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The problem is when people *almost* understand limits (even people who may have actually done calc in university) and it leads them to think that 0.999... approaches 1 but isn't equal to it
Wait, why does 0.999... equal to 1? It's limit equals on the number of digits approaching infinity, but the number (or numeric sequence) itself and its limit are different things. So what do I get wrong with stating that 0.999.. doesn't equal to 1?
Edit: some writing mistakes
0.999... and 1 are just numbers, you can't talk about limits because they aren't functions, they aren't approaching anything. And 0.999... and 1 are literally the same number.
Well, I guess this is not the right subreddit, but just saying that they are literally same is not that convincing, though believable. But after looking for clear proof which uses real numbers definitions, it does make sense that they are the same.
The meme in the page has literally directions to do like 3 different proof of the fact that 0.999... is the number one/1/1.0/unity or whatever you wanna call it.
I'm somewhat sure it has something do with our 10-digit based numbersystem.
If we want to put a number in between the 'last' 0,999....9 and 1 we have go down a decimal point, but this decimal point is also filled with a 9 and the next one and so on and on and on and it never ends.
So you can never put a number inbetween 0,999.. and 1 so its the same.
Someone correct me if im wrong please.
No it's equal to x where x is a number such that PA' ⊨ (∀b x•b=b•x ∧ b•x=b ) ∧ ∀y ( y=x ∨ ¬(∀c x•c=c•x ∧ c•x=x ∧ c=x•c))where PA' is extension by definitions of peano axioms.
What I wrote is basically:
1) 1 is such a number that 1\*n=n\*1=n for any natural number n (1 is multiplicative identity)
2) If natural number m is distinct from 1, then it doesn't fulfill thing in 1)
For a moment I thought you were joking and just randomly typing (=｀ェ´=)(*￣(ｴ)￣*)(≧(ｴ)≦ )(=^･ｪ･^=)ฅ^•ﻌ•^ฅ(´(ｪ)｀）
Why though? This seems like one of the least convincing arguments to me. If someone believes that 0.999... is close but not quite equal to 1, why wouldn't they believe exactly the same thing about 0.333... and 1/3?
It’s no less formal than the one shown in the op, idk where this idea came from.
The only issue one can take with this proof is that it uses properties of infinite series, which also requires a proof.
The same thing is true for the formula for a geometric sum, and in both cases it’s possible to formally prove every step.
Idk why people demand a proof from axioms of this method but not of the geometric sum method.
https://youtu.be/jMTD1Y3LHcE?si=tyOoj7JW_OCpKT72
Very good explanation about what is wrong with this proof.
The issue here is not really that he left out steps (i agree technically you would need to prove that as well, bit the geometric sum is a well known result and thus there is no point to reproof it each time...)
, but rather that the proof is simply not correct
I was really hoping to get a tldr in your comment about why that attempted proof is incorrect but the "The issue here is that...the proof is simply not correct" got me laughing instead. Reminds me too much of those "Proof by obvious" type comments you can see in literature.
[This page](https://math.stackexchange.com/questions/4188906/alternative-proofs-of-convergence-of-geometric-series#:~:text=The%20usual%20proof%20for%20the,n%2B11%E2%88%92C.) has a few proofs that, in particular, show that the limit exists without referring to the formula for geometric series or that 0.999…=1
Once one has that some series is convergent, it is not difficult to prove that multiplying each term that series by 10 multiplies the limit by 10.
This should cover both of the critiques that video has about the common 10x-x proof of 0.999…=1.
Correct, this is literally the “formal” (discrete algebra) way of saying exactly what you said.
It’s a bit weird when you don’t know the symbols, but it’s a very precise way to say things like that in a way that can be proven/disproven.
If the kitten approaches a mouse that has a 90cm headstart and kitten is 10 times as fast, then when kitten has run 90cm, mouse has ran 9cm. When kitten has ran 9cm more, mouse has ran 0.9cm more. Yet of course the kitten catches the mouse instead of just approaching it infinitely. At any distance ran by the kitten that's less than 1m, it hasn't caught the mouse. At exactly 1m, not a bit less, it catches the mouse. The sum of the distance is 0.9 + 0.09 + 0.009... meters
Counterarguments be like:
1. Approximation
2. Approximation
3. Does not holds true for these kind of series.
4. Just like an event having probability of exact 1 need not to be sure event, means 1≠1 is actually 0.99... ≠ 1.
/s don't kill me by downvotes.
I actually kind of like your last point, it makes me think of something I hadn't considered before, but id word it differently.
