The repeating decimal 0.99999.... = 1. They're exactly equal! Let's explore this in five levels, ranging from a quick calculation with fractions, to a more rigorous mathematical proof.
More of the technical details for Level 5: We want to find an interpretation of 0.999... as a nonstandard real number. We'll need a new definition for interpreting this (because if we define it as a real number, 0.999... = 1 as we saw!) One option is to use an ultralimit, defined by Terence Tao. If we interpret 0.999... as having R 9's where R is an infinite hypernatural rank, then we could express this as 1 - 0.1^R. This is like how 0.9 = 1 - 0.1^1, and 0.99 = 1 - 0.1^2. More precisely, we can use Lightstone's decimal expansion for the hyperreals to write this value as 0.999...;...999000... The final 9 occurs at the infinite hypernatural rank R. This value made up of "0." followed by infinitely many 9s, but is also an infinitesimal below the value 1. For short, we could write it as 0.999... and put an R underneath to indicate that the number of 9s is the infinite hyperrank R. This helps to clarify which hyperrank R we are using. But whichever choice of R we make, the value is an infinitesimal less than 1. Because of this construction (and other considerations like learning calculus), some math education researchers have questioned if the intuition that 0.999... is an infinitesimal less than 1 has some merit as it applies to nonstandard settings. (K. U. Katz, M. G. Katz, 2010; also R. Ely 2010). Ian Stewart also mentions this "entirely reasonable" justification in his book (Professor Stewart's Hoard of Mathematical Treasures). At this point, this is perhaps mostly "meta-math", and I'm sure some others may disagree with their sentiments. And just to reiterate, nearly all of the time, we want to just consider 0.999... as a real number. And when we do, it must be precisely 1!
This is not appropriate, because of the transfer principle. You want to make sure that when you define nonstandard analysis, all the first-order properties of the real numbers are preserved. The details are no good, sorry. The "hypernatural rank" interpretation isn't what anyone means when they write .9999..., it ALWAYS means limit as n goes to infinity (past all ranks). The common intuition is simply false, and it must not be coddled.
@@annaclarafenyo8185 you don't actually write 0.9999... meaning a limit in hyperreal aalysis, because the order topology on hyperreals is really poor. You usually mean some but not particular infinite expansion up to some infinite hyperreal rank R. Thus, this is not quite rigorous, but actually an incredible catch on intuition, that 0.999... need not be a 1, because we need not to think of reals when we speak of "infinite sequence of 9s after 0". Transfer principle is not that great, because it only catches a glimpse of all hyperreal arithmetic, tools and even operations that standardify to concepts like continuity, uniform continuity or differentiability. Too many interesting parts of hyperreal analysis are not transferable at all and it's not bad, because in the end, it's not only FO properties that matter.
@@PoweredDragon NO! You mean LIMIT in the hyperreals. Same as always. And it's equal to 1. Same as always. Because it's a first order property, and it's preserved.
@@andytraiger4079 The reason is that decimal expansions are not one-to-one, and they can't be, because an infinite list of digits makes a disconnected space, and the real line is one continuous thing. The ".99999"s are where you glue the real line together from the discrete bits of the digit expansion. It's a different gluing in every base, and it's always at the point where "base -1" turns over to 0.
@@sebastianmanterfield3132 I hope he made this video with source/refernce. But I have an another argument to make I am not sure how we could solve that one. So why we have a representation of something if we are not "able" to use it? 0,999...8, 1 ? 0,999...8, 0,999...9, 1 ?
This video taught me that discreteness in mathematics is completely illusory, and that arbitrary rules are what makes it useful. I could sleep on this for the lifetime of a universe and still not understand the true nature of mathematics, but your video has shined some light onto my ignorance.
Well, real numbers are not discrete they are continuous. But there are branches of mathematics that are done with other kinds of numbers such as integers, naturals and rationals which are absolutely discrete (discreteness is not an illusion in those contexts). The term "real" in real numbers does not necessarilly mean that they are the deepest or most real part of mathematics, although some philosophers argue that we are committed to the existence of real numbers since pi is a real number which appears in some laws of physics which some people think are the literal manifestation of the laws of the universe. As you said, part of the power of mathematics is that they can be adapted to different contexts whether you need discreteness or continuity, mathematics has got you covered.
Math lives in its domain. What's the true nature of a game of chess? Define it as a series of moves without minding metadata such as time taken and name of the opponents. The true nature of a game of chess is it being an abstraction residing in the domain of chess. A game of chess exists that nobody has ever played. And nobody ever will. Or, if I play a game of chess and the cat hits the board it did not change the game of chess. It changed its representation, but you can simply restore the configuration and for what concerns the game itself NOTHING happened. Or, I can play by accident the same game that a chess engine (at low level) played. It's THE SAME game because it does not matter who in the real world did it. Same as math. A line is infinite, our world probably isn't, that's no prob. Same as math. A line is infinite, our world probably isn't, that's no prob.
@@electroborg chess is finite, there is a number of possible actions to take that exponencially grows the more you progress into it up to the middle of possible games being the peak of possibilitys and then it goes the other way around exponentially until the possibilitys end up in draws or wins for either side, it sure feels infinite because well, is not like you would be able to play all possible games of chess in a lifetime, but it sure is finite because it has restrictions and limitations to how much moves can be done because each piece has a limit of posssible moves, the more deeper you go the more possibilitys you find that makes you think that its infinite while its not.
The major problem is you are assuming operations on infinite series or working with Concepts like least upper bound on the number line without proof that these things are valid. How can a person justify going from the finite to the infinite
It's funny how people will bang on about 0.999... never quit reaching 1, but never complain about 0.333... not making it to 1/3; or 0.14285714... staying just shy of 1/7.
@@gabrielandriel8960 they're referencing something called the p-adics, where analysis like this makes sense. In the normal system we're familiar with, not really. But it is fun, so I take it as gospel anyway
Real numbers are dense. That means, between two real numbers, there are infinite other real numbers. And what real number fits between 0,9999... and 1?
@@jacobsmith423920.00000…000001 is not the result from from 1 - 0.999999… This is because for there to be a terminating 1, there must be a finite amount of 9s. Since the 9s are infinitely repeating, there aren’t a finite amount of 9s and thus no terminating 1. The … in 0.0000…0000 is not the same … in 0.99999… It replaces a FINITE amount of omitted 0s, not an INFINITE amount of 0s. If it were an infinite amount of 0s, then the terminating 1 would never be reached. This is until you get into the hyperreals. Then it’s all different
There is another way to define when two numbers are equal. We have proven the density property of the set of real numbers, so for any two different real numbers, we can find some other numbers in between them. Algebraically, if we have a < b, then we can find some real c such that a < c < b; an example is c = (a + b)/2. We then have the statement “If two real numbers are different, then there is some real number in between them”. From this, we deduce the contrapositive: “If there is no real number in between two real numbers, then those two numbers are equal”. Since 0.999… and 1 are real numbers but have no other real number in between them, they are equal.
