Why does 1=0.999...?

A common question, why does 1=0.999...
If you want to know more, Wikipedia has a good article on the subject en.wikipedia.org/wiki/0.999...
I take back the very last thing I said about it being an 'infinitesimal' - that's a 17th century idea that Newton and Leibniz were very keen on, but it was never mathematically rigorous. 200 years later that idea was entirely replaced with the modern definition of a limit, that the partial sums get closer and closer to the limit.

[@TheMrYakobo]

Why no brown paper Oh wait this isn't numberphile

[@nonexistence1836]

lol

[@Anvilshock]

You can clearly see from the excellently still camera work.

[@zachunter1326]

OML yes

[@giacomopamio1191]

Anvilshock and the wind

[@pikilic4481]

Well there is no banana either and it doesnt sing

[@RubbberRabbbit2]

My immediate thought was "this can't be right", but then I realized: There isn't anything you can add to 0.999... to make it any closer to 1, which means there are no other numbers between 0.999... and 1. And you should always be able to find a new number in-between two different numbers, so 0.999... = 1.

[@MuffinsAPlenty]

RubbberRabbbit2 Most people wouldn't think that way. That's really impressive!

[@tweeweekes5309]

RubbberRabbbit2 Your logic is correct but not intuitive. One could ask what number is in between the graph of y = 1/x and the zero axis? Since we cannot think of one number, then 1/x must = 0, but how can this be if x cannot equal infinity in the real numbers. If it were that simple there would be 1 comment instead of the present 5,597 (not counting all the other comments on other sites). In the end, the best understanding is in the defined construction of the real numbers while noting other constructions of number systems (like nonstandard) will give different answers. Unfortunately understanding the construction of real numbers is not simple.

[@tweeweekes5309]

jqbtube 1/x is most certainly a number , just like (1-1/x). For all x. The point is you need to bring in the concept of limits. Look up Hausdorff spaces if you really want to understand more (that is other than just rant).

[@MuffinsAPlenty]

***** I understand what you're trying to say, but it is presented confusingly. While I agree that one cannot fully understand the equality of 0.999... and 1 without understanding the construction of the real numbers, I disagree that people should look up "Hausdorff spaces." There's no reason to bring in heavy duty topology to understand this topic.

[@MCAdrien9606]

+RubbberRabbbit2 I understand what you're sayingbut let me explain what I think and if i'm wrong just let me know. First, you say that 0.999... = 1 because we can't add any number in between so it must be 1. But the thing is, 0.999... goes on *forever*, so obviously you can't add something because there is always something that will come after. But like square root of 2 or pi, we can put it in a finite form. With square root of two for instance, like numberphile showed, a triangle with a 90 degree angle and 2 side of 1 will obviously have a hypotenus = 2^1/2. Well now I'm confused.

[@Asharon]

I always just found it easy to think of it like this: 0,333333 recurring = 1/3 0,666666 recurring = 2/3 0,999999 recurring = 3/3 3/3 = 1

[@justinrenaud6427]

t4r0n yeeeeeeah well if 1doesnt equal 0.(9) then 1/3 will not equal 0.(3) so this proof is not very good

[@shoubhitravi3138]

Justin Renaud 1/3 = .33333, not 0.3, so this proof is definitely valid u scrub

[@IPA300]

@@justinrenaud6427 It's not a proof, it's a way to intuitively explain to people who don't know a lot of math why it's true. If the layman could understand proofs and calculus, then they wouldn't need this example.

[@galaxian2797]

wait if 0.999999…. Is 1, then isn’t 1/3 or 0.33333…. equal to 0.4? edit: bru I’m stupid

[@aayushkothekar]

@@galaxian2797 no it isn't. 1/3 is still 0.33 recurring and nowhere close to 0.4

[@simonenoli4418]

Oh infinity. It always tries to break math.

[@simonenoli4418]

1/9=0.(1) multiply by 9 you find 1=0.(9).

[@pedanticcreeper7822]

Timothy Barth You know what else breaks maths? Dividing by zero. In an algebraic equation, we have this: 2X = 4 To find 'X' we need to divide both sides by 2 X = 2 So consider this: X*0 = 0 Y*0 = 0 Therefore Y*0 = X*0 We can then divide both sides by 0 to get the values Y = X Then let's replace Y with 7 and X with 4 7*0 = 4*0 Divide both sides by 0 7 = 4 BOOM! Maths broken.

[@Geldahar]

PedanticCreeper | a.k.a BritishCriticThat's incorrect. Division by zero creates multiple separations of nothingness, but nothing has no value so there is no way or need to count those separations, which means that the answer would be 0 = 0 which is correct. (7 -/- 0 = 4 -/- 0) =/= (7 = 4)

[@ikasu00]

PedanticCreeper | a.k.a BritishCritic you can't divide anything by zero. Not even zero itself

[@pedanticcreeper7822]

Itsu0 I know. That was an example of what would happen if we could.

[@Andrew..J]

I asked one of my professors this and he said "dont sweat the small stuff, especially not the infinitely small stuff". Pretty sure he knew i was ready to argue.

[@NixinovaMC]

all i needed was that 1 - 0.999... = 0.000...

[@quandovoceleroscomentarios9622]

What a nightmare for my poor little daughter, who is starting to learn math haha

[@Hoerry]

That’s wrong because 1 - 0.999 would equal -0.000...1 you would need to already assume 0.999... equals 1 to say it’s equal to 0

[@alekisighl7599]

Popcornman92017 but the 0's are infinitely many. So you can't put a one in there.

[@tehyonglip9203]

But 0.000... is just 0 there is no 1 at the end of 0, otherwise the series is not infinite

[@edwinmariano5131]

1 - 0.999... = 0.000... is wrong because the end of 0.999... is not established therefore the equation is not manipulable.

[@DiMono]

There's another easy way to clear this up that I'm aware of: if the two numbers are different, then by definition there exists some number between them. But since the 9s never end, that's impossible. Therefor they must be the same.

[@Thepiecat]

I couldn't tell if you were mad or really excited. You're definitely my favorite person on numberphile! Thank you for sharing your love of math with us.

[@dannydewario1550]

*solid proof that 0.999... = 1* *0) define real number X to be equal to 0.999...* X = 0.999... *1) expand the decimal notation into fractional notation by using 9 tenths, 9 hundredths, 9 thousandths, and so on...* X = 9/10^1 + 9/10^2 + 9/10^3 + 9/10^4 + ... *2) factor out 1/10 from each fraction* X = 1/10 (9 + 9/10^1 + 9/10^2 + 9/10^3 + ...) *2.1) focus on what's inside parenthesis* (9 + 9/10^1 + 9/10^2 + 9/10^3 + ...) *2.2) notice we have 9 + 0.999... inside parenthesis, so let's replace it* X = 1/10 (9 + 0.999...) *2.3) replace 0.999... with X inside parenthesis (because we defined X = 0.999...)* X = 1/10 (9 + X) *3) multiply both sides by 10* 10X = 9 + X *4) subtract X from both sides* 9X = 9 *5) divide both sides by 9* X = 1 *6) use law of transitivity to combine steps 0 and 5* 0.999... = X = 1 *7) stating the obvious* 0.999... = 1

[@iwatchwithnoads7480]

For a moment I thought you'd invoke geometric series. This is much nicer

[@TheOvernightcoyote]

This clears up why 33.333...% + 33.333...% + 33.333...% =100%

[@er33t]

+OvernightCoyote Doesnt the true answer equal 99.999999...%?...

[@TheOvernightcoyote]

conman parker and .99999999999...% is equal to 1%, so 99.999...% is equal to 100%

[@er33t]

You will have to explain. Because if infinity TRULY never ends, you will ALWAYS be able to stick another .0000...009 between .9999...99 and 1.000...000. You will ALWAYS be able to find at least one more number that fits between .999...999 and 1 if we are truly talking about an infinite amount of decimal points. Now if we put a finite cap on the sequence, say no more than 1 trillion decimal points once we get a trillion 9s here: .999...999, then we can say that there is literally no number that stands between .999...999 and 1 and using the logic that if you cannot fit a number between two other numbers (if you cannot fit B between A and C), then putting a finite cap while using that logic is the only way it could make sense. However that still is incorrect because 1-.999...999 will equal .000...001. If .999...999 = 1, then 1-.999...999 will equal 0. You will ALWAYS be able to subtract 1.00...000-.999...999 and get a number greater than 0 and you will ALWAYS be able to stick something between .999...999 and 1 (if we are talking about a true infinite amount of decimal points).

[@TheOvernightcoyote]

conman parker Did you watch the video

[@khalididris3859]

+conman parker Still works. 1 - 0.999... with INFINITE 9s = 0.000... with INFINITE 0s. You said it yourself, infinity never ends. The 0s in 0.000... will go on FOREVER and there will NEVER be a 1 at the end, so it is 0. And to say that you can fit a 0.000....9 in the middle of the 1 and the 0.999... is to assume that there's a finite number of 9s in the 0.999..., and that you can fit one last 9 in the end. The 9s are infinite, so by definition, go on FOREVER and you CANNOT fit any 9 in the end.

[@ChrisCuber123]

Another proof: 1/9=0.111... 2/9=0.222... continue 8/9=0.888... 9/9=0.999... right? but as we know 9/9 also equals 1

[@TheAvgCommentator]

That's similar to Vihart's proof, 1/3 * 3 = 3/3 = 1 0.333...*3 = 0.999... = 1

[@FirstNameLastName-tc2ok]

That just uses the assumption that 1/9 = 0.999..., instead of being actually 1/9 of 9.9999... Since in our number system, you can't write a number that's exactly 1/3 of 10

[@robertfrydell499]

ChrisCuber123, That's short and sweet. And, it does have more of a "proof" feel to it.

[@matthewrocky7342]

+Danny Semakov 10/3 is exactly 1/3 of 10.

[@modvind]

ChrisCuber123 no. 9/9 will always give 1 since you devide by itself

[@WarpRulez]

What I find more interesting is the psychological phenomenon of why so many people refuse to accept that 0.999... = 1, no matter how it's explained to them. It's like they *want* for there to exist an impossible number, ie. the number that's the result of 1 minus 0.999... (This fictitious "number" would be the same as "the smallest real number that's larger than zero", which doesn't exist. Curiously, most of these people accept that this number doesn't exist, yet still want for 1-0.999... to exist, even though it's the same fictitious, inexistent number.)