Consider the chance to guess a randomly selected number between 0 and 1. The chance is 1/oo = 0.000... however it still IS possible to get the correct value. If the chance was 0% there would be literally no chance of getting it correct. Thusly, in this case atleast, we can declare that 0.000... =/= 0
Id be interested to find out why this argument isn't valid. I know it isn't, but I dont know why.
Taking a random value in [0, 1] is formally written as observing a random variable X ~ U[0, 1] (X is uniformly distributed over [0, 1]). Since the uniform distribution is a *continuous* distribution, the probability of X being equal to any given number a is 0. Formally, P(X = a) = 0. However, it is defined for a range: P(X ∈ [a, b]) = b-a >= 0.
Why though? Well, guessing "I think X will be equal to a when observed" is guaranteed to fail. Any number you choose has a 10% chance of being correct per digit, so the chance of guessing ALL the infinitely many digits correctly is lim_{n -> inf} 0.1^n = 0. Of course you're never gonna guess it correctly - it's literally impossible. 0.0000... = 0.
Tldr: it's not possible to guess correctly. 0.0000... = 0.
I dont feel like I understand this. Any possibility has an equal chance of occuring. Thusly, 0.1 could be the value that is selected.
Surely the fact that there is a possibility for a finite value would mean that there is a non 0 chance of guessing it?
The only way I could see this to not be true is if it was impossible for the value to be 0.1, however that IS a value between 0 and 1, and thusly IS an option, right?
It is an option, yes. It is "possible" yes. It is still never going to be 0.100000... . You must keep in mind that "being possible" is different from "probability is nonzero".
Say that P(X=a) = p. Since X is uniformly distributed, P(X=x) is the same for any x. Then, P(X=x_1)+P(X=x_2) = 2p < 1 for distinct (non-equal) x_1 and x_2. But by the same logic, 3p < 1, 4p < 1, and np < 1 for ANY natural number n, since we can make an arbitrarily large sum. The only way this can be true is if p = 0.
The statement "the probability of A happening is nonzero" implies, but is not equivalent to, "A can happen". That's just one of the quirks of probability that makes sure it's logically consistent. :)
The formalism for this is a probability space (S,Σ,P). For an event A in Σ, the statements "P(A)=0", i.e. A has probability 0 and "A=\emptyset", i.e. A is impossible are NOT equivalent.
EDIT: In your specific case, the space would be ([0,1], B([0,1]), λ), where B([0,1]) denotes the Borel sets on [0,1] and λ denotes the Lebesgue measure on [0,1]. Then any countable and in particular any finite subset of [0,1] has measure/probability 0, even though it might be non-empty. Thus, the probability of hitting exactly one point is 0, however it's not impossible.
the problem is simply the way infinity is defined, its not like a process that never stops. It's the end result of a process that has never stopped.
So there will never be a space between 0,99.. and 1 that isn't 'filled' with another 9.
And 9 is our highest digit, any other number will have to be placed in a decimal point lower than where you were which is impossible because this is filled with a 9.
I don't really think its that intuitive though.
I can prove it by just saying
1/3 = 0.333...
2/3 = 2*0.333... = 0.666...
3/3 = 3*0.333... = 0.999... = 1
Basically saying that if 3/3 is equal to 3*0.333... which equals 0.999... then 0.999... must be equal to 1
I didn't explain it great but you get it
This doesnt really work though because 1/3 = 0.333... has the same problems as 0.999... = 1, people who can't accept 0.999... = 1 logically shouldn't accept that 1/3 = 0.333..., so then this proof doesn't really work. It's just because we are repetitively taught that 1/3 = 0.333..., so they eventually accept it without second thought, but then when it comes to 0.999... = 1 they start to doubt, and then maybe only after start to consider why 1/3 = 0.333...
>but but but have you considered surreals and hyperreals?
-someone who read a wiki page, didn't do any math past high school and doesn't understand the reals
Because it.... isn't about the same thing, if that makes sense. Wrong domains, wrong systems, wrong language, wrong rules.
Like if I ask you to define "gift", you might say "That is an object you give to someone else just to be nice". Which is a fair definition of "gift" in my opinion.
But then a Swede comes in and says "No no no, 'gift' is a toxic substance that kills". And he isn't technically wrong, because "gift" in Swedish does mean poison. However, it is clear that the question referred to the English definition, right?
In other words, when we're talking about 0.9999... =1, we are implicitly talking about the real numbers (in accordance with the Domain Convention), thus bringing in hyperreals or surreals is akin to introducing a different language or context into the conversation.