Neat story. Except in to actually prove your claim you would have to show the number you call 0.999... truly never ends which you could never do if it doesn't. Your just assuming the 10^10^10^10^1 0^10 digit of that number is there and is 9.
Thank you for your incredible explanations. Honestly it is amazing to see how one can demonstrate common math questions in so many different ways. Thank you for your work.
While I have nothing against the math here, I think the reason why people struggle with this question is because they think 0.99... is a number in itself instead of representation. In this sense 0.99... is not 1.00... but they are different representations of the same number.
Far from the truth. You might as well argue that 0.99 is 1. Because at what decimal do you decide that 0.999 becomes 1? If an Astrophysicist were to do a calculation where time doesn't exist, each number has meaning. So, to calculate "direction" in the entire universe and beyond, the difference between 0.999.. and 1 may suddenly become apparent.
It is the first time that I see that one number or concept represents another. In another million years we would probably argue that 1 = 2. Because 1 and two are the closest to each other when we deal with infinities.
@@sbok9481 Hey, you may argue that the way we use math is wrong, but it is a fact that within the definition of the real numbers and how we represent them in base 10 notation 0.999.. represents "one". Notice that the video introduces another number system in which this is not true. Maybe you like that one better.
oh my i never realized this channel had so little subscribers! I thought this was a popular channel this whole time based on the quality of the videos. Hope your channel blows up more soon!
A medical student here . Though this is very fascinating , i am kind of glad i did not come across this when i was in school . My whole understanding and confidence in maths ( which was weak to begin with ) would have collapsed . Great stuff !!
As a medical student, isn't your area of expertise a system (the human body) that's vastly more complex than what any mind is capable of understanding? Perfect understanding isn't a prerequisite to working with things.
@@paulfoss5385 I was talking about how I would taken this video when I was in school . I hated maths . To me , it very often appeared too inconsistent and too abstract for enjoyment . Around my 10th grade , I started improving my skills in maths and it's abstractness became what I liked . Though still , it was a castle built upon a weak base . Something like this video would have made me feel disillusioned with maths . That's all As for medicine , yeah it's a bottomless pit but still it's far more intuitive for me and
Love your videos, your explanations and format are so easy to follow and make clear sense. And such a soothing voice, I can watch these videos both to learn and to sleep. Thank you ❤
Thank you so much for this. As a layman, you've transformed what seemed at first to me as an odd bit of sophistry into a remarkably insightful primer about different ways of thing about numbers
High level math is just defining axioms and figuring out what you can prove using them. The axiom of choice is rejected by some people and that changes what you can prove.
@@kellymoses8566 So much this. If you say 0.9999 "has a lots of nines in it hasn't it can't be 1" then they go on a rant how dumb you are. But in the end they just believe in the Archimedian principle. Then come other people and say but now i call it epsilon in the Hyperreals and now an epsilon can fitted in there and you are back to 0.999... is NOT equal to one.. Funny how that goes.
You can say that 0.999... = 1 - ε, but so can 0.999... = 1 - nε, or any 1 - f(x)ε where s.t{f(x)} < inf. 0.999... simply fails to be a representation of any specific hyperreal number.
I'm studies maths to degree level (decades ago) and don't recall ever seeing the geometric series formula. So I'm glad you also showed how to solve directly.
Here are two more proofs: 1) Consider relation R between Cauchy sequences of rational numbers: for any two Cauchy sequences of rational numbers a=(a_1, a_2, a_3,...) and b=(b_1, b_2, b_3,...) the relation aRb holds iff lim(a_n-b_n)=0. Any given real number is an equivalence class of such sequences with respect to R. Any given digital representation corresponds to a Cauchy sequence of rational numbers, for example, 0.999... corresponds to (0.9, 0.99, 0.999,...), and 1 corresponds to (1, 1, 1,...). Let's check if (0.9, 0.99, 0.999,...)R(1, 1, 1,...): lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (0.9, 0.99, 0.999,...)R(1, 1, 1,...) and 0.999... = 1. 2) If x is some real number, |x|
@@plresurectzbrokena4997 Why F*K are you here , no one asked you to come to this video, if you are not interested then sh@t somewhere else , he has the right to comment as you have the right to click the video ....
The way it made sense to me was the following: What is 1-0.999...? It's 0.000... infinitely many zeroes and then a 1. But you have infinitely many zeroes, that means an amount of zeroes that never ends, so you can't have a 1 after all the zeroes, there is no "after all the zeros". That's what infinity means. So 1-0.999 must be exactly equal to 0, therefore 1 = 0.999...
Man the very first explanation was so simple and so obvious that I'm amazed we weren't told it in school. I remember being told 0.9999... = 1 but the explanation was just "because there are no values in between" which wasn't enough for me because it's like saying 1=2 in some situation where you're only dealing with integers, such as counting objects etc. As soon as you showed 1/3=0.3333... I could immediately tell where it was going and it made intuitive sense that 1=3/3=0.9999...
Great video - especially the intro to the hyperreals. I taught my 11 year old my favorite method (algebra) and he loved it. I've always been bothered though that other repeating decimals don't have such a nice exact rounding, but I guess that's what happens when you're closest to an integer 😂
My theory could be wrong, you can see as any whole number divided by 9 can give an infinitely long repeating number, for example, 1/9=0.111111... and 31/9 = 3.44444..... There is no possible way to create 0.9999999 in this case while 9/9 =1. Atleast thats how i perceive this
Your assumption is correct (more generally, every rational number can be represented using repeating decimals, and every decimal represents a rational number), but you do need to prove it.
I've been wondering since I was a kid if there was such a thing as infinitesimal numbers as an inverse of infinities. Essentially they do exist, but are non-descriptive of reality similar to infinities.
They are descriptive up to a point. For example plank length and plank time. Normally you don't need more than 3 decimal places, but for say nanoscale you need 9 digits or more and every one of those digits matter.
This one has got me foxed, truly, however I followed the reasoning and your series of mathematical understanding from levels 1-5 is stupendously helpful as is the clear delivery and concise organization per level of explanation. To attempt to add some constructive feedback in the comments might be beyond me other than appreciation expression. Perhaps this is another of the paradoxes of Many = One and vica-versa at the same time as One =/= Many or as another commentator said, it depends? The fractions explanation in Level 1 would drive younger students mad as a way to antagonize them if ever in class they're feeling sleepy and think the world is boring!
