Every PROOF you've seen that .999... = 1 is WRONG

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  • Опубликовано: 20 дек 2024

Комментарии • 6 тыс.

  • @n0ame1u1
    @n0ame1u1 3 года назад +5464

    The proofs are all correct, the justifications are simply left as an exercise to the reader

    • @ujjwalbaranwal6350
      @ujjwalbaranwal6350 3 года назад +75

      Facts

    • @nexovec
      @nexovec 3 года назад +37

      The proofs are;

    • @rephirairis
      @rephirairis 3 года назад +384

      Basically this. To claim that they are wrong is kind of misleading. It's definitely abridged far too much information, but those proofs are usually thrown at kids who are just starting learn the concept of fraction. Highly doubt that series sum is taught to 3rd years.

    • @footlover9416
      @footlover9416 3 года назад

      no

    • @ujjwalbaranwal6350
      @ujjwalbaranwal6350 3 года назад +4

      yes

  • @johnmeo1532
    @johnmeo1532 3 года назад +3438

    My very first day of college the math department head advisor was showing us this proof. Me, having seen the algebraic proofs and trying to make a good impression, went up and showed him a more “elegant” version after the lecture. When he pointed out the fatal flaws, I learned that I didn’t actually know what constituted a rigorous proof.

    • @p_square
      @p_square 3 года назад +73

      can you tell what were the fatal flaws he pointed out?

    • @freedom_aint_free
      @freedom_aint_free 2 года назад +150

      You were luck, they did assume without a proof that "1" is positive, and to prove that is really hard in the analysis field! The for elementary the matter the harder the proof.

    • @NeoiconMintNet
      @NeoiconMintNet 2 года назад +56

      for decimal numbers only one number can be multiplied with a constant to get one unique answer, example :
      10 times x is 10 only when x is 1.
      anything other than a difference of 0 is a difference.
      x=y when x-y=0
      x=1.0...
      y=0.9...
      a difference in value.

    • @NeoiconMintNet
      @NeoiconMintNet 2 года назад +32

      if you don't understand why 0.9... is not equal to 1, after the following info, I know you didn't understand calculus.
      limit, mathematical concept based on the idea of closeness, used primarily to assign values to certain functions at points where no values are defined, in such a way as to be consistent with nearby values. For example, the function (x^2 − 1)/(x − 1) is not defined when x is 1, because division by zero is not a valid mathematical operation. For any other value of x, the numerator can be factored and divided by the (x − 1), giving x + 1. Thus, this quotient is equal to x + 1 for all values of x except 1, which has no value. However, 2 can be assigned to the function (x^2 − 1)/(x − 1) not as its value when x equals 1 but as its limit when x approaches…
      source of quote : encyclopedia Britannia under the subject science-math.
      limit calculation formula :
      first term of series divided by the result of 1 minus the ratio, or the value used to get next term.
      limit of 0.9… = (9/10)/(1-1/10)=1
      limit is not the value.

    • @Chris_5318
      @Chris_5318 2 года назад

      @@NeoiconMintNet LOL. Poor you. Try reading that textbook that you show in your videos. It clearly contradicts your beliefs, as everyone can see.

  • @RandallStephens397
    @RandallStephens397 2 года назад +2459

    "A proof of something is not simply a list of true statements that ends in the one you're looking for."
    Too many people need to hear and understand this.

    • @ianwoodward8324
      @ianwoodward8324 2 года назад +27

      Thanks, that is easy to understand. So how do you get from your description of what a proof is not, to a true proof? Is it the way the true statements relate to each other?
      I'm not trolling, I have no idea what a proof is, but I want to.

    • @ianwoodward8324
      @ianwoodward8324 2 года назад +35

      Ooops, sorry. I watched it again and - "you need to show why each one implies the next".

    • @programaths
      @programaths 2 года назад +14

      @@ianwoodward8324 Or a combination of them. Often you get A implies B, then B and C implies D. And D is also proven somewhere else ^^
      It's like when you use the intersecting chord theorem, wich relies on the relation of angle at the center and inscribed angle AND Thales.
      Then angle at the center relies on the sum of inner angles in triangles.
      Thales relies on similar triangles.
      ^^
      With the intersecting chord theorem, you can sweep a lot of other proofs 😁

    • @mrosskne
      @mrosskne 2 года назад +11

      @@ianwoodward8324 The most reliable method is proof by contradiction. Assume that A is true, and show that this leads to a contradiction, therefore A is false.

    • @insidetrip101
      @insidetrip101 2 года назад +11

      @@ianwoodward8324 That's a mountain of a complicated question, and anyone here trying to answer it is over simplifying the problem to an absurd degree. I'm going to simplify it to an absurd degree as well, but at least I'm not going to pretend like my answer is complete. First, a proof is more or less just a sound argument. Second, a sound argument is just a valid argument where the premises are true. Third, a valid argument is an argument where the form of the argument can eventually be derived from your "base rules" (perhaps the rules of inference, or otherwise) without any gaps.
      How exactly this last point is different from a "list of true statements" is honestly not answered in my explanation--and its not answered in the video either. The important thing to remember is every proof (or argument if you rather), is a rather large example of implication/modus ponens. What we are doing is stacking a bunch of implications together and saying if you can get me this, then I can get you that. An example of a list of true statements, I suppose, would be a bunch of non-sequitors, which is an informal falacy (and all fallacies are more or less the formal fallacy of affirming the consequent, because they assert the conclusion or consequent when the premises or antecedent could be anything at that point).
      I think the truth of the matter is that if you start thinking about this too hard you end up more or less like Russel and Whitehead. You end up trying to form everything on the foundations of practical things on the basis of some theoretical formalism, when somethings just must be assumed. If you take this to an extreme, then really there is no "proof" for anything, but if you're too lackadaisical, then there is a "proof" for everything. I'm not telling you to seek a balance between the two, because that's a lame brain-dead take. I'm telling you that you've actually asked a question that some of the most brilliant minds in history have grappled with and been overcome by. This is a monster that it is unlikely you will be able to tame.

  • @marcrindermann9482
    @marcrindermann9482 2 года назад +755

    7:30 to be totally honest, those algebraic proofs never gave me an 'irky feeling' in my stomach. What did, however, was the constant 'dot 99 repeating' because a simple 'dot 9 repeating' would have been sufficient.

    • @Google_Censored_Commenter
      @Google_Censored_Commenter 2 года назад +36

      Ultimately, symbols are intended to promote a certain meaning. If you want to convey that something is repeating, of course you're gonna succeed in that goal easier, if the symbol has a repeating element in it, in this case, 9.

    • @irrelevant_noob
      @irrelevant_noob 2 года назад +11

      @@Google_Censored_Commenter you're missing the point that it's easier to have a repeating 9 instead of the (equivalent) repeating 99. ;-)

    • @Google_Censored_Commenter
      @Google_Censored_Commenter 2 года назад +27

      @@irrelevant_noob I'm not missing anything. Easier in what sense? To write? Okay, maybe. But the point of writing anything, is to communicate. And that's what MY point is, that you're missing. Visually speaking, a single symbol does not promote the idea of repetition as effectively as two, repeating symbols.

    • @irrelevant_noob
      @irrelevant_noob 2 года назад +11

      @@Google_Censored_Commenter so are you saying that 0.(18) is less "effective at promoting the idea of repetition" than 0.(1818)? 🤔

    • @Google_Censored_Commenter
      @Google_Censored_Commenter 2 года назад +25

      @@irrelevant_noob Yes. I don't imagine you disagree?

  • @andyl.5998
    @andyl.5998 3 года назад +1588

    2:10 For anyone wondering, ...999 = -1 is "true" in the "10-adic" number system, specifically as a ten's complement. The analogy is the two's complement (...1111 = -1), the way computers store negative integers.
    EDIT: I have edited to put "true" and "10-adic" in hand-waving quotations, because people correctly pointed out that p-adic numbers are only well-defined when p is prime.

    • @mCoding
      @mCoding  3 года назад +279

      An excellent example!

    • @qwfp
      @qwfp 3 года назад +210

      So it's just irl overflow?

    • @jenaf372
      @jenaf372 3 года назад +14

      It works similar for the sum of all natural exponents of 2.
      Can propably be generalized from there?

    • @MrRyanroberson1
      @MrRyanroberson1 3 года назад +38

      @@jenaf372 in any base b: let d = b-1. then infinite 0.ddd.... = 1 if math makes sense. Likewise, ...ddd.0 + 1 = ...000.0 (which is the definition of an infinite complement)

    • @GregWeidman
      @GregWeidman 3 года назад +84

      Yes! And, if you are not sure, try adding 1 to ...99999999.0. You will find that you now have an infinite string of zeros going off to the left, which is zero. What number is such that when you add 1 to it you get zero? -1, of course.

  • @marcopollom
    @marcopollom 3 года назад +565

    My multivariable professor called me a "math magician" when I showed him the initial 'proof.'

    • @mCoding
      @mCoding  3 года назад +74

      Like Art Benjamin!

    • @philippg6023
      @philippg6023 3 года назад +19

      Well your Professor knows why you can multiply 0.9(9) by 10 like that ^^

    • @mr2octavio
      @mr2octavio 3 года назад +45

      A mathemagician then

    • @raulf.duarte1856
      @raulf.duarte1856 3 года назад +3

      Is this a compliment or an offense? lol

    • @ahmedaghadi8281
      @ahmedaghadi8281 3 года назад +5

      @Ilya Perederiy what???? Negative number of students?. Hah! Welcome to our universe! Btw here physical laws are somewhat different bro.

  • @robin888official
    @robin888official 3 года назад +835

    "Every proof you've seen is wrong", debunks one proof. (The 1/3 never felt like a proof to me, TBH.)
    My teacher once told me "If 0.p9 does not equal 1, there has to be a third number between them." That convinced me the most.
    I think that's pretty close to the limit interpretation.

    • @mCoding
      @mCoding  3 года назад +292

      Excellent intuition! It's true in some spaces even that do not have this intermediate point property!

    • @EvanOfTheDarkness
      @EvanOfTheDarkness 3 года назад +79

      @ཏྦཱལ་ག་པོ། Not bad just incomplete. They do not contain all the "boilerplate" stuff, just the "essence" of the proof.

    • @EmilMelgaard
      @EmilMelgaard 3 года назад +29

      I have heard that reasoning before but I don't understand it. Is it a property of the real number space? At least it's not true for integers. There are no numbers between 1 and 2 but 1 does not equal 2.

    • @EvanOfTheDarkness
      @EvanOfTheDarkness 3 года назад +93

      @@EmilMelgaard It's a property of Real and Rational numbers. They have no "neighboring" numbers, if A != B there are infinitely many numbers between them. For example (A + B)/2, (A + 2B)/3, etc. And since every rational number can be represented as a possibly infinitely repeating decimal, you'd need to have infinitely many decimal numbers between 0.999... and 1.
      One funny thing with this proof is, that you need to prove that 0.999... is rational to use this proof - and the easiest way is just to prove that it equals 1, so its not really a "useful" proof.

    • @EmilMelgaard
      @EmilMelgaard 3 года назад +4

      @@EvanOfTheDarkness Thanks, that makes sense.

  • @KingBobXVI
    @KingBobXVI 2 года назад +383

    Not a formal proof, but when confronted with this in high school, I figured out this fun fact: you can actually construct any repeating decimal as a fraction by dividing the repeating portion by that quantity of 9's. The obvious being 1/3 = 3/9 = 0.33... but also things like 123/999 = 0.123123...
    And with that established, it then follows that you can construct 0.99... with the fraction 9/9, which of course is equal to 1.

    • @RangerKun
      @RangerKun 2 года назад +17

      More generally, shift the non-repeating portion+1 cycle of the repeating portion to the left of the decimal, then shift only the non-repeating section, then subtract and divide. For example, x=.1121212.., then 1000x-10x = 112.1212..-1.1212.. => 990x = 111 => x=111/990=37/330.

    • @rontyson6118
      @rontyson6118 2 года назад +7

      9/9 is 1. Not .99... . What school taught you math like that?

    • @MrJonnyk000
      @MrJonnyk000 2 года назад +218

      @@rontyson6118 You may have missed the central point of the video.

    • @hawks3109
      @hawks3109 2 года назад +2

      @@RangerKun I wonder if we can use this to construct a better way to store floating point in hardware. Right now we essentially just store as many digits as we can plus the exponent. I bet we could store at higher precision, but we would have to give up speed to computer the other forms. Also you could get into trouble trying to calculate an irrational number.