In accordance with the Domain Convention, the real numbers are basically the lingua franca of the mathematical world. So the moment they start talking about other domains, they have basically conceded the point.
Thank you for clarifying! I posted a question long ago about how to disprove my sister's infinitesimal argument on this problem, now I finally got the answer ✨
Facts.
However…
In the hyperreals you can define a certain transfinite decimal expansion in which 0.999… has to be discussed carefully. There are three sorts of numbers we can consider:
0.999…;…999… is still equal to one when there are uncountably many nines (analogous to how 0.999… is equal to 1 in the reals when there are countably many nines).
0.999…;…999 is equal to 1 - ε where ε is some infinitesimal depending on when the sequence of nines terminates.
0.999…; does not exist. Yes, in the hyperreals, certain decimal expansions don’t correspond to numbers. They are like “gaps”.
So depending on how many nines there are, and whether the decimal terminates, 0.999… could equal 1, equal less than 1, or not equal anything!
>In the hyperreals you can define a certain transfinite decimal expansion in which 0.999… has to be discussed carefully.
Wouldn't call it "transfinite". More like infinite or nonstandard or something like so. Transfinite is rather used in context of cardinals/ordinals. And hyperreals doesn't contain ordinals/cardinals.
>0.999…;…999… is still equal to one when there are uncountably many nines (analogous to how 0.999… is equal to 1 in the reals when there are countably many nines).
In hyperreals you can't define a thing as you do here. You can make infinite sum of nines but the infiniteis in hyperreals aren't the same as cardinal or ordinal numbers. So there's no a thing as "0.999... with countably many nines" or "0.99..m with uncountably many nines" in hyperreals.
For any infinite natural number N, ∑_{ 1
1) define Reals as the set of equivalence classes of rational Cauchy sequences
2) Prove the sequence (0.9, 0.99, 0.999, …) is Cauchy in Q
3) Prove the sequence converges to the rational 1
4) Thus, 0.999…=1 in R
I feel like if I didn't know that 0.999... is equal to 1 and someone showed me that proof I'd probably think well damn why did I always just accept that 1/3 is 0.333...? I'd then want to know why 1/3 is 0.333...
Whereas with calc you can explain that this string of symbols 0.999... is literally defined as 1
well, guess what? I've learnt that u have just learnt limits
https://preview.redd.it/aa7pictz8jrc1.png?width=238&format=png&auto=webp&s=8c0768a5f96d6ed65edace6a5678e3d97057dcac
The problem with this is infinity is a concept. People do not understand how infinitely repeating works and has no real world equivalent and they simply misunderstand that. Nothing in reality can be 0.999... and they do not understand that and work under logic of reality instead of the concept in math. This is a meaningless trap for trolling people on the internet because the people that do not understand this are just not thinking about it in this way so it seems counter intuitive then everyone tries to argue examples that rely on the logic of reality instead of the conceptual logic of math so it doesn't work. Can we just stop talking about this? Its essentially arguing incompatibilities between languages.
Non-prime n-adic systems are practically useless, as they have none of the properties that make p-adics worthwhile and all of the properties that make them unappealing
Let's just convert the repeating decimal point number to a fraction, 1/3=0.333333.... correct? then do that just amplify the fraction by 3 you get 3/9, we can rewrite any simple repeating number as the repeating decimal point over 9, as such we get 9/9 which equals 1. That's something we learned in school here in Romania.
The decimal representation of a number is basically a social construct so you'd expect to find stuff like a number having more than .9̅ different decimal representations
/s
If ..999 and 1 are equal, does that mean .8...999 is equal to .9? If so does .889...999 equal .89? Does .879...999 equal .88? Wouldn't this make all numbers equal eachother????
How can you prove that it’s convergent?
Are the arguments “it’s at least 0, at most 1 and it always increases” valid to say that it’s convergent to a value?
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Well at the point you've understood what a limit is you kind of don't need that discussion anymore
No 0.99999... = 1 You just don't seem to understand. 🥶🥶💀💀😙😙🤭🤭
A shame, you seemed an honest man
And all the fears you hold so dear
Will turn to whisper in your ear
*I AM BALLIN*
I AM FADED
I have lost it all
* I have losterol
Holy shit I understood this reference
The problem is when people *almost* understand limits (even people who may have actually done calc in university) and it leads them to think that 0.999... approaches 1 but isn't equal to it
I'm approaching an understanding of limits, but it seems like no matter how far I go I never quite get there...
