0.999... comes from
0.9 != 1
9 != 10
99 != 100
999 != 1000
...
0.999... * n != 1 * n
>>8110684
>>8110420
Furthermore, you used absolutely no axioms or theorems in analysis or mathematics in general to base that argument. Also you didn't argue why the existence of YOUR stupid number 0.0000......1 and 0.999.... proves 0.999... != 1. That proof is 0/10. If you insist that 0.9999... != 1, then look up surreal numbers and use those instead. The real numbers do not hold decimal expansions to be unique so you're shit out of luck there.
>>8109878
The .999 actually is a simplified notation for the infinite series of .99999999999... you fucking retard. Go take a remedial algebra class to see how that IS a proof. It's a simplified NON Calculus method proof that simpletons like yourself can understand.
>>8109416
Actually, 0.999999999... = -1/12
>>8110600
You do clearly not understand what well ordered means. If A is a well-ordered set then every nonempty subset of A has a least element. Let A = {(0.999...)}, the set A has only one element.
>>8110557
>Try 0.999...5
>Clearly 0.999...5 is more than 0.999... because of the extra 5 at the end
>because of the extra 5 at the end
That's some grade A mathematical arguing right there.
Are we talking about the cardinal numbers or ordinal numbers here? We're talking about 0.9999... and 1 as real numbers, right? 0.999...5 doesn't make any sense in this context.
>>8110563
It has no maximum but it has a supremum and the supremum is 1.
>0.999.... Is obviously within (0,1)
How do you know? Can you build a real number bigger than 0.999.... that's still in that interval? If not, then 0.99... is an upperbound and unless 1 is less than 0.999..., it is the supremum (the supremum is unique) thus they are equal. My arguement is poorly formulated but that's the basic idea.
>>8110724
Other people were saying it's impossible to have something like 0.000...1 because "you never get to the 1". My point was that you can absolutely have a well-ordered set with one value that follows after infinitely many other values, as the 1 comes after infinitely many 0s in 0.000...1. The sequence of place values 0, 0, ..., 1 (or for that matter the values 9, 9, ..., 9 in 0.999...) is the infinite well-ordered set I'm talking about.
>>8110730
0.999...5 = 0.999... + 0.000...5. I don't see what's so difficult here.
>>8110739
0.999... < 0.999...5 < 1. Prove me wrong.
>>8109416
>people still arguing whether or not 0.999... = 1
Mix it up a little: Why not argue that; (10^infinity)-1 = (10^infinity)
its the same question essentially.
>>8110768
What? For one thing, "infinity" isn't really well defined. For another thing, whatever value you assign to 10^infinity, it's obviously going to be larger than that same value minus 1.
>>8110768
But can you prove that 0.000.....0001 is equal to 0?
ITT: /sci/ learns the difference between approximately equal and exactly equal.
>>8110757
>My point was that you can absolutely have a well-ordered set with one value that follows after infinitely many other values
No you cannot when you're talking about cardinal numbers which is what we're dealing with in the real number system. What you're thinking of is an ordinal number.
>0.999...5 = 0.999... + 0.000...5
That is not a true statement. 0.000...5 is not a real number. It doesn't make sense.
>0.999... < 0.999...5 < 1. Prove me wrong.
It just doesn't make sense in the context of real numbers. You are saying some cardinal number is less than some ordinal number which is less than some cardinal number. Usually you compare things that are the same.
1 + x + x^2 + x^3 + ... = 1/(1-x): for |x| < 1. Will you refute that? It is a very elementary result. If you do, we're done here since you're throwing Analysis out the window and have no basis to talk about the real number line.
0.999.... = 1/9(1/10 + 1/100 + 1/1000 + ....) = 1/9[1/(1-1/10)] = (1/9)(9/1) = 1
>>8110582
>in position w+1
So in position w
>>8110809
>No you cannot when you're talking about cardinal numbers which is what we're dealing with in the real number system. What you're thinking of is an ordinal number.
If we're really dealing with cardinal numbers, then what cardinality does 0.5 represent, exactly?
>That is not a true statement. 0.000...5 is not a real number. It doesn't make sense.
Why not? Infinitely many 0s, and then a 5.
>>8110816
That series approaches 1 as a limit at infinity, but there's still 0.000...1 between 0.999... and the actual value 1.
>>8110819
No, [math]\omega+1[/math] is the successor to [math]\omega[/math], meaning they're not equal by definition.
>>8110816
Now in binary!