    • @stephenspackman5573
      @stephenspackman5573 2 года назад +8

      @@hawks3109 Software packages for maths and engineering-and for that matter many programming languages-are perfectly happy to compute directly with rational numbers. Floating point was _invented_ as a performance/precision tradeoff for doing numerical simulations where exact values aren't known, anyway. Irrational numbers can also be expressed directly by using what are called constructive reals (informally, you can think of these as procedures that generate precision on demand, by answering the question “what's the next digit?” if and when it is needed), but these are not exactly the same Reals you are taught about in high school, since they are not totally ordered. In fact, this relates to the topic of the video: if you gain the ability to represent irrationals, what you lose is the ability to say for sure that 1 = 0.9…-though you can still always answer whether they are within any _particular_ given epsilon of each other.

  • @anonymousjohnson8051
    @anonymousjohnson8051 3 года назад +388

    The real difficult thing to understand in these proofs is the idea of a value “approaching” another number meaning that that value is equivalent to that number. Like in an infinite sum that converges to a value, we say that the infinite sum is equivalent to that value, but it almost feels like we’ve just redefined what it means for numbers to be equivalent.

    • @gabemerritt3139
      @gabemerritt3139 3 года назад +57

      Yeah, it's that early concept of a limit that kids struggle with. They can accept it is close to, approaches, or is appropriately 1, but it takes a case to prove it is truly equal.

    • @grandsome1
      @grandsome1 3 года назад +6

      I think 1brown3blue has a fair explanation of limits.

    • @kingthanatos6093
      @kingthanatos6093 3 года назад +55

      Well a function approaching a limit at some input x doesn’t mean that function of x equals the limit. It’s not the same thing.

    • @oscarofastora474
      @oscarofastora474 3 года назад +63

      @@kingthanatos6093 absolutely this. This is why I still think that it's super stupid to say that 0.9 period = 1. If you define 0.9 as a functions as done here, the limit is obviously 1. But saying that because the limit of the function is 1, the number aka function is equal to 1 is like saying a dog is equal to the numers of hairs on its body.
      This would mean that 0.9 period is not a number in itself though, but a short form for a converging function.

    • @benwincelberg9684
      @benwincelberg9684 3 года назад +8

      @@oscarofastora474 converging sequence but yeah

  • @VivekYadav-ds8oz
    @VivekYadav-ds8oz 3 года назад +150

    Video: "Why every proof you've seen is WRONG"
    Me, watching this the 2nd time: *VISIBLE CONFUSION*

    • @asmodeuszdewa7194
      @asmodeuszdewa7194 3 года назад +12

      Look at the description.

    • @prasanna2991
      @prasanna2991 3 года назад +6

      This is a classy joke, does watching this video a second time also mean that either this video itself is wrong, or that the title is wrong? 😀

    • @chri-k
      @chri-k 3 года назад

      @@prasanna2991 it means that you have just altered the rules of logic.

  • @naptastic
    @naptastic 3 года назад +1316

    "A proof is not just a series of true statements that ends in the one you're looking for"
    here's somebody who gets it

    • @surajvkothari
      @surajvkothari 3 года назад +42

      Flat Earthers left the chat

    • @ImNotFine44
      @ImNotFine44 3 года назад +22

      What is a proof then? I’m not saying i disagree but rather than telling us what to do, he’s telling us what not to do. Wouldn’t it be better to teach both?

    • @Lucky10279
      @Lucky10279 3 года назад +99

      @@ImNotFine44 A proof is just an _airtight_ completely logical argument for why something MUST be true. In order for an argument to be valid, it's not enough just for both the premises and the conclusion to be true, otherwise "Paris is the capital of France and 2+2=4, therefore all counting numbers can be uniquely written as a product of primes" would be a perfectly valid proof of the fundamental theorem of arithmetic. Of course it isn't though, because the conclusion doesn't follow from the premises. I can't just say, "A is true and B is true, therefore C is true." I have to explain how I got from A and B to C, why/how A and B imply C. That's what makes something a proof.

    • @ImNotFine44
      @ImNotFine44 3 года назад +12

      @@Lucky10279 ah i see. i kinda assumed that most people would link the points together but then i realised alot people are idiots who want to be right at all costs. the quote makes sense to me now.

    • @Lucky10279
      @Lucky10279 3 года назад +20

      @@ImNotFine44 Most people do. I rarely if ever see something claiming to be a proof that doesn't at least attempt to link things together. I mean, I don't spend my time looking for false proofs, but I don't expect it's a widespread problem, though I could be wrong.
      I don't think the usual algebraic proof that 0.999... = 1 does this though. It's true that it makes a lot of unstated assumptions that arguably should be stated and justified in order to make the proof rigorous, but we never explain _everything_ in a proof. Just like when writing an essay or article, you have to consider the likely audience when deciding what to leave unstated. The average person looking at such a proof isn't likely looking for a detailed explanation of how decimal expansions are defined (there are plenty of explanations elsewhere online if they are).

  • @UltimateDurzan
    @UltimateDurzan 2 года назад +63

    The thing is... .99999 repeating is really just the result of the limitations of dividing a number a base-ten system by something other than 2 or 5. 3/3 is another way of saying 1, but you can't conveniently write 1/3 in base 10 without getting an endless number.

    • @m.h.6470
      @m.h.6470 2 года назад +14

      true. technically 1/3 is 0.333... + ε and if you add up 0.333... + 0.333... + 0.333... you end up with 0.999... + 3*ε and that 3*ε is the missing bit to get to 1.

    • @thewhitefalcon8539
      @thewhitefalcon8539 2 года назад +8

      @@m.h.6470 Nope, it's just 0.3333...
      You can see this because multiplying it by 3 gives 0.9999... which is another way of writing 1

    • @m.h.6470
      @m.h.6470 2 года назад +17

      ​@@thewhitefalcon8539 no, as the video states, 0.9999... is just the limit of 1, while 0.9999... + ε actually IS 1.

    • @m.h.6470
      @m.h.6470 2 года назад +20

      @@coolcatcastle8 and that exactly is the issue: 1/3 != 0.333... it is as I said actually 0.333... + ε.
      ε is the small epsilon, which stands for infinitesimal, which is an infinitely small unknown number and the inverse/opposite of infinity.
      3+3+3 *+1* =10
      0.3+0.3+0.3 *+0.1* =1
      0.33+0.33+0.33 *+0.01* =1
      ...
      0.333...+0.333...+0.333... *+ε* =1
      As you can see, the right part gets smaller and smaller (tends towards infinitesimal) but never gets 0. That is the missing part, that gets suppressed, if you say 1/3 = 0.333...

    • @thewhitefalcon8539
      @thewhitefalcon8539 2 года назад +2

      @@m.h.6470 There's no such thing as ε

  • @johnsavard7583
    @johnsavard7583 3 года назад +372

    You are correct in saying that the common proofs lack mathematical rigor, by failing to say that .999... is a limit. However, people unfamiliar with calculus can do long division, and they can see that dividing one by 3 yields 3 over and over again. So by viewing decimal numbers as common constructed objects, the common proofs are satisfying, and most people don't go beyond that.

    • @otakurocklee
      @otakurocklee 3 года назад +43

      It's satisfying for some people, unsatisfying for others. The others are rightly thinking... is this even a number? What does this even mean? How do we know standard arithmetic operations work here? The proper explanation is using limits. Yes, this is an advanced concept, but I think it's better to say... "you will understand when you learn limits" rather than giving an improper explanation. Or at the very least, the common proof should be accompanied by a strong caveat.

    • @kasroa
      @kasroa 3 года назад +89

      As a non mathematician, the final proof in this video made no sense. So even after watching him dismantle the other "proofs", I am none the wiser. I understood the reasons why those proofs are not enough, but the final section where he gives the real proof only makes sense to people who don't need to see the final section. That's why the other videos exist, because mathematicians are pretty bad at explaining these proofs to lay-people. And that might just be a property of maths, not their fault.

    • @benwincelberg9684
      @benwincelberg9684 3 года назад +9

      @@kasroa That is fair. I would like to take a shot at explaining the final proof so please lmk if I didn’t explain the part you wanted.
      The point is that 0.999... is defined to be the limit of the sequence he mentioned (0.9, 0.99, 0.999, 0.9999, ...).
      A sequence is an infinite list of numbers. We can denote the nth term of a generic sequence by a_n. A limit (we will call it “L”) of a sequence is said to exist if there is a real number such that for any ϵ > 0 there is a natural number N such that for all n > N we have that |a_n - L| < ϵ. That is just a definition but think about what this means. Since we can pick whatever positive ϵ we want, if we look sufficiently far into the sequence all terms will be as close to the limit as we want.
      In this manner, he rigorously showed that the limit of this sequence (i.e. 0.99 repeating) is 1. Like he said, once you understand the definition, you probably won’t feel as much of the proof

    • @chrisbrowy929
      @chrisbrowy929 3 года назад +1

      if that’s your logic then someone dividing 1 by 1 (itself) will never end up getting the 0.999 repeating, so they’ll have to deal with that. it’s not satisfying no matter what

    • @Chris_5318
      @Chris_5318 3 года назад +7

      @@chrisbrowy929 0.999 . . . just exists. It is the starting point, not the result of an operation or algorithm. OTOH there is a Wiki that shows a deferred long division that produces 1/1 = 0.999 . . . but like all such algorithms, it never actually finishes. Even the 1/3 = 0.333 . . . conclusion implicitly used limits as an inspired guess as to where the algorithm was going.

  • @yazanalj1975
    @yazanalj1975 3 года назад +291

    My math teacher who used these demonstrations:
    "Oh? you're approaching me?"

  • @maddo5501
    @maddo5501 3 года назад +35

    I really don't like the comparison between the 0.9(repeating) and 9(repeating). The first is an infinite decimal expansion approaching a finite number, the second IS infinite. Doesn't the error in the 9(repeating) comparison come from treating infinity like an actual number? I don't think the comparison here exemplifies why the first proof is wrong at all. They aren't really analogous.

    • @Jehannum2000
      @Jehannum2000 3 года назад +7

      I'm glad I scrolled this far down the comments and actually found someone sensible enough to question this!

    • @sajanramanathan
      @sajanramanathan 2 года назад +10

      This response is like way after the fact but I feel like your point here is exactly what he was trying to get at. Of course …999.0 makes no sense as a regular number, and if you assume that it is you get weird and bad results. But the same is true for some infinite series (which is how you formally define 0.99…) . It turns out that 0.99… can be well defined but that’s not true for all infinite series (see the alternating harmonic series) and thus cannot be taken as true by default.
      Basically the comparison isn’t “these things are equivalently silly” the comparison is “look at this example that is defined similarly and notice that it actually turns out to be silly! Maybe we need to define things better”

    • @rasmysamy2145
      @rasmysamy2145 2 года назад +1

      @@sajanramanathan you are incorrect. All periodic numbers are convergent series. The justification is trivial. The issue is with adding a number after the period which doesn't make sense and doesn't match the definition of a periodic number

    • @user-lb1ib8rz4h
      @user-lb1ib8rz4h 2 года назад +2

      'the second is infinite" ok but how do we know that the first ISN'T infinite? that's what we're trying to prove. hence the issue with algebraic proofs, we're assuming it converges before proving it!

    • @user-lb1ib8rz4h
      @user-lb1ib8rz4h 2 года назад +2

      @@rasmysamy2145 "all periodic numbers are convergent" okay, prove it using an algebraic proof and never using limits.

  • @justsomeboyprobablydressed9579
    @justsomeboyprobablydressed9579 2 года назад +85

    Good video. But I would say that, although the real numbers are generally presented axiomatically (as a complete, ordered field), there are some treatments of the real numbers that _define_ a real number to be any infinite decimal. So, if you are following that treatment, there is nothing wrong with taking for granted that 0.999... exists.

    • @Pseudify
      @Pseudify Год назад +1

      I don’t know what you’re talking about, but let me ask the obvious question. In the system of which you speak, is the infinite decimal equivalent for 1 equal to 1.0 (infinite repeating zeros) or 0.9 (infinite repeating nines)?

    • @CellarDoor-rt8tt
      @CellarDoor-rt8tt Год назад +13

      @@Pseudify all of the constructions you wrote here are equal to 1. In this construction of the reals, you consider the set of all sequences of rational numbers. You then force every Cauchy sequence to converge. There’s no problem with this as any sequence can be forced to converge by creating a new number system out of Q that includes the number this sequence converges to. The rest of the proof is in showing that forcing the Cauchy sequences to converge still maintains “=“ being an equivalence relation, “

    • @pedroteran5885
      @pedroteran5885 Год назад

      ​@@CellarDoor-rt8ttI really doubt that Cauchy sequences have anything to do with pseudo-defining real numbers as 'infinite decimals', which is something that, as far as I know, was abandoned a century ago.