An explanation I always liked to use was to note that there is no number you can add to .9 repeating to get it to 1 without going past 1, making it the same number.
I like this. My take is slightly different . I believe that 0.999.. is not a number, just a process which lands on different numbers. And approaches 2.
@@Temperans Yes, good point. Yes, but I would argue that 1/x is not a number but a sequence. And even that 0.999.. is really a sequence of numbers , not a number so that there is no way of finding the singular value of 0.999...
@@Temperans My point is more to conceptualize since there is no "infinite-eth" decimal point. Sure, you can approach it and represent that with a function, but you can never actually arrive at it, meaning you never actually add that number. Granted, you can easily argue approaching infinity means little since you are always just as far from arriving as when you started no matter how far you go since there is always an infinite number of numbers/spaces left for you to go.
@@TheMasterPlayer-uo6msLet’s pretend that 1.9999…99998 is a real number (which it’s not). .(9) multiplied by 2 cannot be 1.9999…99998 because 1.9999…99998/2 would be .9999…99994 1.(9)/2 is .(9) because of how infinity works. Most people when they imagine infinity picture something arbitrarily large. When you add an 8 it’s no longer and infinite, it’s arbitrarily large. You could write 10^100 9s every second from now until you die, and never be anywhere close to finishing writing the number .(9) There is no final 9.
@@masalanicholoff3593What are you talking about??? 1.9998 divided by 2 is 0.9999. Just because there is no final 9 does not mean anything in 0.0000...00001 there is no final zero, but you can still add it to 0.999999... and get 1 since 0.0000..00001 would be the theoretical final number.
I think the problem cant be solved by just using 1/9 or 1/3. If we look et the graph we will see it will never touch 1. Infinites are so strong that its hard to comprehend
Awesome video! A thought though; In the "fourth level" explanation though, part of your breakdown to explain that 1 is an upper bound includes defining that 1 minus any portion of the series is always positive. This assertion seems to contradict the later statement that relies upon it when you say 1-x = 0. I understand that a segment of the series is not the same thing as the limit of adding all terms, but by using that explanation to claim that 1 is an upper bound you do actually have to make the assumption that that property of subtracting off more and more terms is logically extended to all terms. One thing I do think would be cool to mention in terms of the limit explanation in part 3 that I just thought of but don't actually know anything concrete about, is that this 0.99999.... series seems to not only approach 1 in the limit, but also to approach 1 faster than any other possible continuous series; No idea if that is actually true, but(despite some of the point of the video being to to NOT do this) that seems like it might be true based on my intuition. I really love how things objects like 0.99999... challenge our understanding of the number line and how to think about what our tools(like algebra, limits, etc.) are really telling us about what's going on. Thanks for the great content!
Forgot to say thank you for this honest and incredibly information rich video. For me Level 5 was where i always where at just didn't use the words Hyperreals and epsilon....
First thing that popped into my head was dedekind cuts for the definition of real numbers. That is 1 and 0.999... share the same dedekind cut of rational numbers as their definitions. Great video!
When I took my first math course in college and we went over the definition of the reals, we defined them as the quotient of a particular set of infinite integer sequences, over the equivalence relation that forced .999… to be equal to 1, along with other similar sequences terminating in infinitely many 0s. Totally valid, and funny in that it just definitionally makes this a nonissue.
Enjoyed your explanation! I am glad I found your channel. I like your approach to teaching. Perhaps you might consider the following for a future video. My wife is an elementary school teacher. In teaching multiplication, I like how there are multiple ways to approach the concept of multiplication of positive integers, including visual and geometric explanations. Is there a geometric interpretation or another visualization for what is happening when multiplying fractions that could be taught to children? Most of what I see focuses on the algorithm for computing the product.
I feel there's a connection with the recently released video from Numberphile on -1/12, which shows how the "-1/12" values is somewhat imprinted (but hidden) in the 1+2+3+4... series (without considering smooth cutoff). Here, the 0.999...=1 is true, but at the same time the converging processing leaves this "infinitesimal" trace.
'Here, the 0.999...=1 is true, but at the same time the converging processing leaves this "infinitesimal" trace' Infinitesimals do not exist in the space of real numbers. The difference between 0.999... and 1 is 0, because 0.999... = 1 exactly, and not approximately.
From a computer science perspective, we experience the challenge of representing fractions since most computer systems use floating point based on binary to store numbers. The issue is that the storage of many values can't be exactly specified in base 2 (binary) including 1/3. The issue with this question starts with how to represent 1/3 (or more specifically 3 * 0.3333). In decimal, we write 1/3 = 0.33333 (repeating). But, the intent of that representation is that 0.33333 = 1/3. Similarly, 3 x 0.33333 = 0.999999. But, just as similarly, 3 x 1/3 = 1, so 3 x 0.33333 = 1 also. Note that this video shows the challenge of representing some rational numbers as decimal numbers. But, the rational numbers have a problem representing real numbers (like PI or square root 2.) The question in this part of math is how to best represent values, and in that light, its should be clear that decimal is not the best way to represent 1/3. But, we can conclude 0.9999 is 3 x 0.3333, which is 3 x 1/3 which is 1.
To follow up on this, the assumption that 1/3 = 0.(3) is true is done for the sake of having a useful number, but the two values are not actually the same. This is why 1/3*3 = 1 but 0.(3)*3 = 0.(9). This is more clearly seen with 2/3 = 0.(6) which is usually represented as 0.(6)7 due to rounding. But 2/3*3 = 2 but 0.(6)*3 = 1.(9)8 and 0.(6)7*3 = 2.(0)1. That is not including the fact that 0.(9)*2 = 1.(9)8 which is decidedly less than 1.(9) and thus breaks the other proof. The entire premise of 0.(9) = 1 is the result of people rounding because it is more practical.
@@Temperans For the 2/3 example, 2/3 is written either as 0.(6) (where parens mean repeating,) OR is written as 0.6667 (with some number of sixes in there.) If we use this rounding convention, then 0.3 would be written as 0.333330 with 0 after it, non repeating, which is not 0.(3). In either case. 2/3 is represented in decimal as 0.(6). And, 3 * 0.(6) is 1.(9) and we've already shown 0.(9) is 1 so 3 * 0.(6) is 2. It all comes to representing fractions in decimal. 0.(6) is more accurate than 0.66667 but if 0.(6) is changed to 0.6666667 that latter is meant to mean the former and that 0.(6) is the rep for 2/3.