0.111...= 1(1/2+1/4+1/8+1/16+...) = (1)(1/1) = 1.0
>>8110820
I'll admit I don't know anything about Cardinal, Transfinite Cardinals etc so I'll stop talking about that.
>Why not? Infinitely many 0s, and then a 5.
Numbers like that are not defined in real analysis. It's like working with integers and using fractions as an argument.
>That series approaches 1 as a limit at infinity, but there's still 0.000...1 between 0.999... and the actual value 1.
No there isn't. A limit is defined such that for ANY value greater than zero, you can extend the series between 1 and that number. Literally in the definition it says, there is NO number between 0.999... and 1 when you take the limit.
>>8110832
Good one. Gave me a nice chuckle
>>8110827
Arithmetic operations on infinity result in infinity. Don't be so trash.
>>8110832
>forgot a step
0.111...= 1(1/2+1/4+1/8+1/16+...) = 1[1/(1-1/2)] = (1)(1/1) = 1.0
>>8110850
I'm not sure what the point is that you're trying to make. I know you're trying to say:
0.9999... = 0.9 + 0.09 + 0.009+ .. = 9/10 + 9/100 + 9/1000 + ... = 9(1/10 + 1/100 + 1/1000 + ...)
is not true but I assure you it is.
>>8110842
A limit is defined using delta/epsilon formalism. I'm not sure what you mean by "you can extend the series between 1 and that number". Certainly there's nothing within the definition of "limit" itself that says there's no number between 0.999... and 1.
>>8110849
Only for infinite cardinals. But these are clearly ordinal numbers because they're indexing the decimal places of a real number.
>>8110860
If you're going to use "avanced" math like ordinals don't simultaneously claim that .(9) != 1
>>8110860
If you're using functions, yes. For sequences you use epsilon/N (almost the same so I will use that to illustrate my argument). The lack of formalism will hurt me a little bit it doesn't seem like you care anyway. 0.999... is a sequence of 9's and consider any 9n the nth 9. For any N > n there is an epsilon, you will always have |0.999....9n - 1| < epsilon. Choose ANY number and you can do that. Since you can do that for ANY number, there is no number between 1 and 0.999... when you take n to infinity thus they are equal. That is the essence of the series argument.
>>8109416
Let's never talk about this shit tier,
>https://answers.yahoo.com/question/index?qid=20070620111810AAF5EZC
The number in between 0.999... and 1 is 0.000...1
Thusly.
0.999... != 1
QED.
>>8110876
But what that's actually showing is that 1 is the limit of the sequence. Yes, the LIMIT is exactly equal to 1. But that's not the same as the value 0.999..., which is not a limit but rather a real number.
>>8110614
It's always going to work. The "debate" will never end.
>>8110884
Then how would you describe 0.999... in a mathematical way? How else do you describe infinity? If you don't want to talk about infinity, then you have 0.9999.....9 which is in fact not equal to 1. I'll give you a hint about how you talk about infinity: limits and sequences.
>>8110884
0.9999... is greater than every term of the sequence and less than or equal to 1
What this implies is crystal clear.
>>8110876
I fudged this up a bit too.
>For any N > n there is an epsilon, you will always have |0.999....9n - 1| < epsilon
It should be:
For any epsilon > 0 there is an N > n such that |0.999....9n - 1| < epsilon
>>8110891
Limits and sequences aren't the best. They're unintuitive and you run into problems with undefinable real numbers and similar things. Instead we can say that 0.999... has 9s in all decimal places corresponding to positive integers. Much simpler, and it still allows for usual arithmetic operations. From these, it can be easily shown that 0.999... < 1.
>>8110898
Yes, 0.999... is bounded between the terms of the sequence and 1. So are 0.999...5, 0.999...1, and plenty of other numbers.
According to measure theory, there is a difference between 0.999999... and 1
see the difference between surely and almost surely.
>>8110909
But the sequence converges to 1 as you yourself have admitted
>>8110909
>They're unintuitive
They're fairly intuitive. Regardless math isn't supposed to be intuitive, it is supposed to be logically sound.
I'm going to stick with Analysis where you use limits to talk about infinity. You can stick with whatever it is you made up.
>Yes, 0.999... is bounded between the terms of the sequence and 1. So are 0.999...5, 0.999...1, and plenty of other numbers.