[@applessuace]

Well, the number is only fictitious in the realm of the reals, we can conceive of these types of numbers/objects pretty easily (but it is NEVER 1-0.999..) and there is nothing wrong with the idea. It's just not 1-0.999... But in regards to the actual point you were making - I agree. I too find it interesting. I think it all stems to the terrible way mathematics is taught in gradeschool. Manipulation of abstract symbols, with a complete negligence of the ideas behind them, make students see math as being itself, the notation. It is obvious that the symbol 0.999.. and 1 are wildly different, so the student assumes the ideas that they represent must be as well - not so.

[@phrygianphreak5408]

In addition to what +applesauce said, although this very pure, basic example of infinitely recurring series does not leap out as intuitive, it does not mean things like this don't happen in real life. In real life, there are many problems which require infinitely recurring series to achieve a solution, and things like 0.999... are useful in some solutions to simplify answers. In fact, there is a whole division of mathematics devoted to infinity and infinitely recurring series and how they can be used to solve previously thought unsolvable problems: calculus.

[@32266ms]

I think for most people the psychological hurdle is in the understanding of infinity. People think it doesn't matter how many nines there are repeating 0.999... is always less than 1. They fail to make the connection that the nines never stop repeating. They keep thinking there's some number of 9's that are repeating - sure it's a really big number of 9's but it doesn't matter how many 9's there it's always less than 1. They think infinity is a really big number. They fail to realize that infinity isn't a number and are unable to deal with it - I guess.

[@clarkkent701]

Michael Stein That's my problem, but I'm an open minded guy. Can you read my first comment about the meaning. Because I talk about that. I said pretty much that, that we come close to 1, but never reach 1. Can you read that and help me out...or someone else, kindly please?

[@phrygianphreak5408]

It has to do with how mathematicians see it. What .9999-> is doing is it's getting closer and closer to one. The next step is knowing that looking at an infinitely recurring series and what it is getting close to is just one out of one hundred ways to solve an infinite series, this technique is just one of the oldest and most useful ways. What the mental hurdle is that you are trying to apply finite logic to infinite numbers/series. You can't use the logic "closer and closer but never there" because that logic only works on finite numbers. In infinite series and numbers, closer and closer but never there is, by the definition of infinity, there. And this concept isn't just some arcane artifact of an oversimplified system: in physics, escape velocities are dependent on infinite sums. We coax the answer out of the problem with nifty tricks like looking at what the series is getting closer to; in college level calculus, all you learn is tricks to solve infinite series and sums. This conclusion, that 0.9999-> is equal to one comes out of applying the same tricks and equations we used to solve real world infinite problems to this question. In any situation in real life where the problem can be boiled down to 0.9999->, then treating it as 1 has always yielded the expected answer. Heck, most scientists are so comforted by those 9-s that in particle physics, a certainty about a theory being correct to just 4 9-s, or 0.9999, they call it 1, or true/correct; in the real world, just four 9-s is enough for most professions to call it a 1. So the short answer: unfortunately, you are using the right logic in the wrong place. Infinite questions in real life like escape velocities and derivatives are solved with special tricks, and those special tricks are what tells us 0.9999-> = 1.

[@xenon9562]

Is this James Grime's account?

[@singingbanana]

+Xenon Yup.

[@xenon9562]

I didn't expect you to comment o_o

[@dantdx787]

+singingbanana Try 4 digits with 4 digits.

[@jonasvanderschaaf]

Xenon how can you react, you're a noble gas...

[@fdnt7_]

lol u r genius

[@givemeyourmoneynao]

another way to show this is 1/3 = .3333... multiply that by 3 you get 3/3 = .9999...

[@MathTravels]

But why is 1/3=0.33333.....

[@AlDunbar]

@@MathTravels divide 1 by three and it should become obvious, as none of the digits produced in the result will be anything other than 3.

[@sofarky]

You look triggered in the thumbnail

[@johnprice233]

finally, someone who found out about the fact that the thumbnail looks so weird.

[@wesso27]

AHJDUEJAJDJEIAJDUHAEUAHRUHAEUAHURAH

[@rosikhunafikriassafari2660]

People: "1 can't be equal to 0.999..." Mathematicians: [This Thumbnail!]

[@orange_leaf4913]

LMAOOOOOO@@rosikhunafikriassafari2660

[@eddyeddyd]

lol

[@SteveMcRae]

Before anyone posts that .999... = 1 is wrong, and that Dr. James Grime is wrong I would like you to first try to answer these questions: 1) ALL rational numbers by definition can be expressed in the form of a ratio of two integers p/q as ℚ = {p/q | (p,q)=1, p,q ∈ ℤ where q ≠ 0}. Since .999... is rational by definition as all repeating decimals are rational. It has to be able to be expressed in the form p/q which is the ratio of two integers. If .999... is rational by definition what is p/q = .999...? 2) .999... = lim k→∞ Σ 9/10^n, n=1 to k by the definition of a limit of a sequence. lim k→∞ Σ 9/10^n, n=1 to k = 1. If both .999... and 1 = lim k→∞ Σ 9/10^n, n=1 to k please explain how they are not equal. 3) 1- .999... = x. What is x? infinitesimally small values are not allowed in ℝ due to the Archimedean property so given no infinitesimals can exist in ℝ please give the real value of x if x ≠ 0. (If you claim x = 0.000...1 then please construct 0.000...1 using the definition of a Cauchy sequence ε > 0 there exists N such that if m, n > N then |am- an| < ε and please give the sequence and show it is convergent upon 0.000...1) 4) By the trichotomy property you can only have xy. If x y then there exists (x+y)/2 between them else as the reals are dense else x=y. If x=1 and y=.999... what is (x+y)/2 Yes Virginia, using REAL math 1=.999... is absolutely true.

[@nmarbletoe8210]

i am convinced. 1= 0.999... this sounds similar to Hilbert's Hotel.

[@SteveMcRae]

N Marbletoe There is a relation there when using a the common method for converting repeating decimals to their equivalent fractions actually:) Such as even a simple multiplication of 10 * .999... = 9.999... it does directly relate to Hilbert's paradox of the grand hotel. Nice! :)

[@nmarbletoe8210]

Steve McRae hoop! maybe i heard of the grand hotel in a vid on quantum field condensates. hmm.

[@danl2729]

You're right... Because 0.999... = 1

[@___________2204]

+Steve McRae Dude, please teach me math, youre awesome

[@StyxTBuferd]

I never liked the S and 10S proof because it sort of implies there's a zero at the end that is being subtracted from, so it still gives you that idea that there's a 1 at the end. You can prove it for sure using series, which is the way that works most intuitively for me, HOWEVER I think the best informal proof I've found is this one: Consider 1/3. 1/3 = 0.333... Then 1/3 + 1/3 + 1/3 = 0.333... + 0.333... + 0.333... Every place value is getting three 3s added together- there's no denying that, there's no place that's being left not added- so you should get 0.999... At the same time, we know 1/3 + 1/3 + 1/3 = 1. Thus, 1 = 1/3 + 1/3 + 1/3 = 0.333... + 0.333... + 0.333... = 0.999... This 1=0.999... This isn't a hard proof, but it's one that sort of clicks best with high school kids in my experience. The series proof is more fun I think and I recommend you check it out even if you don't quite know series notation yet.

[@singingbanana]

StyxTBuferd But the proof why 1/3 = 0.333... is the same as why 1 = 0.999.... So it's a bit circular.

[@StyxTBuferd]

Yeah, and in that way I know it's not a formal proof. But if a student already accepts that 1/3 = 0.333... but doesn't grasp that 1 = 0.999..., then that's a sort of good way to have them approach the idea, and in my experience most kids do accept that 1/3 = 0.333... at face value. Asking for a proof of 1/3 = 0.333...is, I think, a good follow up question for someone who might accept 1/3 = 0.333... at face value but asks for a proof of 1 = 0.999... Also, since I have your attention right now, I want to say how much I love your work and numberphile. As someone who's very soon going to be a formal high school teacher, I think you guys play with ideas that are both fun and accessible to people regardless of how much math they've seen, and I plan on using some of those ideas in my lessons.

[@singingbanana]

StyxTBuferd Hey, thanks. I agree with your comment, it would be a good approach with some students, especially if you can then show them why 1/3 = 0.3333.... Incidentally, if 1 =\= 0.999... then 1/3 =\= 0.3333.... and worse, there would be no decimal expression for 1/3. So the choice is have no decimal expression for 1/3 or accept two forms for the number 1 (and other terminating decimals).

[@SteveMcRae]

StyxTBuferd I often find it is more adults (even those with a decent amount of mathematical education) that for some reason think of .999... (and even .333...) as some type of on going process, as if the number is approaching some value (such as the asymptomatic behavior of a function) rather than a fully formed number. Kids who haven't generally been exposed to these types of concepts are more like as you said to accept that 1/3=.333... as unlike the aforementioned adults they are far less likely to conflate various mathematical concepts if they have yet to be exposed to them.

[@SteveMcRae]

singingbanana Could you elaborate on a question for me if possible....In my understanding of set theory, .999... being an equivalence class to 1 that .999... must be an element of Z and of N as it is just another representation of the element of 1 and would be represented in Z and N as 1. In other words both 1 and .999... represent the same real abstract number in both of those sets are are the same element, but notationally different.

[@randoomnik]

It makes sense if you think of it as a limit. The "recurring period" notation ("...") should be interpreted as defining the limit of an infinite sequence.

[@EmperorZelos]

There is no limit in the notation or it.

[@randoomnik]

***** ?

[@mikeguitar9769]

randoomnik, exactly. I agree. And it ought to be fairly obvious conclusion, but the "mathematicians" are being stubborn about it for some reason. They are "infinitesimally" close. I'd like to write that: 1 - δ = 0.999... , where δ is an "infinitesimal", or some such thing. Anyone who is okay adding δ to the right hand side (by in effect equating 0.999... to 1), will also be okay with adding δ to the left hand side as well to get 1. This operation also triggers the ripple-carry on the RHS that is needed to truly satisfy the equality and make it a correct equation.

[@Jay-kk3dv]

Exactly. This is an issue of Limits, which deals heavily with these kinds of numbers. In calculus 1.999 recurring is definitely not 2.

[@DowzerWTP72]

Thank you so much for explaining this to me. I'm a second year student, and have been at Uni for 3 total, and I do a lot of maths for my course, and this has finally cleared it up for me! Brilliant!

[@pangainoon17]

to continue the previous post if that point I described does not exist then does that imply that the number line has gaps or holes in it? if not how not? and where can I find an explanation?