@@TemperansOne last way to look at this is is to consider rounding. If we round 0.(6) to some number of decimal points, we'd choose a point so the rounding error is insignificant. Say we did it with money, and we only cared about cents. We could round 0.(6) to 4 decimal places, giving 0.6667. Then, when we multiply by 3, we get 2.0001, which is a rounded answer. If we care about 2 decimal places, 2.0001 is exactly 2.00. Again, its not as much about limits or other explanations as it is about representing fractions in decimal and how to use those representations.
@@BrianStDenis-pj1tq 3.0.(6) is not 1.(9) but 1.(9)8. The 8 shows up where ever you decide to stop adding digits. You can see it arimethically by seeing how 3*0.6 = 1.8, 3*0.66 = 1.98, 3*0.666 = 1.998, etc. Similarly 2/3 is not actually equal to 0.(6) but a number that lies somewhere between 0.(6) and 0.(6)7 those numbers are just the only way to represent 2/3 in decimal form. The issue stems from the fact that base 10 cannot properly represent fractions that are not a multiple of 1, 2, 5, and 10.
@@BrianStDenis-pj1tq that we can agree with, the whole thing is just a matter of what is more convenient for the problem and for most problems going beyond 9 decimal places is too precise. Which is why we use 3.14 for Pi and 9.8 for earth's gravitational constant.
It's nothing but a representation of the property "rational numbers are dense". The latter one is defined as: between any two different rational numbers, there are infinitely many rational numbers. If the same principle is used with real numbers, it's nothing but a representation of the property "real numbers are dense".
For Level 4, here is how I would rigorously prove that 1 is the least upper bound of that set. The set is defined by {1 - 0.1^n : n is a natural number} (this could easily be proved, but I don't feel like writing that). For the sake of contradiction, assume that an upper bound for this set t exists such that t < 1. Thus, 1 - t > 0. Note that lim n->inf [0.1^n] = 0 (one could write a rigorous proof for this too, but I don't feel like writing it out). By definition of the limit, since 1 - t > 0, there exists a natural number n such that 0.1^n < 1 - t. Thus, 1 - 0.1^n > t. By definition of the set, 1 - 0.1^n is in the set. However, this is a contradiction, since t is supposed to be greater than or equal to all numbers in the set. Thus, for all upper bounds t, 1
Exactly what you said. There isn't (as far as I know) a "closed" numerical representation of it, because of the Archimedean property (or if you prefer the density of the reals). However, that is not a property only of the integers. What is the next smallest number to π?
@@trumpet_boooi No. '-0' is a concept from computing. It has nothing to do with real numbers. There is no such thing as 'the next smallest number to 1' in the space of real numbers.
There isn't any next smallest number. If you choose any two real numbers you can always find another real number between them. You can keep doing that forever.
Might be a dumb question but I need to ask it, does this work for everything that ends in ".999..." ¿?, like for example take "61÷3", it equals "20.333", and if u multiply it by 3 you get "20.999....", is "20.999..." also equal to "21" ¿? And second, let's pick a random number, "14.8" for example, is "14.79999..." equal to "14.8" ¿?
My problem with the last one is, if 1-epsilon is an upper bound, what about 1-2epsilon? Why is that _not_ an upper bound? If it _is_ an upper bound, then 1-epsilon is not the least upper bound, since 1-2epsilon is less than it. The obvious ad hoc solutions don't work for equally obvious reasons. How do you solve this? How does one define classical limits in the hyperreals? The classical epsilon-delta definition no longer works. Do you know any resources I could read on this?
As Sebastian said, the answer is basically that he lied in that last part. Even in the hyperreals, there is no way to consistently say that 0.99999… = 1 - ε. Indeed, if we define ε to be the class of the sequence (1, 0.1, 0.01, 0.001, 0.0001, …) we have that ε² = (1, 0.01, 0.0001, 0.000001, …). It’s clear that 1-ε and 1-ε² both “should be” 0.99999… and yet they are distinct from each other. There are ways to represent these values with “decimal-like” notation, but 0.999… will never be one of them.
"How does one define classical limits in the hyperreals?" - You're going to hate it, because the answer is mind-bogglingly stupid. The answer is that you don't need classical limits anymore. These "numbers" literally exist now. You don't need to think of them in terms of limits anymore. For a concrete example. You don't need to think of the derivative as the limit as h->0 of [f(x+h) - f(x)] / h. The derivative in the context of nonstandard analysis is literally just: [f(x+ε)+f(x)]/ε. The number ε literally exists now. So you don't need to reframe it inside a limit. With the caveat, that in nonstandard analysis, you would also "sort of round this number to the nearest non-infinitesimal value". So if you were to look at the value [f(x+2ε)+f(x)]/2ε, it's going to essentially just round to the same number as the former. Similarly, if you have a function f(x) that converges to L as x->a, we would just say: Limit of f(x) as x->a = L iff x-a=ε, implies f(x)-f(a)=L.
I think that you should have added a section on the construction of real numbers. We know that fractions have many representatives. For example, 1/2=2/4. This comes from the fact that swapping 1/2 for 2/4 "does not change the result of a calculus" (equivalent classes inside). Now what is a real. We can construct that as Cauchy sequences of rational numbers. This formalizes the level 3 step: the sequences 0, 0.9, 0.99... and 1, 1, 1... "converge" to the same value. And as for factions, real numbers have a ton of representatives. Note that, in this regard, Dedekind cuts are much better. Each real number correspond to precisely 1 such cut.
I could follow and understand all this, makes perfect sense, great explanations. But do I really understand it if I can't repeat the final two explanations to an audience.
Was having a day long discussion about this with my shop. Whenever someone came in that day to use our copier we asked them if .9 repeating equalled 1. Best answer was a guy who thought about for a bit and answered "close enough".
thank you! me and some of my friends like to think we are math nerds, and on of them keeps on insisting 0.9 repeating is not 1! this will be a nice proof.
@4:58 I agree that x must be an upper bound of the set, but why do we assume x must be the LEAST upper bound? where does this more restrictive assumption come from?
it kinda sounds like that's one way to define 0.99... but I don't really like it either; however, if you ignore that x=0.99... and just define x as the sup (which exists because in the real numbers any non-empty set with an upper bound has a supremum) the video still proves that x=1, the rest follows quite easily: since 1 is the sup and 0.99... is an upper bound, we know that 1≤0.99... and if we agree that 1≥0.99... it follows that 1=0.99...