Just note ff 0.9999...5 is indeed a number between 0.999... and 1, then for |0.999...9n - 1| < 0.99...5 there is a real number epsilon = 0.99...5 such that there is no N >n satisfying that inequality. Either 0.99...5 is not a real number or you don't have limits. I know what I will choose
>>8109891
So since you can't perform a the Linear Identity to get the same value back, one is the the vector space of another
>>8110927
Not* stupid phone
>>8110912
>almost surely
P=1, with possibly nonempty complement in the sample space of measure zero
>surely
P=1, complement is the empty set
Measure is an extraneous notion here though, and certainly does not contradict a basic property of the reals.
>>8110914
He would have a point if "integers x and y are equal if and only if there is no integer z such that x<z<y" was an axiom of or equivalent to the axioms of the integers. Which it's not.
The guy is clearly just confused about the properties of R, trying to represent them with transfinite decimal expansions
>>8109455
the only true answer
>>8110913
Yes, the limit of the sequence is 1, meaning it slightly "overshoots" the value of 0.999....
>>8110923
>Just note ff 0.9999...5 is indeed a number between 0.999... and 1, then for |0.999...9n - 1| < 0.99...5 there is a real number epsilon = 0.99...5 such that there is no N >n satisfying that inequality.
I have no idea what you're trying to say here. The definition of a limit: for any given positive real number (such as 0.999...5, sure) there is a natural number N such that [math]|x_n-x| < \epsilon[/math] for all n > N. Choose 0.9...9 with n 9s as your xn and 1 as your x, and this absolutely does hold for an epsilon of 0.999...5. Literally any positive N will work.
>>8110939
I learned long ago that .999... = 1 due to how the Reals are constructed, though it must be fun to work with a new set allowing stuff like .999...001...223 and the like, when Im advanced enough I'll screw around possibly there ahaha but who knows
>>8109464
Are you trolling?
>>8110958
Exactly. So that means 0.9...9 with n 9's is closer to 1 than 0.999...5. That's what the absolute norm measures: the distance between two real numbers. How can 0.9...9n be closer to 1 than 0.999...5 when 0.999...5 is supposedly bigger than 0.999... with INFINITE 9's?
>>8110958
>meaning it slightly "overshoots" the value of 0.999....
What the fucc are you smoking?
>>8110970
Replace 0.999...5 with 1 in your argument. Clearly |0.9...9n-1| < 1 for any positive choice of N, so by your definitions 0.9...9n is closer to 1 than it is to 1. But this doesn't make sense, so there must be something wrong with your definition of distance.
Hi /sci, in calc we are doing infinite series. Just came across the problem in my textbook.
The pic is the problem and the answer given in the back of the book. What gives
>>8110988
Jews trying to force the .99999999 meme to brainwash pure pedigree white males into believing the incorrect maths
>>8110984
What the fuck?
|0.9..9n - 1| < 1 says that the distance between 0.9...9n and 1 is less than 1. |1 - 0.5| < 1 says the distance between 0.5 and 1 is less than 1...
The correct answer was given many times, you cannot compare infinities like this
0.999... is an illegal operation, because 0.999... is an infinite sum and that's the end of it.
Infinities cannot be treated like normal numbers that work with the same arithmetic operations that you are used to, this is the error in your thinking.
>>8111015
>0.999... is an illegal operation
It's not an operation at all. It's a real number.
>because 0.999... is an infinite sum and that's the end of it.
No, 0.999... is the unique least real number to be an upper bound on every interval of the form (0,a) with 0<a<1. This puts it in every set of the form [a,1], which are each clearly compact. The intersection of all of these is simply the set {1} and the equality follows (nested sphere theorem)
dumbfuck
>>8111036
0.99...5 = .999... = 1
>>8109791
>Not that guy but they LITERALLY do.
>>8111039
That is the point I'm trying to make with that
>>8111032
>No, 0.999... is the unique least real number to be an upper bound on every interval of the form (0,a) with 0<a<1.
This is wrong, you're implying that the set of uncountable infinity in 0<a<1 has a "final last step", which is definitely wrong.
0.999... is not a real number as it doesn't terminate.
The rest of your post falls apart here.
>>8109479
[math]1-.999...=\lim_{c \to 0}c[/math] is what you're saying
>>8111059
>real numbers have to terminate
>>8111036
0.999... doesn't "approach" anything. It's a constant real number with a fixed value.
>>8109925
they forgot the infinitsemal math..
>>8110356
you have the entire set of fractions between 0 and 1
they go like 0.999... +p/q, where p/q is an infinitely small fraction
it's necessary because you can't map between fractions and decimals
0.444444444444444.....