    • @CellarDoor-rt8tt
      @CellarDoor-rt8tt Год назад

      @@pedroteran5885 I mean Cauchy sequences are the system that replaced this defining of infinite decimals, but these systems aren’t that dissimilar.
      The idea is that we take as an axiom in the reals that any set of real numbers that is bounded above has a supremum (a least upper bound). You can prove that this axiom is equivalent to saying that every Cauchy sequence in R converges in R.
      The point is that infinite decimals aren’t that dissimilar. 0.9… as a repeating decimal is basically just short hand for the sequence, (0, 0.9, 0.99, 0.999, …). It can be proven that this sequence is Cauchy and therefore, by the completeness axiom, it must converge. We then just have to show that it converges to 1. It can even be shown that this method of defining numbers always works i.e the more decimals you take the closer to the number you get.
      The reason why this had to be abandoned had nothing to do with numbers like 0.9… repeating. The reason why it was abandoned was because this doesn’t work very well for irrational numbers. Since these have decimals that don’t ever repeat, you need a process for generating each of the irrational number’s digits and that’s not trivial depending on the number. It’s far easier, at that point, to just provide a sequence that you can prove converges to the number. For example, it’s possible to show that pi = sum(n from 0 to inf, (-1)^n * 4/(2n + 1)) or 4 - 4/3 + 4/5 - 4/7 + 4/9…
      You could even do this in reverse by giving pi this as it’s definition and then prove that this number you defined has all of pi’s geometric properties. This is how we do it more now

    • @empathogen75
      @empathogen75 Год назад +2

      The problem with explaining this to most people is that they’re introduced to real numbers without any foundation for understanding the logical foundations of what real numbers actually are. Which is fine, you don’t need to know what a Cauchy sequence is to use pi in a formula or calculate the decimal expansion of a rational number, but it does make it hard to explain something like this who thinks they know what a real number is, but don’t.

  • @BurgerFred1
    @BurgerFred1 3 года назад +322

    Algebra teacher: "Prove that .99...=1"
    The kid who knows about geometric series: *pushes up glasses* "My time has come."

    • @naptastic
      @naptastic 3 года назад +16

      That's how I did it in 10th grade. The algebraic proofs just *seemed* wrong, so I did (apologies for the notation; I speak perl now)
      Σ(1..∞) {x += 9/(10**n)} which converges to 1.
      Since then, I have learned the quirks of IEEE-754 floating point "math", which is, I believe, the worst possible way to represent numbers. In floating-point math, the only numbers that exist have a denominator of 2^x where x is an integer between -23 and 24 inclusive. So x = 0.1 cannot be represented.
      As I type this, I'm 2:30 into the video... looking forward to how mCoding does it.

    • @brainletmong6302
      @brainletmong6302 3 года назад +25

      @@naptastic We tried our best to trick rocks to thinking, but after we put the lightning in them, they've been very uncooperative and refuse to acknowledge the existence of decimals, or of anything aside from 1 and 0 really.

    • @TheEnmineer
      @TheEnmineer 3 года назад +2

      @@brainletmong6302 There either is or isn't lightning there, I don't understand what you flesh-bags mean by "some" lightning.

    • @pussy8229
      @pussy8229 3 года назад

      @@TheEnmineer kek

    • @user-lb1ib8rz4h
      @user-lb1ib8rz4h 2 года назад

      proof of convergence is essentially the same as just using the limit directly tho

  • @julianfogel5635
    @julianfogel5635 2 года назад +54

    Something worth adding is that the reals are usually defined axiomatically, using formal logic and set theory, rather than as an infinite sequence of decimal digits as taught in high school.
    The decimal representation of a real can be shown as valid because every axiomaticallly defined real has such a representation and likewise every decimal representation refers to a unique real number.
    There is the minor caveat that some reals have two decimal representations, one ending in repeating 9s and the other in repeating 0s. For example, 1.4329999999... is the same real number as 1.43300000... This is a consequence of the axioms for reals and the way decimal notation is defined.

    • @zandaroos553
      @zandaroos553 Год назад +5

      In fairness in low-intermediate level analysis the axiomatic definitions are shifted to the Natural numbers and arithmetic operations, with Z and Q being constructions built up from there and the Reals being defined like how our lovely presenter showed via Cauchy spaces constructed through sequential ep-delts

  • @TrimutiusToo
    @TrimutiusToo 3 года назад +63

    Well mathologer didn't do mistakes that you mentioned. He actually did do proper infinite sum in his proof (second half of his proof) in second half of the video he defined how to multiply infinite sum by 10 instead of assuming it can be just done, so it solved that problem with arrow you talked about...

    • @Chris_5318
      @Chris_5318 2 года назад

      No, Mathologer did not use limits. He had assumed >>without proof

    • @TrimutiusToo
      @TrimutiusToo 2 года назад

      @@Chris_5318 hence why I mention second half of the video, where he did do the limits. Structure of his videos is often like that, he does something without proof and then in second half does the proof for the most keen of us

    • @Chris_5318
      @Chris_5318 2 года назад

      @@TrimutiusToo Perhaps we are talking about different videos. The one I'm referring to is titled "9.999... really is equal to 10" and starts of with a Total Drama Island scene with a bloke in a barrel of nastiness. That one does not refer to limits at all.

    • @TrimutiusToo
      @TrimutiusToo 2 года назад

      @@Chris_5318 nah i was just confused which comment section this was... Indeed proper rigorous proof of infinite sums was done by him only in his -1/12 debunking video... In 9.999 video he is a little bit hand wavy but he still didn't fall for any of the pitfalls this here video says, but didn't do full rigorous proof. But it was an old video before he started to be more rigorous

    • @Chris_5318
      @Chris_5318 2 года назад

      @@TrimutiusToo His 9.999... video was intended or children.

  • @jhawley031
    @jhawley031 2 года назад +53

    My favorite "proof" was from one of my professors
    He pulled down the chalk board and got a new stick of chalk like he was about to write out a long proof then simply wrote: "It is trivial to see that 0.9̅ = 1 ▪︎" i should mention that this was just a joke after class

  • @michamichalski8633
    @michamichalski8633 3 года назад +5

    It’s not really that the proofs are incorrect. It’s about imprecisely formulated question. Proofs are always in the context of assumptions. You have just provided a proof with different set of assumptions which doesn’t make it more or less correct. The question doesn’t specify what you can take as a given so you can operate under any assumptions you choose.

    • @gianglai7346
      @gianglai7346 2 года назад

      Exactly. Most of my acquaintances don't have the knowledge about Analysis, so I just cannot pull off formal definitions of limits which would only confuse them more.
      It's easier to make assumptions which are familiar to them and work with the simple demonstration using fractions as proof.
      The goal is to convince people and not to confuse them.

  • @patatopie
    @patatopie 3 года назад +31

    Not wrong just not as rigorous.
    Usually within the context those proofs are brought up, the assumptions used are already accepted as true (even if just because someone more knowledgable has said so), so are fine to use as axioms without extra justification. Not stating them as such is a bit messy to be fair.

    • @NyscanRohid
      @NyscanRohid 3 года назад

      Then they aren't proofs; they're intuitions. A proof _must_ be rigorous. Else, you absolutely cannot be sure that it's true. The calculus and infinite series "proofs" you've been doing in class are called intuitions. The actual proofs used to justify them are much much longer and take a lot more time than your professor has to teach you with.

    • @patatopie
      @patatopie 3 года назад

      @@NyscanRohid If you're coming from the perspective of a college-level analysis class then I fully agree those wouldn't be acceptable as proofs.
      But words are context dependent even in maths, and the line between assumption and axiom always has to come somewhere

    • @thekenanski8789
      @thekenanski8789 3 года назад +1

      I guess what really irritates me is that the assumptions they make aren’t true in general-they just happen to be true here. It’s one thing to omit a proof of something that’s always true, but quite another to omit a proof that some necessary condition is satisfied. (Eg sure proving that addition of real numbers is well-defined for the construction of the reals by Dedekind cuts is probably overkill here, since it always is.)
      And then tons of people see that we make these assumptions about how series behave and can be manipulated, but don’t realize that they’re assumptions that depend on convergence, and so they get all confused about 1+2+3+...=-1/12 or some stupid thing like that because they assume that the assumptions made here are true in general.

    • @patatopie
      @patatopie 3 года назад +1

      @@thekenanski8789 That's fair- imo what makes it alright here is that the original thing being proven is stated in terms of infinite decimals not infinite sums, which are obviously a subset- but a subset that's more commonly accepted and used at high-school level than the whole set of convergent infinite sums. (e.g. wouldn't expect a student to prove pi exists before using it in a proof)
      But obvs adding the nuance isn't a bad thing at all, and you make a good point that for any students who're aware of series it probably is worthwhile making those steps more clear (at the very least making it clear that there *are* assumptions being made)

    • @Takkunda
      @Takkunda 3 года назад

      @@patatopie words have specific meanings. A proof in math is a proof, a theory in science is a theory.
      We are not laymen.

  • @alvideiectiones5097
    @alvideiectiones5097 2 года назад +4

    "Every proof you've seen that .999... = 1 is wrong"
    Me: watches the video twice

  • @hxbel
    @hxbel Год назад +7

    Me telling the cashier that she can't make me pay one extra cent for the 0.99 drink:

    • @Chris-5318
      @Chris-5318 Год назад +1

      What's that got to do with the fact that 0.999... = 1?

    • @ElusiveEel
      @ElusiveEel 10 месяцев назад

      @@Chris-5318 equality not being transitive is the issue, unless also 0.99...8 = 1

    • @Chris-5318
      @Chris-5318 10 месяцев назад +2

      @@ElusiveEel Equality is transitive. Not that I have any idea why you mentioned it. 0.999...8 is a terminating decimal. In fact 0.999... - 0.999...8 = 0.000...1999... (= 0.0002) where the 1 (and the 2) is in the same decimal place as the 8 was.

  • @Daniel_VolumeDown
    @Daniel_VolumeDown 3 года назад +36

    4:41 "Why are you so okay with one-third being exactly equal to 0.33 repeating?"
    It's not just because of what I think is called "long division" in English? (I'm not a native speaker)

  • @shohamsen8986
    @shohamsen8986 3 года назад +141

    One could argue that at 2:12, u have a classic infinity=infinity expression. In that case, u cannot just take x to one side and then cancel stuff to get x=-1

    • @maxmustermann3938
      @maxmustermann3938 3 года назад +32

      There is a reason why we never see this "repeating" thing in front of the comma, which is related to this. Because we can't "calculate normally" anymore of we allow that. Which is why the whole point of this video is moot anyways, it only matters if someone doesn't actually have a grasp of what this "repeating" thing means, but you don't really need that, because the regular transformations still just work fine.

    • @shohamsen8986
      @shohamsen8986 3 года назад +5

      @@maxmustermann3938 What do you mean by "Because we can't "calculate normally" anymore of we allow that"? What do you mean by regular transformation. Anyway, my point is any number repeating before the decimal point should be thought of as infinite which is what the algebra actually gives us.

    • @Rastafa469
      @Rastafa469 3 года назад +2

      @@shohamsen8986 "any number repeating before the decimal point should be thought of as infinite" Hate to be a smart ass but what about repeating 0 before the decimal point? :D

    • @shohamsen8986
      @shohamsen8986 3 года назад +4

      ​@@Rastafa469 Its been some time since I typed the above comment. So, I cant exactly recall the context, but if you go through the algebra, you get
      x=0000000000000000000000,
      10x=00000000000000000000,
      9x=0,
      is x=0.
      Which seems okay.

    • @NerdyStarProductions
      @NerdyStarProductions 3 года назад +4

      I think that's basically the point that mcoding was trying to get across. You can't assume you understand what the actual value of 0.9 repeating is; you need to see what it's value will be when you actually use its definition and see the limit it approaches. In 0.9 repeating's case, it ends up being a finite number, so all of the statements that were written in the "algebraic" proof end up being valid. But for 9 repeating, it has no finite limit, and so many of the algebraic statements made with it afterwards are nonsensical.

  • @fn3200
    @fn3200 3 года назад +12

    I think 99999...9.0 is a contraddiction in itself, you cant write a signed number (saying "it goes like this forever") amd then putting there the dot (saying "it stops somewhere").
    0.99999.... is different because there is no contraddiction in the definition of the number itself

    • @3snoW_
      @3snoW_ 3 года назад +2

      write the number from right to left and it's fine

    • @amjadsiddig2085
      @amjadsiddig2085 3 года назад +2

      Laughs in ordinal numbers

    • @fn3200
      @fn3200 3 года назад

      @@3snoW_ yeah but the problem is not writing the number, it's how the bar that you put on top of the number works

    • @fn3200
      @fn3200 3 года назад

      @@amjadsiddig2085 explain the meme pls😂

    • @amjadsiddig2085
      @amjadsiddig2085 3 года назад +1

      @@fn3200 I'm not an expert by any means but in set theory, you can basically have numbers after an infinite series of numbers, that's the joke. you can have a number that comes after an infinite string of numbers. I think that's the gist of it but I may have butchered it 😂

  • @KissyfaceJenkins
    @KissyfaceJenkins 2 года назад +16

    As someone who finds math fascinating but was never all that good at it, thank you for explaining things in an easy to understand way.

