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!

I like that you made level 5 'it depends'. This is the natural progression as you learn Maths at higher levels.

A physicist would just say the difference between .999 and 1 can be explained by experimental error.

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.

my dumbass would say that it's because you can round off or estimate it to the nearest ones 💀

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.

Loving this channel so far, friendly for all levels

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.

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?

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.

Man youre explaining voice is so soothing to listen too. Would love too attend a lecture just for that.

I'm at this point convinced that high level math is just fanfiction.

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!

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 ❤

You are great bro, thanks. Thats the channel i've been looking for!

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 !!

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...

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

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.

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.

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

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...

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!

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|

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.

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 😂

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!

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.

Another very easy way of doing this is by saying: x = 0.999… 10x = 9.999… 9x = 10x - x = 9.999… - 0.999… 9x = 9 x = 1

this was the exact video I was looking for!

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....

1/9 =0.111..., 2/9 = 0.222..., ... 8/9= 0.888..., 9/9 = 0.999...., 9/9 = 1, so 0.999... = 1

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!

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.

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.

I love this video bro!

I like to use the 1/3 * 3 argument but demonstrate it using a different base, maybe 12.

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.

dude already at level one you blew my mind. I think I need to study up

Fire video doc!!

I cant say i fully understand, but i will say this is fascinating and well explained

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.

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.

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.

The way I've always thought about it was numbers are equal if you can't find a number between them. There is no number between .9999 repeating and 1. No idea if this actually makes sense or is correct but it works for my smooth brain. Edit: Looks like your #5 discounts what I thought lol. Darn scary maths.

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

Would you mind making videos on less understood formulae? Things like the Archimedes principle and why it works.

Thank you . very good explanation

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.

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" ¿?

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".

What is the fraction form looks like in hyperreal though?

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.

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.

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.

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.

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.

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)

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.

"U dont have1 cookie, I have 0.9999999999999.... cookies" 😂

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.

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.

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?

I'm surprised you didn't do some sort of delta-epsilon proof. I assumed that would be the rigorous proof for the reals.

what's the next smallest number to 1?

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.

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…

In level 2(simplified aproach) you are assuming the same as in number 1

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?

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! 🤓

@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?

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.”

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

One of my favorites is: Assume 1>.(9) or 1b, then (a+b)/2>b and if a

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.

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.

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.

My 1st rational was you can spit an infinite into 2 , so .3 repeating can be .3 repeating point 3 repeating and turn it into .3 repeating point 1/3 when the 2nd one times 3 = 1 and caries over to the first and carries over all the 9's, but then I thought of .3 repeating point 2 repeating and things get confusing.

My way of thinking of this is that .9 repeating is just the left decimal representation of 1

What's the difference between hyperreals and surreals?

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"...

So in Hyperreals, 1/ε is possible?

.999... would still have to be equal to 1 in the R* (Hyperreals) due to the transfer principle.

surreal numbers also have infinitesimals (they are different though, right?)

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

Hmm. For level 4, it feels like the part doing the heavy lifting is the Archimedean property, but you didn’t go through that here.

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

I have a point regarding the concepts VALUE and REPRESENTATION. In maths a number's REPRESENTATIONs are DEFINED to be equal to its VALUE -- AND VICE-VERSA. BY DEFINITION: 1) "1/3" represents the value of "1" divided by the value of "3" in the form of a fraction. 2) "0.333..." represents the value of "1" divided by the value of "3" in decimal form. So both "1/3" and "0.333..." represent the exact same VALUE. So when said VALUE is multiplied by the number's VALUE that the number "1" was originally divided by (ie the VALUE of "3"), the result is the VALUE of the number represented by "1". The POINT is this: Is the-above a PROOF of the concept of "INVERSE"? Or is it a PROOF of the equality of "1/3" and "0.333...", since "INVERSE" has a DEFINITION? Basically this is about MAKING USE of "value representation", but then suddenly DISallowing it.

It confuses me how people don't get this honestly.

The problem with the hyperreal approach is that only very few of the non-real hyperreals can be represented in such a way, so even then this notation is not very useful for this purpose.

What about this one: 7/9=0.7777777 8/9=0.88888888 9/9=0.99999999=1 ?

Would that stand to reason that the number one, as an actual number, isn't an actual number and instead, just a concept, like 0?

Can you do a video on 5 reasons why n choose 0 is equal to 1? Because I still believe it shouldn't exist - that is, that is - I believe that there are an infinite amount of ways to choose 0 or no items. Because there is nothing to choose, is why - it is all the various ways not to choose anything.

In hyperreal numbers, each standard number is surrounded by a monad of nonstandard numbers(including the standard number that is surrounded by this monad) that does not intersect with the monads of other standard numbers. Thus, if we assume that 0.(9) is not equal to 1, then the equality 0.(9)=1-ε cannot be true, because this would imply the intersection of monads, and if we assume that 0.(9) and 1 are equal, then the equality 0.(9)=1-ε is also incorrect, because this would mean the equality 1=1-ε. In addition, there are an infinite number of infinitesimally small numbers and it is unclear why you chose ε and not any other infinitesimal number greater or lesser than ε. The limit of the sequence {0.1, 0.01, 0.001,...} is equal to zero in both standard analysis and nonstandard analysis. The proof of the equality 0.(9)=1 is possible on the set of hyperreal numbers and does exactly the same as in standard analysis: through the theory of series. The difference is only in the definition of the limit, but the results of a nonstandard analysis are equivalent to the results of a standard analysis

Most (if not all) of the proofs using elementary methods are based on "assuming" what 0.999999... is. People trying to show an elementary proof usually don't realize this. To give a rigorous proof, the symbol 0.999... must be defined first. This requires the construction of real numbers which is at least level 4.

@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.

So too anyone who watched the Infinity ball theorem. I would imagine the fact that the smallest number between 9/x and 1 is 9/infinity, and there are an infinite number of sets that correspond too 9/infinity, so would that also make an intuitive point that 0.9999 = 1

A curiosity from another comment on KZfaq. 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?

I thoughts are like this: you can't come up with a number that is in between 1 and 0.999... therefore there's no gap. No gap means they're equal.