Every Proof that 0.999 equals 1 but they get increasingly more complex

[@ThoughtThrill365]

Have suggestions for a future video? Leave a comment below! Thanks :)

[@Nicolegusso_nfg]

I really like mathematical topics about geometry, like fractals and non-Euclidean geometry... your video is amazing!

[@jsalsman]

0^0=1

[@jsalsman]

or how about why 1/0 can't be {+∞, -∞}?

[@juliavixen176]

The monotonous AI generated voice makes it difficult to follow the arguments. There are no pauses or emphasis to distinguish what is important and what is meaningless filled words.

[@jsalsman]

@@juliavixen176 this is clearly not text-to-speech narration.

[@d4n737]

My favorite is the Engineer proof, which goes like this "It's not gonna make a difference, just round it to one"

[@Chris-5318]

Why would you round 1 to 1?

[@shadowd9810]

@@Chris-5318 pi = e = g = 10

[@Chris-5318]

@@shadowd9810 I hope that you aren't an engineer.

[@heh_boaner]

​@@Chris-5318 obviously not. Engineers round pi and e to 3, not 10.

[@Chris-5318]

@@heh_boaner Exactly :)

[@rennangandara7697]

The best part is that there's an entire video about proofs that 1 = 1

[@Rabbit_Stewer]

Nah 1 ≠ 1

[@Charles-pf7zy]

@@Rabbit_Stewer sigh. 1. suppose 1 ≠ 1

[@callmetefa7871]

This is like building an spaceship to cross a road instead of just walking.

[@gustavsturksteinwall4027]

Except it turns out that the road doesn’t exist and beyond it is but an infinite void of paradoxity

[@darrelmasterson5850]

While it may be like that, it is not that, and indeed it is most accurately exactly what it is, which is straightforward and does not require comparisons to things it is not in order to help us gain insight as to what it is.

[@kenzou776]

Welcome to math😂😂

[@santerisatama5409]

@@kenzou776 Gladly genuine math is not what a horde of postmodern academic crancks from the school of Cantor's crackpotterism teach as "math". :)

[@BloodRoseRecords]

@@kenzou776 There are no "numbers" between (0.9 repeating but with the "last" value an 8) and (0.9 repeating), therefore they much be the same. There are likewise no "numbers" between (0.9 repeating but with the "last" value being a 7) and the same with an 8, therefore they are also the same thing. You could repeat this pattern ad-infinitum and conclude that all numbers are the same number, even though that is obviously not true.

[@jaroslavsevcik3421]

The first proof is already axiomatic since you claim that 1/3 = 0.333.... All remaining steps are redundant.

[@andishyti7664]

Yep, sadly.

[@darinpringle5611]

This is untrue. Divide 1 by 3 and you will get 0.333 ..... 3 +(1*10^-n)/3 this is equal to 0.333 ..... 3 + (1/3)* 10^-n then substitute the 1/3 for the above derived value you get 0.333 ..... 3333 ..... 3 + (1/3)*10^-m Clearly this is a repeating proces with the result 0.333... with no end therefore 1/3 = 0.3333 infinitely repeating. He added no proof because it is immediately understood and nobody could disagree, not because it is an axiom

[@VvxVxvV-j2y]

@@darinpringle5611 But people who argue that 0.999... isn't equal to 1 wouldn't agree that 0.333...=1/3. They'd argue that while it's a very close approximation, you'd still be left w/ some space at the end.

[@ryan41748]

1/3 having infinite decimals is axiomatic to a base 10 number system. We don't need to do any work prove that, that's just how we choose to represent 1/3.

[@santerisatama5409]

@@darinpringle5611 I strongly disagree. 1/3 is not "division", it's a non-reducible coprime fraction. It's very important to understand the difference. I also don't believe that decimal numbers exist as a coherent part of pure mathematics.

[@EricDunn-ci3ks]

My favorite proof, not included in the video, is that any two real number, if they are not equal, will have an infinite number of numbers in between them. There exists no numbers in between 0.999... and 1, ergo, they are equal. Maybe that's what you were getting at on the last one, but I didn't really follow that one too well. edit: I tried to add this proof below, but for whatever reason, the algorithim keeps deleting that comment ... so I'm going to add it here: The real number line is continuous (That means there are no gaps in it). This is by definition. Therefore, any two numbers that are not equal MUST have some numbers in between them. If a and b are not equal, there must be some numbers in between a and b. For instance (a+b)/2 [the average of a and b] would be in between a and b. Let's set a = 0.999... and b = 1 What is (a+b)/2? a + b = 1.999... Now divide a + b by 2 0.999... _____________ 2 | 1.999... 0 [2 goes into 1, 0 times] _ 19 [bring down the 9, 2 goes into 19, 9 times] 18 [2 * 9 = 18] __ 19 [19-18=1, bring down the next 9 ... and so on] So (a+b)/2 = 0.999... [which is just a] Thus (a+b)/2=a [now solve for b.] a+b = 2a [multiply both sides by 2] b = a [subtract a from both sides]

[@Asango]

Yes, the term for that is "densely ordered" or the slightly stricter "linear continuum". The set of real numbers is a linear continuum, meaning there is an (in this case uncountably) infinite amount of real numbers between any two distinct real numbers, meaning 0.999... and 1 can't be distinct and must therefore be equal.

[@briant7265]

The trick is to prove there is nothing between. Using the mean makes it simple. For a, b, m=(a+b)/2 Also, m-a = b-m = (b-a)/2 Plug in a=0.999... and b=1 and m=1.999.../2 Do long division m=0.999... m-a = 0, so b-m = 0, so a = b.

[@LaTortuePGM]

that's literally the archimedean property argument bro lmao

[@EricDunn-ci3ks]

@@LaTortuePGM Which one was that in the video? I found some of them in the video were explained very poorly.

[@WhateverIwannaupload]

this is wrong because 0.999 never ends and there will always exist a number between 0.999... and 1 or else its not infinitely increasing via a infinitely small number. the concept that 0.9999... and 1 have no real numbers in between means that you have reached the end of 0.999... and even as a concept itself, it does have something missing its just 0.0000....1 and that will never be reached bc 0.999... never stops

[@sararamaiah1915]

Reading people arguing about this is giving me brain damage

[@adamheuer8502]

The problem is that he is using an irrational number alongside rational operations, division, addition, subtraction, etc. While you can do this it only creates more irrational numbers. Otherwise we could say 2 divided by infinity is 0 and since 0 times infinity is still zero. 2 divided by infinity times infinity equals 0 while also saying that 2 infinity divided by infinity is 2. If we do this we end up with 2=0. This guy is just wrong.

[@Arthur-io4ey]

​@@adamheuer8502No you are the wrong one. 1) he doesn't bring irrational numbers, 1/3 belongs to the rational number. And 2) your "demonstration" is a part of why division by 0 doesn't work. You cannot just use operation like that with infinity. I don't know how to foemalize that in better words but I promise that your reasoning has flaws

[@bruxasemdoce]

​@@adamheuer8502 wtf are you saying by "this guy is just wrong"? You really think that he just put all these proves from mind?

[@user-jq2xp8xx8e]

@@adamheuer8502 0, 1, 0.(9) and 0.(3), none of those numbers are irrational, go back to middle school

[@Adamin_The_III]

No intro, no "hello guys, and welcome to our new prove video" just straight to the point, that's what I'm talking about

[@joaomigueloliveiratavares6579]

I was gonna say that 0.333 was 0.34 but then i proved myself that im stupid

[@Chris-5318]

0.34 = 0.33999...

[@shalev3800]

@@Chris-5318 So every number equals every number? If the LSD doesn't matter then we can do it infinitely and get that 1 = 10000. 0.34 = 0.33999... -> 0.3399...9 = 0.3399...8 -> 0.3399...8 = 0.3399...7 and so on

[@Chris-5318]

@@shalev3800 I have no idea how you came to that conclusion. What on Earth is 0.33999...9 etc., supposed to be? (There are no such decimals). 999... does not have a LSD. "Infinite" means "endless" not "big".

[@ChiefMakes]

@@Chris-5318 I think he meant 1=1.0001 1.0001=1.001 etc. to 1=1000 by logic since 1= 1.00….1 etc.

[@Chris-5318]

@@ChiefMakes There is no such decimal as 1.000...1. You can't have a digit at the non-existent end of an endless string of digits. If you meant there to be an unspecified finite number of 0s, then 1.000...1 > 1 by 1/10^n where n is the decimal place at which the last 1 is.

[@minirop]

if the reciprocal of 1/ɛ is ω which is a 90° rotation, therefore the reciprocal of 1/8 is ∞

[@GabrielMirandaLima-hv7oe]

I would like to add then the immediate theorem which states that 1/0=0

[@wicowan]

and the lemma 1/N=Z

[@2070user]

Actually, the reciprocal of x is 1/x, by reciprocal we mean 1/... So the reciprocal of 1/E would then be just 1/(1/E) = E, same for 1/(1/8) = 8. Your joke works better if you drop of the words "the reciprocal of" or turn the 1/E and 1/8 into just E and 8.