  • @thedoublehelix5661
    @thedoublehelix5661 3 года назад +18

    The proofs aren't really "wrong". That's like saying every proof by euler or gauss or any proof of a calculus fact before real analysis was created are all "wrong". While technically true, it's a bit pedantic in my opinion to say they're strictly wrong. The main idea is correct and leads to a correct proof with very minimal additions.

    • @mCoding
      @mCoding  3 года назад +8

      I'm certainly not saying euler and gauss are wrong (though some of their proofs were, of course, wrong, it's just part of life). The point was that a proof is for a particular audience, and if your audience does not know calculus, then to them it is not a valid proof, while to someone else it may be a valid proof. As for the idea being correct with minimal additions, consider that this this proof uses only algebraic manipulations. If it were correct, then it would be true for all algebras. However!! It is false in Z/11Z, for example, where .999 repeating is NOT equal to 1! This is the danger of proofs like this. Your audience needs to realize that the details left out of the proof can change the answer!

    • @jenaf372
      @jenaf372 3 года назад +1

      A proof beeing "wrong" doesbt mean the dervied statement is wrong. Only that we cant know the truth value of that statement from that proof.
      Its propably better to label such proofs as incomplete or invalid, just to make the difference more clear?

    • @thedoublehelix5661
      @thedoublehelix5661 3 года назад

      @@jenaf372 I meant that proofs that are wrong do to really technical details can still clue us in on the correct proof. A proof can be correct for the cases that mathematicians considered in the past; now we have more precise boundaries for our domain.

    • @jenaf372
      @jenaf372 3 года назад

      @@thedoublehelix5661 well wr should amend suvh proofs wich only apply to a restricted domain with ckear demarkations on where it is valid. Or else really really sneaky "bugs" and pathological cases may pop up.

    • @thedoublehelix5661
      @thedoublehelix5661 3 года назад

      @@jenaf372 I agree

  • @matsim0
    @matsim0 2 года назад +5

    1:55 - sorry, I don't understand the disproof. How come 10x = p90.0 is equal to 10x + 9 = x?
    For the equality to hold, 9 must be added to both sides, so suddenly p90.0 + 9 becomes x? If anything, x is p9.0? So it should be _smaller_ than p90.0, and if anything p90.0 + 9 should be _bigger_? (yes, I know it is inf, so it makes no sense speaking of bigger or smaller, but still...)
    What am I missing?

    • @luphaestus
      @luphaestus Год назад

      10x = 9 recurring 0,
      10X + 9 = 9 recurring 9 which is the same as just saying 9 recurring which is the same as x therefore 10x+9=x

    • @luphaestus
      @luphaestus Год назад

      ​@Ok-se6mkwhat do you mean by exactly?

  • @Biggyweezer69
    @Biggyweezer69 Год назад +9

    I still believe that the best way to understand this problem is thinking about our notation for numbers more abstractly. Numbers are not strictly fractions like 1/2 or decimals like 0.5. Numbers exist independently of notation, 0.5 is simply a notation we use to represent numbers base 10. Switch to base 9 and all the sudden 1/2 is 0.444... Our representation of numbers is a choice.
    It's then we have two options: We either decide that the only numbers that can be notated in this way are numbers who's decimal expansion is finite in length, or we decide that all real numbers can be written with this notation. The obvious choice is the latter as it allows us to see things from a more familiar viewport. Under this system its clear that many of these real numbers will have to be written as a limit of a sequence of decimals. Examples are 1/3 , sqrt(2) and pi. We chose to define 0.3... as the limit of the sequence that converges to 1/3. Writing 0.3... = 1/3 is a nice shorthand, imagine having to write sigma notation every time you wanted to do math with 1/3 written in decimal. 0.9... = 1 is an artifact of our extension of notation. This is true for all numbers, all infinite decimals are sigma notation compressed and they all equal the limit of the sequence they are notated as.

  • @DarinBrownSJDCMath
    @DarinBrownSJDCMath 2 года назад +4

    There's absolutely nothing wrong with the algebraic proof, given a rigorous definition of real numbers and their decimal representation. In fact, this is exactly how I teach beginning algebra (even pre-algebra) students to convert repeating decimal numbers to fractions.
    If we took the attitude of this video to heart, then we should never teach anything about irrational numbers to students. We should not even mention the square root of two, or e, or pi, because after all, we have not shown that those numbers exist. At least not until we teach them the Cauchy sequence or Dedekind cut construction of the real number system. Come to think of it, we should stop teaching fractions, because when we "prove" 3*(1/3) = 1, we are assuming that 1/3 exists as a number, and when do we ever define the number 1/3 to elementary students? (Hint: we don't rigorously define it for them.)
    Clearly this is absurd, because for one thing, mathematicians worked with irrational numbers and real numbers long before Cauchy and Dedekind were born. And humans were using fractions 2,000 years before their definition as a quotient field of equivalence classes of ordered pairs.
    The algebraic proof condemned here is a gem, and has been described by many mathematicians as elegant and poetic. It made a powerful impression on me as a young child, and the fact that its rigorous formulation cannot be fully understood until much later in life should not keep us from exposing students to its beauty and charm.

    • @stephenolan5539
      @stephenolan5539 2 года назад

      And it is the word "repeating " that is confusing the "not equal" people.
      The 9's are already there. You are not repeating them.
      So every math proof is missing why people don't agree.

    • @DarinBrownSJDCMath
      @DarinBrownSJDCMath 2 года назад +1

      @@stephenolan5539 People generally have trouble getting an intuition for infinite processes or handling infinity as a mathematical concept.

  • @Txominde
    @Txominde 3 года назад +9

    Sorry, but I had to downvote your video. I normally like your videos (I subscribed for a reason).
    Reasons:
    1. The title was very click-baity and you only showed one "false proof" (I put these words in quotation marks because by definition a proof is always right) which is only (maybe?) taught in school. If one attends a university/college with a math lecture, they will see a "right proof".
    2. Your proof has a huge gap. The real work is to show that 10^-n < 1/n for n large enough but you just state this fact as it is absolutely clear - which it is not. You more or less only state the definition (besides the inequality stated in the previous sentence).
    Greetings from a mathematician

    • @mCoding
      @mCoding  3 года назад +4

      Understandable, thanks for letting me know. My reasoning was as follows: 1) is there any title of a video about this topic that could NOT be construed as click-bait? 2) 10^-n < 1/n for all natural numbers n for which the expression makes sense (i.e. all n
      eq 0), not just for all n large enough. Rearranging this is just saying n < 10^n. While it is true that any proof that does not literally go down to the axioms contains a "gap", a proof is constructed for a given audience in order to allow gaps of acceptable size, and I think that the average person watching this video is not being swindled by a basic algebraic inequality like n < 10^n in the same way they were being swindled by unknowingly using calculus in 10 (.99...) = 9.99... I can see how, if this video were meant for children, more explanation would be needed. But this video is not really meant for people who are not familiar even with algebra (note 10 is an integer so no calculus is needed). I do understand your concern though, but I'm sure you know it is not worthwhile to cater to everyone and that was where I decided the line was. I will keep your concerns in mind for future videos though. Greetings from a mathematician!

    • @Txominde
      @Txominde 3 года назад +4

      @@mCoding Thanks for the response.
      1. Maybe a title like "Why the standard proof from school for ... is wrong" or something like this would be more appropriate but never mind.
      2. Yes, you are right that the inequality is more or less clear. (In my head I always think about q^n for |q| < 1 and hence the additional "for n large enough".) You are also right that you have to assume some background knowledge (if one does not want to go back to the axioms).
      Thanks again for the discussion and keep up the good work.

    • @DajesOfficial
      @DajesOfficial 3 года назад

      ​@@Txominde so, 1 hour earlier you stated that proof is always right and now you propose to name the video using the statement "proof is wrong"?

    • @Txominde
      @Txominde 3 года назад

      @@DajesOfficial Yeah, your are right. My suggestion is really not that good. Maybe "proof" (with quotation marks) or something like that. I better stop suggesting any titles^^

    • @rohitchaoji
      @rohitchaoji 2 года назад

      @@Txominde Unfortunately, clickbait titles are useful, even for bigger channels, if they want their videos to show up on people's recommendations and home page. Just how RUclips algorithm works. It's obviously not necessary, but it is a valid advertisement tactic, if it isn't meant to be sneaky. I'm honestly okay with clickbait titles like these, rather than "You wouldn't BELIEVE what happened NEXT! Watch until the end" type titles because they're not meant to mislead, but rather intrigue.

  • @mvryan
    @mvryan 2 года назад +39

    I feel really weird, I have math degree and only knew about proving as a limit and the start of the video had me asking myself why I’d made it so much work for myself 😅

  • @tricky2014
    @tricky2014 3 года назад +6

    The x = 0.9999 example is actually correct and your counterexample isnt. It becomes extremely obvious when you write 0.999 and 9999. as infinite sums. You can do this because 0.9999 is convergent while 99999. is divergent. With the sum notation you see that 9.99999 - 0.99999 goes to 9 at the limit while 99990 plus 9 is not x. You should always be very careful doing math with divergent sums. This is the same mistake that leads people to conclude that the sum of all natural numbers is -1/12. Bad infinite series math.

    • @mCoding
      @mCoding  3 года назад +1

      This was exactly the point I was making, showing you what goes wrong if you manipulate a series that is not known to be convergent. However, this is not "bad math". 999... DOES converge some number systems, e.g. Z/10Z or the 10-adic numbers, in which case it is equal to -1, as the proof showed.

  • @cola-z9219
    @cola-z9219 2 года назад +1

    If the problem is assuming .999 can exist, why do you assume that an infinite sum could exist. The problem is it’s not a finite number, yet even you use an infinite proof. If you can assume to have an infinite proof why not assume to have an infinite number.

  • @Krunschy
    @Krunschy 3 года назад +60

    I'm not so sure about these other proofs confusing even more. The way they explain it might not be a 100% waterproof proof, but neither is most reasoning that people get for the mathmatical truths they learn at school. What I'm trying to say is that this proof fits considerably better with what idea many have of maths. One could argue that everyone should be learning formal maths from the get go, but I'm sure you'd lose people way earlier this way in school.

    • @eliezra83771
      @eliezra83771 2 года назад +1

      We studied limits in high school .-.
      Yeah last grade but still, most schools in my state have them in their program. How well they are explained and studied is another matter... but in my class, we definitely learned them well 👍

    • @holomurphy22
      @holomurphy22 2 года назад +1

      Many (most) people only have an idea of how big or small a number is, how addition and multipliction works with actual numbers, etc.
      How would those many people define 0,123456...414243... ? Why would this be a number if it is infinite ? Most people that don't do math dislike when infinity shows up.

    • @Roescoe
      @Roescoe 2 года назад +1

      I think the issue actually lies in that fractions to exact decimals isn't something that should be taught without calculus limits. Since there's a lot of stuff hiding in "infinite repeating decimal"

    • @programaths
      @programaths 2 года назад

      @@Roescoe Well, it can be done when the teaching is done in spirals.
      When introducing 1/3=0.333333...
      You can already explain that in reality, it's: 0.33333... and something so small it can be ignored.
      Because if you do the Euclidean division, you would get: "1.0=0.3×3+0.1"
      And you quickly find out it can be writen as: 1.0000...0=0.3333...+0.00000...1
      So, that 0.0000...1 is the samellest number or the closest to zero from the right. (it's a limit, but you don't name it)
      It's also so small that we don't write it, but one should mind that it's still there.
      And then, you tell that the more serious explanation is coming when you will explain limits.
      That way, pupils know you are not pulling that straight out of your ass and just skipping material that is too hard to learn.
      It works better if you already did that on shorter time frame, so that student get used to work with partial (but working) knowledge. This by itself is an invaluable skill in real life. As an example, as a programmer, you may be using an API on something you don't know, like a machine learning model to discriminate cats and dogs. You don't have to be good in ML to just use it and it would be a waste of time if all you want to to is classify pictures for a pet shop.
      And one fact is that it already happens during the curriculum. Most students can do algorithms to add and divide in radix 10, but can't in radix 7.
      They can do an Euclidean division in Z (integers), but not in P(polynomials). Yet, that's the very same mecanisms. So, they already work with incomplete knowledge without minding it.
      Mind that in the real world with real problems, you often have to use knowledge of many fields at a level above your own as one can't specialize in everything. So, you defer parts of your problems on others (even if it's looking it up in StackOverflow, Google or company data bank) and trust it. Even if with time, you may acquire more knowledge and specialize in more fields.
      And that's something a new teacher will strugle with and drown students ^^
      It's why we do didactic transposition for 3 years, to learn how to shake knowledge in a way it's manageable for pupils.