The statement is that x=0.999... is the least upper bound for the set S={0.9, 0.99, 0.999, ...}. Assume for the sake of a contradiction, that x is not the least upper bound for the set. Then there exists a value "y", such that y0. Let z = x-y. Note: y = x-z (use algebra). Since z is a positive number, not equal to 0, by the Archimedean Property, there exists some natural number N, such that 0 x-z = y. But what does x-(1/(10)^N) look like? This is 0.999... - (0.000...9000..), where the 9 is at the Nth position, which would be 0.999...0999.... Therefor x-(1/(10)^N) is less than some value inside the set S. Namely, the (N+1)th element of set S. So x-(1/(10)^N) is greater than y, and the element S_n+1 is greater than y, but y is an upper bound for the set S, so S_n+1 is both greater than y, and less than y. This is a contradiction. Therefor our initial assumption that x is not the least upper bound for the set is incorrect. So negating this, x must be the least upper bound for the set S. Take a few minutes to really read through this carefully. It's easy to get lost. Let me know if you have any questions about the proof.
You asked if there's another way I (and anyone else) think about. You didn't need to say it this way, but the way I heard it was that you can't add another number in between .99... and 1. Using basic knowledge of decimals you and the number, you can always think of another number. Then the person knew explained the number line, which I already knew. Finally they explained the difference is infinities. That the infinity of decimals is bigger than fractions. Finally it is then proof, since no number can sneak its way in between 1 and .99..., that it is in fact 1. I wish I could remember the channel name! My most vivid memory of her , besides the video I mentioned, is drawing in sharpie (permanent marker!) on Derek's head from Veritasium. That for me, is all I remember of her. The video I mentioned and Derek's forehead marked for days.
One can further ponder the question, if .999.. is equal to one, how does it compare to the same value in a different base? .111111 in base 2 approaches 1 at a different speed. If we look at it another way of 1-.11111111111 vs 1-.9999999 on the scale of infinity both of these numbers are infinitely close to 0. Can you say they are equal though?
Possible issues with levels 1 and 2: Level 1: You start with the statement that 1/3=0.333... The problem here is that many people who reject 1=0.999... will also reject 1/3=0.333... Your starting assumption implies that '0.333...' represents the result of the division of 1 by 3 in base 10. Indeed, this would appear to be the premise behind the assertion that 1/3=0.333... Disbelievers might argue that before Simon Stevin's L'Arithmetique published in 1594, the consensus of opinion was that fractions such as 1/3 could not be represented in base 10. In those times it was accepted that the fraction 1/3 could only be converted to a decimal in base 3 or 6 or 9 or any base that has 3 as a factor, because in these cases the conversion process would complete and thus provide a constant as a result. Stevin simply asserted that they, along with all irrationals, COULD be represented by the use of unending decimals. However, some people still rejected this assertion on the basis that for an infinite decimal to be not changing, and thus be a constant, it requires a completed infinity which appears to be a self-contradictory notion. It is easy to accept why a terminating decimal can be called a constant, but if we have unending non-zero trailing digits then it seems impossible to reach a position where the value is no longer changing. A constant requires a static collection of non-zero digits and yet these strange 'infinite' decimals claimed to have unending non-zero digits. How could unending non-zero terms form a constant???!!! At best this appeared to be a fuzzy idea that could never be clearly demonstrated and at worst it appeared to be a blatant contradiction. Level 2 (part 1): Here you provide a meaning for '0.999...' as being a 'geometric series'. You say that the geometric series formula tells us that this infinite sum will be equal to a/(1-r). Disbelievers might object to this claim on the same basis that they objected to your 'level 1' argument, in that infinitely many non-zero terms cannot be said to sum to a constant. They might say that it is true that for some geometric series, the increasing n-th sum appears to tend towards a constant. But the idea that infinitely many non-zero terms can somehow complete and mysteriously become equal to a constant is problematic. Indeed, if we consider two simple geometric series such as 1+1+1+... and 1+2+4+8+... then the formula for the infinite sum produces a divide by zero error for the first, and a negative value (-1) for the second. The disbelievers might argue that if this was not proof enough, consider an alternating series like 1-1+1-1+... Here it is blatantly obvious that this can't somehow end with a mysterious completed infinity of terms. To claim that such a series can somehow sum to a constant is obviously absurd. Disbelievers might consider it farcical that mathematicians still would not conceded that this was further evidence of the blatantly obvious fact that there cannot be a completed infinity of non-zero terms that somehow sums to a constant. Mathematicians devised different categories of geometric series. Then they simply defined the sum to be a constant value for cases where the sequences (of the partial sums of the geometric series) were found to be 'absolutely convergent'. To a disbeliever, these arguments are just extra layers of definitions designed to try to justify the blatantly absurd notion of a completed infinity of non-zero digits. In terms of this particular argument for 1=0.999..., the assumption that these are equal is hidden within the formula 'a/(1-r)'. Therefore the disbelievers will claim that by using this formula you have already assumed what you are attempting to prove. Level 2 (part 2): You say "if you don't want to use the geometric series formula we can solve for this directly" and then you proceed to demonstrate the most popular algebraic proof of 1=0.999... The disbelievers might claim that this is slippery beyond belief. Rather than being a valid proof, they might claim that this is just really bad arithmetic. They will say that you have made an arithmetic mistake and, instead of realising that you've made a mistake, you have presented the argument as being a valid proof. They might argue that if we start with a simple statement of 'x=0.ddd...' (where d is some digit) and then we apply some numeric manipulations that amount to multiplying both sides by 1, then we should always end up with exactly what we started with. Any other result is a clear indication that we have made a mistake. There is an assumption in your argument that 0.999... minus 0.9 is equivalent to 0.999... divided by 10. The important thing to consider here is how to properly and correctly perform arithmetic operations on geometric series. In particular, if we were to remove the constant '9/10' from the geometric series '9/10 + 9/100 + ...' then would it be arithmetically equivalent to dividing the series by 10? Whatever arithmetic manipulations we perform, it is important to maintain integrity with respect to the n-th sum of the series. Therefore by working with n-th sum expressions we can apply normal arithmetic operations to any geometric series, regardless of whether or not it is classified as having a sequence that converges. The n-th sum expression in this case is '1 - 1/(10^n)'. And so if we remove '9/10' from this we get '1/10 - 1/(10^n)' However, if we simply divide by 10 then the n-th sum expression of the resulting geometric series would be '[ 1 - 1/(10^n) ]/10' = '1/10 - 1/(10^(n+1))'. And so this highlights that the n-th sum expression of 0.999... minus 0.9 is different to that for 0.999... divided by 10. Thus they are not the same since they produce different values up to any specified n-th term. For arithmetic correctness the n-th sum value needs to be maintained. Therefore this is the arithmetic mistake in your logic.