[@nickacoffin]

Here is the actual why: how many numbers are between 1 and two? An infinite amount. Or between 1.5 and 2, there is also infinite. between 1.9 and 1 there again, is an infinite number of numbers between the two. That is what makes two numbers not the same, you can cut the space between them an infinite amount of times. Between 1 and .99999999… however, by definition, there are no numbers between them and that is why they are the same.

[@loupax]

That must be the best answer I've read on the subject. +1 because it didn't say "because the mathematicians said so, so don't bother asking and just accept it"

[@evelyningraham8627]

But what about 0.999...8 assuming the ... has infinite 9s there is no number closer to 0.999... than 0.999...8 so are they equal to? And by logic isn't 0.999...8 equal to 1?

[@quantoris4848]

@@evelyningraham8627 There is no number like .999...8. It is just not defined

[@justinrenaud6427]

How does this prove that 1=0.(9) ? The only thing you showed is that there is no number between 1 and 0.(9) that doesn’t mean that they are equal it only shows that they are very close. For example , if we remove all decimal (1.xxxxxx we remove x’s) then we count like 1 2 3 4 5 but there is no number between 1 and 2 but that doesn’t mean that they are equal.

[@NatePrawdzik]

@@justinrenaud6427 The decimals don't stop existing just because you're ignoring them.

[@sarahast8437]

I prefer this : By definition 0.999... is defined by : the limit of the sequence ( 0.9 , 0.99 , 0.999 , 0.9999 , 0.99999 , ... ). And 1 is the limit of the sequence (by the definition of a limit).

[@EmperorZelos]

And most morons do not understand this.

[@aka5]

@@thehumblepuppy4488 They mean the mathematical definition of a limit. lim {n->infinity} (sum {i, 1->n} (1/10^i)) = 1

[@stoikes]

singingbanana can you do a vid explaining an example as to when you would want or need to use the 0.999 recurring equivalent, or is this just something that we know to be true mathematically?

[@benjaminbrady2385]

I saw the link in the description and wondered if the Wikipedia link went on forever, it didn't.

[@fft2020]

"there is no pot of gold at the end" hilarious :)

[@kye4840]

Question: what is 0.249999... to one decimal place, is it 0.3 or 0.2?

[@EmperorZelos]

Either because it is 0.25

[@TheDiamondGames]

To clarify ether, because 0.249999... is around 0.2 and if 0.24999... is 0.25, then it is 0.3. But look at the comment above by Marcus Anderson, he has made valid points and this whole idea is very disputable.

[@EmperorZelos]

The Diamond Games - Channel closed WHAT!? he isn't talking about 0.2444.... or the likes nad it'd be rounded to 0.2!

[@TheDiamondGames]

***** Edited, thanks

[@thefran901]

To only one decimal place, it is 0.3. If you take all the numbers with two decimals, half of them should be round down, and the other half should be round up. So from 0.20 to 0.24 it is rounded down to 0.2, and from 0.25 to 0.29 it is rounded up to 0.3: 0.20 = 0.2 0.21 = 0.2 0.22 = 0.2 0.23 = 0.2 0.24 = 0.2 0.25 = 0.3 0.26 = 0.3 0.27 = 0.3 0.28 = 0.3 0.29 = 0.3

[@jonahansen]

This is a perfect explanation - Dr. Grimes starts out with the heart of the matter, that decimal representations are just representations and are not one to one with actual numbers. I've seen so many videos showing the proof, but not even mentioning this essential fact that really explains why.

[@estebanmartinez4803]

I've thought about "proof" for those who still think this isn't right. I hope someone believe it useful. Real numbers are densely ordered, that means that for every x,y belonging to R with x

[@burlapsack7759]

Best way to show this. 1/3 is .333... so, add 1/3 3 times. You get 3/3 = .999... Obviously, 3/3 is 1

[@nagyandras8857]

expect 1/3 is not equal to 0.333....

[@dannydewario1550]

Nagy Andras what....explain

[@nagyandras8857]

Mitchell Dobbs 1/3 is not equal to 0.3 1/3 is not equal to 0.33 And so on. Eatch time you put 3 at the end you get closer to 1/3 but you never reatch it. If its infinate long you just get infinately closer. Might as well say it converges. But never reaches.

[@dannydewario1550]

Nagy Andras That would make sense if we said 1/3 could only be translated into decimal form using a finite amount of 3's. But the very definition of 1/3 implies the use of an infinite amount of 3's beyond the decimal point (because 1 can't be evenly divided by 3). The proof (I guess) that 1/3 requires an infinite amount of 3's is to start by asking "What's 1/3 of 100?" Clearly the answer is roughly 33.3. But then ask what's 1/3 of 1000. Well that's 333.3. We can keep this going forever. At no point do we run out of 3's when increasing our question by 10. So this must imply that in order to write 1/3 in decimal notation, that would require an infinite amount of 3's. Hence the proof for using 3/3 (or 0.999...) is equal to 1.

[@nagyandras8857]

Mitchell Dobbs If it makes sense for you then its cool. It makes sense to me that it only converges. Like 1/x is never gona be zero. You can write larger and larger numbers in place of x but you never get zero. For me its the same case here. You write more and more 3 s there, but you only can get closer and neverhit 1/3.

[@Armiteus]

James, an interesting thought: Graphically, you can represent a repeating decimal with shorthands (a bar over the repeating number, or an ellipsis after, or whatever form we deem acceptable). In reality, one could never stop calculating this number. Even if you represent that infinite smallness with a bar or three dots, once you've stopped calculating or writing the number, even if you wrote it out for very long time, in essence you'd eventually truncate it by stopping the calculation and moving onto the next part of the proof. We should never even be able to get to the subtraction part at all, still trying in vain to reach the infinite end of 0.999999... How possible is it that shortening the number to do the proof is itself leading to a problem?

[@EmperorZelos]

You assume that one must do digit by digit, which is a superflous activity in mathematics. I don't need ot do that.

[@rosepinkskyblue]

You’re not shortening it, you’re just realising that all of them are nine and so they’ll all cancel out even if there are infinitely many of them Writing it shorter is just a representation of the number that we use for convenience

[@Armiteus]

@@rosepinkskyblue i totally forgot about this video, and my comment definitely shows me how much better I understand math now than I did 5 years ago. I think I know what I was trying to say, but it’s pretty silly and you’re absolutely right.

[@rosepinkskyblue]

@@Armiteus ah sorry I didn’t notice it was old but you’re right a lot can happen in five years

[@kidbuu8025]

Hi James, will u make a video on Zeno paradox, with a physicist?

[@TheDaniel2718]

Have you done any videos on the mathematically rigorous system of infinitesimals developed by Abraham Robinson?

[@singingbanana]

+Daniel Evans I did a video about infinitesimals on numberphile - I think I mentioned the maths was made rigorous in the 1960s but didn't go into details. I think was another video as well on numberphile with someone else talking about details.

[@majorkatzmann2240]

applessuace said pretty much what I would have said, I only say couple words on intuition in maths. Many mathematical results are extremely counterintuitive and often things that seem obvious at the first glance are wrong or very difficult to prove. Moreover mathematicians often work in fields that are so abstract that humans have no intuition at all about them. Sure you can use your intuition when you're solving a mathematical problem but you mustn't forget that it might be hiding from you the most ingenious solutions. But what you absolutely cannot do with intuition is using it AS a proof. Even the most evident statements (if they aren't axioms) can't be considered true if they aren't proved logically. The remark about math majors isn't just pointlessly rude but also completely wrong. I know many mathematics students and they are probably the smartest people I've met. They didn't forget how to use the intuition, they've just learnt to use it in less naive way.

[@SteveMcRae]

Didn't Peter finally say he understood why .999... = 1? It rather seems now he is just fixated on that which he may just not have the ability to understand. He needs to realize that mathematicians don't sit around and think why .999... = 1 , as they know as apple points out ...that is how our mathematical system works. It is axiomatic and seems more he is undergoing an exercise in futility to wonder why they are the same after being explained why a number of different ways.

[@applessuace]

How boring maths would be if we never found results that conflicted with our intuitions. How pointless maths would be if when such results arose, we'd throw them away in favor of our intuitions. What is the point of searching for truths if we're going to let our "intuitions" win out in the end? Studying mathematics tempers our intuitions to be more inline with the logical. That's why math majors are valuable in unrelated fields, that's why everyone is required to at least dip their fingers in it. Our intuitions are worthless without learning how to become logical thinkers, which is exactly what math teaches us.

[@applessuace]

Furthmore, I'm trying to understand how, even if you rely on intuition, that .999... is merely "very close" to 1. This only seems reasonable if you mistakenly view infinity as an "extremely large number." I don't see how an informed mind could have this intuitive view at all.

[@Arkalius80]

Maybe someone can explain the phenomenon I see in any context where this topic is discussed. You always see the professional mathematician or educator explain this particular identity, often showing various proofs that it is true. Then, in the subsequent discussion, you get tons of people who are not mathematicians attempting to argue vehemently why its wrong, as if this idea completely offends the core of their being and they cannot endure a world where this is true. What is it about this simple identity that gets some people so up in arms?

[@singingbanana]

Arkalius80 People have studied that from an educational point of view.

[@Arkalius80]

***** Simple rules of multiplication. When multiplying a decimal number by 10, you simply shift the decimal point one to the right. If 10 x 0.999... is not 9.999... then 10 x 1.000... is not 10.000... which means 10 x 1 is not 10. Weird math you have there. What is 10 x 0.999... then? Since you think you've solved the challenge of identifying a number x such that 0.999... < x < 1 (thank you for identifying the flaw in your solution for us), try this one. Since any repeating decimal is a rational number, you should be able to tell us the integers a and b such that a/b = 0.999... and (since you insist this isn't 1), where a is not equal to b. Remember, all integers have a finite number of digits.

[@Arkalius80]

"the simple rules of multiplication do not apply to infinities" 0.(9) is finite. There are no infinities here. "I would like you to pull up a proof that "all" repeating decimals are in fact a representation of a fraction." Easy. A repeating decimal represents a geometric series with rational elements and a ratio less than 1. Any such series has a finite, rational sum. Therefore, any repeating decimal represents a rational number. "For example: 0.33333.. ≠ 1/3" Incorrect. The geometric series that decimal represents has a sum, which is 1/3. Thus those values are equivalent. "As far as I can tell, the only numbers that can actually be written in decimal is the set: {2^a x 5^b | a,b ∈ Z}" Those are the only numbers with finite expansions in decimal, yes. We can still write infinite repeating expansions with parentheses or ellipsis notation. They are not approximations, because they give us all the information we need to understand the precise value of the number, since they describe geometric series with finite sums. If a repeating decimal is just an approximation, it should be impossible to mathematically determine the precise value a/b (a and b are integers) that produces it. But I can do that for any repeating decimal you give me. Applying that same process to the value 0.(9) will net the answer 1.