    • @Roescoe
      @Roescoe 2 года назад +1

      @@programaths I had a professor whose mantra was "Decimals are a disaster, fractions are your friend" I tend to agree with him, especially since decimals don't really exist anywhere except in the digital. Fractions are ratios which you can see in the physical.
      Stating that decimals are approximations is a good way to teach if you must, but I guess I would stay away from it.

  • @kiiometric
    @kiiometric 3 года назад +74

    Despite them not being proofs, they are a very fast and easy way to teach to high-schoolers

    • @naptastic
      @naptastic 3 года назад +23

      If correctness doesn't matter, I can make teaching arbitrarily fast and easy.

    • @imoomi4067
      @imoomi4067 3 года назад +24

      @@naptastic that may be true but I think it's missing the point that young people might not have the skills yet to understand some of the more complex aspects of proofs like these. Also while they may not be technically correct, these exercises do teach young people the skills for algebraic proofs and a way of viewing algebra that they hadn't considered yet. I remember doing proofs like these as a teenager and they were what got me excited about maths and pushed me to study it further.

    • @dekkonot
      @dekkonot 3 года назад +12

      @@naptastic shortcuts and approximations aren't exactly the same as arbitrary; it's like how you when teaching division, you don't open with the concept of fractions and instead use remainders. It isn't arbitrary or wrong, it just leaves out a few steps.

    • @markasiala6355
      @markasiala6355 3 года назад +2

      The real issue comes when those high schoolers enter college and they learn the rigorous methods for this or any other mathematical concept. Students too often take this shortcuts as fact and nothing else is needed. Hence the college instructor has to "unteach" some of these shortcuts before you can teach them the more rigorous methods. It is a difficult balance that can be painful for instructor and student if not done well.

    • @prumchhangsreng979
      @prumchhangsreng979 3 года назад

      @@markasiala6355 so many people that do well in math in highschool start to hate math in college

  • @Tomyb15
    @Tomyb15 3 года назад +58

    the reason people are comfortable with 1/3 = 0.333... is that at some point in their lives, a math teacher made them do long division on 1 by 3 when teaching long division with decimals, and easily saw that the numbers to put after the decimal point would never stop coming and that the numbers were always '3'. They then trust the teacher that there is an infinite number of decimals in the actual result.

    • @panjak323
      @panjak323 3 года назад +3

      Ehem, calculators... You know, the simple ones... Without fractions...

    • @yyeeeyyyey8802
      @yyeeeyyyey8802 3 года назад +45

      it's not about trust, it's a fact that you will keep getting 3's indefinately and that's provable. i'm pretty sure most people recognize the pattern and why they will keep repeating indefinately, rather than simply trusting their teachers.

    • @federicovolpe3389
      @federicovolpe3389 3 года назад +8

      You can easily prove 1/3 has an infinite decimal expansion.

    • @leonardovalente9772
      @leonardovalente9772 3 года назад +14

      not only you can trust the teacher but the teacher is also right

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 года назад +3

      @@marvinmallette6795 You are speaking nonsense. Decimal notation is not broken. It is a demonstrable fact that 0.333... = 1/3.

  • @Elektrolite111
    @Elektrolite111 2 года назад +1

    The best proof for me is to define a real number as a Cauchy sequence of rational numbers. Two real numbers are equal if the first sequence subtract the second tends to 0.
    1 is the sequence 1, 1, 1, …
    0.999… is the sequence 9/10, 99/100, 999/1000, …
    Subtracting gives the sequence
    1/10, 1/100, 1/1000, … so all thats left to show is that this tends to 0.

  • @isodoubIet
    @isodoubIet 3 года назад +56

    First let me say that I think this is a good video, and that anyone who's confused about the algebraic proofs and/or the statement that 0.999.. = 1 would do themselves a favor by watching it. Your exposition of the definition and proof is clear and accurate.
    But I don't agree with the titular claim that the algebraic proofs are wrong. The key fact that is assumed by such proofs at the outset is that you're working in a real number system and that you have a well-defined positional representation. Those facts are plenty enough to rectify any shortcomings in the algebraic proof! In particular, by merely saying the words "real number" you're already saying (for example) that you believe in the Cauchy sequence construction of the real numbers, of which the positional representation is a concrete example. That construction guarantees that any (finite or not) sequence of digits with a finite number of digits left of the decimal point corresponds to an existing real number, and it further guarantees that multiplication by 10 is a valid operation.
    So I don't think it's fair to say the proofs are wrong. The purpose of a proof is to convince the reader of a proposition by making logical inferences. In any proof there is a set of facts that can be assumed known, and facts that are to be demonstrated. In the algebraic proofs, it is assumed (for instance) that the reader knows that there exists such a thing as the set of real numbers, that listing their digits is a valid way of specifying them, and that the multiplication algorithm they learned in school is correct. Given those assumptions, the proofs are absolutely correct, and they _do_ in fact prove what they set out to prove, provided you believe/understand those assumptions.
    By the standards advanced in this video, I could also argue that the proof presented here is wrong: you didn't, for example, establish that the equivalence classes of sequences of digits with the same limit points can be usefully regarded as a number system, that the limit technology you used actually works for them (that well-defined limits exist even for other digit sequences), etc. I could keep going back, and ask for constructions of rationals, of integers, and so on all the way to ZFC. But I don't think that's the case. I think the proof in this video is correct, and I think it's useful for everyone to see it at least once. But that doesn't make proofs that start from stronger assumptions wrong.

    • @CountSolo
      @CountSolo 3 года назад +10

      I was about to post an explanation that tries to clarify certain distinctions that I felt should’ve been made in the video, and then I saw this comment. It echoes *exactly* what I wanted to write.
      I do agree that the video is well-intentioned. It can especially help students have a better understanding of (or at least be introduced to) the foundations of mathematics, and take a deeper dive into mathematical concepts that they may have taken for granted.

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 года назад +2

      *The key fact that is assumed by such proofs at the outset is that you're working in a real number system and that you have a well-defined positional representation.*
      Yes, and the problem is that these assumptions are being made, and they should not. To anyone that is not convinced of 0.p9 = 1, the statement that the real numbers are a well-defined system, and that there is a well-defined positional representation for real numbers, is far from obvious, and is in fact what is being challenged by the 0.p9 = 1 deniers to begin with.
      *In particular, by merely saying the words "real number," you're already saying (for example) that you believe in the Cauchy sequence construction of the real numbers, of which the positional representation is a concrete example.*
      No, not necessarily. There exists multiple constructions of the real numbers, and the second-order axiomatization of the real numbers does not rely on any given construction. A person can feasibly believe that one construction is valid, without believing that the others are valid. Yes, I know the constructions are all isomorphic, but what I am getting at is the hypothetical person I am talking about may not believe the constructions are isomorphic. As such, to dispel that belief, you need to at least attempt to explain that the constructions are all isomorphic, not shove it under the rug. And I am not saying you need to be all rigorous about it. I am saying it is inappropriate to not explain any of it at all. But also, you are missing the point: many people really do not believe that the Cauchy construction is valid. Most people think that numbers are sequences of digits in themselves, they do not believe sequences of digits are merely a form of representation. This is why, when people talk about using base 2 or base 12, they describe it as a different nunber system (it is not a different number system), rather than as merely changing the notation. So, yes, the positional representation already entails the Cauchy construction, but a person who denies 0.p9 = 1 would disagree with you on that point. So, shoving an algebraic proof at them without properly addressing that grievance is annoying, and simply incorrect.
      *That construction guarantees that any sequence of digits with a finite number of [nonzero] digits left of the decimal [radix] punctuation corresponds to an existing real number, and it further guarantees that multiplication by 10 is a valid operation.*
      Again, those who believe 0.p9 = 1 is false would disagree with the premise that any such sequence of digits corresponds to a real number. You are missing the point. You have to prove the premise. Not just assume it. Also, even if multiplying by 10 is a valid operation in that instance, a person may not be convinced that it is correct to conclude that upon performing the multiplication, that to the right of the decimal radix punctuation, there still remains an infinite string of the digit 9. This also must be shown.
      *So, I don't think it's fair to say the proofs are wrong. The purpose of a proof is to convince the reader of a proposition by making logical inferences. In any proof, there is a set of facts that can be assumed known, and facts that are to be demomstrated.*
      No, it really is not that simple, and this is where you are wrong. A proof has to consist of premises that can be _proven_ to be factual, or otherwise and at the least, the parties being addressed to with the proof can agree are factual. The problem with the proofs the video calls "wrong" is that they contain premises that not all parties agree are factual (even if, objectively speaking, they are factual). If the party being addressed to is still convinced by the proof, then it just means they are contradicting themselves and not thinking it through properly, and are accepting the conclusion not on the basis of understanding the proof, but on the basis of being impressed by its elegance, so much so that they choose to ignore the fact that their personal grievances with 0.p9 = 1 have not been addressed. This kind of impressionability is inevitable, but the fact that people are willing to exploit it with mediocre proofs is part of the problem.
      *In the algebraic proofs, it is assumed (for instance) that the reader knows that there exists such a thing as the set of real numbers, that listing their digits is a valid way of specifying them, and that the multiplication algorithm they learned in school is correct.*
      And this is wrong. The problem is that you are conflating "accepting an assumption" with "understanding an assumption." Most people accept the real numbers are real, but their beliefs on what the real numbers _are_ just happen to be completely wrong, and in direct contradiction with all the assumptions that go into these proofs. I already have given examples of this.
      *By the standards advanced in this video, I could also argue that the proof presented here is wrong: you didn't, for example, establish that the equivalence classes of sequences of digits with the same limit points can be usefully regarded as a number...*
      But the difference is, he is not required to do that at all for his proof to work. In other words, this is not actually an assumption of his proof. All he really has to do is say that if d(m) represents the mth digit after the radix punctuation, then the string 0.d(1)d(2)d(3)... is simply _defined_ to be the limit of (0, 0.d(1), 0.d(1)d(2), ...), and then prove the limit exists. An equivalence relation between digital strings does not need to be established at all. If you want to motivate the definition, then yes, you would talk about the fact that strings of digits are merely a representation system, and that two distinct strings may represent the same number, in so far as they are equivalent. But motivating a definition is not actually necessary in the proof. Stating the definition, however, IS necessary.
      *I could keep going back, and ask for constructions of rationals, of integers, and so on all the way to ZFC.*
      No. This is an invalid analogy. To begin with, 99% of the people who deny 0.p9 = 1 have no actual grievances with ZFC itself, or the majority of consequences from it that build up to real analysis. Also, real analysis is itself a formal theory, and is, as such, a relatively self-contained formal system that is only a small fraction of ZFC. You are not actually required to trace a mathematical theory back to ZFC. You always can (though perhaps not elegantly so), but you are not required to. If you are working with a theory of toplogical spaces, you are not required to start with ZFC. You do, however, start with the axioms of topological spaces. Not starting there would be wrong. If you are doing real analysis, then you have to start with the axioms of real analysis. In that regard, you can simply assert that (R, 0, 1, +, ·) is a field, and that (R,

    • @isodoubIet
      @isodoubIet 2 года назад +3

      @@angelmendez-rivera351 "es, and the problem is that these assumptions are being made, and they should not. "
      What is "should"? "Should" is always a conditional particle: you should do X _if_ you want Y, even if the Y is often implicit. The Y in question here are the specific pedagogic goals of the exposition. For some audiences this is perfectly ok to do algebra tricks -- it's a corollary of the construction of real numbers, taken for granted. For an audience of math undergraduates it's probably not ok and you want a more detailed proof. It all depends greatly on who your audience is and what you're trying to convince them of.
      " To anyone that is not convinced of 0.p9 = 1, the statement that the real numbers are a well-defined system, and that there is a well-defined positional representation for real numbers, is far from obvious,"
      Who? Every person I've seen deny such proofs was completely satisfied that the positional representation is valid, but incorrectly assumed that it is also unique (in the sense that there's a one-to-one correspondence between representations and real numbers. That is the key error, and the algebraic proofs really do show that this is an error.
      "A person can feasibly believe that one construction is valid, without believing that the others are valid."
      What's the overlap between the set of people who believe 0.999... != 1 and the set of people who are even aware that you need to construct the real numbers, let alone in multiple (equivalent) ways? Vanishingly small, surely. But if you do encounter such a person you can go ahead and prove what you need to convince them -- it still doesn't make the algebraic proofs wrong.
      "Again, those who believe 0.p9 = 1 is false would disagree with the premise that any such sequence of digits corresponds to a real number. "
      I've never seen anyone dispute that a sequence of digits is a valid way to specify a real number. I've seen people disagree with the premise that a given sequence of digits doesn't _uniquely_ specify a real number.
      "No, it really is not that simple"
      It really is that simple. Every proof has a set of facts that are assumed known and facts that are to be demonstrated. That's the fundamental structure of logical proofs dating back to classical antiquity.
      "And this is wrong. The problem is that you are conflating "accepting an assumption" with "understanding an assumption." Most people accept the real numbers are real, but their beliefs on what the real numbers are just happen to be completely wrong, "
      Those incorrect beliefs will remain unchanged upon watching a proof like this, because most people aren't going to sit down and learn real analysis, but such people _can_ gain a better understanding of their structure from working through an algebraic proof which is vastly more accessible than a rigorous construction of the real numbers.
      "But the difference is, he is not required to do that at all for his proof to work"
      But by your own standards, someone who's not convinced that real number constructions make sense would also not be convinced that this construction has any relevance to real numbers.
      " To begin with, 99% of the people who deny 0.p9 = 1 have no actual grievances with ZFC itself, "
      99.99999% of those people have no idea what ZFC is.
      "In that regard, you can simply assert that (R, 0, 1, +, ·) is a field, "
      A field is a set.
      " You are not actually required to trace a mathematical theory back to ZFC. You always can (though perhaps not elegantly so), but you are not required to. "
      Thanks, that's what I'm saying.