>>8110415
it's one of the rules of algebra. if you multiply 0.999... by itself, you get back 0.999...
You actually get the same no matter what you multiple 0.999... by, 0.999... x 18 = 17.999... and so on, it works as an identity, and identity must be unique. that's the proof that convinced me at least, even though i didn't understand it in algebraic terms for years.
>>8110416
From sheer logic. I use my massive intellect to think hard about these problems and always come to good conclusions. People have confused fractions and decimals for centuries. It's time to enter a new paradigm where we accept they're only approximately equal.
>>8110572
it doesn't terminate, you just keep adding in terms in the middle instead of the end.
If 0.999... = 1
then 0.333... must logically be 0.333...4, which is impossible, as a non terminating real number cannot have a last step.
>>8109416
.99 repeating is equal to 1-h (infinitesimal)
it is 1 for all practical purposes
>>8110785
this, 0.333... only approximately equals 1/3
>>8111083
0.333...4 * 3 = 1.000...2 =/= 1
>>8110988
ivory tower mathematicians are stuck in their old paradigms, and fight to keep that lush government grant money. the truth will win out anon.
>>8111064
0.999.. = 9(1/10 + 1/100 + 1/1000 + ...) = 9[1/(1-1/10)] = 1. 1/10 + 1/100 + 1/1000 + ... looks like an infinite sum to me. An infinite sum is just a limit of partial sums. If you don't like that then how else would you describe a sequence of numbers?
Either, way, I'll go back and explicitly ask you the question from the point of view of the limit we were talking about. If 0.99...5 is less than 1 and 0.99...9n approaches 1 as you go to infinity, then 0.99...9n approaches 0.99..5. So either, 0.99..5 is the limit where it follows 0.99..5 = 1 or you have an n such that 0.99...9n exceeds 0.99....5. Can you explain that?
>>8111083
>If 0.999... = 1
then 0.333... must logically be 0.333...4
Show that and then we can start talking.
>>8111090
Thanks for proving my point... I guess?
>>8111074
does not equal 4/9.
>>8111099
Guess again.
(0.333... * 3) = 0.999... = 1
(0.333...4 * 3) = 1.000...2 =/= 1
(1.000...2 / 3) =/= (0.999... / 3)
0.333... =/= 0.333...4
(0.999... / 3) =/= 0.333...4
hexes for 0.999... =/= 1
>>8111084
>everyone ignores my answer which is correct
you guys need to retake calc 1
>>8111059
>you're implying that the set of uncountable infinity in 0<a<1 has a "final last step"
Yeah? Where? Do you not know what "upper bound" means?
>protip: no, I'm not.
>0.999... is not a real number as it doesn't terminate.
1. That doesn't preclude a number from being real
2. Simply writing down "0.999... " is not an unambiguous construction. If we specify that 0.999 is the real number as described, it necessarily equals 1. If we define it similarly as a real number less than one, it does not exist. If you're not defining it as a real you must state which number system you're defining the scrawl "0.999..." to belong to and what its properties are. It has no meaning absent a definition.
0.999999999999999999...FOOT DOCTOR
>>8111112
shut up nerd, the numbers do what i say they do
>>8111101
>There are plenty of numbers less than 1 which 0.9...9n does not approach. 0.999...95 is simply one of them.
Well fuck me right? How can 0.9...9n approach 1 but not 0.99....5?
>>8111124
maybe you should go back to /s4s/
>>8111126
all of the high population boards care about gets
notice how I stated "I suppose it doesn't really matter that much on this board"
>>8111111
rigorously and unambiguously define the relevant hexes for 0.999...
>>8111111
The true irony would be if you had gotten a 8111112 get
>>8111129
>all of the high population boards care about gets
that's because they've been colonized by /s4s/ immigrants who have forced this aspect of their board culture on the other boards. it's board-cultural imperialism.
>>8111123
Well, because 0.999...95 is a distinct, smaller number than 1, obviously 0.9...9n can't approach both 0.999...95 and 1 at the same time. And none of your proofs have proven anything other than that 0.9...9n has a limit of 1.
if 0.999...9n >= 0.999...5
9n >= 5
n >= 5/9
if 0.999...9n =< 0.999...5
9n =< 5
n =< 5/9
0.9 + 0.09 + 0.009 + 0.0009 ... + 0.000...9 = ???
0.000...9 = ???
??? = ???
???