[@YunxiaoChu]

@@GabrielMirandaLima-hv7oedafuck?

[@santerisatama5409]

Butt in horizontal position ɛ and then waking and sitting up to cough the morning cough in horizontal position ω is indeed a quarter rotation. But I do suspect there is also something else to math besides Greek letters making their butt jokes.

[@Bodyknock]

I've never heard people say "repeating 9", I've always heard it as "9 repeating". Is it actually phrased differently in different parts of the world?

[@ThoughtThrill365]

Yes, we had the discussion internally and did some research. There isn't much literature online discussing the pronunciation or geographical differences, and but in the end sided with the conclusion from a person with an Asian background.

[@n7x]

I always heard it referred to as “.9 recurring”

[@kingcobraarchie]

I noticed that while watching. In a similar way whenever a fraction was multiplied he would say times quantity and then whatever the fraction was. For example in the infinite series part he said 9 times quantity 1 over 10. It just stood out to me

[@dommorris8163]

Yeah i learned "[digit] recurring" in school. Pretty sure ive also heard "[digit] recursive". The word always came after the relevant digit though.

[@justdavelewis]

I also used “x-recurring” - uk based

[@jrgamma]

The most rigorous proof is as follows. Real numbers are equivalence classes of Cauchy sequences of rationals, where sequences x,y are equivalent iff |x_n - y_n| -> 0 as n -> infinity. Take x to be a sequence defined by x_n = 1 for every n, and define y as in the video, where y_n is 0.999…9 with n 9’s. Then we see that the sequence y represents 0.99999… since y converges to 0.99999…, and clearly |x_n - y_n| approaches 0. So by definition the reals 1 and 0.999… are in the same equivalence class and so they are equal. QED

[@Taubenkill0r]

Yes, I think that proof is the one that really gets to the core of the matter. A proof that doesn't start with a definition of what the real numbers are leaves something unsaid. Once one understands that the real numbers are groupings of many "objects" into one number it becomes really clear.

[@Censored_Truth_Addict]

This isn't convincing to me. At best it just proves 0.9999 is not a real number, or that we were mistaken to assume it approaches 1. (it was 1 all along, as you conclude)

[@Relrax]

​@@Censored_Truth_Addictyes. technically neither 0.99999... nor 1 are real numbers, they are merely representatives of the same equivalency class (and thus the same real number) Like in the comment above, if you use the definition of the real numbers, showing that 0.9999... and 1 are in the same equivalency class is trivial. This entire issue only arises because schools don't teach what the real numbers are, and why the field of real numbers is our desired object to work with.

[@lantzevongkorad4084]

Very nice, but you seem to have “hand-waved” some parts.

[@archon8255]

When your math exam says "Prove your answer" so you make up random shit

[@gustavsturksteinwall4027]

When you know your answer is incorrect but you still need to motivate it

[@teldd4930]

yeah except this isn't random shit 👍

[@felipejonas7655]

But this is right

[@chickennuggetman2593]

how​@@felipejonas7655

[@HunsterMonter]

​@gustavsturksteinwall4027 When the answer is right but you made a mistake and refuse to admit it.

[@ghostlore3660]

these proofs aren’t what people need to hear to understand it, I think. The real key is understanding that infinity is NOT A NUMBER. it can’t be quantified, and it simply does not behave like a number. there is a whole series of paradoxes that can be easily answered with the simple truth that infinity is a concept, not a number.

[@greatthodric2936]

Exactly! Infinities cause paradoxes. Paradoxes are illogical. Therefore 0,999... = 1 is an illogical statement. 1/3 = 0,333... Is also an illogical statement. It's a problem with base 10. In base 12 we don't have this issue. 1/3 = 0,4.

[@michaelpristin9944]

My favorite (most complicated but simplest) proof is the topological proof - the real numbers are Hausdorff, meaning that if 0.999… =/= 1 then there exist two disjoint open sets containing 0.999… and 1, which is not possible.

[@YouFydes]

"which is not possible"... but why?

[@michaelpristin9944]

@ good question. The definition of an open set U in R is that any point in U has an epsilon-neighborhood completely contained in U. In particular this implies that there exists some nonzero positive number epsilon such that 0.999… + epsilon is in U. But if you add any nonzero number to this 0.999… you get a number larger than 1. Hence 1 is in U and so U cannot be disjoint from an open set containing 1.

[@YouFydes]

@@michaelpristin9944 so you should prove that 1 - 0.999... < epsilon for all epsilon ... which is the proof using the limit 😅

[@piglava]

@@michaelpristin9944this feels like a really roundabout way of saying the limit of 0.9999… equals 1, in which case “0.999… = 1” by definition of (convergent) infinite series. Still a cool approach though. (and of course these ideas are equivalent in R, anyhow)

[@faner8653]

What is hausdorff?

[@bakamitai6654]

I fully know and accept the fact that 0.9999… is equal to 1. That being said however, I’m still fucking pissed off that’s the case.

[@xinpingdonohoe3978]

I think it's cool. It's like a byproduct of the base we choose. In base 2, 0.111...=1 In base 3, 0.222...=1 In base 4, 0.333...=1 In base 10, 0.999...=1 In base 16, 0.fff...=1

[@michaelpristin9944]

A little interpretation that made this less infuriating for me is the fact that infinite decimals don’t really exist on their own from base axioms - you have to formally define what they even mean which is why the result follows. For example 0.33… is meaningless unless you suppose that this number is defined to be the limit of its partial sums, which turns out to be 1/3

[@Ghork1]

Tbh, I don't agree at all. It's functionally the same. But it's seperated by 1/infinity

[@bakamitai6654]

@ as annoying as it may be for me 1/inf is not a real number, and thus cannot be a difference of two numbers

[@xinpingdonohoe3978]

@@bakamitai6654 1/∞ can be a real number. Look up Hölder's inequality. That's a context where we sometimes evaluate 1/∞, and it evaluates to 0.

[@marcelob.5300]

X-elent.

[@golem778]

See what's always annoyed me about this one is no math teacher ever explains that in our system of math infinitesimals don't exist, so when people try to explain 0.9.... = 1 using proofs like this, it doesn't actually answer the real reason it feels wrong, so it just feels like you're being gaslit

[@Chris-5318]

Decimals cannot represent non-zero infinitesimals.

[@golem778]

@@Chris-5318 Well I don't know the exact nomenclature or representation standardly used for infinitesimals, I'm just saying the whole 0.9... = 1 thing is based on a system of math that discludes infinitesimals, and that if you were using a system that included infinitesimals then the two would not be equal.

[@Chris-5318]

@@golem778 Robinson's hyperreals have infinitesimals but the equivalent of 0.999... = 1 remains true because of the transfer principle. Exception, some authors use an abysmal redefinition so that 0.999... is infinitesimally less than 1. But that breaks decimal arithmetic. For instance, with that redefinition, there is no decimal for 1 - 0.999... or for 10 * 0.999... - 9. That redefinition corresponds to the Lightstone extended decimal 0.999... ; ...99900... = 1 - 1/10^H (where H is a hyernatural) rather than the more obvious 0.999... ; ...999... = 1 A natural surreal (due to Conway) equivalent of 0.999... is { 0.9, 0.99, 0.999, ... | 1. , 1.01, 1.001, ...} and that is also 1. The surreals include infinitesimals.

[@johnlabonte-ch5ul]

​​ K Chris only wants to debate college grads. This discussion is open to anyone. From a middleschool math pov I see the problem is in how we represent numbers. There is no numeral that can be constructed between ".99..." and 1 in base 10 place notation of numbers even if infinite digits are allowed. That still doesn't mean that ".99..." is 1 or is even next to 1. I can't debate his last reply to you, but I can discuss how I have been educated on the basics of math.

[@golem778]

​@ Well again only if you assume infinitesimals don't exist. In a system of math with infinitesimals yes you very much can have a number between those two. Since we don't use that system though there is no number between and the numbers are exactly equal to eachother. My problem is just that we use this rulebook and don't tell students that we're using these rules, and then use proofs that also assume these rules as if the proof means anything to those unaware of the rules

[@FusionReborn95]

5:26 Secure, Contain, Protect

[@PAGMAOnline]

Engineers: See, exactly the same measurements.

[@ishkanark6725]

Somebody shared a story where their prof. Rounded pi to 10.

[@Glaster6]

@@ishkanark6725 bro how 😭

[@DrakeDenney-nd3go]

I knew it was going to get extremely complex when I started to get confused three minutes in

[@dragoonsunite]

I like to imagine this videos origin was someone arguing that .9999 _ didn't equal 1 and this video was created to post as a response to that person.