[@SteveMcRae]

***** "You cannot start a proof that 0.999..x 10 = 9.999.. by stating that 0.999.. x 10 is 9.999. That is a null statement. When you say simple rules of multiplication, the simple rules of multiplication do not apply to infinities. Not without a proof." Well, actually you can since when doing a proof it is not necessary to prove something (such as multiplication with repeating decimals) that has already been proven or more specifically how repeating decimals are constructed. A repeating decimal such as .999... as Arkalius points out represents a geometric series such that if the common ratio is |x| < r then - 1 < r < 1 you can use the formula a/(1 - r) which gives you .999... = (9/10)/(1-(1/10) = 1 You don't need to prove that .999... is a rational number in a proof as all repeating decimals are rational, nor would you have to show the construction of rational numbers from abstract algebra. Could you? Sure, you could define rational numbers as a ratio of two integers such that ℚ = { m/n | (m,n) = 1 and m,n ∈ ℤ and n ≠ 0 } and go on to construct a more well ordered field from there. Since .999... is defined as rational it behaves as a rational number. Therefore 10 x .999... works just like any other multiplication by a power of 10. The decimal point shifts over one and you have 10 x .999... = 9.999... this was also the proof used by Euler in his work "Elements of Algebra" and is an established proof (trivial) and not required in math to prove again when used.

[@FernieCanto]

One of the factors, I guess, is people's overall animosity towards mathematics. Many times, math is shown as something deliberately obscure and difficult, and any unusual and surprising result therefore may be interpreted as math being deliberately frustrating. Yeah, for many people, math has a personality. Also, in many instances I've seen, 1 = 0.(9) is presented in a pretty smug and condescending manner, which immediately causes resistance. Some people fail to show math as something exciting and fascinating, and instead go on an ego trip to show how smart and awesome they are, and that's pretty irritating. That can permanently "taint" things in people's mind. Also, when people think they are being fooled, denial is an immediate reaction, and 1 = 0.(9) is often shown like some kind of dirty trick. It should be something that people can figure out on their own, but instead, that truth is shoved down their throat. That may explain the delusional and irrational "explanations" to why the identity is false, like that "there is a zero at the end of infinite nines", or that you can somehow change the number base and act like it's no big deal. In the end, I think it's mostly due to poor education. If kids were taught to be fascinated and thrilled by mathematics, a lot of that wouldn't happen. But a lot of people see mathematics as a pretentious, snotty hipster who constantly wants to make you feel stupid.

[@NoriMori1992]

I love saying your channel name out loud with your accent. It just sounds cute, and it's fun!

[@KongofPip]

If we of the ≠ clan are not using the real number system, then what system are we using and what number system are you of the = clan using?

[@KongofPip]

***** There are people who know things and know why; there are people who know things and don't know why; there are people who don't know things and know why; finally there are people like you, who don't know anything and don't know why. QED

[@KongofPip]

*😔💤*

[@SteveMcRae]

"The answer to the question posed in the title of this video:" - aafdafd Yea, what would a PhD in Mathematics know about math verses aafdafd's not even high school education. A person who thinks (2)3^2 = 36 and that x^2 is referred to as implied multiplication is going to refute a PhD mathematics? # EPICfacepalm

[@EmperorZelos]

"A humerous fact about 0,(9) for all is this 0,(9) if you convert it to ANY radix, base, or whatever, binary, ternary, dozenal etc, is the only fraction that will always result in a 0,(b-1) constantly, where b is the base/radix, in decimal case b=10, binary b=2 and so on. 0,(9)_10 = 0,(1)_2 = 0,(2)_3 = 0,(B)_12 and so on :) take a guess why! as such it is the one and only fraction that no matter what base you convert it to will not get a terminal decimal expansion 1/17 = 0,1_17 for example, but 1/17 = 0,repeatingstuff in decimal" Posted it :P Whatcha think?

[@Chris_5318]

@singingbanana, Gestalt Ishtarie is another alias of Neoicon Mint, Logicartistist Whitefalcon etc. He has already called you a fool a/o a liar. Please add add him to your block list.

[@Chris_5318]

@singingbanana, thank you.

[@alvinlepik5265]

so 10^(negative infinity) is 0 - is it? So if I would divide by this 10^neg.infinity I would get infinity - where as if I divide by zero I get something out of the real numbers' domain. Am I correct to deduce that +/- infinity and the result of dividing by zero belong in the same domain and therefore eventually make these conclusions of 0.9(9) being 1 etc?

[@alvinlepik5265]

Yes, you can divide by zero/infinity - the result simply doesn't belong in the real numbers - that's why we say undefined or empty set or something like that.

[@alvinlepik5265]

How can you say that .(9) = 1 And consequently 1 - .(9) = 0 then? If there are infinitely many 9s then the value will be infinitely close to being 1 and the counterpart subtraction would be infinitely close to 0 - how come can we do limit operations with the argument approaching 0 when in standard math division by 0 is not permitted? It sounds like a maths-philosophy to me by now - I just can't get around that something infinitely close to a value is that value - it doesn't make sense.

[@alvinlepik5265]

Oh okay, you're right - thanks for the eye-opener. The algebraic proof seems not entirely accurate altho the limit proof looks solid.

[@wmheric]

I like that you said the non-uniqueness of decimal representation was one thing you wished that you'd been told early on. Many who teach prefer to jump directly to "let me show you" instead of guiding students to think more conceptually.

[@ferrishthefish]

My personal proof that 1 = 0.999... is that for any two real numbers X =/= Y, there must be at real number expressible by (X+Y)/2 that is in between them. 1 + 0.999... = 1.999... 1.999... / 2 = [1.8 + 0.18 + 0.018 + ... ] / 2 = 0.9 + 0.09 + 0.009 + ... = 0.999... In other words, the average of 1 and 0.999... is not a number that is in between those two numbers. The only way that the average of X and Y could be equal to Y is if X and Y are equal.

[@emoemolga8842]

It may be me missing something but your math doesn't look right. It says 1.999.... = 1.8 + 0.18 + 0.018 + and so on, but 1.8 + 0.18 - - - 0.98 + 0.018 - - - 0.998 + and so on No matter how far its goes it's always putting an 8 at the end and 1.999... doesn't have an 8 anywhere in it.

[@Paul-iq8bu]

x=0.999 10x=9.999 10x-x=9 9/9=1

[@rbdoppler7841]

The problem with you seeing it like that is that it is an infinite string of addition, and you cannot place an 8 at the end of an infinite string of 9s. It's just like you can't have an infinite string of 0s with a 1 at the end. 1.8 + .18 + .018 + ... is the same as writing 2 * (0.9 + 0.09 + 0.009 + ...) ferrishthefish is completely correct in his math.

[@matthewrocky7342]

If I assume orange and banana are real numbers, then orange = banana because I can't find one that is 'between' them?? Every proof for this system is shakey at best, but maths comes down to what you 'accept' as the principles. I personally don't accept infinite addition as a single number. I can accept that the limit is a number; no problems there, but the infinite series is not automatically a limit. I provided an answer for your question elsewhere. It is redundant where the addition of a terminating decimal and an infinite decimal relies on the same principles of why 0.999... = 1 in the first place, thus, it's not a proof. Sorry to burst your bubble dude.

[@ferrishthefish]

Matthew Rocky Uh, do you even algebra? Do you understand that when I compare X and Y, I am comparing the numbers they represent, not the letters I've chosen to represent them with? By your logic, if 'orange' = 5 and 'banana' = 5, then 5=/=5 because oranges are different from bananas. Anyways, my answer has to do with Dedekind cuts, not "infinite additions" or "the addition of a terminating decimal and an infinite decimal." In layman terms, the 'typical' Dedekind cut argument is that, if 0.999... =/= 1, then a cut (A,B) at 1 would produce a B with a smallest value (i.e. b=1), and an A with a greatest value (i.e. a=0.999...), and this is contradictory because it implies the existence of c = (a+b)/2 which is not contained by either A or B. I just worked it backwards because it seems more intuitive that way.

[@radiumbreon2109]

You can also follow a pattern. 7/9= 0.777..... 8/9= 0.888..... To continue the pattern, you would think that the next term is 0.999......, but 9/9 is 1. But that's just how I interpret it.

[@EdgeOfEntropy17]

But 7/9 = 0.7777777777777778

[@jetison333]

@@EdgeOfEntropy17 no, it doesn't. that would be strictly greater than 7/9. the sevens don't stop.

[@uforob5601]

I knew all this but first time I find this two explanation together and clear in a short video

[@CrniWuk]

Is 0,999...=1. True or not? I always thought, it depends on the Math you're doing. Infinitesimals work with limits and there 0.999 ... does NOT equal 1. The concept of Hyperreal numbers. That's of course only if you want to be a real smart ass about it.

[@singingbanana]

+CrniWuk Sorry to disappoint you, 0.999... = 1 in the hyperreals too.

[@CrniWuk]

singingbanana Thx for the clarification, and hey! It's not a disappointment :), learning something new is awesome. But I am curious, I hope you can answer me this, is there absolutely no situation in math where 0,999... is not 1?

[@singingbanana]

+CrniWuk The transfer principle of the hyperreals says that any proposition that is true is the reals is also true in the hyperreals. This 0.999...=1 is a property of real numbers and a consequence of limits, so it will still be true in the hyperreals and any other system containing the real numbers and that allows limits.

[@tweeweekes5309]

+singingbanana This is true as the real number 0.999... = 0.999...;...999... hyperreal which is equal to 1, but what about all the other hyperreals like 0.999...;...999 which we know has a nine in the last spot shown here, the infinite spot, and is strictly less than one? That is , 0.999...;...999 < 0.999... = 1 Why only see the one hyperreal where the equality holds and miss the infinite many others on our journey to and past infinity (actually getting there with hypernatural indexes and not having to take the limit)? You may have heard of the natural string evaluation.

[@OnamKingtheKing]

+CrniWuk what is true for reals is also true for hyperreals, so 0.999... is indeed equal to 1

[@keithcamilleri1497]

plus if you divide both by 3 you get the same answer

[@charwolf1]

How about this for the doubters. We know that 1/3 = .333333333.... so multiply both sides by 3 and we get 3/3 = .99999999999... how can you refute that kind of logic?