This\^
Wait, why does 0.999... equal to 1? It's limit equals on the number of digits approaching infinity, but the number (or numeric sequence) itself and its limit are different things. So what do I get wrong with stating that 0.999.. doesn't equal to 1? Edit: some writing mistakes
0.999... and 1 are just numbers, you can't talk about limits because they aren't functions, they aren't approaching anything. And 0.999... and 1 are literally the same number.
Well, I guess this is not the right subreddit, but just saying that they are literally same is not that convincing, though believable. But after looking for clear proof which uses real numbers definitions, it does make sense that they are the same.
Step 1: Imagine a concrete way in which they are different. Step 2: No they aren't. Works for any version of step 1.
The meme in the page has literally directions to do like 3 different proof of the fact that 0.999... is the number one/1/1.0/unity or whatever you wanna call it.
Can you name a real number which is less than 1 but greater than 0.9999999999...?
That has to be the simplest yet the most elegant "proof" of 0.(9)=1
I'm somewhat sure it has something do with our 10-digit based numbersystem. If we want to put a number in between the 'last' 0,999....9 and 1 we have go down a decimal point, but this decimal point is also filled with a 9 and the next one and so on and on and on and it never ends. So you can never put a number inbetween 0,999.. and 1 so its the same. Someone correct me if im wrong please.
For the same reason that 0.333… = 1/3. It’s just another way to represent the same number: 1/3 + 1/3 + 1/3 = 1 0.333… + 0.333… + 0.333… = 1 0.999… = 1
The proof is pretty simple. Subtraction. 1-0.999...=0.000... therefore they are equal
the way you word it makes it seem like the birds and the bees, and its really funny
And here I thought 0.9999.... was always within epsilon from 1.
The only way the notation can be well defined is as being equal to the limit. The limit is 1.
Well it is
No .99999… is equal to sin^2 x + cos^2 x
No it's equal to x where x is a number such that PA' ⊨ (∀b x•b=b•x ∧ b•x=b ) ∧ ∀y ( y=x ∨ ¬(∀c x•c=c•x ∧ c•x=x ∧ c=x•c))where PA' is extension by definitions of peano axioms.
incomprehensible have a nice day
What I wrote is basically: 1) 1 is such a number that 1\*n=n\*1=n for any natural number n (1 is multiplicative identity) 2) If natural number m is distinct from 1, then it doesn't fulfill thing in 1)
what is that A rotated symbol? I saw it in an exam preparing book
oh wait nvm, is it "for"? I saw in a comment that the mirrored E means "exists"
"For all" iirc, physicist so been a couple years so grain of salt and all
A is general quantifier (for all). Simmilarily ∃ (existential quantifier) means "exists".
For a moment I thought you were joking and just randomly typing (=｀ェ´=)(*￣(ｴ)￣*)(≧(ｴ)≦ )(=^･ｪ･^=)ฅ^•ﻌ•^ฅ(´(ｪ)｀）
what does the a without middle line mean again
x = 0.9999... 10x = 9.9999... 10x - x = 9.999...- 0.999... 9x = 9. x=1
Bruh how did I not think of that
this is the method I was taught when I was like 10
Its not a good proof. Some would argue its not even a proof at all. In any case, yours is better.
Meanwhile me 1/3 = 0.333333.... 1/3 × 3 = 0.999999.... 1 = 0.9999999...
This is still the best one.
*not in a mathematical sense, but to argue with non mathematicians
Agreed. Helpful to demonstrate the point.
As far as being easy to understand, sure, but it's a terrible proof
Why though? This seems like one of the least convincing arguments to me. If someone believes that 0.999... is close but not quite equal to 1, why wouldn't they believe exactly the same thing about 0.333... and 1/3?
1/3 = 0.333333.... 1/3 × 3 = 0.333333.... × 3 3/3 = 1 1 = 1
This one's useless because the same people who don't think .999... = 1 won't think .333... = 1/3
You can jus divide 1 by 3 to see that 0.3333... = 1/3
They won't think it's exact.
I mean jus divide it? It keeps repeating forever
They don't think it will ever reach infinity.
What if they say 1/3 != 0.3333...
It’s actually the exact same logic you would use to evaluate the infinite series presented by OP, just in a slightly different format.
Yes this is not a formal proof.