[@cshairydude]

The duplicates are far from being a quirk of decimal expansion. In fact, _any_ representation of the real numbers by infinite strings over a finite alphabet that can be used to compute computable functions _must_ have multiple representations of some numbers. If it didn't, either some real numbers are unrepresentable and so it's not really a representation at all, or functions that should be computable can't be computed using that representation. "Computable" means "computable by a halting algorithm" for a function whose result is finite; when the output may be infinitely large, such as an infinite decimal expansion, we mean instead "computable by a productive algorithm", where "productive" means _the next character of the output_ can be computed by a halting algorithm. Since reading the input takes time, this implies that each digit of the result depends on a finite amount of the input. Therefore for real numbers the functions that are computable in this sense are precisely the uniformly continuous ones. A representation maps each finite prefix to an interval. For the representation to be complete and unique, the interval can't be open (because it would leave out the end points), nor closed (because the intervals would overlap in at least one point, so representations wouldn't be unique). But if they are half-open (as those who deny that 0.99.. = 1 believe), some uniformly continuous functions become uncomputable, because when approaching the open end of an interval, a finite amount of input can get you arbitrarily close to the boundary but cannot give you the boundary point itself.

[@briant7265]

For any a < b, the mean, m=(a+b)/2, is between them, a < m < b For a=0.999... and b=1, m=1.999.../2 Perform long division, m = 0.999... m = a, thus b = a, thus 0.999... = 1

[@vincentiusthegreat]

O.9 repeating isn't a number, so it cannot be represented as a variable, so therefore your premise is invalid

[@tfx9223]

@@vincentiusthegreat how is it not a number? Because it has an infinite amount of places? Pi is a number with an infinite amount of places, and I don’t see many people refuting that.

[@vincentiusthegreat]

@@tfx9223 we don't do math or use pi with infinite digits, for the reason that such a thing would be impossible. Our numeric approximation of pi is serviceable, and we never use pi in the same way as 0.999 repeating is used here. The problem with using 0.999 repeating here is also not because it has infinite digits, but that you're trying to calculate using those infinite digits.

[@Mufasa4142]

for a = 0.999.... = 9/10 + 9/100 + 9/1000 + ..... = lim n --> inf 1 -1/10^n and b = 1 m = (a + b )/ 2 = lim n --> inf (1-1/10^n + 1)/2 = lim n --> inf 1 - 0.5/10^n Now the question here is when a number is divided by 10 infinite times, will it be equal to zero. Well no, because by the principle of mathematical induction we can show that 0.5/10 0. If 0.5/10^n 0, then definitely 0.5/10^(n+1) will not be zero (a process of division can never result in zero). So Principle of mathematical induction shows that 0.5/10^n will never be zero for any natural number n. This states that m will never be equal to one. When dealing with infinity, we have to understand that just as there is no number which can be attributed equal to infinity, there is no number whose inverse can be attributed equal to zero. Just like infinity is never ending and we can go on and on, 1/infinity is never ending and it goes on and on closer to zero, but never equal to zero. Example lim x --> inf 1/x approaches to zero. while lim x --> inf 1/x^2 also approaches to zero. Both are approaching to zero at different rates and their values will be different. The difference can be so much that one number divided by another can yield another infinity. lim x--> inf (1/x)/(1/x^2) --> inf

[@EthanOhlendorf]

I never expected to see a flawless parabola of mathematical difficulty in my life as flawless as this video 😂

[@aepokkvulpex]

My favorite fractions test is pointing out how ninths work. 1/9=0.111..., 3/9=0.333...(aka 1/3), etc. And so obviously 9/9=0.999..., and 9/9 also = 1

[@super.heraut.officiel]

that is false. 9/9 = 1. 0.999999... = (9/9)-0.000...0001(this 1 being at infinite position).

[@tzorfireis425]

@super.heraut.officiel That’s not how that works There’s no such thing as an infinite position. Every single possible digit in a number is a finite position. It’s like saying “infinity + 1 is bigger than infinity” That’s just not how it works

[@super.heraut.officiel]

@@tzorfireis425 i know that theory. but if people can pretend that 3x1/3 = 0.9999999... then i SHALL conceptualise the infinite position

[@tzorfireis425]

@@super.heraut.officiel That's not how that works either. Nobody's pretending that 3x1/3 = 0.99999..., that's a side effect of how the decimal system works because no power of 10 is divisible by 3. If we used base 12 (0,1,2,3,4,5,6,7,8,9,A,B), 1/3 would be 0.4, and 3x1/3 would equal 1, because 12 is divisible by three. And then we'd have to deal with .AAAAAAA... being equal to 1 for all the same reasons .99999... is equal to 1 And you can't just *say* there's an infinite position because there's no where for it to fit. For every finite number, there is another finite number exactly 1 less than that, and another finite number exactly 1 more. There is no finite number that is exactly 1 less than infinity. Meaning that every single digit position is necessarily a finite position, no matter how far you go. That's what infinitely repeating actually means. Unless you want to tear math down all the way to base axioms and simply define some finite number the absolute biggest number and force anything that would go above that to be infinite automatically, an infinite position *can't* exist. Flat out.

[@heh_boaner]

​​@@super.heraut.officiel if you divide 1 by 3, you don't get 0.333... 2 thirds of the time and 0.333...34 the last third. They are all equal. If you are uncomfortable with one of the last digits not being 4, then you can just split up the difference between the three parts.

[@AnimeStories_tr]

The infinite numbers between 0 and 1, you can add infinite amounts of 9's behind 0,999..., it will always come closer, but it will always come a little shorter than 1. Infinity itself is not a number. And that goes for the infinite amounts of 3's as well. Therefore, you can't take 0.333... into any calculations because you will never reach the last digit. What this means is that, you can't multiply 3 with infinity. That's why you have to make the calculation with 1/3, not with 0,333...

This is like an overflow problem in real life

[@ГеоргийСамбуров-ф8ь]

Same level as arguing that gojo's infinty completely stops the object somewhere close to him

[@black_m1n825]

Assume 0.999... and 1 are not equal. Which means that there is a number between 0.999... and 1. However such a number doesn't exist, since we already filled up 0.999... with as many 9's as possible (infinitely). Therefore 0.999... and 1 are equal.

[@xinpingdonohoe3978]

Yes, there's no way to make a decimal starting with 0 that is bigger than all 9s.

[@Ghork1]

That would be like saying that 1 and 2 are the same number when looking at intigers because there is nothing between them. It's just infinitely close to 1

[@xinpingdonohoe3978]

@Ghork1 no, the integers don't form a continuum. They are a discrete set. The real numbers are a continuum. They are not discrete.

[@Ghork1]

@@xinpingdonohoe3978 yes, and being infinitely close shows it to be a continuum

[@teldd4930]

​@@Ghork1 Nope, the proof works, because R is not a discrete set; it describes a continuum of values. Because of this, if you take two different real numbers, I can ALWAYS find a number between them. This means that, if I can't (such as with 0.999... and 1), then they must be the same number. But if we take only whole numbers, that is not a continuum: that's a discrete set. Which means our earlier statement, where I said I could always find a number between any two numbers you give me, does not apply. So, no; saying that 0.999... = 1 does not in fact correspond to saying that 1 = 2

[@rqmxn.]

My favorite proof of this (since it is the most rigorous) uses the Dedekind cut construction of the real numbers in order to show that 0.999...=1. The proof essentially amounts to that the two Dedekind cuts are the same, so the two numbers are in fact the same.

[@ccbgaming6994]

💯

[@funnyfish1982]

2:38 You can also use this function, replace n with infinity and add (1/2^k), this means 1 devided by 2 to the power k. That should also give you one.

[@Ang3lUki]

The hyperreal numbers one is so silly, like it’s just mathematicians declaring “if you believe this, you’re breaking the rules, and the number you believe in is 0”

[@retroshredmc]

yeah it’s like imaginary numbers like they couldntbfigure out how it exists so they just said “it’s not real fuck you”

[@SolutionInnPro]

This was such a creative way to showcase the depth of math! 0.999... = 1 has never felt more justified. Having SolutionInn as a study companion makes tackling even these proofs a rewarding experience!

[@jdlech]

What convinced me was going backwards into it. 1-0.000... = 1. Why? Because there is no 1 at the end of 0.000.... It's simply zeros forever. No digit, no matter how insignificant, ever gets to be anything but zero. And what is zeroes forever? 0, of course. So if 0.000... = 0, then 1-0.999... equals 0. Because there is no end to the 9s just like there is no end to the 0s. Thus, 1-0.000... = 1-0 = .999... = 1. It's all the same thing expressed 4 different ways.