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 года назад +1

      @@isodoubIet *What is "should"? "Should" is always a conditional particle: you should do X if you want Y, even if the Y is often implicit.*
      I am no aware of any such grammar rule of the English language.
      *The Y in question here are the specific pedagogic goals of the exposition.*
      Okay, so why you are being purposefully obtuse and needlessly pedantic? This is dishonest. Frankly, I should already stop taking you seriously and dismiss the rest of your response, just from this alone, but out of respect for the fact that you seemingly put some effort into the reply, I am going to move on.
      *For some audiences this is perfectly ok to do algebra tricks -- it's a corollary of the construction of real numbers, taken for granted.*
      I already explained what the problem with this. There is no point in you repeating it now, and there is no point in me repeating the objection.
      *For an audience of math undergraduates it's probably not ok and you want a more detailed proof. It all depends greatly on who your audience is and what you're trying to convince them of.*
      I know as much, and I made as much clear within my response. Who are you trying to convince: me, or yourself? Because you cannot convince me of what I am already convinced of.
      *Who? Every person I've seen deny such proofs was completely satisfied that the positional representation is valid, but incorrectly assumed that it is also unique...*
      Okay. I have met people that deny the validity of the positional representation. Many of them. More than I can feasibly keep track of in my head. To be clear, I have not met anyone that denies the validity of representations such 53 or 5.456 or 0.010001011 in binary. However, non-terminating strings as part of this representation system are denied by many people. If you do not believe I have met such people, then it will have been a waste of my time to have this interaction.
      *...in the sense that there's a one-to-one correspondence between representations and real numbers. That is the key error, and the algebraic proofs really do show that this is an error.*
      The algebraic proofs do not show that error. The algebraic proofs show that if the assumptions being made about the positional representation system are true (and they do not show that they are true), then the assumption about the uniqueness of representation is false. However, you have to assume parts of the conclusion of the proof as premises in the proof, hidden as part of the definition of 0.p9 to begin with.
      *What's the overlap between the set of people who believe 0.999... != 1 and the set of people who are even aware that you need to construct the real numbers, let alone in multiple (equivalent) ways? Vanishingly small, surely.*
      This is a baseless assertion, and it is also irrelevant.
      *But if you do encounter such a person you can go ahead and prove what you need to convince them -- it still doesn't make the algebraic proofs wrong.*
      This is a non sequitur, since the existence or non-existence of such people is not relevant to the correctness of the proofs, hence my previous sentence.
      *I've never seen anyone dispute that a sequence of digits is a valid way to specify a real number.*
      Such people exist in the comments section to the very video we are commenting to. These people are by no means rare. They exist in just about every neighbourhood of the Internet you can think of.
      *I've seen people disagree with the premise that a given sequence of digits doesn't uniquely specify a real number.*
      So have I, but they are not the only family of deniers in the subject matter, and the fact that you think it is means you are making a bold implicit assertion.
      *It really is that simple. Every proof has a set of facts that are assumed known and facts that are to be demonstrated. That's the fundamental structure of logical proofs dating back to classical antiquity.*
      No, this is false. There are no facts you assume known in a theorem. You either prove the facts in question, or the parties being addressed confirm, of their own volition, that they know the facts. You can cite well-known results, you can cite a theorem and cite a source where a proof is discussed, if you are addressing a non-replying audience. This is what people have been doing for centuries. Most laypeople will refer to a source that discusses the fact in question, rather than proving it themselves, because they lack the sufficient knowledge to prove it. If the fact is extremely elementary, then perhaps you skip that, but none of the facts in these algebraic proofs are so elementary to warrant that. In fact, they are so non-elementary, that many people get them wrong when explaining them, or outright deny them. The kind of "skipping" you are talking about is not the grade school skipping where students avoid proving 2 + 2 = 4 whenever they solve a problem because virtually everyone knows that 2 + 2 = 4 is true anyway. What you are doing is taking the very point of contention in question, and saying "don't worry about that, look at this other thing instead," and hoping that the distraction is sufficient to create the false illusion of understanding, and purely by a matter of statistical fluctuation, it just so happens that a number of people fall for it and believe themselves convinced that they understand why 0.p9 = 1, without actually understanding, because the point of contention that lead to the discussion to start with was shoved under the rug. It is misinformative.
      *Those incorrect beliefs will remain unchanged upon watching a proof like this, because most people aren't going to sit down and learn real analysis,...*
      This is false. I know it is false, because I am a tutor, and I have experienced first-hand that this is false. Your assumption that people have to learn real analysis to be to intuitively grasp and conceptually understand what is going on with a proof such as this is one is a completely erroneous assumption that I debunk on pretty much weekly basis whenever I give tutoring sessions. The algebraic proofs do require a lower effort, yes, but at the cost of being effectively circular. However, how much the effort is lowered is not so much to warrant the kind of defense you are seeking for.
      *...but such people can gain a better understanding of their structure from working through an algebraic proof which is vastly more accessible than a rigorous construction of the real numbers.*
      These algebraic proofs do absolutely nothing to clarify the algebraic structure of the real numbers, nor do they create any understanding as to how non-terminating decimal representations work, which is what you should be aiming to do in this context.
      *But by your own standards, someone who's not convinced that real number constructions make sense would also not be convinced that this construction has any relevance to real numbers.*
      Nowhere in the fragment that you quoted and responded to did I say that such people would consider the construction relevant. In fact, I explicitly stated that the construction is not relevant to mCoding's proof, and only relevant to motivating some of the definitions, which is, strictly speaking, not a requirement. I have no idea why you misrepresent my argument.
      *99.99999% of those people have no idea what ZFC is.*
      I know as much.
      *A field is a set.*
      No, it is not. A field is an algebraic structure. An algebraic structure can be _interpreted,_ in the formal sense of the word, as a set _with additional structure,_ but it need not be interpreted as such. In fact, the surreal numbers form an ordered field, but they do not form a set. They form a proper class. As far as category theory is concerned, the objects of the category of fields not even be classes at all. We only tend to interpret them as classes, because the mathematical consensus is to do so, so as to take advantage of set theoretic ideas. This is beneficial, because there exists a forgetful functor from the category of fields to the category of sets. The theory of algebraic structures is relatively self-contained and it need not rely on set theory at all. In fact, set theory in first-order logic is, in the informal sense, "modelled" after the theory of lattices, which is, ironically, the theory that we use to study deduction systems, which include the various families of logic, classical and otherwise, you and I are familiar with.
      *Thanks, that's what I'm saying.*
      No, that is not in fact what you are saying, because your statement is about proofs, not about tracing back theories to other, more fundamental theories. In fact, your claim is that any proof whose conclusion follows from premises that at least one person on Earth knows how to prove should be presented as correct, even when those proofs completely fail to address skepticism of the conclusion being denied to begin with, and even when the premises themselves are what the deniers are challenging. This is very, very different from what you are asking mCoding to do, which is, inexplicably, to trace every theorem back to the Zermelo-Fraenkel axioms and the axiom of choice.

    • @isodoubIet
      @isodoubIet 2 года назад +3

      @@angelmendez-rivera351 "Okay, so why you are being purposefully obtuse and needlessly pedantic? This is dishonest"
      You can disagree with me all you like, as you can tell by how I responded to you politely before, but that accusation is both rude and uncalled for. I draw a hard line at spurious imputations of malice. If you wish to have an actual conversation, let me know, but otherwise, I have stuff to do.

  • @amirh6712
    @amirh6712 3 года назад +35

    Man i went to shower and you had 14.7K subs
    I'm back and now you have 15K subs
    Your channel is grown really fast
    Keep up the good work!

    • @mCoding
      @mCoding  3 года назад +5

      I appreciate it!

    • @programaths
      @programaths 3 года назад +1

      You have to be very clean now!

  • @Norhther
    @Norhther 3 года назад +16

    6:26 I think this would lead to confusion. Here you have to be familiar with limits. This is claiming that lim x->oo x_n is EXACTLY one

    • @mCoding
      @mCoding  3 года назад +11

      Yes, the number the sequence APPROACHES is EXACTLY one. This is not saying any term in the sequence is ever equal to 1, just that you can get as close to 1 as you want (epsilon > 0) by going far enough out.

    • @gideonmaxmerling204
      @gideonmaxmerling204 3 года назад +1

      well that limit is exactly one.

    • @jenaf372
      @jenaf372 3 года назад

      Wait until someone points out that the value of the limit towards some "number", (evrtn if it exists from "all" "directions") need not be the value "at" that "number".
      Thats one reason why I kinda hate the shorthand of sums wich leave out the limit.

    • @gideonmaxmerling204
      @gideonmaxmerling204 3 года назад

      @@jenaf372 what shorthand of sums?

    • @jenaf372
      @jenaf372 3 года назад

      @@gideonmaxmerling204 if you write something like "sum from n=1 to infinity of f (n)".
      That an infinite sum only makes sense in rare cases, and most of the times its "the limit of c approaching infinity of (sum of n=1 to c of f (n))".
      But the first is most often used as a shorthand for the second.
      While the limit often does exist, the first one, if read litteraly, rsrely makes sense.

  • @pectenmaximus231
    @pectenmaximus231 2 года назад +1

    Tl;dr:
    Repeating decimals are notation signalling that infinity is being invoked
    Algebra is not strictly on its own equipped for infinity
    Therefore the real proof has to come from the maths of infinity, eg calculus
    Post scriptum: infinity in this context is something constructed via series, and also we require specialised definitions for how to approach such objects (no pun intended)
    Infinity also has different conceptions and uses in maths, and calculus is also a matter of convention (as it has changed over time)

  • @chrisjduplessis
    @chrisjduplessis 3 года назад +7

    "In mathematical terms, we say that it is the limit of the sequence but it's ok if you're not familiar with limits."
    *Proceeds to use the definition of the limit for the proof.*
    Those were still good points in the vid.
    Maybe a rigorous proof that more people would understand would be to use the geometric series formula (the theorem itself describes the conditions for convergence so we are not even assuming existence either).

    • @mCoding
      @mCoding  3 года назад +1

      I think this is a good idea, even though it uses calculus that people may not be familiar with, it is at least very explicit about it. That way, anyone interested in filling in the details can lookup the proof of the geometric series formula

  • @yogeshchandrasekharuni4054
    @yogeshchandrasekharuni4054 3 года назад +8

    At 2:15 you said there are number systems where things like ..99999.0 = -1 are true. Can I get an example number system for that?

    • @mCoding
      @mCoding  3 года назад +6

      The simplest example is that this is true in Z/10^nZ for any integer n>0, i.e. integers modulo some power of 10. Consider Z/10Z. Then 10^k = 0 for all k >0, so ....9 = 9 = -1 in this number system, the infinite sum collapses to a finite sum of just one number and then all zeros.

    • @mCoding
      @mCoding  3 года назад +6

      (This is also why your computer represents -1 as 11111111111111111111... in binary, it's the same example but mod 2^32 or 2^64 or whatever your int size is, because 1 in mod 2 is like 9 in mod 10.)

    • @Maurycy5
      @Maurycy5 3 года назад +1

      @@mCoding How about number systems in which 0.(9) does not equal 1? Can we get an example of that?

    • @rsa5991
      @rsa5991 3 года назад +1

      There is actually a way to make that true even in R. You just need to redefine what "convergent series" means. There is some definitions of convergence, according to which 9 + 9*10 + 9*100 + 9*1000 + ..... = -1.
      Also, fun math exercise: you can try to add 1 to ..99999.0, and all digits will roll over, giving you ....00000.0 ! It works with other numbers too: adding them to ..99999.0 will give you a number one less. So, in some way, ..99999.0 does indeed "act" like -1.

    • @rsa5991
      @rsa5991 3 года назад

      @@Maurycy5 The obvious one is Z (integers). In Z, number 1/10 doesn't exist, therefore 0.(9) (which depends on 1/10) doesn't exist either.
      If we want to make 0.(9) exist, but not equal 1, that is much harder. We need a system in which 9+1 =/= 10. And we still need the whole thing to converge. Besides the trivial case, where we just butcher all arithmetic (by replacing 9 with 4, for example), I cannot think of such system.