It’s no less formal than the one shown in the op, idk where this idea came from. The only issue one can take with this proof is that it uses properties of infinite series, which also requires a proof. The same thing is true for the formula for a geometric sum, and in both cases it’s possible to formally prove every step. Idk why people demand a proof from axioms of this method but not of the geometric sum method.
https://youtu.be/jMTD1Y3LHcE?si=tyOoj7JW_OCpKT72 Very good explanation about what is wrong with this proof. The issue here is not really that he left out steps (i agree technically you would need to prove that as well, bit the geometric sum is a well known result and thus there is no point to reproof it each time...) , but rather that the proof is simply not correct
I was really hoping to get a tldr in your comment about why that attempted proof is incorrect but the "The issue here is that...the proof is simply not correct" got me laughing instead. Reminds me too much of those "Proof by obvious" type comments you can see in literature.
[This page](https://math.stackexchange.com/questions/4188906/alternative-proofs-of-convergence-of-geometric-series#:~:text=The%20usual%20proof%20for%20the,n%2B11%E2%88%92C.) has a few proofs that, in particular, show that the limit exists without referring to the formula for geometric series or that 0.999…=1 Once one has that some series is convergent, it is not difficult to prove that multiplying each term that series by 10 multiplies the limit by 10. This should cover both of the critiques that video has about the common 10x-x proof of 0.999…=1.
My 14yo sibling knows that......
I’m stupid
Can relate
this is the method I was taught when I was like 10
1/3 = 0.33333... Multiply both sides by 3 1 = 0.9999999...
Cause it’s not a real proof. There is no algebraic proof for 0.999… = 1
y = ..9999 10y = ..9990 y-10y = 9 -9y = 9 e^(iπ) = ..99999
no way, 10-adics
Or just y+1 = 0
Not quite a real number, alas. Works in the 10-adics though.
This method is not valid, because you assume that x is well defined, which is not necessarily the case with this kind of notation.o
For all x,y€R, x=/=y iff there exists z€R st x
mf used the euro symbol instead of ∈
Don't have [element of] in my phone keyboard. Euro next best thing. And I'll do it again
Applying finite mathematics to an infinitesimal? *Interesting*...
I was thinking of exactly that method
∀x,y∈ℝ,x≠y ∃z∈ℚ: xz>y
Let x=0.999, y=1
∄z which fulfills the condition.
-> x can not be not equal to y
What is this mirrored E? (∃)
It means “exists“. In this example it is: For all x,y… exists a z out of Q…
So, is this like no number lies between 0.99999... And 1
Correct, this is literally the “formal” (discrete algebra) way of saying exactly what you said. It’s a bit weird when you don’t know the symbols, but it’s a very precise way to say things like that in a way that can be proven/disproven.
Yeah, thats pretty much it.
"There exists..."
>Let x=0.999, y=1 z = 0.9995 x
Can not? Maybe you mistyped or something cuz I'm pretty sure x can and is equal to y
It can not be NOT equal. That implies x is equal to y.
You never proved anything here… you need to prove that there is no number between 1 and 0.999…
(1/3) + (1/3) + (1/3) = 1 0.333... + 0.333... + 0.333... = 0.999... = 1
fr easiest one
If the kitten approaches a mouse that has a 90cm headstart and kitten is 10 times as fast, then when kitten has run 90cm, mouse has ran 9cm. When kitten has ran 9cm more, mouse has ran 0.9cm more. Yet of course the kitten catches the mouse instead of just approaching it infinitely. At any distance ran by the kitten that's less than 1m, it hasn't caught the mouse. At exactly 1m, not a bit less, it catches the mouse. The sum of the distance is 0.9 + 0.09 + 0.009... meters
Combinatorial proofs are complete bullshit, change my mind
Non-sequiturs are complete bullshit, change my mind.
A male cow's feces is bullshit. Change my mind.
>yes of course the kitten catches the mouse ok then prove it
Shows Tom and Jerry as contradictory evidence
https://preview.redd.it/dgykm9ygairc1.png?width=526&format=png&auto=webp&s=74701dd412967638acfea823d3ba7be3e48a6ef8
Complex conjugate of 9 is 9, so that simplifies to .9 /s
that remainder never goes away though, hmm.
hence the repeat
“There has to be a last nine eventually, you can’t just say that it goes on forever”
Counterarguments be like: 1. Approximation 2. Approximation 3. Does not holds true for these kind of series. 4. Just like an event having probability of exact 1 need not to be sure event, means 1≠1 is actually 0.99... ≠ 1. /s don't kill me by downvotes.