Could there be a level 0. A level of first impression. (Based on how we have been taught the basics. And the axioms and definitions we retain) We represent numbers in base 10 either as fractional number or decimal using digits and the decimal point (radix point in all number bases) In decimal, any digits, other than 0, right of the radix point is the fractional part (similar in other base representations. ) Basic number groups. Natural, Whole, Integers, Rational and Irrational based on a real number line. (Other sets are then added. Depending on your basic concepts.) The definition of an integer is a number with no fractional or irrational parts. (0's to the right of the decimal point) A Rational is a number that can be represented by the ratio of two integers. At this level you can't prove that .999...=1 [Does modern math rebuild the number system without the counting numbers and the number line? Maybe that is why ".99..." is 1. I believe we can make ".99..." as close to 1 as we desire, except equal to 1 or even next to, by just considering any amount of 9's, even an infinity of 9's. (Which is what Stevins proposed. ) Not very exact!]
@KarmaPeny, you've been peddling your nonsense for at least 10 years. It's about time that your learnt some math. Most people learn it as fast as they are taught it.
Math saying that ".99..." is 1, is like a builder constructing over the n'th floor decides he no longer needs the building beneath for support and goes freestyle.
Geometric series is more of a calculus argument tho. The algebraic argument was the equation you later explained. Plus, the second and third levels are basically the same argument, since a geometric series is exactly defined the limit of that sequence, that is the sequence of partial sums.
Long ago, there were countless pointless debates about this on message boards. I tried explaining it every which way, and there are certain people who will never grasp this because they don't really understand what infinity means. To them, infinity means a really large finite number. It's just a quirk of our number system that some numbers can have more than one notation.
Thank god someones got my back on this. I saw this on a confidently incorrect post and I’ve not been okay with it at all. I will die on my infinitely small hill.
A curiosity from another comment on RU-vid. Divide any integer, except all 9's, by a integer of all 9's with the same number of digits. You get the same original integer repeated infinitly after the decimal point. Is there any connection to ".99..." not fitting the definition of a rational number?
@KarenTheBonehead, let 0.(N) represent the n digit string N being repeated. Then 10^n * 0.(N) = N.(N) = N + 0.(N) => (10^n - 1) * 0.(N) = N => 0.(N) = N/(10^n - 1) Note that 10^n - 1 = 999...9 (n 9s) and the latter has the same number of digits that N has. Also note that N and 10^n - 1 are natural numbers, so 0.(N) represents a rational number. It trivially follows that every eventually repeating decimal represents a rational number. Examples 0.(2) = 2/9 0.(9) = 9/9 = 1 0.(999999) = 999999/999999 = 1 0.(285714) = 285714/999999 = 2/7 There are no exceptions to that rule. Why on Earth do you think there are any exceptions? That formula also follows from the more general geometric series formula a + a*r + a*r^2 + a*r^3 + ... = a/(1 - r), if |r| < 1 That formula doesn't magically go wrong if a = (1 - r) Note that 1 = 2/9 + 7/9 = 0.(2) + 0.(7) = 0.(9), so we have consistency despite your insistence, elsewhere, that we don't.
You will note that KC starts with a infinite number and ends with a finite number. That is mathic. Interesting that starting with finite and ending in infinite is more natural in math and a sign that an error occurred in physics. Starting with a infinite number and forcing it into a finite number is unsafe, incomplete, inconsistent and imprecise.
@@johnlabonte-ch5ul KarenTheBonehead says: "You will note that KC starts with a infinite number and ends with a finite number. That is mathic. Interesting that starting with finite and ending in infinite is more natural in math and a sign that an error occurred in physics. Starting with a infinite number and forcing it into a finite number is unsafe, incomplete, inconsistent and imprecise." LOL. I couldn't make this stuff up. You must be making your speshul needs teacher work hard for her money. 0.(N) is a finite number you muppet. It is N/(10^n - 1). The represented number is not proportional to the number of digits in its decimal representation. If it was, then, according to you, 0.1111111111 would be larger than 0.9 because the former has more digits.
Yes I do have another way to interpret this. And it goes back to the explanation in level 1. 1/3 = 0.3333… 2/3 = 0.6666… 1/3 + 2/3 = 0.3333… + 0.6666… = 0.9999… = 3/3 = 1 However, here’s the big issue with this. It’s infinite(whatever that means). Using null hypothesis, “What if 2/3 is not 0.6666… forever? What if at infinity, 2/3 = 0.6666…7?” In that case, you have a value that is 100% equals to 1 and there’s no more argument to it anymore as it makes the whole repeating decimal equals to 1. But then again, how do we even prove that? How do you prove that at infinity(assuming we somehow know what happened at infinity and infinity has an end finite value but the human brain cannot see all the mid point but we somehow see the beginning and end point), there is a value bigger than the previous decimal value by 1 or smaller than the previous decimal value by 1? After asking this question, I also start to question whether we are doing math properly or not. Yes it does sounds like we are just trying to convince people rather than doing math as in mathematics “1/3 + 2/3 = 1” is easily accepted with the use of the epsilon-delta proof of a limit. Since I don’t have a math keyboard, I’ll write my proof in latex form: \begin{align*} \frac{1}{3} + \frac{2}{3} &= \sum_{n=1} ^{\infty} \left((3)10^{-n} ight) + \sum_{n=1} ^{\infty} \left((6)10^{-n} ight)\\ &= \sum_{n=1} ^{\infty} \left((3)10^{-n}+(6)10^{-n} ight)\\ &= \sum_{n=1} ^{\infty} \left(\left(10^{-n} ight)3+6 ight}\\ &= \sum_{n=1} ^{\infty} \left(\left(10^{-n} ight)9 ight)\\ \text{Where, } &\frac{1}{3} +\frac{2}{3} = \frac{3}{3} = 1\\ \text{Therefore, } &1 = \sum_{n=1} ^{\infty} \left(\left(10^{-n} ight)9 ight) \end{align*} Writing this in an infinite sum gives me the freedom that I do not need to compute the whole thing and show that at the end of 2/3 might have a 7 at the end. Which leaves the proof to the operation rather than the computation.