[@ryangosling239]

you cant do 1-0.(9) though

[@jdlech]

@@ryangosling239 sure you can. Since .999... = 1 and 0.000... = 0, then 1-0 = .999... and 1+0.000... = 1 Law of substitution.

[@HouseHoldAdventures]

But it's not 1, it's .99 forever which isn't 1

[@jk-2053]

​@@HouseHoldAdventures That's like saying 1+1 isn't 2, or 0.5 isn't 1/2, or 1/1 isn't 1. They're equal, hence why you can use the Law of Substitution, so they're the same (e.g. 1+1+1+1 = 4 is the same as 2+2=4, it's written differently but I fail to see how that's significant).

[@jdlech]

@@HouseHoldAdventures Except that it is.

[@bunnycarrot9937]

The problem with the first example is that during the division you have to throw away the 1 to reach infinit precision

[@tsume_akuma8321]

Have you considered using a background that is less of a flashbang?

[@fshi1525]

my guy its the colour white, do you live in complete darkness? if so, understandable

[@tsume_akuma8321]

@fshi1525 No, but I watched it in the evening with the light turned down. Got flashbanged.

[@fshi1525]

@@tsume_akuma8321 think faster chucklenuts 🤷

[@ThoughtThrill365]

By popular demand, we will produce a future video with a dark background.

[@TheColonThree]

need this!

[@johnlabonte-ch5ul]

As a middleschool math, I agree with the Real Number line. (now Rn=Real Number) The line is infinite, and infinitely dense forming a continuum. (The concept of infinity is dangerous.). Each number on the line is unique and precise. Fraction numerals exist on the Rn line in the form #/× where # and × are usually integers. There is also an operation on Rn called division with a symbol #/×. To differentiate, #//× will be used to indicate "divided by".(If both # and × in #//× are integers, the symbol #/× is rational. RATIOnal numbers are the RATIO of 2 integers.) 1//3 does not equal ".33..." as the operation is incomplete and the value inconsistent at each step in the operation. On the other hand, ⅓ as a symbol is unique and precise. ".33..." does not equal ⅓ exactly. ".33..." has a limit of ⅓ as each step of the division is completed. The valuable concept of infinity is dangerous, incomplete, inconsistent and imprecise.

[@Chris-5318]

You are a muppet school math master. Your entire comment is balderdash written in gibberish. That's because you are a troIIing muppet. I've dealt with every one of your false claims before. e.g. 0.333... does not have a limit. You've been at this for over a year, yet you still don't know the difference between sums and limits or series and sequences.

[@johnlabonte-ch5ul]

Good to know, I guess. lol

[@Chris-5318]

@@johnlabonte-ch5ul It would be helpful if you actually bothered to learn the terminology. Being at this for over a year and still not knowing what sums, limits, series or sequences are is incredible. You are very speshul.

[@Chris-5318]

@ThoughtThrill, you really ought to change the video title. It's 0.999... that equals 1, not 0.999

[@VividBoricua]

I'm glad there's a number system out there that deals (or rather attempts to deal) with the absurdities and inconsistencies of infinity and infinitesimals. I've always been of the opinion that our number system must be lacking in some way to define 0.999... = 1. There is a number between 0.999... and 1. It's 0.999... . Admittedly, my interpretation does not work with real numbers. Luckily we have improved with hyperreal numbers. lol

[@selphy_geumja]

If i ask to someone how much is 0.9999999... missing to reach 1? Everybody should answer: 0.000000...

[@mitchhudson3972]

0.0...1

[@alfred7846]

@@mitchhudson3972 how many 0 before you reach the 1?

[@JustifiedNonetheless]

This is a perfect example of why conceptual, linguistic, and propositional differences (descriptors) do not imply a difference in the referent.

[@AffanKarimullah]

My math teacher always used to say that fractions like 1/3 is not = 0.333 because it is just an approximation. He made a rule that we are not allowed to use decimals in his class and instead use fractions because it's more accurate. He was the best math teacher I ever had

[@godfreypigott]

He would have said that 1/3 is not 0.333. He would NOT have said that 1/3 is not 0.333...

[@AffanKarimullah]

@godfreypigott oh yeah, you're correct. It was a typo from me. Sorry for the misinformation but he did say it just how you said

[@Nikola_M]

in the schools i went to in Austria we marked infinitely repeating decimals by putting a dot/line over all repeating decimals, so 1/3 would be 0.[3 with a dot above it, i didn't find a way to display it with unicode]

[@paytondettmer7315]

I feel like it’s one of those things where it’s just close enough. I can’t think of many scenarios where basically 1/infinite will make a difference, plus it makes math easier for a 10 based system where we have continuous repeating numbers due to fractions.

[@Chris-5318]

1/oo = lim n->oo 1/n = 0

[@fdileo]

It would be interesting to know more about hyperreal numer

[@paultapping9510]

numberphile and mathologer are both very good and have probably covered hyperreals

[@RSLT]

😂 Definitely, non-standard analysis shows that all of these are false.

[@RSLT]

Definitely, non-standard analysis shows that all of these are false. Furthermore, in standard analysis, it also makes little to no sense to claim that .999... equals 1.

[@ethanbottomley-mason8447]

You sound like you have no idea what you are talking about. In standard analysis, it is perfectly sensible (and true) to claim that 0.999... is exactly 1. You need to define 0.999... in a sensible way (it is a Cauchy sequence in a natural way, as are all real numbers). From there proving that it is equal to 1 is trivial and follows very quickly from the definition of R as equivalence classes of Cauchy sequences in Q.

[@RSLT]

@ Let me answer on behalf of @fdileo. First, you’ve probably heard of contradiction (I know you have, but I can never know for sure). If one system says 0.999... = 1 and another system says they are not equal, then one system must be correct. Or, you would really have to have some very special case that allows both systems to coexist without contradiction. What he meant in a nice way is this: Are you kidding me? It sounds like you've never heard anything about hyperreal numbers. If you had, you'd know that they cannot be equal. When you make these claims straight out of a real analysis textbook, it feels to a math PhD like a kid saying x^2 + 1 has no roots because of what they learned in high school, without understanding that in higher-level math, we have complex numbers. I have a playlist called 'The Naked Emperor,' where I have fun with a collection of videos about 0.999... and 1. Just to give you an idea of how silly it is to claim that an infinite-digit number is a real number-it’s like saying pi is a rational number just because we can divide two infinite-digit numbers and get the pit. So, if you accept that infinite-digit numbers are real numbers, then by the same logic, pi should be a rational number.

[@Сасичлен666тотсамый]

The main thing here is that 1/3 is a number in R. I mean the confusion here is about 0.(9) being so close to 1. But in that way 0.(3) is also so close to something. And by being so close to something it's naturally define itself not as a number or a dot to say, but as a function in a way that it's form a subset that is nearly close to that thing. And the problem here is that in rational numbers 1/3 IS NUMBER and to say it's also a dot. And we just for a sake of rational numbers say that 0.(3) is a number and moreover that all subsets that are really nearly close to some point are those points. In that case 0.(3) is a number and a dot. And 0.(9) is a number and a dot, and it's a 1. And yes, by saying that all nearby subsets are equal to what they are nearby we also say that all nerby subsets to those subsets are equal to them as wel. And you can try to imagine what will happen if you try to say otherwise. And this axiom also include deep connections with axioms of separability.

[@Chris-5318]

You really need to learn the difference between a numeral and a number. Numbers are pure abstract - they don't have digits or dots. 0.(3) = 1/3 and 0.(9) = 1. You can choose a number to be as close to another number as you like. But you can't get closer then being the same number.

[@Сасичлен666тотсамый]

@Chris-5318 you mean that numbers can't be described as a dots? Simple to say i used more of a visual part of this thing, bc if you glance at 0.(3) for a first time as a dot on a line you might not see a dot, but dots. And to describe it you will eventually use some sort of a function. So I just tried to say, that 0.(3) is a number not a function, or so called dot but not dots on a line, and that 0.(9) is also a number or a so called dot just like 0.(3). And that they are not only numbers but they also equal to what they nearby.

[@Chris-5318]

@@Сасичлен666тотсамый This is getting confusing. You hadn't previously mentioned the number line. I thought you were talking about the decimal point. You have to realise that the drawing of a number line and points on it is just an illustration. The number line is actually a pure abstract concept. It is not a line drawn with a pen on a piece of paper. 0.(3) is a numeral, not a number. It can be indicated as a point on the real number line. It is the same point that indicates 1/3. 0.(3) is a concise notation for a constant formal series that has a sum. It definitely is not a function. Its sum is also represented by the rational numeral 1/3. Of course in everyday use, numerals usually called numbers because it is understood that that is what is meant. However, when we are looking at, especially whether or not infinite decimals represent a number and how they can do that, we need to be more careful and spell out exactly how we are to associate a numeral with a number. It should be obvious that most people are not really aware of how that is done. Hence the arguments on pages like this.