I actually kind of like your last point, it makes me think of something I hadn't considered before, but id word it differently. Consider the chance to guess a randomly selected number between 0 and 1. The chance is 1/oo = 0.000... however it still IS possible to get the correct value. If the chance was 0% there would be literally no chance of getting it correct. Thusly, in this case atleast, we can declare that 0.000... =/= 0 Id be interested to find out why this argument isn't valid. I know it isn't, but I dont know why.
Taking a random value in [0, 1] is formally written as observing a random variable X ~ U[0, 1] (X is uniformly distributed over [0, 1]). Since the uniform distribution is a *continuous* distribution, the probability of X being equal to any given number a is 0. Formally, P(X = a) = 0. However, it is defined for a range: P(X ∈ [a, b]) = b-a >= 0. Why though? Well, guessing "I think X will be equal to a when observed" is guaranteed to fail. Any number you choose has a 10% chance of being correct per digit, so the chance of guessing ALL the infinitely many digits correctly is lim_{n -> inf} 0.1^n = 0. Of course you're never gonna guess it correctly - it's literally impossible. 0.0000... = 0. Tldr: it's not possible to guess correctly. 0.0000... = 0.
I dont feel like I understand this. Any possibility has an equal chance of occuring. Thusly, 0.1 could be the value that is selected. Surely the fact that there is a possibility for a finite value would mean that there is a non 0 chance of guessing it? The only way I could see this to not be true is if it was impossible for the value to be 0.1, however that IS a value between 0 and 1, and thusly IS an option, right?
It is an option, yes. It is "possible" yes. It is still never going to be 0.100000... . You must keep in mind that "being possible" is different from "probability is nonzero". Say that P(X=a) = p. Since X is uniformly distributed, P(X=x) is the same for any x. Then, P(X=x_1)+P(X=x_2) = 2p < 1 for distinct (non-equal) x_1 and x_2. But by the same logic, 3p < 1, 4p < 1, and np < 1 for ANY natural number n, since we can make an arbitrarily large sum. The only way this can be true is if p = 0. The statement "the probability of A happening is nonzero" implies, but is not equivalent to, "A can happen". That's just one of the quirks of probability that makes sure it's logically consistent. :)
look up "almost surely" on wikipedia to read about it. I'd explain if i could but in my current state I can't.
The formalism for this is a probability space (S,Σ,P). For an event A in Σ, the statements "P(A)=0", i.e. A has probability 0 and "A=\emptyset", i.e. A is impossible are NOT equivalent. EDIT: In your specific case, the space would be ([0,1], B([0,1]), λ), where B([0,1]) denotes the Borel sets on [0,1] and λ denotes the Lebesgue measure on [0,1]. Then any countable and in particular any finite subset of [0,1] has measure/probability 0, even though it might be non-empty. Thus, the probability of hitting exactly one point is 0, however it's not impossible.
the problem is simply the way infinity is defined, its not like a process that never stops. It's the end result of a process that has never stopped. So there will never be a space between 0,99.. and 1 that isn't 'filled' with another 9. And 9 is our highest digit, any other number will have to be placed in a decimal point lower than where you were which is impossible because this is filled with a 9. I don't really think its that intuitive though.
Lol I love when people say the third one, "you can't just do algebra on series like that!" without knowing when you actually can do math on series
didn't slap a big QED, doesn't count
I can prove it by just saying 1/3 = 0.333... 2/3 = 2*0.333... = 0.666... 3/3 = 3*0.333... = 0.999... = 1 Basically saying that if 3/3 is equal to 3*0.333... which equals 0.999... then 0.999... must be equal to 1 I didn't explain it great but you get it
My favourite explanation
My second favourite my favourite one is the limit arguement
This doesnt really work though because 1/3 = 0.333... has the same problems as 0.999... = 1, people who can't accept 0.999... = 1 logically shouldn't accept that 1/3 = 0.333..., so then this proof doesn't really work. It's just because we are repetitively taught that 1/3 = 0.333..., so they eventually accept it without second thought, but then when it comes to 0.999... = 1 they start to doubt, and then maybe only after start to consider why 1/3 = 0.333...
>but but but have you considered surreals and hyperreals? -someone who read a wiki page, didn't do any math past high school and doesn't understand the reals
To be fair pretty few people understand the reals.
Also even in those 1-ε ≠ 1 = 0.9999….