[@applessuace]

The general refutation is "but .9999.. is scary."

[@roninnib6635]

@@applessuace I don’t like this proof, as it just makes people believe that 0.333... =/= 1/3

[@AlDunbar]

@@roninnib6635 but it is. 1/3 is not at all equal to 0.333 but it is equal to 0.333... You can prove this to your own satisfaction by dividing one by three. You will note that the only digits produced in the result are all three. How far you take it before convincing yourself that no digit will ever be produced that isn't a three is something I will leave up to you to determine. Please report back when you know how far you have decided you'd have to go to be sure of tge result being no different digits.

[@roninnib6635]

@@AlDunbar I know it’s true that 1/3 = 0.333… It’s just that when you use this kind of explanation, I have found that a lot of people instead of agreeing that 0.999… = 1 Instead now believe that 0.333… =/= 1/3 as well.

[@MisterrTheEditor]

if something was length pi (or any repeating number) could it have a definite length?

[@EmperorZelos]

Of course

[@roninnib6635]

Pi isn’t repeating

[@MisterrTheEditor]

@@roninnib6635 I meant infinitesimal

[@roninnib6635]

@@MisterrTheEditor how Did Infinitesimal auto correct to pi

[@MisterrTheEditor]

@@roninnib6635 replace the repeating lol

[@hittoman3219]

I have given this some more thought to this lately and considered also the option that perhaps 10 simply can't be divided 3. If you do you will get infinitely continuing number that is close but not quite the answer but adding anything to it you will get something that is even more wrong. If you can't give an exact answer no matter how you try then shouldn't that mean that there just isn't any? What do you think? Am I being totally illogical here?

[@Araqius]

If you are trying to say 0.333... is less than 1/3 or 0.999... is less than 1, just find a number between them or find the difference of the two numbers. If two numbers are not to be equal, there must exists a number between them.

[@EmperorZelos]

+Hitto Man You are because in mathematics you can divide anything with anything (zero being the only exception) and get exact answers.

[@sonaruo]

another easy way to do it change base in the number 1/3 in base 12 will be a nice number 0.4 and of course 0.4+0.4+0.4=1 all base 12 which is 1/3+1/3+1/3=1 people need to understand that 0.333333333... is a representation limit of how we represent number since we can not write infinite decimals we need to make a cut somewhere on what we see on the paper but 0.3333... is the same as 1/3 is the same as 0.4 base 12 ITS THE SAME NUMBER

[@Powersd451]

@@NeoiconMintNet 0.3... is the same as 1/3 by definition. You're simply fundamentally wrong about the topic.

[@DragenmanX]

I really hope this isn't just a joke because I'm going against my Maths teacher, because she says that 0.(9)=/=1.

[@aimuser3111]

Your maths teacher is unfortunately wrong. Please stay curious and involved in maths!

[@andrewzhang8512]

did you win?

[@JKrollling]

the 10s - s = 9 equation was forgotten for so long in my brain, i knew there was something like that, but just thanks (both the but and the just are needed in this clause, for reasons...)

[@confuciuscalculus388]

James Grime hi there, Ive got a great idea; why dont you guys (you or numberphile) make a video about milenium problems of clyde mathamatics ensitute ? (I mean you've made videos about the poincare conj. And riemann hypo. I think if you would make something like that it would be better.) (Please?) :)

[@SteveMcRae]

Confucius Calculus Better yet, have them actually solve one of the millennium problems, then have them share the video and we all can submit it for publication and split the prize money. Sounds good to me. Think Dr. Grime would go for that? LOL :)

[@confuciuscalculus388]

Steve McRae a kind of it ;)

[@VisionaryFire]

James Grime is a cool guy! I really like his videos. I have watched many videos that prove 0.999999... = 1 but I still have my reasons for saying it is not exactly 1 for similar reasons he mentioned at 1:45

[@mikeguitar9769]

I agree. It is "infinitesimally" close. I'd like to say that: 1 - δ = 0.999... , where δ is an "infinitesimal". Anyone who doesn't have a problem adding δ to the right hand side (by in effect equating 0.999... to 1), will also have no problem adding δ to the left hand side as well. This operation also triggers the ripple-carry that is needed to truly satisfy the equality and make it a correct equation.

[@VisionaryFire]

To me the only difference in and infinitesimal and zero is that infinitesimal can be operated with, such as multiplied by infinity or even by any number. Zero, no matter how much you multiply by, or how many times you add it, the value will never change.

[@mikeguitar9769]

excellent point , and I think that is the issue. For the sake of operational-ability , it's better to leave the infinitesimal in place and denote it with some explicit, unambiguous symbol. And only after all the operations are done should it be replaced by the limiting value. We might all cringe at this, but interestingly, this is sometimes how "significant digits" are handled in calculations, I suppose because in a way it is the same sort of issue.

[@tanishmishra9152]

They are the same, just divide both sides by three

[@jextra1313]

__Quick Proof__ 1 / 3 = 0.333 recurring 1 = 3 x 0.333 recurring 1 = 0.999 recurring

[@emoemolga8842]

Again, I'm no mathematician, but is 1 / 3 = 0.333... truly accurate? Look at how the decimal is made. 1 / 3 == (1 / 3 = 0 carry 1) →0.(10 / 3 = 3 carry 1)→0.3(10 / 3 = 3 carry 1) →0.33(10 / 3 = 3 carry 1) → 0.333(10 / 3 = 3 carry 1) →and so on So it's not an infinite amount of 3s so much as an endless iteration of (10 / 3 = 3 carry 1). As such, wouldn't it be right to somehow acknowledge that? I mean, no matter how many 3s you pile in there you will never get out of iterating (10 / 3 = 3 carry 1) over and over again because if you did the amount of 3s that would eventually result would become finite rather than infinite.

[@nagyandras8857]

expect, 1/3 is not 0.333 recurring.

[@gearcomNPG]

I don't have questions. I just love to hear you talk about numbers.

[@simonecipriano]

Mi piace tutto di questo video. L'argomento in sé, la spiegazione del perché, ma soprattutto l'entusiasmo con cui lo spiega e il sorriso finale

[@Araqius]

Caper keep confusing himself. One day he say as n = infinity. Another day he say n can't equal infinity. One day he say ... doesn't reach infinity (That's also means 0.000... has finite 0s and doesn't equal 0.).

[@Murdervator]

Can someone show me a number multiplied by 10 that doesn't end with a zero? .999...x 10 = .999...9990. Then the algebra falls apart. Maybe I'll represent my twos as threes from now on.

[@SteveMcRae]

Murdervator What? 999...x 10 = .999...9990. ??? (999...) * 10 = (9.999...) =10 There is actual mathematics, then there is nonsensical mathematics...try working the the former more than the latter.

[@FernieCanto]

"Can someone show me a number multiplied by 10 that doesn't end with a zero?" Yes: pi * 10. Also, e * 10. Also, sqrt(2) * 10. Also, amazingly, .(9) * 10.

[@kerseykerman7307]

Steve McRae If you do .(9) * 2 the answer is theoretically 1.(9)8. If you would write an infinite amount of 9s and an 8 at the end, you wouldn't ever write the 8, but its still there. (Theoretically). So, by Murdervators logic, .(9) * 10 = 9.(9)0 I really like to work with that 'type of logic', even if its not true. It makes things harder and more challenging.

[@AlarmClock65]

When you multiply a number by 10, in base 10, each digit moves up a decimal place. So, if there are only finitely many digits, then there will still be zero, or nothing, to the right of the rightmost digit. But with infinitely many digits, as with .(9), the nine in the tenths place will move into the ones place, the nine in the hundredths place will move into the tenths place, thousandths to hundredths, etc. There will still be an infinitely long trail of nines to the right: the new nine in each negative decimal place will come from the place directly to the right. As another example of when you can multiply by 10 without having a 0 at the end, take .3 times 10. 3 isn't usually written with a 0 at the end, although it could be, as in 3.0.

[@beepster4096]

I remember a long while ago people arguing this video is wrong. Somehow the arguments continue.

[@AndresAmelieEsp]

The world needs more math's teachers like u, Felicidades ;) Greetings from Mex

[@Skwisgar2322]

another way I like to explain it is 1/3=0.333... if we multiply both sides by 3 we get 3/3=0.999... or 1=0.999...

[@tweeweekes5309]

I think the calculators are the reason people are not quite getting what some are saying. If you take 1 and divide it by three you get (long division) goes into it three times with a remainder of 1. Then you take that 1 (ie. 10 when long dividing) it goes in three times with a remainder of 1. Then you take that 1 and divide it by three and you get (long division) that it goes in three times with a remainder of 1 etc .... to infinity. The repeating 3 to infinity is the same as the repeating remainder ... to inf.ie. never quite get to exactly one third with a decimal since we always end up with this remainder of 1 which then leaves yet another 3. To get rid of the remainder means we got rid of the 3 repeating. So saying .333... is the same as one third is the same as saying .999 is the same as 1 which we already do not agree. So let's just cut to the chase and look at .999... The reason people say it is not equal to 1 is because there is exactly 0.000...1 between it and 1. Now the author of the videa says those are the same (0.000...1 and 0 are not the same in my opinion). My reasoning is a computer program runs better actually if its time step is 0.000,,,1 and the computer program cannot run at all if the time step is zero. Thus I do not agree that an infinitesimally small number (precisely 0.00...1) is the same as zero. I think the author of the video should really say something and not keep going by what he was taught at an early age.

[@Skwisgar2322]

I know it doesn't hold up as a proof, but it makes it easier for people to understand. most people understand that 1/3 in decimal is 0.333... 3*3=9 and that 3/3 =1. so explaining it as(1/3)*3=(0.333...)*3 seems to help people understand.

[@Skwisgar2322]

***** The flaw in your reasoning it that there is not "0.000...1 between it and 1" because it is an infinite process. it has no end, and since every time you go one decimal place further out it gets smaller, that makes the limit of 1-0.999... as the number of 9's approaches infinity = 0. at infinity this becomes 0 so at infinity 0.999... and 1 have a difference of 0 and thus 1=0.999... and your computer analogy fails for the same reason. a time step of 0.000...1 is not equal to 1-0.999..., that is equal to 1-0.999...9, a finite number. infinity breaks alot of things like this.