[@johnlabonte-ch5ul]

In terms of sets, it looks like ".99..." is an open set, (.99..9,1) where .99..9 is an unimaginable large finite number of 9s following the decimal point.

[@Chris-5318]

@@johnlabonte-ch5ul 0.999... is not a set (unless you are talking in terms of a Dedekind cut). No matter how large the natural number of 9s is in 0.999...9, your open set has an uncountable infinity of numbers greater than 0.999... and less than 1. It does NOT represent 0.999.... Your set is the same set as (0.999...9, 0.999...) and, FWIW, does not include 0.999.... As you have introduced intervals, consider the following infinite union of closed intervals: [0, 0.9] U [0, 0.99] U [0, 0.999] U ... =[0, 0.999...) = [0, 1). Hence 0.999... = 1. That's because none of the subintervals contains 0.999... or 1. Infinitely many of them contain your 0.999...9 though.

[@mahmoodayesh6706]

People believing that this is not true and believing that 1+2+3+....=-1/12 is crazyy😂😂😂

[@XDGB97DX]

its all fun and games until the mathematicians ask "but why?"

[@matthewsouthall5021]

What's the average of 0.999... and 1? If the average is equal to either number, then those numbers must be equal.

[@cubed.public]

It can be even more easily explained as to find the average, you need to find the middle of the difference. What is the difference between 0.999... and 1? It must be 0.000..., so therefore there is no number between them

[@DracoSuave]

The one I like is based on the definition of the value of an infinite series of digits. The value of an infinite series of digits (for a positive number) is equal to the supremum of the set of finite partial digital representations. So, 0.999... is represented by the set {0, 0.9, 0.99, 0.999, ....} and so on. Every value is less than 1, so the series is bounded above and 1 is an upper bound. Because this is a set of real numbers bounded above, there MUST be a supremum. And because any arbitrary number less than 1 is less than some member of the set, 1 must therefore be the supremum of the set, and therefore, must be the value of 0.999...

[@jsalsman]

All are spectacular, I can't choose!

[@TheBluePhoenix008]

It was interesting to see my highschool journey in this video. It first starts with class 8 methods, and slowly goes to highschool maths, then to what I'm studying right now. I love weird maths.

[@ccbgaming6994]

I feel the same lol

[@R.B.]

When I learned this concept, I came up with my own "proof," which might not stand up to academic rigor, but it satisfied me that I reached a similar conclusion based on what I was being told. Use long division to solve 1/1. The answer should be 1, but instead assume it is 0. The next place would be 10/1. Were you to solve it as 10, you'd carry the 1 when adding and feet the answer you seek, but instead guess that it is 9. Subtract and you'll have a remainder of 10/1 for the next place. Using long division, you effective solve that 1/1=0.99... solving each place with this intentional underestimation. If at any point you solved 10/1=10 for any place, then the tens 1 would carry and ripple up each higher place until you get 1/1=1.0... or extended out an infinite number of terms, it would be that 1/1=0.9... It is then that I tried other long division problems like 4/2 and would intentionally underestimate or overestimate each term and observe that whenever you apply the correct answer for one of the places, the correction will resolve to the correct answer, so understanding each term doesn't provide an incorrect answer when fully carried out to a remainder of zero. This was before I had any exposure to limits or calculus, but it was enough to convince me that what I was being told could be accepted as fact until I had more substantial proof.

[@johnlabonte-ch5ul]

You don't guess in long division. Also you verify your answer by reversing the division.

[@R.B.]

@johnlabonte-ch5ul maybe "guess" doesn't explain it fully. As you are building the quotient, you are trying to find the value which when multiplied by the divisor which gives you the smallest positive remainder for that unit. 16/4=4*10^0, but also 16/4=2*10^0+20*10^-1. The answer is the same when you combine terms, 4 and 2+2.0. As such you don't need to be precise with each unit as long as you are going to reconcile things. This can actually be a useful technique for solving a division since you have a chance to break the problem down into what may be easier to solve sub-problems. There's nothing which says you need to under "guess" those terms either. 16/4=5*10^0+-10*10^-1 is also a valid solution. Once you see that you can be incorrect on that term because you will eventually consolidate the terms and find that it gives you the same answer, then the following makes sense: 1/1 = 0*10^0 + 9*10^-1 + 9*10^-2 + 9*10^-3 + ... When you extend this series out to infinity because of associativity and commutativity, you reconcile that 1/1 = 0.999...

[@Nekoma_is_flexible]

@@R.B. "When you extend this series out to infinity..." - it never ends, so there's no answer

[@ChadFaragher]

You can derive it in reverse for those of us that learned long division. You set up 1)1̅ and you write the first digit as a 0, leaving remainder 1, then bring down a 0 from the tenths column to form 10. Then write 9 above the tenths column, then 9 times 1 is 9, and 10 minus 9 is 1 and you continue bringing down 0's and writing 9's forever. I find the procedure is as compelling as the bar notation itself (to embody the notion of an infinitely repeating digit).

[@johnlabonte-ch5ul]

That is not long division in reverse. You use long division in reverse to prove your long division is correct. What is long division in reverse, multiplication I called, what you described, anti-math. It also works for 1/3. You could use the formula x/9 where x is a digit, to get "xx..." it has a counter example. If x =9 then 9/9=1, not ".99..."

[@ChadFaragher]

@johnlabonte-ch5ul That's not a counterexample. You're mistaken if you think 9/9 does not equal 0.999... repeating.

[@johnlabonte-ch5ul]

It is really interesting what 9 can do in base 10 numbers. Also there are more correct ways to show unusual properties. You have to first invoke this videos discussion that ".99..." is 1 before you can say, since 9 divided by 9 is 1, then 9/9 is ".99...". You don't guess in division without verification. In math you should always verify. Middleschool basics.

[@manolosardo3661]

Op 76, 2 jean Sibelius in background

[@darkarchon2841]

This mathematical fact was one of inspirations for me to do my Bachelor of Mathematics thesis (I don't recall how it is called exactly, so I am using this word) on non-standard analysis.

[@nikolakosanovic9931]

If it ever comes in conversation I would say what is the difference between them. They will probably say 0. infinitely many 0 then 1. Then I will ask them if there are infinitely many 0 what is the value

[@xinpingdonohoe3978]

Explain the flaw. Explain how there can't be "then 1". There isn't an ∞th decimal place. That doesn't make sense. Just one decimal place for each natural number n, and each of those has a 0 in it.

[@nikolakosanovic9931]

@xinpingdonohoe3978 I know that ∞th decimal place doesn't make sense. That is argument. If difference is is ∞ zeros then 1 what is in practice (0) numbers have to be the same. Ask anyone is that correct

[@xinpingdonohoe3978]

@@nikolakosanovic9931 precisely, but when you mention "then 1", you'll get it into people's heads that somehow there is a 1 there, when there isn't. There's only 0s.

[@nikolakosanovic9931]

@@xinpingdonohoe3978 I am pretty sure that they will say that difference is ∞ 0 then 1

[@jonathanparsons461]

Unfortunately if you have someone who is waiving even one of the proofs here, they are most likely disingenuous and will most likely say "Nu-uh!" to every single one of them. For example, they most likely will not accept the "limit definition" as being true even if shown otherwise. My favorite argument for why 0.9999... = 1 ends up being the fractions test for this reason. The way we are taught in grade school is to arrive at an answer, you have 3 shake hands with every digit of the number you are trying to multiply (i.e., 0.33333...). When 3 shakes hands with a digit, that being 3, it becomes 9, and you shake hands over and over again until you are done. Similarly with fractions, you have 3 shake hands with the numerator of the fraction (i.e., 1/3) and it becomes 3/3. This is the best working model of how to perform multiplication in math. Until you find a model that provably does it better with no error, you don't have a model and you must accept the best working model.

[@anonym5160]

A: 10÷3×2+10÷3 3.33‾×2+10÷3 6.66‾+10÷3 6.66‾+3.33‾ =9.99‾ B: 10/3×2+10/3 20/3+10/3 =10

[@Why553-k5b_1]

And?

[@HunsterMonter]

​@@Why553-k5b_1 It means 0.999... = 1, which a lot of people seem to have a hard time to accept for some reason.

[@DamienHaga]

I love the thousands of random people thinking they know more about math then all mathematicians. Simply exemplary

[@wesleydeng71]

If 0.999... is different from 1, then there must a number between them. But what is it?

[@Bruh-bk6yo]

(1+0.(9))/2 Problem, professor?

[@massterlomasster]

@@Bruh-bk6yo it will be 0.999......999995, which less than 0.(9)

[@Bruh-bk6yo]

@@massterlomasster who said it will not be a periodic number?