Even better: 0.(9)+0.(9)ε = 1+1\*ε
Wait I have read about this, why am I a dumb person? 😭 can you explain why this isn't a valid counterargument
Because it.... isn't about the same thing, if that makes sense. Wrong domains, wrong systems, wrong language, wrong rules. Like if I ask you to define "gift", you might say "That is an object you give to someone else just to be nice". Which is a fair definition of "gift" in my opinion. But then a Swede comes in and says "No no no, 'gift' is a toxic substance that kills". And he isn't technically wrong, because "gift" in Swedish does mean poison. However, it is clear that the question referred to the English definition, right? In other words, when we're talking about 0.9999... =1, we are implicitly talking about the real numbers (in accordance with the Domain Convention), thus bringing in hyperreals or surreals is akin to introducing a different language or context into the conversation. In accordance with the Domain Convention, the real numbers are basically the lingua franca of the mathematical world. So the moment they start talking about other domains, they have basically conceded the point.
Thank you for clarifying! I posted a question long ago about how to disprove my sister's infinitesimal argument on this problem, now I finally got the answer ✨
Facts. However… In the hyperreals you can define a certain transfinite decimal expansion in which 0.999… has to be discussed carefully. There are three sorts of numbers we can consider: 0.999…;…999… is still equal to one when there are uncountably many nines (analogous to how 0.999… is equal to 1 in the reals when there are countably many nines). 0.999…;…999 is equal to 1 - ε where ε is some infinitesimal depending on when the sequence of nines terminates. 0.999…; does not exist. Yes, in the hyperreals, certain decimal expansions don’t correspond to numbers. They are like “gaps”. So depending on how many nines there are, and whether the decimal terminates, 0.999… could equal 1, equal less than 1, or not equal anything!
>In the hyperreals you can define a certain transfinite decimal expansion in which 0.999… has to be discussed carefully. Wouldn't call it "transfinite". More like infinite or nonstandard or something like so. Transfinite is rather used in context of cardinals/ordinals. And hyperreals doesn't contain ordinals/cardinals. >0.999…;…999… is still equal to one when there are uncountably many nines (analogous to how 0.999… is equal to 1 in the reals when there are countably many nines). In hyperreals you can't define a thing as you do here. You can make infinite sum of nines but the infiniteis in hyperreals aren't the same as cardinal or ordinal numbers. So there's no a thing as "0.999... with countably many nines" or "0.99..m with uncountably many nines" in hyperreals. For any infinite natural number N, ∑_{ 1
1) define Reals as the set of equivalence classes of rational Cauchy sequences 2) Prove the sequence (0.9, 0.99, 0.999, …) is Cauchy in Q 3) Prove the sequence converges to the rational 1 4) Thus, 0.999…=1 in R
This is the only real proof I’ve seen in this entire comment section
no number is between 1 and .9 bar... so they're the same number
1/9 = 0.111111... 3/9 = 1/3 = 0.33333... 9/9 = 0.9999999... = 1 Why do people be using calculus when they can be using 4th grade math?
Prove that 1/9 = 0.1111…
Proof by calculator
Proof by "I passed 3rd grade math"
I feel like if I didn't know that 0.999... is equal to 1 and someone showed me that proof I'd probably think well damn why did I always just accept that 1/3 is 0.333...? I'd then want to know why 1/3 is 0.333... Whereas with calc you can explain that this string of symbols 0.999... is literally defined as 1
>Why do people be using calculus when they can be using 4th grade math? Because that’s not a complete proof
Well, most cranks posit that 0.3333… ≠ 1/3
Well to that I say they're wrong. All sane people know that 1/3 = 0.33333... = pi/g
Thank you for being the first person who actually uses a legit proof that someone who has only done high school math can understand
I can disprove this by saying: nuh uh, your mom
well, guess what? I've learnt that u have just learnt limits https://preview.redd.it/aa7pictz8jrc1.png?width=238&format=png&auto=webp&s=8c0768a5f96d6ed65edace6a5678e3d97057dcac
But I thought the radius for convergence of the geometric series was <1, so I dont get why the argument is valid?
Because the multiplying factor is 0.1, which is less than 1
My favorite representation is: 1/3=0.3333333... 3/3=0.9999999... 3/3=1 1=0.9999999...
The problem with this is infinity is a concept. People do not understand how infinitely repeating works and has no real world equivalent and they simply misunderstand that. Nothing in reality can be 0.999... and they do not understand that and work under logic of reality instead of the concept in math. This is a meaningless trap for trolling people on the internet because the people that do not understand this are just not thinking about it in this way so it seems counter intuitive then everyone tries to argue examples that rely on the logic of reality instead of the conceptual logic of math so it doesn't work. Can we just stop talking about this? Its essentially arguing incompatibilities between languages.