  • @SafetySkull
    @SafetySkull 3 года назад +20

    What's wrong with using assumptions people have that they haven't necessarily proven in a pop math youtube video? You used addition in your proof even though the general public hasn't seen a rigorous definition but has a general understanding of how it works.

    • @thekenanski8789
      @thekenanski8789 3 года назад +8

      I guess the big difference is that, in general, addition behaves exactly as our intuition says it should, so not much is gained by studying formal definitions of it. On the other hand, people’s intuition about infinite series (including repeating decimals) is often totally wrong precisely because they don’t understand issues of convergence and the fact that, in general, one cannot simply manipulate series as if they were numbers (eg if a series only converges conditionally then addition in that series is not commutative because rearranging the terms allows you to change the value the series converges to).
      The other issue is that the algebraic proofs just over complicate things-they don’t show why 0.999 repeating is equal to 1, which is a problem, since all of the confusion about why that’s true stems from people not understanding what we mean when we say “repeating” (ie that “repeating” means taking a limit).

    • @SafetySkull
      @SafetySkull 3 года назад +2

      @@thekenanski8789 infinite decimals behave intuitively though. Every sequence of integers between 0 and 9 times 10^-n converges when summed

    • @phremic5867
      @phremic5867 3 года назад +5

      I agree with you saying that these proofs are not "wrong". They just assume that the series/sequence 0.999... converges, which is true in the real numbers (the real numbers can actually even be defined as the limits of rational Cauchy sequences), therefore simple multiplication by any number is possible.
      However I think this video is good for understanding what the question "Is 0.999...=1 ?" even means. It is not obvious for many people that "0.999.." is just a short way to describe the limit of the series "Sum(9*10^-n)", so they just see it as any number and believe that - since every element of the series (0.9, 0.99, 0.999,...) is smaller than 1 - it must be smaller than 1 too. But you usually only learn at university how to work with sequences and series and because of that most people don't really understand the concept of it.

    • @Lucky10279
      @Lucky10279 3 года назад +1

      @@thekenanski8789 That's a valid point. I think the main reason it's so counterintuitive that 0.999...=1 is because most people don't understand that decimals are just shorthand for addition of fractions and infinite decimal expansions are shorthand for sums of infinitely many fractions, which, as you said, are typically defined as limits of partial sums. Once I realized that that 0.999... is just shorthand for the limit of the sequence (0.9, 0.99, 0.999,...) all my confusion disappeared.

  • @jamirimaj6880
    @jamirimaj6880 2 года назад +2

    4:42 it is because dividing 1 by 3 using elementary division method will give you 0.33333..., thus the notation of 3 with a bar above it is created.

    • @Chris-5318
      @Chris-5318 2 года назад

      You cannot get 0.333... using the long division algorithm. You can only generate the start of the sequence 0.3, 0.33, 0.333, ...

    • @bgdgdgdf4488
      @bgdgdgdf4488 Год назад

      ​@@Chris-5318yes but the pattern is repeating

    • @Chris-5318
      @Chris-5318 Год назад

      @@bgdgdgdf4488 I know it is. OK, I'll concede, at least for the purposes of argument, that you can prove that it generates 0.333... if you use an inductive argument. Now, why do you think that that proves that 0.333... = 1/3. Hint: you haven't considered what is happening to the size of "remainder" after each step, and you need to prove that it is approaching 0. That involves a limit based argument, and that is a large part of what James is saying is needed. But really, even that is not good enough - you have to introduce all of what James is saying is needed.
      The formal definition is: the sum of a series is the limit of the sequence of its partial sums (if the limit exists).
      I'll apply that to 0.999... because it is slightly easier to read. The sequence of partial sums of 0.999... is 0.9, 0.99, 0.999, ..., whose n th term is 0.999...9 (n 9s) = 1 - 1/10^n. So:
      [the sum of] 0.999... := lim n->oo 0.999...9 (n 9s) = lim n->oo 1 - 1/10^n = 1.
      Back to the long division algorithm for 1/3. We have:
      1/3 = 0.3 + 1/30 = 0.33 + 1/300 = 0.333 + 1/3000 = ... = 0.333...3 (n 3s) + 1/(3 * 10^n) = ...
      I'm calling the second addend, at each step, the remainder. That does not get to the limit case of 0.333... + 0, and there is no [pure] inductive argument that can prove that the limit of the sequence of remainder terms is 0, even though it obvious that is approaching 0. Whatever, the fact that the remainder approaches 0, intrinsically involves a limit argument.

  • @baptiste2b31
    @baptiste2b31 3 года назад +8

    Actually, even if the series 9999...,0 diverges, you can rigorously give it acreal value and if you do so, that value has to be -1 (if we want to conserve "good" properties from the sum operator). We can do that using "super summation" (the idea being to create a linear opperator S that is a continuation from the usual "Sum" that deals with converging series).
    One can convices themself that it has some sens with the idea of "overflow" in computer science : working base 2, with 4 bits as an exemple you get that 1111+1=0 therefore 1111=-1, that's an overflow. That idea generalised to the base 10 and with an infinite number of bits gives that 9999...,0 = -1.
    You say (2:08) that we "shouldn't belive it", that "there are number systems where this kind of things is true" but "the real numbers is not one of them".
    The "number systems" you think of may have been the modular ones.
    However, I disagree with the last statment : it can work with real numbers because 0.999... (=1), can be defined as the image of a (converging) sequence by "the" sum operator. Then, in the same way, we can, define 999...,0(=-1) as "the image of a (non-converging) sequence by [a continuation of] "the" sum operator. Defined that way, it is equal to -1 so it's a real number.
    I know it doesn't change the message form your video that deals with the idea that "a statment being true or false is irelevant to the validity of it's proof". I just wanted to point it out.
    Anyway, it was a good video.

  • @taufiqutomo
    @taufiqutomo 3 года назад +20

    The conventional proof is not rigorous enough to be mathematically valid, but is good enough to shut people up and move on to more important topics.

    • @cinemarat1834
      @cinemarat1834 3 года назад +1

      I'd argue that teaching people about proofs IS the important topic especially in the long run. after all maths is fundamentally about PROOFS. I'm fine with pseudo-proofs but people should always give disclaimers when they offer pseudo-proofs cuz It can create a false sense of what proofs are really like.

    • @enzzz
      @enzzz 3 года назад +2

      My main problem is I don't see how could an infinite number possibly exist and in which case 0.(9) can't possibly equal 1, because something that does not exist can't equal something other that does exist. And I'd say 0.(3) does not equal 1/3, it's a close approximation.
      For how long we can go and check along with infinite number whether it equals 1, it never does, so how can we from this conclude that it must equal to 1?
      I don't see how any of those proofs could convince or satisfy anyone, except unless you try to "hack" it to make sense and then you tell yourself, that yes, this does make sense.

    • @prumchhangsreng979
      @prumchhangsreng979 3 года назад

      @@cinemarat1834 nah. Thats becuz u think people learn math just for the sake of math. What if some people only wanna learn math to solve simple issue, or to become biologist or physicist. These kind of stuff should leave out for those that are really interest in math. Despite how much thing we cut off from math, there are so many people alr not interest in math, can u imagine how many ppl would ignore math completely in highschool if u were to go rigorously from the start? Leaving off some detail early on isnt necessary a bad thing.... it's the same as telling first grader that zero is the smallest number and slowly reveal to them that it is wrong later on.

    • @scubasteve6175
      @scubasteve6175 3 года назад +1

      @@enzzz math breaks when it comes to infinity. I agree with you. I feel like it's a little desperate to assume we can make logical sense of it when we cannot comprehend unending sequences whatsoever. Not to mention mathematics aren't necessarily structured to support equations involving infinite numbers

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 года назад

      @@scubasteve6175 You are so wrong it is not even funny. Firstly, we DO understand unending sequences. Set theory is all about that. Secondly, mathematics DOES have the structure to handle equations with infinite quantities. John Conway's surreal numbers are proof of this. Nonstandard analysis is another example of this.

  • @astroknight5
    @astroknight5 3 года назад +5

    1:33 it's not very similar though.
    0.9999... is finite
    ...9999.0 is infinite
    You can't do simple algebra with infinity

    • @branthebrave
      @branthebrave 3 года назад

      That's the point

    • @user-lb1ib8rz4h
      @user-lb1ib8rz4h 2 года назад +3

      okay, prove that the first number is finite.

    • @astroknight5
      @astroknight5 2 года назад

      ​@@user-lb1ib8rz4h using the representation at 5:25, we can represent 0.9999... as a limit of a series. Apply Cauchy ratio test (0.1 < 1). The series is convergent ie. the limit exists and is finite. You can try the same for the second number but the ratio is 10 > 1, so the series is divergent.

    • @user-lb1ib8rz4h
      @user-lb1ib8rz4h 2 года назад +2

      @@astroknight5 good job! you understood the video which literally said that 0.99... is a limit and you can't just use algebra to assume that it's finite. ergo the algebraic proof is begging the question here.

  • @LaurenceRietdijk
    @LaurenceRietdijk 2 года назад +11

    "The other proofs are not good because they require the reader to understand something that is understood via calculus, which they may not know. So here is my proof, which is 10x more complicated so that the reader is even less likely to understand." - well done sir...

    • @mCoding
      @mCoding  2 года назад +2

      Although a correct proof may be more difficult to understand than an incorrect proof, you must keep in mind that the simple incorrect proof can be so simple precisely because it is incorrect.

    • @LaurenceRietdijk
      @LaurenceRietdijk 2 года назад +15

      @@mCoding But by "incorrect" you actually mean "incomplete" in that one statement does not satisfactorily follow from the previous statement without some assumed knowledge. But, by that standard, pretty much all proofs are insufficient, since they tend to assume basic arithmatical computations (such as 1 + 1 = 2) without revisiting 'Russell's Principia Mathematica'. And there is a reason for that. A kindergarten student is taught by rote that 1 + 1 = 2 and indeed, understands how to conduct the calculation, but is not capable of understanding 'Russell's Principia Mathematica'. Similarly, a highschool student knows that multiplying something by 10 means shifting the decimal point, regardless of whether they can understand calculus. Assuming that x10 means moving the decimal point is no more "incorrect" than assuming that 1+ 1 = 2

    • @ethandavis7310
      @ethandavis7310 2 года назад +5

      @@LaurenceRietdijk Yeah idk why this video resonated so much with people who actually know math. It's just moving the goalpoast in a non-constructive and clickbait manner.

    • @pedroteran5885
      @pedroteran5885 Год назад

      ​@@LaurenceRietdijkIt's hard to believe that the point the video is making can be misconstrued in such an absurd way.

    • @Davs211
      @Davs211 Год назад +1

      @@LaurenceRietdijk 100% agree. I don't think anyone doubts the covergence of such series even if not proved (which is trivial tbh).

  • @SantoshKumar-ku1mq
    @SantoshKumar-ku1mq 3 года назад +88

    The effort you put into thinking there will be people in the comments saying, 'now the title is invalid' and including a disclaimer in the description, really shows how much you think of a viewer. Thank you for the content, keep up the good work.

    • @mCoding
      @mCoding  3 года назад +8

      Thanks for that

    • @roberthorne9597
      @roberthorne9597 2 года назад +5

      But the thing is, even at university or in a textbook, we don't need to do this type of proof, the rigousness is only there to allow for more rigorous definitions elsewhere. And assuming 0.99 is a member of the reals is hardly a stretch.
      It is a clickbaity title, and worse it makes maths seem like it needs to be complex to be really useful. It's classic Hilbert (he was a formalist right?)

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 года назад

      @@roberthorne9597 No. The title and the video do not imply maths have to be complex to be useful (though maths are indeed complex, and Gödel did prove that maths have to be complex fo be useful).

    • @roberthorne9597
      @roberthorne9597 2 года назад +1

      @@angelmendez-rivera351 Mmm, it claims every proof you've seen is wrong.. which misses the point of actually, what Godel hinted at... There is not absolute set of axioms to go with... which implies that proof rigorousness is controlled by the math we are trying to solve and link. Point is that these proofs are perfectly fine.
      Also, not to nitpick too much, but on: "though maths are indeed complex, and Gödel did prove that maths have to be complex to be useful"... I am not super familiar with his work, but I do think the point was to undermine that maths can be complete and self consistent and provable at the same time, nothing about complexity of the proving system chosen?