[@EmperorZelos]

caperUnderscore26 You're god damn circular argument again, you wouldn't know what circular mean even after ebign stuck in a roundabout for 75 years. 1/3=0,(3) is just from division algorithm it assumes nothing of 0,(9)=1

[@tweeweekes5309]

***** ahem, I am trying to point out that your argument is circular. Debating whether 1/3 is the same as .3333... is the same as debating .999... is 1. So let's let's just deal with why .999... is not 1.

[@CharlesSprinkle]

If 1=0.999... and 0.000...(1) = 0, then what have we learned in calculus? The whole revolutionary idea in differential and integral calculus is that we deal with things that are either infinitely close to, or infinitely big or infinitely small. In limits, we argue that we approach a value of x, infinitely close, but never touching. To touch would be to equal, which invalidates the definition of a limit. When we do differentials, perhaps the slope of a curve at a point. We say that we let the change in x and change in y be infinitely small, but not zero. if we let it be zero, it's undefined because it would just be 0/0. But the infinitely small changes, dy/dx give us the slope. In integrals, we sum up infinitely many, infinitely small pieces. If each of these infinitely small pieces were really zero, well then the integral is useless, because you're just adding up zero repetitively to nothing more than zero will ever be. There's no such thing as calculus if 0.999... = 1, and if something infinitely small was numerically equal to zero. We can always treat infinitely small things as zero, while they are not actually numerically zero, and we can treat things that are infinitely close to another number, as the actual number, even though it isn't numerically equal to.

[@SteveMcRae]

CJ Sprinkle This is because both you can Caper seem to confuse the term LIMIT with Limit of Sequences which are the summation of partial sums. A limit of sequences that converge is DEFINED to be that value. They are not the same thing. If I took the number 1 and broke it down to .9 + .09 +.009 +.0009 ... and repeated this ad infinitum , regardless of what people think about "remainders" I could break 1 down as .999... and the sum of .999... will ALWAYS still be equal to the whole. 1=.999... by absolute definition of limit of sequences...not by the limit.

[@SteveMcRae]

caperUnderscore26 Again, the "LIMIT" and "Limit of Sequence" are different terms. You are equivocating the terminology. ,999... is DEFINED as ≡ Σ 9/10i , i=1 to n as n goes to infinity. This you know to be true, but you disagree with "definitions".

[@SteveMcRae]

caperUnderscore26 The limit of the SEQUENCE defines it as .9, .99, .999, ,9999, .99999 ... that converges to 1 as the limit of the sequence is .999... That is the definition of limit of sequence for Σ 9/10(i).

[@SteveMcRae]

caperUnderscore26 A sequence is such as .9, .99, .999, .9999 ... the sum of the sequence is the series of partial sums as sn = a1 + a2 + a3 + ... + an which is the same as the limit of the sequence for an infinite series. For .999... all the numbers of the sequence are already there, it is the summation of all of them to infinity which makes it equal to 1. Sequence: .9, .99, .999, .9999 ... Series of Σ 9/10(i) = limit of the Sequence. = 1 We don't have to reach .999... as all the numbers in the sequence are already there and exactly by summation are equal to 1. It is the same for any real number, as all of them can be expressed as a limit of sequence the same way and If you use a geometric series progression to describe .999... = 9(1/10) + 9(1/10)^2 + 9(1/10)^3 + ... = [9(1/10)] / 1 - (1/10) = 1

[@SteveMcRae]

caperUnderscore26 "I haven't a clue what he is trying to do." obviously not...no where did I say series of Σ 9/10(i) was the exact same thing as the actual sequence. I said, the LIMIT of the sequence of .9, .99 , .999 ... is EQUAL to the series of Σ 9/10(i) = .9 +.09 + .009 + ..., which is what the definition of the a series means! The definition of a series is the sum of the sequence.

[@shaihulud4515]

WTF? I ve been looking vids about this topic for the last hour - and this guy explains it easily and comprehensive in 1:58 minutes!!! Brilliant!

[@joi314]

gosh this was short but extremely interesting i love maths so much

[@ThexxCODxxFATHER]

I still don't understand why all of these people are telling the mathematician that he's got the maths wrong...

[@nagyandras8857]

common sense.

[@Koisheep]

Have another mathematician, then: kzbin.info/www/bejne/raSyiamHo5h9q6c The thing is, this is not a formal proof, just a way to get the message across. That being said, you can actually PROVE the equality by using recursive sequences, but the process involves proving a property of sequences which requires the formal definition of limit. And that's not friendly.

[@jyrikgauldurson8169]

Konhat Lee Sakurai LOL, why are you citing an April Fools video

[@nekrataali]

Dunning-Kruger in full effect for this comments section. I can only assume it's a bunch of middle schoolers upset their teacher gave them a 1 / ∞ on their homework.

[@Jack-vo1bf]

I watched a pretty interesting video by Vi Hart in which she tries to show why the proofs for 1 = 0.999... are invalid. The video is here: kzbin.info/www/bejne/raSyiamHo5h9q6c

[@singingbanana]

+Jack That was a joke. It was uploaded on April the 1st.

[@Jack-vo1bf]

+singingbanana Oh, that makes sense. Thanks

[@rbdoppler7841]

+singingbanana Don't you mean April the 0.999... ? :D

[@mikeguitar9769]

Yeah it was actually the last day of March, slightly before midnight.

[@AnteP-dx4my]

γRB Doppler april of the 0.99.... st * now its correct :P

[@AlexGF008]

theres an easier explanation: 1/3 + 2/3 = 0.33333... + 0.666666.... = o.9999999..... as your calculator rightly states when u do it, but also it is: 3/3 = 1 and this are absolutely true facts and absolutely compatible, people usually get mad about a number is equal to another aprently different number but as he said its a problem of notation

[@Martin-pb7ts]

When I was at school we were taught that you could indicate "recurring" by putting a dot above the number. So 0.999.... recurring would be written as 0.9 with a dot above the 9, is that notation still valid?

[@Martin-pb7ts]

That's right. I forgot about the series of repeating digits. :-)

[@urahi830]

the comment section of this video is astonishingly depressive

[@nekrataali]

Seriously it's full of people who haven't gotten to Algebra 1 lmao.

[@martijnvanweele6204]

If you're saying 0,000... is infinitessimal, then aren't you contradicting yourself? Isn't an infinitessimal a value that is as close as you can get to 0 without it being 0? If that is true, and 0,000... is infinitessimal, that means that de difference between 1 and 0,999..., which equals 0,000..., is infinitessimal, and therefore not 0, and if the difference between two numbers doesn't equal 0, they do not equal each other, right?

[@EmperorZelos]

No, infinitesimals isn't "as close as you can get", any infinitesimal isn't closest, there are always closer elements.

[@martijnvanweele6204]

***** I thought that was the point. An infinitessimal is like the opposite of infinity. Of course, just like how ever large a number gets, there will always be a larger one, however close to 0 a number is, there's always a closer number. The point is that both infinite and infinitessimal are theoretical concepts made to represent respectively the largest something can get and the closest something can get to 0 even though those don't actually exist. Or am I getting that completely wrong?

[@EmperorZelos]

martijn van weele You are wrong, again, there will never be anything "largest" or "smallest"

[@martijnvanweele6204]

***** You seem to misunderstand; I never said there was anything largest or smallest. I'm just saying that infinity and infinitessimal are simply nonexistant, hypothetical concepts that represent a largest or a smallest *if* such a thing existed...

[@EmperorZelos]

martijn van weele They exist as mathematical objects like all numbers.

[@BigDBrian]

decimal notation represents a sum. e.g. 123.45 = 1*100 + 2*10 + 3*1 + 4/10 + 5/100 therefore 0.999... = 9/10 + 9/100 + 9/1000 +... which is a geometric series. The sum of which is n/(1-r) where n is the first term (9/10) and r is the ratio between consecutive terms (1/10) so we get: 0.999 = (9/10)/(1-1/10) = 1

[@PH03N1CS]

It can be done as reoccurring digits in decimals can also be written as geometric progression. When we do GP to write 0.99999… it gives exact 1.

[@exowelle]

well I've seen lots of these explanations and read the comments and stuff, but I still feel very uncomfortable about saying 0.999... is the same as 1, because the number "before" 1 indeed is 0.999... so you can't just say 1=0.999... because if you would do that 0.999... would equal 0.999...8 (so the number "before" 0,999...) and if you would continue that you would say 1=2 and stuff like that so I am really pretty uncertain about that

[@Araqius]

You misunderstood something. There is no number right next to another number as there will always be a number (or numbers) between them. This is explained by Dedekind Cut. Also, there are no smallest number (that is not 0) and there are no largest number. 0.999...8 doesn't exist since there are infinite 9s. Infinite 9s means the 9s never end so you can't put 8 there. You can't put something at the end of something that has no end.

[@exowelle]

Araqius I understand all of your opinions but still I don't like to agree that 0.999... is 1 :(

[@GeorgeSorrell]

You can't have numbers one after the other because the real number line is continuous. In fact, if two numbers are different, you should be able to find infinitely many numbers between the two. If you can't find a number between 0.999... and 1, then they must be equal, and this is in fact the case.

[@Araqius]

Exowelle It's not my opinion, it's logic. Real number line is dense, there will be infinite amount of numbers between two different numbers. The same goes as 1/3 = 0.333...

[@exowelle]

Araqius but that underlines my opinion even more for my sense... if there is an infinite amount of numbers between 0,999... and 1 how can it be the same? between 1 and 2 are also infinite numbers and they aren't the same

[@evalsoftserver]

Approximation of interger notation basically conserves prime even number factorization to a constant minima maxima

[@filipsperl]

Does that means that limits really equal the value their series tends to? I always thought we just defined it to equal the value, even though it's never quite it.

[@EmperorZelos]

the sum is defined as the limit of the partial sums.

[@julienr3374]

Or 1= 3*1/3 = 3*0.3333...= 0.9999...

[@frankyjayhay]

It seems to hinge on the two infinite series of 9s. Suppose S = .99 then 10S = 9.9 and so 9S = 9+(.9-.99) or 9-.09 which is 8.91 so that S=.99 as we started with. It seems to me they've sneaked an extra 9 on to the end of the infinitely long series of 9s after the decimal point when multiplying by 10 to make the two infinite lengths exactly the same (so to speak) so they can be subtracted exactly. It seems to me more of an "acceptable fiddle" than a rigorous poof, more like infinitely close rather than equal. You're not adding much but you are adding something.