[@xinpingdonohoe3978]

​@@Bruh-bk6yo what's the decimal expansion of (1+0.999...)/2 It's 0.999... But that means 1+0.999...=2×0.999... 1+0.999...=0.999...+0.999... 1=0.999...

[@Schnorzel1337]

@@massterlomasster YOU can not state. I put a zero a dot and infinite amount of 9 and then 5. I will forever dance and sing. After that I will be silent. Does that sentence make sense?

[@hey...4702]

The way i explained it to a kid who asked (whom i babysit) was that 1-0.9999... equals 0.000000...1, which is basically 0. Prob not the best proof, but he's only 7 so doesn't rlly understand fractions yet

[@Chris-5318]

There is no such decimal as 0.000...1. You can't have a 1 at the non-existent end of an endless string of 0s. 1 - 0.999... = 0.000... = 0. If it was OK, then 1 - 0.000...1 would be 0.999...9, not 0.999.... OTOH, surely 0.999... + 0.000...1 = 0.999...1. What do you think 0.000...1 * 10 and 0.000...1 / 10 are?

[@mattsgamingstuff5867]

Not quite there but pretty close to getting there. You can assume they are non equal and demonstrate that if they are not equal the difference would be an infinitesimal. And since there are no infinitesimals in the real numbers you have a contradiction that the difference between two real numbers is not a real number. Therefore you conclude that the false assumption (10.999...) is wrong and the inverse (1 = 0.999...) is true. Proof by contradiction is probably a bit much for a kid though.

[@Chris-5318]

@@mattsgamingstuff5867 You are confusing the series 0.999... with the sequence 0.9, 0.99, 0.999, .... That's the thing that doesn't get to 1. It also doesn't get to 0.999... either. But it does get arbitrarily close to both 0.999... and 1. It cannot be getting arbitrarily close to two different numbers, therefore 0.999... and 1 are the same number.

[@danielpazekha9758]

​@Chris-5318 Yes there is. It's called in infinitesimal. Clearly no one here has ever done calculus here

[@Chris-5318]

@@danielpazekha9758 Decimals cannot represent non-zero infinitesimals. See my first comment as you clearly don't understand what a self-inconsistency is. Something that is endless doesn't have an end - what part of that aren't you able to grasp? Clearly you do not know the definition of decimal notation. Every decimal place is indexed by an integer. Which integer indexes the 1 in your wannabe infinitesimal 0.000...1. Be aware that infinity is not an integer, or a number of any kind. Some of us have studied up to Calculus 7 at university. Some of us know that Calculus 2 (and all standard Calculus) does not deal with infinitesimals. Some of us have knowledge of nonstandard analysis based on Robinson's hyperreals. Some of us know that the numeral you want is 0.000... ; ...0001000... using Lightstone's extended decimal notation. Note carefully that that is NOT 0.000...1.

[@teh_kaczuch]

and here comes nonstandard analysis with infinitesimals

[@RSLT]

Definitely, non-standard analysis shows that all of these are false. Furthermore, in standard analysis, it also makes little to no sense to claim that .999... equals 1.

[@Lightning_Lance]

Imagine you had a large aquarium, and you filled it with increasingly smaller buckets. Each bucket fills the remaining emptyness by 90%. So the first bucket fills the aquarium to 90%. The next bucket fills it to 99%. Then 99.9%. Then 99.99%. Clearly, the aquarium will never be completely full. You will just edge closer and closer to it being full without ever getting there. This is why your intuition is telling you that 0.9repeating does not equal 1. But if you had infinite buckets and you spent eternity filling the aquarium. Then what would happen at the end of that? Well, there is no end to infinity so the question doesn't make sense. But IF there was an end, then the aquarium would actually be full. The takeaway isn't that you're wrong to feel weird about it. The takeaway is that infinity is weird.

[@johnlabonte-ch5ul]

The valuable concept of infinity is dangerous, incomplete, inconsistent and imprecise

[@Chris-5318]

@@johnlabonte-ch5ul LOL. Only in your hands.

[@Chris-5318]

Your aquarium will never get to 99.999...% full either. The end of eternity does not exist. You can never finish filling the aquarium. There will never be a case where you pour the last bucket. (Obviously the water is mathematical water that isn't made of molecules). Be clear, the complete infinite sequence 0.9, 0.99, 0.999, ... does not include 0.999... or 1. The limit of that sequence is both [the value of] 0.999... and 1.

[@thuroria7631]

"But if you had infinite buckets" But you don't. Infinity does not exist. Infinity is not infinite, it is UNDEFINED. "Then what would happen at the end of that?" There is no end, but you can choose an arbitrary stopping point. "But IF there was an end, then the aquarium would actually be full." Meaningless statement. If my mom had balls she would be my dad. "The takeaway isn't that you're wrong to feel weird about it. The takeaway is that infinity is weird." The takeaway is that infinity does not exist and assuming that it exist is the cause of this weird behavior. Logic like this is what has blessed up with logarithm of 1 being undefined and division by 0 being undefined. Because it's faulty mathematics.

[@Chris-5318]

@@thuroria7631 You do not know what you are talking about. Infinity exists in mathematics. How is it possible to be over about 5 years old (wild guess) and not know that? Starting from the first bucket, the tank would become filled to 0.9, 0.99, 0.999, .... If you could complete that sequence, you still would not have 0.999... because 0.999... is NOT in that sequence.

[@Nurlan270]

1:08 you can multiply equalities?? so i can 5(y = 7) = (5y = 35) this feels cursed

[@Razorcarl]

Yea, this is called the Multiplication property of equality. As long as you multiply the same number on both sides, they remain equal.

[@iosefka7774]

Imagine two people who are the same height. If we then make both of them twice as tall, they will not suddenly be different heights from one another.

[@HPTopoG]

It can be formalized in various ways, but yes. I like to think of it using the Leibniz substitution rule.

[@fedoraguy6774]

Yes but this is the most cursed way to show it

[@StormySensei]

You can do it, but the notation I would use is different. In America we would say "multiply both sides by 5" and show it as " y=7 5*y=7*5 "

[@pacattack2586]

fun fact assuming N>=2 (n-1)/n+(n-1)/n^2+(n-1)/n^3...+(n-1)/n^∞=1 (for example 1/2+1/4+1/8+...=1, 2/3+2/9+2/27+...=1 etc)

[@mongooseV]

My favourite one is just to ask: 1 - 0.9999999... = ?? Which is inevitably an infinite amount of zeroes

[@ishkanark6725]

That's simple, 10^-infinity (Can't find the real symbol)

[@คนน่ารักrassdd]

@@ishkanark6725if you said 0.9999999…=1 ,1-0.99999999… should equal 0

[@shiinondogewalker2809]

I've tried this and got someone to insist that it's 0.00000... which is different from 0

[@worldball503]

​@@shiinondogewalker28090.00000.. is equal to 0

[@Existence111-qj5mw]

“Any number that does on a whiteboard basically rounds down to zero”

[@RandomDude236]

Using pretty stupid and flawed logic, I just thought of it like this: Let's think of the difference between 0.999(...) and 1. The difference must be 0.000(...) followed by a 1. However, you can't exactly follow an infinitely repeating string of numbers with anything, as there are always more zeroes. Thus, there's no possible place to put the 1, leading to the difference simply being 0. As such, 0.999(...) = 1.

[@cmyk8964]

That’s pretty similar to the simplest (IMO) proof. If two values a and b aren’t the same, then there is at least one value where a < c < b (or b < c < a, depending). Try as you might, you can’t find a value between 0.[9] and 1.

[@AravindKarthigeyan]

You can have an infinitely repeating string of something! Look up the p-adics! …111111 in the 2-adics is equivalent to -1!

[@irinaseif9691]

​@@cmyk8964great, now do a proof by contradiction that this value doesn't exist. (What my professor would propably say)

[@ryangosling239]

​@@irinaseif9691 ⬜suppose it exist, lets divide it by 10 again and again, writing it as a decimal number. It must be less than 1, so it would start with 0.9. If every digit is equal to 9, then its = 0.(9), contradiction. If the nth digit is not 9, then this number is less than 0.(9), contradiction. ⬛

[@briant7265]

​@irinaseif9691 Take the mean, 1.999.../2. Simple long division gives 0.999.... Contain that in a little algebra fluff and you're there.

[@thesugareater8607]

Every post like this has so many people who refuse to believe it

[@arthurgames9610]

0.99999... = 1-ε, where ε is an infinitesimal

[@RSLT]

Well, Said.

[@steamdiary9526]

Just as you can't have a number x, which is an infinite number, you can't have an epsilon which is infinitesimal. Both concepts can't be represented with a real number.

[@АлександрМарач-с8ы]

Ok... Than 1 - 2ε = ?