'It’s infinite(whatever that means). Using null hypothesis, “What if 2/3 is not 0.6666… forever? What if at infinity, 2/3 = 0.6666…7?”' That 'null hypothesis' is really easy to check. 0.666... = 6/10+0.666...*1/10 => 0.666... = (6/10)/(1-1/10) = (6/10)/(9/10) = 6/9 = 2/3. Why didn't you just check that? 'How do you prove that at infinity(assuming we somehow know what happened at infinity and infinity has an end finite value but the human brain cannot see all the mid point but we somehow see the beginning and end point), there is a value bigger than the previous decimal value by 1 or smaller than the previous decimal value by 1?' Just pick up an introductory calculus/real analysis textbook and read a few chapters.
Infinitesimals are basically a way of saying that the DEFINITION "representation value" is not valid. Which can ALSO be CONCLUDED from the definition of infinitesimals, that they are NOT-REAL numbers, ie NOT real numbers.
If you extend 2's/10's/n's complement in the style of the p-adics (but not necessarily using the p-adic metric), and then tack a real fractional part on the end, you get a number system in which ...999999.99999... = 0 (for base 10)
Possible issues with levels 3 and 4: Level 3: Disbelievers might complain that you have apparently abandoned the definition of '0.999...' as being a geometric series and instead you now define it as the limit of its corresponding sequence of increasing partial sums. This dual definition might be ok if we had already proven '1=0.999...' by the first definition, but this is not the case. Indeed, it seems clear that if '0.999...' is a geometric series then it cannot possibly equal 1. They might also point out that this is not a 'proof' in any way shape or form. You have simply defined '0.999...' to be the limit of the corresponding sequence. Also they might say if we start with the assumption that 0.999... is a constant that may or may not equal 1 then why can't the corresponding sequence be approaching 0.999... rather than approaching 1? They might also point out some of the bizarre consequences that necessarily result from such a definition. Since all decimal values can be considered as having infinite decimal places, even if they are just trailing zeroes, then all of them can supposedly be defined as being limits. And so when we write down '0.123' this can no longer be interpreted as simple base 10 place values. It conveys the value of the limit of its corresponding infinite sequence. But if we try to write down the value of that limit as a base 10 decimal value we can't do it. This self-reference means that base 10 place values no longer exist since all representations are now limits. If we were to allow them to be both limits and base 10 place values then not only do we have a lot of ambiguity and self-reference going on, but we would presumably have to allow 0.999... to be defined as being a geometric series. And as we already seen, as such it is not equivalent to 1. Also there is the apparent inconsistency in that the limit of 0.333... is itself whereas the limit of 0.999... is apparently not itself but 1. Perhaps 0.999... can have two limits, both 0.999... and 1, but then does this mean that 1 has a corresponding sequence (of 1,1,1,...) for which the limit is 0.999...? By what logic is this demonstrated? Level 4: Similar to the previous argument, a disbeliever might ask why do you insist that 1 is the least upper bound. Why can't 0.999... be the least upper bound? They might infer that your real argument here is that there is no gap between 0.999... and 1 and therefore they must be the same number. This is confirmed by your 'Archimedean property' argument. But again they might complain about the starting assumption that 0.999... is a constant. They might point to other geometric series such as 1-1+1-1+... and 1+1+1+... and claim that these objects clearly do not 'sum to a constant'. Therefore why would you assume that other series of unending non-zero terms can somehow sum to constants? They might insist that this is an obvious flaw in your reasoning.
Question: In base 10, there may be no number between .999... and 1 but what about in higher base number systems. Is the similar .(b-1)(b-1)(b-1)... closer to 1? Satisfying the Archimedean property.
@KarmaPeny. You: "Level 3: Disbelievers might complain that you have apparently abandoned the definition of '0.999...' as being a geometric series and instead you now define it as the limit of its corresponding sequence of increasing partial sums." The sum of a geometric series is the limit of the sequence of its partial sums. The same applies to all [convergent] series (with real terms). You've been peddling your nonsense for at least 10 years. It's about time you learnt some math. The "geometric series" Wiki is a particularly good place to start. You: "Level 4: Similar to the previous argument, a disbeliever might ask why do you insist that 1 is the least upper bound. Why can't 0.999... be the least upper bound?" Both the sum of 0.999... and 1 is the LUB of the sequence 0.9, 0.99, 0.999, .... I bet you've been told that hundreds of times.
This cracks me up because the fifth level (assumingly the most advanced of the maths) says what the little kid things “0.9 repeating is just barely smaller than 1.”
Answer is easy. It’s just an artifact of the decimal system. 1/3 is no different to 1/4. 4 times 1/4 is without any doubt 4 times 0.25 which is 1. But 3 times 0.3333… is also 1.
1 divided by 1, long division. Make the dividend 1.0. Put 0.9 above to construct the quotient. You get the remainder 0.1. Drop a 0 down to get 0.10. Quotient becomes 0.99. Remainder 0.01. Continue ad infinitum. 1 equals 0.9999…
Im not joking when I say that in like 11th grade I protested to my math teacher with the level 5 explanation when I found out about the concept of an infinitesimal. And now I find out that I accidentally instinctively tapped into higher math than anything I am usually capable of. Nice! 🤓
I've always just figured like this: What is 1 - 0.999...? The first intuitive answer is that it's something of the form 0.000...001. But how many zeros should you put in after the decimal place but before the 1 in the difference? An infinite number. After each zero, there will always be another zero. There is no zero that is not followed by another zero in the difference, because there is no nine that is not followed by another nine in the subtrahend. In other words, the "rule" for writing out the difference for a subtraction of the form 1 - 0.999...999 seems to be: For every nine that is followed by another nine in the subtrahend, put in a zero in the difference; for every nine that is NOT followed by another nine in the subtrahend, put in a one in the difference. So it follows that if there are an infinite number of nines that are followed by another nine in the subtrahend, there will be an infinite number of zeros in the difference; and if there are no nines that are NOT followed by another nine in the subtrahend, there will be no ones in the difference.
This is my go-to, though I state it simpler. The cool thing is when you present the argument you can say, "So the difference between 1 and .9 repeating is literally zero." which means they are the same. Very pleasant.
It kinda blows my mind how this initially 'counter-intuitive concept' of 0.9999... = 1 is actually so logical that it can be easily proven in 100 ways using simple numbers.