[@EmperorZelos]

Acctualyl infinitely long strings of 9s are always equally long, INFINITE, or more preicely, aleph-0 No one has added anymore or removed any 9s

[@SteveMcRae]

857frank If there is an infinite string of 9's then ANY value of 10 you multiply it by will never change the number of 9's to the right of the decimal. As EmperorZelos points out, it is INFINITE , aleph-0. S=.99... , 10S = 9.999... 10000000x = 9999999.999... It doesn't matter what you times it by, 10, 100, 1000, or a trillion...there will ALWAYS be the exact same number of 9's to the right of the decimal. All you are effectively doing is changing the position of S on the number line to a value that is 10 + S , so that .999... shifts from .999... on the number line, to 9.999... when you multiply it by 10. (i.e. x.999... the .999... is always the same infinite string regardless the value of x)

[@Araqius]

The number of 9s is infinite and we all know that "infinity + 1 = infinity". Basically, adding another 9 won't change the number of 9s.

[@frankyjayhay]

Araqius That's the sort of "fiddle" I was getting at, not saying it's wrong but that proofs involving infinity can be on shaky ground. It's true that infinity + 1 = infinity but then you could deduce that 1 = 0. Similarly I can see it's true, above, that 10x the infinite string .999... can't really add a 9 when there's no real end to add it to, just a figure of speech on my part. Perhaps it's just a manipulation with infinity that we can get away with on this occasion because it makes intuitive sense in the first place.

[@SteveMcRae]

857frank "It's true that infinity + 1 = infinity but then you could deduce that 1 = 0" Why would you think this type of deduction would be true? Any finite process done on an infinite set and the set still remains infinite. Proofs are either valid or invalid...and are based upon axioms we accept. There is nothing "shaky" when you accept how the reals are constructed. No 9 is added when we multiply 10 * .999... = 9.999... = 10 as nothing changes on the right of the decimal: 10 * .999... = 10 * 1 = 10 (as 1=.999...) and that 10 * .999... = 9.999... = 9 + .999... = 10 (again 1=.999...) If we have 9.999... or 99.999... or 999.999... the number of 9's to the right of the decimal for all of them are all the same infinite string of 9's. How we got there by multiplication or addition makes no difference. 9.999... = 9 + .999... and 10 * .999... = 9.999... = 9 + .999... the .999... in both of the above lines are complete the same.

[@firstnamelastname-oy7es]

I know why .9 recurring is equal to one, but it seams counter intuitive to think about it that way because maybe there is more use to the gap between 1 and the convergence area to one that we think there is.

[@holycrapitsachicken]

How does this equality function in regards to limits? More specifically 1+. 0.9999... can represent 1- and 1, but isn't it impossible to consider it 1+?

[@EmperorZelos]

what does 1- and 1+ even mean?

[@holycrapitsachicken]

The left and right limits of a number. So 1- would be the left side of the number one and 1+ is the right side of number 1. It's important for asymptotes since the left side could be something like a value negative infinity while the right side of the same number could be going to positive infinity.

[@EmperorZelos]

In real numbers they are both 1. You are thinking of functions.

[@holycrapitsachicken]

Well 0.9999... is a real number since fractions are part of real numbers, and this represents 3/3. I just find it strange that 0.9999... can represent 1 because it implies that there can be right side of the number 0.999... We can take limits of irrational numbers like pi after all. I just can't see the right side of 0.999... being anything other than 1. 1+ is 1 + infinitesimally small number. But 0.9999 is 1 - infinitesimally small number. So 1 - infinitesimally small number + infinitesimally small number should be 1. Even if the small numbers are different. 1+ is something like 1.00...001. I don't see how the right side of 0.999... can be 1.00...001 because it passes through 1. Unquestionably, 1 - an infinitesimally small number = 1 so 0.999... = 1. However, the representation of 0.999... is usually known as 1- since that's how we are able to state that a limit approaches an infinity. I reflected on this a bit myself and came to the conclusion that it's because it's an infinitely repeating stream of nines which means it can represent a perfect fraction 3/3. While 0.999...+ would become something like 3/3 + infinitesimally small number. Since 0.999... in limits is known as 1 - infinitesimally small number. I guess that means 3/3 = 0.999... + infinitesimally small number = 1. So strings of repeated digits essentially need to add an infinitesimal to become an actual fraction. My theory is that 0.333... does not equal 1/3. Rather 0.333... + infinitesimal = 1/3. So then when it is multiplied by 3 it becomes 0.999... + infinitesimal which definitely is 1. This is simply due to the nature of the remainder. If you divide 1 by 3. You will get .3 then you add .03 then you add .003. If you want to stop at 0.333 you would need to say the quotient is 0.333 + 1/3000. However, if you stream this all the way to infinity then the remainder is 3/infinity which is simply an infinitesimal. So that means that all fractions that have infinitely repeating decimals must have an infinitesimal remainder added to it.

[@SteveMcRae]

Once again CAPER is being QUITE disingenuous as he uses a Wikipedia reference with out validating the whole picture taken WELL out of context: "Oh look: Dr. Fred Richman ! " *Perhaps* the situation is that some real numbers can only be approximated, like the square root of 2, whereas others, like 1, can be written exactly, but can also be approximated. So 0.999... is a series that approximates the exact number 1 . Of course this dichotomy depends on what we allow for approximations. For some purposes we might allow any rational number, but for our present discussion the terminating decimals---the decimal fractions---are the natural candidates. These can only approximate 1/3, for example, so we don't have an exact expression for 1/3." Also should be included: "The above approach to assigning a real number to each decimal expansion is due to an expository paper titled "Is 0.999 ... = 1?" by Fred Richman in Mathematics Magazine, which is targeted at teachers of collegiate mathematics, especially at the junior/senior level, and their students.[17] Richman notes that taking Dedekind cuts in any dense subset of the rational numbers yields the same results; in particular, he uses decimal fractions, for which the proof is more immediate. He also notes that typically the definitions allow { x : x < 1 } to be a cut but not { x : x ≤ 1 } (or vice versa) "Why do that? Precisely to rule out the existence of distinct numbers 0.9* and 1. [...] *So we see that in the traditional definition of the real numbers, the equation 0.9* = 1 is built in at the beginning."[18] A further modification of the procedure leads to a different structure where the two are not equal. Although it is consistent, many of the common rules of decimal arithmetic no longer hold, for example the fraction 1/3 has no representation; see "Alternative number systems" below."* Although the real numbers form an extremely useful number system, the decision to interpret the notation "0.999..." as naming a real number is ultimately a convention, and Timothy Gowers argues in Mathematics: A Very Short Introduction that the resulting identity 0.999... = 1 is a convention as well: *However, it is by no means an arbitrary convention, because not adopting it forces one either to invent strange new objects or to abandon some of the familiar rules of arithmetic*.[47] *One can define other number systems using different rules or new objects; in some such number systems, the above proofs would need to be reinterpreted and one might find that, in a given number system, 0.999... and 1 might not be identical. However, many number systems are extensions of -rather than independent alternatives to- the real number system, so 0.999... = 1 continues to hold.* Fred Richman was writing an expository paper on possible ALTERNATIVE number systems to eliminate repeating decimals, which would lead to complications and basic rules of arithmetic. It was a paper written to TEACHERS and EDUCATORS to get their students to THINK why 1=.999... and not just accept it as fact. Also, yes in any possible ALTERNATIVE NUMBER SYSTEM then 1 may not equal .999... but in the REALS 1=.999... which is the WHOLE point of his paper. Nothing in his paper suggest that for the REAL NUMBER system 1 doesn't =.999...

[@cperez1000]

what is hard for me to understand is not that 0.999...=1, but the fact that there is people here with no expertise in math (not even basic knowledge) saying that 0.999... does not equal one, despite any demonstration and math doctorates showing them otherwise.

[@nagyandras8857]

quite simple. we think otherwise.

[@MrCmon113]

Those demonstrations are null and void without proper definitions.

[@jonahansen]

Thank you so much. You cut to the heart of the matter up front: the decimal notation represents the number and isn't unique; it's not the number. So many other videos just keep proving it various ways but never really mention this, the crux, of the misunderstanding.

[@oxylepy2]

If .9999... = 1 then 1 minus an infinitesimal = 1 does that mean that 1 minus any finite number of infinitesimals is equal to 1?

[@johnwhite30]

caperUnderscore26 _"Σ[n = 1 to ∞] n= 1 to n = ∞ ?_ _That is fail."_ That's why you have to properly define this notation: Σ[n = 1 to ∞] a_n := lim(N → ∞) Σ[n = 1 to N] a_n Your statement _"0.999... = 1 - 1/10ⁿ as n tends toward infinity."_ is simply wrong. The correct way to say it, is 1 - 1/10ⁿ = 0.9(n-times) *tends* towards 0.999... as n tends towards infinity. 0.999... does *NOT* equal to 0.9(n-times)! Later in your post you say _"f(n) = 0.(9)"_ where _"f(n) is also defined as 1 - 1/10ⁿ for n ∈ N (all natural numbers)"_ which is even more laughable. Now you say, 0.9(n-times) is equal to 0.(9) for *all natural numbers*?!? Since when did 0.(9) become a function of n? Can you give a single source, that defines 0.(9) like that? I guess not. _"At no point does f(n) = g(n)"_ f(1) = 0.9 ≠ 0.(9) f(2) = 0.99 ≠ 0.(9) f(3) = 0.999 ≠ 0.(9) f(4) = 0.9999 ≠ 0.(9) ... Guess what, for all n ∈ N, *at no point does f(n) = 0.(9)* either! So much fail in a post, that ironically started with _"That is fail"_. Back to school with you! Oh and btw. disabling replies, makes your garbage look even more pathetic.

[@EmperorZelos]

Yepp, he doesn't know how to go about things at all.

[@luigiionascu9797]

..your calkcul is correct ..!.callculus implcated 0,(9)different1 ..

[@lol93716]

You could also theorise it: Between every 2 real numbers there is an infinite number of other real numbers For example: between 0 and 1 lies 0.5, 0.25, 0.125, 0.75 etc This is not true for 1 and 0.999... Because you cannot imagine a number between those 2

[@709zzy]

yeah i can, its called infinitesimal, we use it in calculus all the time. dx, dy, dz, they are absolutely real. And if you say they don't exist or they = 0, you break calculus.

[@blykgod]

Yes,I can just take the average of two

[@nagyandras8857]

depends on the decimal system you use.

[@alanturingtesla]

As Nagy says, it is imaginable in system with more than 10 units.

[@cotronixcoolson3999]

That does not make 9.999... equal to 10.

[@InverseSwagster]

If you know your sums of infinite series you can prove this as a basic ratio series’s beginning at a=9 with a ratio of (1/10)^x between each item in the series. 0.9 + 0.009 + 0.0009..... = 0.999999... = Which by the ratio rule on their series described above is equal to 1.