[@arthurgames9610]

​@@steamdiary9526I get wt ur tryna say. In real analysis and normal arithmetics we can't indeed have "infinetly small" or "infinitely large" numbers (being infinite, in real analysis, is a property of certain sets meaning there is an injective function from it to the natural numbers). But there are other kinds of arithmetics, field extensions of the real numbers. Search about dual numbers, which have a real part and an infinitesimal part (their geometry is rly cool btw, and their sometimes used in differential equations). A video that helped me a lot understanding arithmetics with infinite is "The Search for the Longest Infinite Chess Game" from the channel Naviary, there you can understand clearly the difference between ω, 2ω, 3ω, ω^2, ω^3,... I believe that 0.99999... can be seen as a dual number

[@arthurgames9610]

​@@АлександрМарач-с8ыBro, this is a really interesring question, and I had to sit and think. I think I have come to a conclusion, but I might be wrong: Let us take a look at 0.1111... in binary base. It seems to me at least quite intuitive that 0.222... in base 3 is closer to 1 than the first number, even tho both of them are infinitely close to 1. Furthermore, if we kind of did the subtraction, we would thee that 1 - 0.111... 0.000..1 (base 2) and 1 - 0.222 = 0.00...1 (base 3). Notice that, sice the first difference is in base, this "last 1" represents half of a figure, while the "last 1" from the second difference is 1/3 of a figure. Therefore, it seems right to me to define 0.111... (base 2) to be 1 - ε/2, 0.222... (base 3) to be 1 - ε/3, and so on, up to 0.999... which is 1 - ε/10. Hope this helped you, reply to me if this idea made sense to u.

[@chauphongchau5565]

This is how the Lawyer will convince the Judge at the trial….

[@felipejonas7655]

The lawyer would be right then

[@tlllllllllll]

If youre using decimal representation of real numbers then by DEFINITION 0.999(9) = 1. By definition, in decimal representation, a trail of 9s is replaced so that every number has a unique representation.

[@oliverly897]

One way or another, we will always get stuff like 0.9999... = 1 because in reality the countability of anything composed (aka the totality of the universe) doesn't care about our expressive notation of it all. The number 0.333 will always be 1/3. But in a different way of writing it down. Getting attached to the way we speak of and write about things doesn't change them for a fact. Our base 10 mathematical notation simply doesn't work for when writing about the third part of a whole. The first proof already contemplates that well.

[@juliavixen176]

The monotonous AI generated voice make it difficult to follow the arguments. There are no pauses or emphasis to distinguish what is important and what is meaningless filled words.

[@xinpingdonohoe3978]

AI? Just sounds kind of Asian to me.

[@sweetdannyandlisa6537]

This doesn't sound a lot like AI to me

[@TallinnCav]

Definitely not AI. Rude lol

[@Am_Cookie2436]

I'm gonna need this at some point even tho it's something simple made complex

[@DrJulianNewmansChannel]

Interesting video. But I don't think it's true that in the hyperreal numbers, 0.999... is an infinitesimal less than 1, unless perhaps your sequence of 9s is a terminating transfinite sequence. The usual interpretation of the word "recurring" is that there is a 9 in every finite position after the decimal point, but no notion of digits at infinity; under that interpretation, nought point nine recurring will either be undefined or exactly 1, depending on your scheme for evaluating decimal expressions.

[@denki2558]

No. Hyperreals operate with the transfer principle. All the properties of it's subalgeba of reals are retained. So 0.9(9) = 1 still. This is just a limitation of base 10 representation. Also, you wouldn't write a hyperreal "One minus an infinitesimal" with the notation 0.9(9). There's infinitely many infinitesimals so you can't pick which one 0.9(9) correspond to even if you accept that it's "one minus an infinitesimal". It could be 1-ε, 1-2ε, 1-ε², or any 1-f(ε) where f is a real polynomial.

[@maksymisaiev1828]

@@denki2558 well, the last assumption is also kinda wrong. By definition hyperreal is between any real and 0, so 1-0.(9) should give or hyperreal, or 0. It is directly stated that 0 is assumed as a solution. Problem mostly with repetition number is that they are kinda irrational. At least, for 0.(9) you can't represent it properly, as you bump into weird cases. Like yeah, 3/3 may be 0.(9), but by the same logic, 4/4 is also 0.(9), but 1/4 is then 0.2499...975, because we know that at some moment 75 will be the end of this number despite having infinite amount of 9s before that, but we can easily prove that there is a number between 1/4*0.(9) and 0.25, so 0.(9) can't be equal to 1.

[@jefflim3559]

Slice a whole circular cake into 3 equal parts. Each slice is 1/3. When you piece them back, you should get 1 whole cake. But if u look closely, some crumbs of cake have fall off or stick to the knife while slicing, that is your 0.0..01. So with this example, some may argue 0.99...9 may not equal to 1.

[@Chris-5318]

There is no such decimal as 0.000...1 or 0.999...9. Also when you cut a mathematical cake into 3 pieces, there are no mathematical crumbs on the mathematical knife.

[@johnlabonte-ch5ul]

KC's reply is correct, yet some still say ".99..." is not 1.

[@Chris-5318]

@@johnlabonte-ch5ul You aren't competent to judge if my reply is correct or not. You say that millions of degreed mathematicians are wrong. That's because you are a deIusionaI troIIing muppet.

[@VvxVxvV-j2y]

The proof at 4:44 implies that division by 0 is possible. If 1/x=0, 1/0=x. If 1/∞=0,1/0=∞. As we know, dividing by 0 isn't allowed. That doesn't mean that 0.999...=1, just that I'm not sure if this proof supports that fully.

[@thiagolucas4651]

you're not dividing 1/∞, but 1/(a very, very big number). Infinity is not a number, so when we do a limit x -> ∞, we're not saying that x *is* infinity, but a real number that is infinitely close to ∞, but never being actually ∞, because that's impossible

[@xinpingdonohoe3978]

Even so, later on we do do 1/∞, and it does equal 0. For instance, Hölder's inequality requires two numbers p and q, where 1≤p,q≤∞ and 1/p+1/q=1 When, for instance, p=1, we have q=∞.

[@TGamer1359]

whos gonna tell them what 0 * x is

[@kyleliao4445]

the second and third one are nearly identical, the proof for infinite geometric series sums is very similar to the second proof

[@Seagaltalk]

Axiomatic and using infinity is already problematic

[@Chris-5318]

he foundation of math is axiomatic set theory. Why do you claim that infinity is problematic (that's news to me)?

[@Seagaltalk]

​@@Chris-5318 Axiomes are a choice that now predetermines the result. how is infinity not problematic? It's not a number, it's not measurable, it's not physical, it creates logical inconsistencies and paradoxes. A number or quantity being infinitesimally close to 1 actually 1 is in the realm of philosophy at the end of the day. But if one pays full price for Infinitesimally less than 1 full pizza I think they should get an infinitesmal amount of money back. Or maybe Infinitesmals don't exist????

[@Chris-5318]

​@@Seagaltalk The axioms have not been chosen to predetermine the result that 0.99... = 1 as you seem to be insinuating. As nobody with a clue has found a problem with infinity in mathematics, I dispute your assertion that infinity is problematic. You claim that infinity creates logical inconsistencies and paradoxes, yet you haven't given as much as one example. All you have is vague assertions. I don't care about paradoxes, they are usually easily dealt with. A number that is infinitesimally close to 1 is not 1 in mathematics. 0.999... is not infinitesimally close to 1, it is exactly equal to 1. I am ignore the hyperreal argument because that involves an abysmal redefinition of what the symbol 0.999... means. Infinitesimals definitely exists in mathematics. Robinson's hyperreals and Conway's surreals include them. Both contain numbers hat are infinitely large. Neither contain infinity, oo, though, because oo is not a number.

[@Seagaltalk]

@@Chris-5318 Zenos paradoxs are of course the most famous. But in general math is just a tool even the idea of anything including a numberline being continuous is problematic and likely needs to be dealt with at some point to use it properly in physics and other sciences. And mathematicians thoughout the centuries have been grappling with infnities so it's incredibly odd you seem to thing that it no longer and issue or has never been one? anyone with a clue would know. If there is Infinitesimals than 0.9999.... is different from 1 by an Infinitesimal amount. you can't have it both ways. The axiomes have been chosen not just to make 0.999... equal to one but certainly in part, even the axiom of infinity is a choice. When someone says 0.9999.... repeating is equal to 1 what they are really saying is in my chosen mathematical formulation I have 0.9999...=1. math however ism't all there is. The concepts of a whole or 1 element and part of said element are concepts that while are can be described mathematically go beyond mathematics in to the realm of philosophy.