Well this was my idea of it: Any repeating decimal can be expressed as a fraction. For example 0.3333… is 3/9; 0.252525… is 25/99; 0.812812812812812… = 812/999 From this you can see that to express a repeating decimal as a fraction, the numerator should be the digits that were repeating while the denominator should be 9s equal to the length of the numerator. For example if the numerator has 3 numbers the denominator should have 3 9s. So according to this, 0.999999… can be expressed as 9/9(as there is only 1 digit repeating, thus the denominator should be only a single 9) and 9/9 is equal to 1
I like the intuitive explanation that there can't be a number inbetween 0.999.... and 1 because any number you say will be smaller than the infinatly repeating 9s (and because of the properties of real numbers you should be able to construct a number that is inbetween two non-equal numbers) So since that number can't exist they must be equal
As far as fractions, I always liked to think of it as following the pattern of x/9, which for single digits all equal 0.x repeating! And 9/9 is of course equal to 1.
By definition, fraction of integer divided by integer, are rational real numbers. A discussion is why the contradiction when; x= integer digit, and x//9 [See NOTE below] x//9=".xx..."? (?"at infinity" another related discussion) 9//9= 1 not ".99..." I can show 1//1=".99..." by anti-math (changing basic long division). Math says ".99..." is 1 by definition. (The sum of the infinite series, represented by the infinite sequence of ".99...", is the real number limit of the partial sums, 1 and they can't find a real number that equals 1-".99...". I suggest 1-".99..."=epsilon. ) [NOTE // is the result of × divided by 9, not the fraction x/9, in this reply. Equivalent but different like ".99..." and 1]
@@johnlabonte-ch5ul LOL. You pinched // from Terence Tao. There is no difference between the numbers represented by a//b and a/b, i.e. a//b = a/b. You contradicted yourself. You said that you can show that x//9 = 0.xxx... but then said 9//9 != 0.999.... You failed to explain how you came to that ridiculous conclusion. Worse, take, e.g., 1 = 2//9 + 7//9 = 0.222... + 0.777... = 0.999... . Are you also going to say that x//9 + (1-x)//9 i not allowed for some reason that you have disclosed? Is it OK to say 2//9 + 5//9 = 0.222... + 0.555... = 0.777... = 7//9. Why is it only fractions numerals that add to 0.999... that are not allowed? Don't bother answeing, the questions are rhetorical. I know that you haven't got a clue about anything. 1 - 0.999... = 0 and I have proven that to you many times. So your claim that "they can't find a real number that equals 1-".99..."." is BS. You also say " I suggest 1-".99..."=epsilon". That's fine, because epsilon = 0. The thing is, you have completely failed to understand the role of epsilon in the definition of limit. You continue to use the phrase "at infinity". You have never explained what it is supposed to mean. (Hint: it is gibberish).
one of many mathematical concepts that seem paradoxical and happen to require some aspect of infinity... similar to such "infinity-related paradoxes," there likewise exist many "zero-related paradoxes"...
We could define a sequence: a(n) = 0.9+0.9/(10^n) . But my knowledge on series is a bit rusty. Someone could find an upper bund sequence that converges to 1 . By theory that would make a(n) converge to 1 too.
Decimals are just integers divided by something. They're alternative representations of fractions. In real world, when a number of objects, compounding some set, is divided (or fractionalized) by some integer value larger than 1, one gets a consequence set of objects that has a larger number of elements than the previous one. That said, if the number of elements in the primary set is infinitely near the number of elements in the consequence set, they rigorously have the same number of elements. So, the divider involved in that assumption must be 1. Every number divided by 1, and only 1, must result oneself.
What I want to know is if there is a number between 0.9999....(9 repeating) and 1. Edit: What if you had a digit that ends with an 8 but there are an infinite digits of 9, like that 0.9999...(infinite 9s)...8 The digit is infinitely repeating with 9s but there is an 8 at the end. If you have difficulty imagining, take 0.8 , add .18, you get .98 then add .018 for .998, then .9998, etc... till .999..(infinite 9s)...8 does that also equal 1 or not?
'What I want to know is if there is a number between 0.9999....(9 repeating) and 1' 0.999... = 1, so there aren't any numbers 'between' 'them'. 'What if you had a digit that ends with an 8 but there are an infinite digits of 9' That would just be nonsense. Among other things, decimals represent series, like 0.999... = 9/10+9/100+9/1000+... What position would the '8' be in? What would its corresponding term be in that series? 'like that 0.9999...(infinite 9s)...8' That would be a terminating decimal with finitely many '9's. 'If you have difficulty imagining, take 0.8 , add .18, you get .98 then add .018 for .998, then .9998, etc... till .999..(infinite 9s)...8 does that also equal 1 or not?' It does not, because that's not a valid decimal representation of any number. The sequence (0.98, 0.998, 0.9998,...) does, however, converge to 1.
There is a real number between 0.999… and 1 if you can increase any of the digits in 0.999… or add a digit to the end. But all the figures are 9 and there are infinitely many of them. So you can’t do either those things. There is no real number between 0.999… and 1. They are the same number.
@Dr Sean, I've only just noticed your level 5. You have redefined 0.999.... Decimals are not suited to representing hyperreal numbers. For instance, using your redefinition, there would be no decimal for 1 - 0.999... or 10 * 0.999... - 9.
@@johnlabonte-ch5ul That depends on what you mean by an infinite number. If you mean transfinite hyperreal numbers cannot be represented by decimals (or other bases), then that'll be the third time that you have got something right in a week. That must be a record for you. How is your proof that 0.222... + 0.777... cannot equal 0.999... coming along? I'm still WAAAAAAAIIIIIIITINGINGING... I'm sure everyone will be fascinated to see how a snot nosed school kid has beaten the mathematicians at their own game.
@@thesugareater8607That would mean that means the limit of an expression would always be equal to the expression (at least when the limit approaches infinity), which is false.
@@f.b.i6889 Not necessarily, if you graph the sum .9 + .09 + .009... the graph will approach 1 and will be equal to 1 at an infinite distance. However, if you consider a function such as 1/x, it will approach x=0, but it will never be the case that x=0.
@@thesugareater8607 Sorry, I meant to say "when the limit point of the limit approaches infinity," not the limit itself. For example, the series you are talking about can be represented as the limit as x approaches infinity of 1-0.1^x. However, this would imply that an expression like (1-(1-0.1^x))/(1-(1-0.1^x)) when x approaches infinity does not only equal 1 as a limit but its value is also 1 despite the fact it would simultaneously equal (1-1)/(1-1) making it undefined, which is why if the limit as x approaches n of f(x) equals k, it does not imply that f(n)=k. What I'm saying is that your claim would require more proof than a limit.
Maybe you could also have proven it by a taylor - series. If you take a series that expresses 9/10 + 9/100+ ... And aproximates f=1, Then you could conclude that for each eduration the remaning term is smaller for each n of the function, and so for n-> inf the rest term goes to 0, the series is equal to the function it aproximates, which is f= 1