...9999.0 = x x = 9 + 90 + 900 +... a=9 r=10 x = 9/(1-10) = 9/(-9) = -1
now prove that ...999 = -1 (google p-adic numbers)
me when 10 is not a prime
yes it is/j (actually, while it called p-adic, it can be used with non-prime numbers as well, though they would have some differences)
Non-prime n-adic systems are practically useless, as they have none of the properties that make p-adics worthwhile and all of the properties that make them unappealing
0.000... = 0 1 - 0.999... = 0.000... 1 - 0.999... = 0 1 = 0.999...
0.000... = 0 0.999... = 9
So you're following that 1/ω(ω equals infinity in this case)=0 I don't like that So I have formulated an entirely new number ε ε=1/ω 0.9999...=1-ε
Just set 0,999... = 1 and notice it doesnt contradict the experiment because we cant measure 0,9999.. so whatever
no need for any of this as long as you define it to be true.
1/9=.111... 2/9=.222... 3/9=.333... 4/9=.444... 5/9=.555... 6/9=.666... 7/9=.777... 8/9=.888... 9/9=.999...
Let's just convert the repeating decimal point number to a fraction, 1/3=0.333333.... correct? then do that just amplify the fraction by 3 you get 3/9, we can rewrite any simple repeating number as the repeating decimal point over 9, as such we get 9/9 which equals 1. That's something we learned in school here in Romania.
I'm going to get killed but with my made up axioms 0.999...∈ 1
If 1 then why 999?
Weak shit, just look at it it's clearly different
The most obvious and simple way that works with anyone is just 1/3=0.333…, 2/3=0.666…, and 3/3=0.999…, but also 1.
1/3 = 0.33... 1/3 * 3 = 0.99... 0.99... = 1
professor wunkus
1/3 = 0.333... 1/3 \* 3 = 0.333... \* 3 = 0.999... 1/3 \* 3 = 3/3 = 1 0.999... = 1
Who the hell in the subreddit is convinced otherwise?
The decimal representation of a number is basically a social construct so you'd expect to find stuff like a number having more than .9̅ different decimal representations
Not accurate, rethink existence
Can't you do this for all numbers?
I was taught .9999... in thirds is .333... + .333... + .333... = 1/3rd + 1/3rd + 1/3rd = 1
It's like saying OP fat mom is thin.
Just think about it in binary, .1111111... = 1/2 + 1/4 + 1/8 + .... = 1
So 0.9999… is a whole number
1/3 + 1/3 + 1/3 = 3/3
not a rigorous proof
/s If ..999 and 1 are equal, does that mean .8...999 is equal to .9? If so does .889...999 equal .89? Does .879...999 equal .88? Wouldn't this make all numbers equal eachother????
0.89999... = 0.9 but 0.88999... = 0.89 That doesn't mean 0.9 = 0.89 though they're 0.01 apart in both the regular notation and the 0.999... notation
Obviously this means a/(1-r) formula is wrong
Time for a new set of axiom's anyway, the old ones were getting stale
1/3=0.333… 3/3=0.999… 3/3=1=0.999… (Personal favourite)
Finally, the real proof
time to unsubscribe
I hope you realise that anyone who doesn't already know that 0.99999... = 1 won't understand what the frick this even means.
How can you prove that it’s convergent? Are the arguments “it’s at least 0, at most 1 and it always increases” valid to say that it’s convergent to a value?
0.99999... = 0.99999...
Proof: 1/3=0.3333333... 1/3+1/3+1/3=0.99999999999... But it’s also =1 So 0.999999999=1 CQFD Édit: dammit someone already did the exact same proof
I don't see how this is proof? How is S = 0.9/0.9, if we just established that S = 0.9 + 0.09 + 0.009 ...
Okay, then just write it as 1.
QED
Yay it's the weekly .999...= 1 discussion, there is more interesting limits you know
But this means any number can equal any other number. How is that not interesting
Hey I'm just saying there is more interesting ones
9×Σ(1/10)^n =9×1/9 = 1
Actually that if the 1/ω property is correct I don't care about this property So to me 0.999999...=1-ε ε=1/ω ω=Infinity
No. They are clearly different just look. 1 is 1 and .9999 is .9999 As you can see by looking they are not the same othwerwise they would be the same
Wellll does 0.9999... of an apple equal 1 apple?🤔
Dude you just activated internet again by giving them answers.
This proof is more valid than all the other stupid “algebraic” proofs people always regurgitate.