[@Chris-5318]

​@@Seagaltalk All of Zeno's paradoxes have been resolved. Continuous number line: There you go again with your vague and unfounded assertions (opinions). Also, what on Earth has it got to do with physics? Mathematics doesn't have anything to say about physical reality. It is pure abstract. (Please don't tell me about the history of mathematics). What physics do you think that 1+ 1 = 2 claims to describe? Don't tell me about mathematical models of reality or that 1 apple + 1 apple = 2 apples because that isn't physics. I didn't say that nobody ever struggled with the concept of infinity. Besides, that is irrelevant. If you go back further in time, we didn't even know how to count. You have NOT provided a single instance of infinity causing an INCONSISTENCY, i.e., an unresolved contradiction in math. 0.999... = 1 exactly. The same symbol was used to represent a different thing in part 6 of the video. i.e. the meaning of 0.999... was redefined. ALL of math is built from axioms. The statement that 0.999... = 1 is true using those axioms. Of course math isn't all there is. What possessed you to mention it? Why do you keep blathering on about philosophy?

[@soyoltoi]

17:01 This would have been a good opportunity to introduce the standard part function

[@TherealBENNYBOI]

I liked my own comment

[@Seagull-sc9gp]

I liked your comment

[@Rudy137]

I disliked your comment

[@Road-to-100-x]

Like beggar

[@pridhvi4789]

Simplest algebraic proof using patterns that convinced me before l learned any calculus. Take a number with n digits (let's use 45) Divide this number with n digits of 9 (as 45 is two digits, we'll divide by two digits of 9. this gives me 45/99) Using a calculator, notice that the decimal is the number repeated. (45/99 is 0.4545....) If you test each number from 1 through 8, you will see the pattern emerge for each number. 1/9 = 0.111... 2/9 = 0.222... 3/9 = 0.333... until 9/9 = 1, but following this pattern, we can also claim that 9/9 = 0.999... Simple, possibly flawed, proof that .9 repeated is equivalent to 1!

[@Mega-wt9do]

999th like LETS GOOOOO!!!!!!!!!

[@pimphatwaggoner1655]

If 0.999... = 1 - ε and 0 < ε < R then the difference is up to ε which can include another infinitesimal (ζ) with a size 0 < ζ < ε < R, and therefore 0.999... != 1. The world doesn't care about real numbers.

[@Chris-5318]

LOL. Your initial premise is wrong, because 0.999... = 1.

[@pimphatwaggoner1655]

@@Chris-5318 "0.999... = 1 because 0.999... = 1.....its just true because it....le IS OKAY!!!"

[@Chris-5318]

@@pimphatwaggoner1655 LOL. "0.999... isn't 1 because 0.999... = 1 - ε and and 0 < ε < R because I say so and I also say that the mathematicians are wrong." Simple proof that 0.999... = 1: 10 * 0.999... = 9.999... => 9 * 0.999... + 0.999... = 9 + 0.999... => 9 * 0.999... = 9 => 0.999... = 9/9 = 1 That proof uses the fact that 0.999... does not have a last 9. Formal(ish) proof: The value of 0.999... := lim n->oo 0.999...9 (n 9s) = lim n->oo 1 - 1/10^n = 1.

[@pimphatwaggoner1655]

@@Chris-5318 Now you're not even being coherent. You didn't even state my argument! Try harder & do better!

[@Chris-5318]

@@pimphatwaggoner1655 LOL. You didn't like the taste of your own medicine. Learn math and do better. In simple words you are wrong and the mathematicians are right. I'll add that your claim that the world doesn't care about real numbers is insanely wrong. They are easily the most important real world numbers. Very few people even know what infinitesimals are, yet alone things like Robinson's hyperreals or Conway's surreals. You are a nutjob, so I probably won't respond further because I am not a psychiatrist.

[@Golden_Official100]

When you desperately want to prove yourself right, to the point of going on things that are way to complex for something as simple as your proof:

[@Jack-O_hedgehog]

NO The title is INCORRECT ZERO* POINT NINE NINE NINE IS *_APPROXIMATELY_* EQUAL TO ONE

[@1vanmax]

did u watch the video

[@uncopino]

no. it is exactly equal, dummy

[@Chris-5318]

It seems that neither ivanmax6039 nor uncopino have read the title of the video.

[@uncopino]

@ oh… you mean because there are no dots the 0.999 in the title? i’m not sure op was joking about that

[@Jack-O_hedgehog]

@@uncopino Where is the ellipsis

[@cupiodissolvi9942]

What I find most interesting (which is obvious when you understand the equality) but wich is never said and is confusing for those who don't understand exactly what the equality is about, is that there is no real number whose integer part is zero and whose fractional part is an infinity of 9s

[@Chris-5318]

To be clear IntegerPart[0.999...] = 1

[@johnlabonte-ch5ul]

In math, there is only one exception in base 10 notation of integers, that infinite 0s follow the decimal point. Note that math says that ".99..."=1.000... but 1.00..1 |=1 where 1.00..1 is finite Of course the difference is infinite. Infinity is dangerous, incomplete, inconsistent and imprecise.

[@Chris-5318]

@@johnlabonte-ch5ul You forgot to take your meds again.

[@cupiodissolvi9942]

@ that's why 0.999... is not the real number whose integer part is 0 and whose fractional part an infinity of 9s

[@AaronAcademyOfficial]

Please note that 1/3 is APPROXIMATELY 0.333… 2:05 Edit: I should stop this before it gets out of hand

[@Chris-5318]

Nope. 10 * 0.333... = 3.333... = 3 + 0.333... => 9 * 0.333... = 3 => 0.333... = 3/9 = 1/3. Also 1 - 0.333... = 0.666... => 0.666... + 0.333... = 1 = 0.999... a) dividing 1 = 0.999... by 3 => 1/3 = 0.333... b) using 0.666... + 0.333... = 1 => 3 * 0.333... = 1 => 0.333... = 1/3 You might want to try to find 1 - 0.333... by yourself. Just remember, there isn't a last 3.

[@tfx9223]

1/3 is approximately 0.333, 1/3 is actually 0.333…

[@wiilli4471]

No it is not.

[@WattsonorSomething]

If Douglas Adams was a mathematician instead of writing books

[@de_oScar]

around 10-11 minutes in: I wish the delta epsilon thing was explained to me that way before

[@CubeCatGaming]

1st proof: I understand that 2nd proof: Yeah, I mostly get it 3rd proof: What the fu-

[@cnone3785]

I like useing 3rd's when measuring. The .00000000003 buffer helps take the blame when I may be a bit off .. + when yell out a measurement in 3rd's the new guy spends multiple minutes looking for the mark on the tape ..

[@novaprime2297]

This is always one of those mathematical phenomena that the philosophical part of my brain struggles with. I see the cold calculated math, I agree with it, but the 3 A.m shower thoughts still persist. Thank you for helping me with this. I’ll move on to other thoughts. Haha

[@DaviRangelSLima]

Every time you put a "9" in the number, you make the difference 10 times smaller. If there are infinite nines, the difference tends to be infinitly small, so it simply doesn't exist.

[@macksnotcool]

I think the best way to think about it is zero point zero repeating one (0.0000... etc... 1) would be an infinetly small number. What is infinitely small? Well that is zero. So 1 minus zero is 1 and 1 minus 0.000... etc... 1 is 1.

[@Chris-5318]

Although 0 is an infinitesimal, your wannabe infinitesimal 0.000...1 is not 0. In fact that is an invalid decimal. Your idea that there is a final 1 is completely wrong. It suggests that you really think 0.999... is 0.999...9 and that is wrong (and not a valid decimal either). Also wouldn't 0.999... + 0.000...1 = 0.999...1 !?

[@johnlabonte-ch5ul]

KC, math seems to be saying that you can add a 0 to "9.99..." to make it divisible by 10. 09.999...0/10=0.999...

[@Chris-5318]

@@johnlabonte-ch5ul LOL. It's 9.999.../10 = 0.999... Your 0.9.999...0 isn't a valid decimal. If it were valid, then 09.999...0 / 10 would be 0.999...9

[@Chris-5318]

@@johnlabonte-ch5ul PS the term "is divisible by" is for use with integer. e.g. 10 is divisible by 5, but it is not divisible by 3 (it goes 3 time with a remainder of 1). Geddit!?!

[@abbe1255]

You could also try to think of a number x such that 1>x>0.999..., you can't; hence 1=0.999...

[@davidcotham1939]

I have NEVER seen the first proof and I am shocked. I'm quite familiar with most the other proofs. I have a distinct memory of arguing with my second grade teacher that 3/3 = .99... She marked it wrong because it didn't say "1." Looking back, I was a genius because I was technically right (I was a terrible student).

[@dewaard3301]

Why does this equality fascinate us so?

[@moustacheskinautique8951]

It's cheaply provocative. But it doesn't bring any great perspective on anything, except maybe on how using unsatisfying notations brings unexpected and apparently ludicrous conclusions. I feel like people who know the proof are just glad to be able to play a little "gotcha" on their unsuspecting friends.