>>8111142
>Well, because 0.999...95 is a distinct, smaller number than 1
How about you formally construct this new number system you're inventing?
Is it a metric space?
>>8111142
How does 0.99...9n approach 1 without approaching 0.99...5?
>>8111161
I know right?
>>8111163
How does 1 + 1/2 + 1/4 + ... approach 2 without approaching 1.99? They're different numbers, that's how. But 0.999...95 is much closer to 1 than 1.99 is to 2, in fact so close that it's larger than any element of the 0.9...9n series (though still smaller than their limit).
>>8109655
you fucking retarded faggot
>>8109766
>If 0.00...001 is a nonzero real number what is its reciprocal?
ez. that would be 10000000000 ... 1
you math faggot. /s
>>8111142
what is the value of 1/0.999...95 ?
of 1/0.000...1 ?
>>8110359
>>8111179
No, that is not how. In 1 + 1/2 + 1/4 +... you have an nth term such that 1 + 1/2 + ... (1/2)^n is larger than 1.99, that is how. You still haven't explained yourself.
>But 0.999...95 is much closer to 1 than 1.99 is to 2, in fact so close that it's larger than any element of the 0.9...9n series (though still smaller than their limit).
Using your logic I can just say 0.99999....95 has infinite numbers and then 5. 0.9999....99 has infinite numbers and then 9 . Since 9 > 5 you have 0.99999....99 > 0.999.....95 but then `obviously' 0.9999.....99 = 0.999....
>>8110420
>If you can have infinite 9s followed by a 9 then you can have infinite 0s followed by a 1.
HAHAHAHAHAHAHAHAHAHAHAHAAHHAAHHAHAHAHA
0.000...1 / 0.999... = 0
>>8111188
1/0.999...95 = 1.999.../2.
1/0.000...1 = 100...0.
>>8111199
You're getting mixed up because of the infinite decimals. 0.999... only has 9s through decimal place [math]\omega[/math], while 9.999...5 has 9s through decimal place [math]\omega[/math] and then a 5 in place [math]\omega+1[/math]. Of course you could also have 0.999...9, which has 9s through decimal place [math]\omega+1[/math], but that's also a different number than 0.999... by itself.
In summary, 0.999... < 0.999...5 < 0.999...9 < 0.999...95 and so on.
>>8111215
>1/0.999...95 = 1.999.../2.
Whoops, meant to say 0.999...5 = 1.999.../2 so 1/0.999...5 = 2/1.999...
>If you can have infinite 9s followed by a 9 then you can have infinite 0s followed by a 1
ok. Let's create this number:
>>8110359
>the ω decimal point
This does not exist in the real number system. Kill yourself.
All rational and recurring numbers can be given in the form a/b where a,b are integers. Prove to me that you can show 0.999... in this form?
Protip: you can't.
>>8111220
That is not an element of the complex numbers.
If you want to "create a number" and you want us to call it complex, then you need to prove to us that it is a complex number. This is easy to do for all known complex numbers. It is not possible to do for yours.
What you have just written makes as much sense as the number 0.0...fish
Both are undefined until you give it a precise definition.
>>8111229
.99999999 is irrational faggot
>>8111229
Let a = 1 and b = 1. Wow a/b is fucking 0.999....
That wasn't hard.
>>8111229
0.999... = 999...999/100...000. It's rational.
>>8111234
Then it doesn't equal 1 =)
>>8111234
I clearly said rational OR recurring.
0.999... is recurring.
>>8111215
>0.999...9, which has 9s through decimal place ω+1, but that's also a different number than 0.999... by itself
Ya okay
>>8111254
It is, though. 0.999...9 - 0.999... = 0.000...9 > 0.
>>8111237
>infinity is an element of the integers
You dumb fuck
>>8111229
Sorry for confusion, my intentions of this post were to prove that 0.999... could not be expressed differently than 1.
Hence, 1/1=1=0.999...
>>8111264
No, "infinity" isn't a precise term that corresponds to a single number nicely. 100...000 just has an infinite series of 0s following the 1, but preceding the decimal point.
>>8111270
Not an integer. Integers are finite. Buddy :-)
>>8111270
>has an infinite series of 0s following the 1, but preceding the decimal point.
Not an integer
>mfw I fail students who get this wrong
You better not assume 0.999.. = 1 because you'll get a fail for that task
>>8111255
So you're telling me there's more digits in 0.99....9 just because you insist? Do you understand how infinity works?
>>8111274
>teaching objectively incorrect maths
absolutely AMERICAN