[@Sydrooo]

Very interesting to be honest. Nice video. :)

[@KongofPip]

What if God was 0.999... of us?

[@xZak-A-42]

Underrated comment.

[@cryptexify]

1/9 = 0.111... 2/9 = 0.222... 9/9 = 0.999... 9/9 = 1 Simple as that.

[@Baerchenization]

+cryptexify I want to see you sitting down with pen and paper and perform 1/9, and then 9/9. You do not get 0.999... if you do that, you get 1. The only reason you could add 9/9 = 0.999... to your list is with hindsight, after having learned that this result is true through other ways of looking at this issue, i.e. other explanations. 0.999... is NOT what you get if you sit down to calculate it. That would be 1. You do not get 9/9 = 0.999... any more than you get 6/2 = 2.9999... righties? ;) What you have offered is true - but it is not a proof for the fact.

[@user-pj8kp4vl9n]

Baerchenization x = 0.9999.... 10x=9.9999.... 9x=9 x=1

[@kenmolinaro]

@@Baerchenization 9 x (1/9) = 9 x (0.1111...) (9/9) = (0.9999...) You made an incorrect assumption about how each step was reached in the sequence given by crypexify. They simply shortened up the post by leaving out the multiplication step that occurs on each line to generate the next line.

[@applessuace]

@Ian Switzer So you end with: "For example, the formula works like it should." But started with: "The formula is a mathematical illusion." Thanks for playing.

[@jacquesjenseen7355]

So, by making the above statement, especially involving limits, isn't this really about semantics, then?

[@restlessimagination765]

This is a fallacy in reasoning resulting from the confusion in mixing unreal infinite concepts with defined and definite mathematical values. 0.999... is nothing but just another form of 0.999...9 wherein the "..." symbol represents a never-ending ( infinite) repetition of the digit 9. It would never equal to 1 because no matter how you expand the value of the number it would always have a 9 at the end, therefore it is always short or less than 1.0 by an amount represented by 0.000...1 ( which is an ever decreasing number that would never approach zero) I agree however that 0.9999... ( or 0.9999...9 ) is APPROXIMATELY equal to 1.0 in our practical ( within acceptable tolerance) world application, but not exactly equal to 1.0 in pure and exacting mathematical concept.

[@MisterrLi]

No, they are exactly the same. The two numeral symbols 1 and 0.999... are defined as equal in the real number system. This is because the real number system is limited to finite values and no infinitesimal values (except for zero). If you end up with 0.000...1 and similar you are dealing with infinitesimals and these are all zero in the reals, = 0.000... .

[@hollth6770]

You're slightly misunderstanding the infinite part. .99999... is not an ever increasing value. It is not a dynamic thing. For 1-.999... Because there is no end to the 9's there is no 0.00..1 remainder. There is only 0.00..1 if .9999... stops at some point.

[@AlarmClock65]

There is no nine at the end. There is no end. Numbers don't increase or decrease. They just exist. Functions can increase or decrease, but 0.999... is a number, not a function.

[@I_leave_mean_comments]

this is not an explanation.

[@selfcentered3406]

It is proof.

[@mrpablodelosmarbles7966]

I am not entirely sure whether it makes sense to even think about what numbers with infinite decimal places are, since they don't exist in the physical world. Therefore, in principle, they could be anything you want them to be.

[@Mmmm1ch43l]

By the same token, no numbers exist in the real world. All numbers are just what we want them to be. It's just that if you define the decimal notation in any meaningful way, then you'll get 0.99999...=1

[@abdullahyosof6041]

from what I taught, subtracting these two equations from each other resulting in 9s=9 is invalid because those two equations should be linearly independent (not multiplied or divided by any real number)

[@EmperorZelos]

That is not neccisery at all at all, equals subtracted by another equals is still equal.

[@abdullahyosof6041]

+EmperorZelos but will not lead to a valid proof!

[@EmperorZelos]

Abdullah yousef It is perfectly valid, the proof given here is 100% valid and accurate.

[@Araqius]

+Abdullah yousef Assume x = 2 4 - x = 4 - 2 What is wrong?

[@mikeguitar9769]

Abdullah, that's a good thought, but as I recall, what that rule says is you need at least one linearly-independent equation for each variable you want to solve for. Since there is only one variable here, only one equation is needed, if I'm not mistaken.

[@florinburian7291]

1:09 this is wrong. you have broken math.

[@florinburian7291]

9s is not equal to 9, its 8.9999999.. say you have n digits after 0 : after multiplying with 10, you will always get n-1 digits after the decimal point even when n tends to infinity say for example, n=5 so 0.99999*10 =9.9999 9.9999- 0.99999 _______ 8.99991

[@Araqius]

florin burian. You have broken your head. There are infinite 9s. There is no last digit. "after multiplying with 10, you will always get n-1 digits after the decimal point" This means you are so stupid, infinity - 1 equals infinity.

[@SteveMcRae]

florin burian. 8.9999.... = 9 my mathematically challenged friend. You are also quite confused as you are expressing finite iterations when .999... is formed from an infinite amount of iterations. .999... * 10 =9.999... was a trivial proof by Euler himself or are you a better mathematician than Euler? If so please publish your paper and collect that Fields Medal! The only thing here that is broken is your understanding of infinite convergent series and basic mathematics.

[@LassekLorenz]

florin burian. "even when n tends to infinity" n does not tend in infinity. It IS infinite!!

[@SteveMcRae]

Nee Roo He thinks .999... is some on going process where there is some type of asymptotic behavior taking place rather than understanding that .999... is a complete, finite, rational number..or as a completed process of an infinite convergent series. It is a all to often misunderstanding of maths.

[@BenRK90]

I never understood why 1 - 0.9999... had nothing at the end. I get that it would simply be impractical. In my mind I figured that if one were to take an infinitesimally small chunk out of an object, you wouldn't notice a difference no matter how well you examined the object. Thus you took out practically nothing, or 0. But surely you did take SOMETHING away from the object. Eh, I just don't understand why we just automatically assume 0.999999... is equal to one when it clearly is equal to 0.999999... I get why it is practical to do that however. Fun fact, my friends an I used to call the solution of 1 - 0.9999... the length of a point.

[@kenteato]

0.999... is a infinitely repeating number. You will never reach the end of the number. Think of it this way. A finite number divided by infinity. If you had 1 trillion dollars and lose 1 dollar, the 1 dollar lost is almost nothing to you right? But if you lose 1 dollar out of infinity dollars, you are losing nothing because there is no "finite" source to take from. 0.999.... has an infinitely repeating decimal (remember infinite in a concept, you can't apply finite metaphors to prove a correlation with infinity). 1-0.999... = ? You will never reach the end of 0.999.... You will get 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000............................. towards infinity. You may say that at the very end of the number will be 1, but that's like saying a infinitely filled glass of water must be able to be empty at some point. No, infinity will never stop no matter how deep into you go. You will just reach more .0000000 the farther you go into infinity, you can never reach the 1 at the end of a finite source in an infinite source. 1-0.999... = 0, if it was 1-0.999999, you would get a number, but since it is a infintely repeating number you can't.

[@BenRK90]

Kenneth Hua Trust me, I get the practicality of it 0.99999... being 1, and the practicality of 1 - 0.9999999 = 0, I wont argue that. Though wouldn't it be more accurate to say 0.999999... approaches 1, and that 1 - 0.999999... approaches 0?

[@kenteato]

If you are assuming 0.999... = 1 + x lim (0.999...) = 1 lim (1) = 1 lim (x) = 0 lim (0.999...) = lim (1) - lim (x) 1=1-0

[@BenRK90]

It still feels like we're saying 99 = 100, to oversimplify things.

[@Araqius]

Ben RK Nope. 99 =/= 100 but 99.999... = 100 "wouldn't it be more accurate to say 0.999999... approaches 1, and that 1 - 0.999999... approaches 0?" 0.999... is a number, no number approach another number.

[@777Dunkelheit666]

what about lim [n->inft] 1^n-(0.999...)^n ?

[@EmperorZelos]

it's 0 :)

[@duckles426]

A way which might help someone, in binary an equivalent of 0.9_ would be 0.1_ (using _ as recurring) which equals a half plus a quarter plus and eight and so.on which sums to one, and it's the same for 0.9_ Or noting that 0.9_ is 3 × 0.3_, and in dozenal (base 12) three thirds is 1.

[@Chris-5318]

In dozenal (AKA duodecimal) 3/3 is 0.bbb . . . as well. Some people think that 0.999 . . . (base 10) is a bigger 1 that 0.111 . . . (base 2). Amazingly, they have almost identical DNA to ours (or so I have been led to believe).

[@deannarobles2700]

Doesn't make sense. Isn't 0.999 right before one?

[@SteveMcRae]

Deanna Robles 0.999 ? You mean .999... which is .999repeating? And no, .999... is not before one as there is "no number right before one" or a highest number that is less than 1 but not equal to one because the real numbers are dense. .999... is just another way to represent the number 1 as they are completely equal in value. All rational numbers can be expressed in the form of an infinite repeating decimal. 1=.999... or 2=1.999... or 64.86=64.85999... ect.

[@freddieorrell]

Take any single number from the infinite set of 9.999..., for example 9.99, 9.99999999999999999999 or 9.99999. There are an infinite number of them, but none of those numbers is equal to 1. Subtract any of those numbers from 1. The answer could be 0.01, 0.00000000000000000001 or 0.00001, but never zero. [Erratum: I meant 0.9-etc throughout, not 9.9-etc, of course. My own typo and careless proof-reading]

[@SteveMcRae]

Freddie Orrell Allow me to introduce to you the "archimedean property". You confuse finite iterations with infinite iterations. 1- .999... = 0 because the archimedean property does not allow for any infinitesimals in ℝ (other than zero).

[@freddieorrell]

I'm appealing to logic, not notation. You can say 'this notation means that equals the other thing' and always be right.

[@SteveMcRae]

Freddie Orrell If you appeal to logic then please use predicate logic in a proof that shows I am incorrect. The reason we can say notation equals what it does is because in mathematics things are clearly defined. In math the definition of a limit of a sequence means that the value a sequence converges upon is the value for that sequence and are equal. Mathematically this is expressed as .999... = lim n →∞ Σ 9/10^k, k=1 to n which happens to equal 1. If you "appeal" to logic in math you first need to have an understanding of the mathematics involved else you fall into logical fallacies as you have.