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 года назад

      @@roberthorne9597 *it claims every proof you've seen is wrong...*
      Which is correct. The proofs are wrong, simply for the reason that they fail to address that, in order to prove a statement about 0.(9) is true, you must use the definition of 0.(9) in the proof, and those so-called "proofs" fail to do this, hence they are not actually correct, even though they do hint at true statements.
      *...which misses the point of actually, what Godel hinted at... There is not absolute set of axioms to go with...*
      Gödel never hinted at that... because the idea of axiomatizing mathematics with several alternatives to do so long preceded Gödel's time. Before Gödel even published anything, mathematicians were already well-aware that different axiomatizations of mathematics with equal consistency existed. Even so, the fact that there is no absolute set of axioms is irrelevant to what the video is claiming.
      *...which implies that proof rigorousness is controlled by the math we are trying to solve and link. Point is that these proofs are perfectly fine.*
      No, they are not. There is more to proof systems than just axioms. Also, I really would love for you to expand on what you meant by "proof rigorousness is controlled by the math we are trying to solve and link," since, as far as I am concerned, this is is an incoherent sentence.
      *Also, not to nitpick too much, but on: "though maths are indeed complex, and Gödel did prove that maths have to be complex to be useful"... I am not super familiar with his work, but I do think the point was to undermine that maths can be complete and self consistent and provable at the same time, nothing about complexity of the proving system chosen?*
      If you are aware that you are not particularly familiar with Gödel's work, then you have no business trying to nitpick my alluding to his work as if you even understand what you are talking about, let alone do so with the confidence that you are doing it. You may think I am being rude in telling you that, but I think your attempt at nitpicking me while knowing you are not qualified to do so is rude to begin with, and arrogant. Hopefully, you understand that if this is how you wish to approach the conversation, then this will become unproductive extremely soon.
      Gödel's Incompleteness Theorem states the following: for any sufficiently expressive proof system of arithmetic (and this can be generalized to broader families of proof systems), if the proof system is consistent, then it is incomplete. In the theorem, this idea that the proof system must be sufficiently expressive is important. This is because there are proof systems that are consistent and complete, such as Baby Arithmetic, but this proof system is not sufficiently expressive. In particular, in Baby Arithmetic, there are no statements with universal quantifiers. You can never speak about an infinite subset of the natural numbers. Thus, for discussing arithmetic and number theory, this proof system is extremely limited, almost to the extent that it is worthless and useless for its purpose. This is because most statements about arithmetic you would wish to make and study do not exist in the language of Baby Arithmetic, hence why it is not sufficiently expressive. The idea of sufficient expressiveness for a proof system literally codifies the fact that the system is capable of expressing enough of the ideas we care about for it to be useful. With sufficient expressiveness comes complexity, and comes Gödel's Incompleteness Theorem. You could almost rephrase Gödel's Incompleteness Theorem as "For any useful formal proof system of number theory, if the system is consistent, then it is incomplete." What does it mean for a proof system to be consistent? It means that the proof system does not prove a contradiction. What does it mean for a proof system to be complete? It means that for every formula φ in the proof system, the proof system proves φ, or it proves -φ (if it is also consistent, then it never proves both).

  • @cool_scatter
    @cool_scatter 3 года назад +11

    I don't know, is it really fair to say a proof is "wrong" just because it doesn't explicitly justify all the steps it takes? By that logic, your proof is wrong too, and so is every one that doesn't reproduce all of Principia. It seems especially odd to criticize it when those steps taken - sure, they don't work for the same reasons, but they do work the same as terminating decimals. Plus, like you said, if you define 0.999... as you should be, it's very obvious that it equals 1. "Proving" that's true is about as useful as proving 5/5 = 1. The proofs of 0.999... = 1 are aimed at people who don't already understand that intuitively. It's pop math. Which is fine, but I think if you're writing pop math, understandability to the layman should be prioritized over rigid correctness.

    • @EvanOfTheDarkness
      @EvanOfTheDarkness 3 года назад +2

      No its not fair. The only thing this video does is highlight his stupidity.

    • @pedrofuster9161
      @pedrofuster9161 3 года назад

      @@EvanOfTheDarkness This video was worth watching for me though, so it certainly can do a little bit more for at least some people ⚽️🏀🎖🏆🎏🎎🎃

  • @bluerojo
    @bluerojo 3 года назад +14

    engineers: 0.99999 = 0

  • @Errenium
    @Errenium 2 года назад +2

    eh, it seems to me a bit of a reach to say both those arguments are wrong. if there's an issue, it's labelling them as proofs and not any particular step taken in them. as well, i think faulting them for the assumptions of existence and the unstated nature of the starting definitions ignores that they are done in a particular context where these assumptions and definitions are natural and pointing them out risks seeming patronizing or irrelevant.
    i do also think that arguing based on limits would be a pedagogical mistake since a recurring theme among the target audience is misunderstanding of infinity. they may deny such a concept outright, or they may think that there could still be a "last" digit even if they nominally accept the infinite nature of the number's decimal representation.

    • @Chris-5318
      @Chris-5318 2 года назад

      Saying "wrong" was wrong.

  • @cobbcoding
    @cobbcoding 3 года назад +46

    thanks for clearing this up, was keeping me awake at night thinking about this.

    • @mCoding
      @mCoding  3 года назад +4

      Any time!

    • @MichaelBrown-gt4qi
      @MichaelBrown-gt4qi 3 года назад +1

      This was literally bothering me, too. But now that I've seen it, I feel like I should have known all this time.

  • @_tsu_
    @_tsu_ 2 года назад +8

    idk whatever you said is right but this video feels like a nothing burger

  • @CjqNslXUcM
    @CjqNslXUcM 2 года назад +8

    In balanced number systems, you can end up with 1/2 being one of two infinitely repeating fractional numbers, where none is 0 repeating. It gave me a new perspective on .000... vs .999...

    • @irrelevant_noob
      @irrelevant_noob 2 года назад +1

      Wait, what?! o.O
      But first, before i ask for any further details on those "two infinitely repeating fractional numbers"... let me ask: what's a "balanced" number system? 🤔

    • @CjqNslXUcM
      @CjqNslXUcM 2 года назад +3

      @@irrelevant_noob A number system where there are the same number of negative digits as positive.
      Here's an example: Z, T, 0, 1, 2 are the digits, where Z and T represent -2 and -1 respectively.
      The second column is multiplied by 5, as there are 5 digits.
      decimal : 5-based balanced
      1 : 1
      2 : 2
      3 : 1Z
      4 : 1T
      5 : 10
      6 : 11
      7 : 12
      8 : 2Z
      9 : 2T
      10 : 20
      Intuitively, it jumps to the nearest 5s column as you cross the midway point, just like telling the time using phrases like "quarter to 10". This eliminates the need to round numbers entirely, as they are always perfectly rounded, no matter how many columns you change to 0.
      Here are some fractional numbers:
      0.4 : 0.2
      0.6 : 1.Z
      0.48 : 0.22
      0.52 : 1.ZZ
      0.496 : 0.222
      0.504 : 1.ZZZ
      What happens at exactly decimal 0.5? Either 1.ZZZ... or 0.222...

    • @irrelevant_noob
      @irrelevant_noob 2 года назад

      @@CjqNslXUcM oh, that's so interesting. Also, the trick is that there is *_no_* 0-based representation for "1/2", it's like if the decimal 1/3 would have 0.3333... and 1.SSSSS... as representations (where S = -6 in a poorly extrapolated pseudo-base-10). Thanks for explaining these introductory notions.

    • @CjqNslXUcM
      @CjqNslXUcM 2 года назад +1

      @@irrelevant_noob I'm not sure you understood. Your representations are correct, but a system starting with -6 and ending with 3 would be tedious and pointless. If the base is not an odd number, you don't get the useful properties. Try converting FFF (F representing -5) to a positive number in your system. Try rounding 1S to the nearest 10.

    • @Muhahahahaz
      @Muhahahahaz Год назад +1

      Interesting… This really highlights the fact that the two representations arise due to taking a limit from above vs below (a non-increasing series that starts off greater than or equal to the number in question, vs the opposite)
      In the traditional case, one of those series just happens to be exactly equal to the desired real number the whole time, while adding a bunch of inconsequential zero terms
      But with a balanced number system, one series is strictly above the limit (with the other below), and they both meet at the same number in the middle

  • @alexismiller2349
    @alexismiller2349 2 года назад +1

    I suppose that algebraic proofs are "doomed" to be unsatisfying, since the claim is one about analysis and so you can never really make an argument that doesn't rely on calculus.
    It's like how the d'Alembert-Gauss theorem doesn't have any purely algebraic proofs (despite attempts) and they all use calculus to some degree, probably since the real number (and therefore the complex numbers) are constructed analytically...

  • @johnnygodoy8329
    @johnnygodoy8329 3 года назад +14

    Worth noting that every short proof of a sufficiently complicated statement will have a few implicit assumptions to simplify. A few follow up questions one might get after showing this proof is:
    * What does it mean for a sequence to approach a number? Doesn't that sequence also 'approach' 2 (in the sense that 2-x_n is a decreasing sequence)? This, and all related questions get cleared up by looking at the formal definition of limit
    * What if the sequence has more than one limit? This is a really good question, because if there is more than one limit it means you can't say the limit 'equals' any particular value, and the definition of .999.. wouldn't make sense. This gets cleared up by proving that in the real number's usual topology, the limit is unique.

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 года назад

      *This, and all related questions get cleared up by looking at the formal definition of limit.*
      You do know he used the formal definition in the video, right?

  • @xuepingsong5329
    @xuepingsong5329 2 года назад +6

    Yea this is pretty much exactly why I've been confused in high school, when I ask why, I just get an answer of "of course this works", but in a proof you are not supposed to assume anything unless it's been explained before. I guess it's a common theme of learning math, for me it's probably best to just memorise the steps of the proof before really understanding it to not be stuck as I've wasted so much time on proofs.

    • @HappyBeezerStudios
      @HappyBeezerStudios Год назад +1

      @falastin2658And it feels so good to actually understanding instead of memorizing.
      I remember when I stumbled over Thales' theorem. I noticed that there seems to be a pattern with right-angled triangles. So I looked deeper into it (which basically means drawing a ton over triangles on a sheet of paper) The moment it clicked was a true eureka moment.
      When you understand the why and not just the how it feels for a moment like your mind expanded.

    • @bernardovidigal1306
      @bernardovidigal1306 9 месяцев назад

      youre right, I hate when teachers expect me to belive in what they say because they say so even without proof

  • @williambaillargeon5266
    @williambaillargeon5266 3 года назад +10

    I’m not sure about this, the way I see it:
    The value of 9 periodic (_9) should be 9.99999....
    Not 9999999...9.0
    The periodic symbol is used for representing right side continuity of the number, not left side; it is used to show a number with a better accuracy, not a number that grows forever.
    Hence
    If x = _9
    10x = 99.999...
    By subtracting
    9x = 90
    x = 10
    Proving that _9 = 10
    (9.999... = 10)
    The original proof is still valid.

    • @WukongTheMonkeyKing
      @WukongTheMonkeyKing 3 года назад

      The video was like nails on a chalk board because of this.
      Add in the pedantic nature of the arguments and the result was aggravating.

    • @Excalibur0126
      @Excalibur0126 3 года назад

      Wouldn’t the 9s being added be before the decimal point either way?

    • @vertihippo1274
      @vertihippo1274 3 года назад

      @@WukongTheMonkeyKing Small details matter in mathematics, so nitpicking is necessary for accuracy. The only issue is the slight logic errors present that lead to a faulty result.

    • @globulin
      @globulin 2 года назад +2

      He's not using 9 periodic as an extension of the repeated decimal concept, he's introducing a new concept where the repetition continues to the left to demonstrate that we need to be careful when new concepts are introduced.

  • @Kram1032
    @Kram1032 3 года назад +6

    1:47 "there are number systems where this is true"
    that is, as far as I know, in fact the logic of p-adic numbers.
    Of course, there are still two problems with that:
    - as you point out, real numbers are not p-adic numbers so you are now talking about an entirely different number system and the result won't apply to reals
    - more importantly, p-adic numbers are only properly definable for a p that is prime. 10-adic numbers aren't really a thing.
    I'm not sure if there is like a weakened version of p-adics that would make this kind of thing valid for any base. Like, clearly you can proceed that way, and so in a sense it works, but what is the object you end up with if you go "well it's basically p-adics but for a non-prime base"?

    • @MuffinsAPlenty
      @MuffinsAPlenty 3 года назад +2

      You can make sense of "n-adic integers" for any nonzero integer n. You can complete the ring Z with respect to the (n)-adic topology, which is equivalent to taking the inverse limit of Z/n^kZ. Doing this with n = 10 produces a completely valid topological ring in which ...999.0 = -1.
      If you want your "numbers" to be part of a field, then sure, you can't get a _field_ of 10-adic numbers, since 10 is composite. For a composite base, you can cook up zero divisors in a construction like this, so you can't form a field of fractions of the ring of 10-adic integers. (For p a prime, the ring of p-adic integers is an integral domain, so you can take the field of fractions of the p-adic integers, and this is isomorphic to the construction of the p-adic numbers you get when you take the field of rational numbers and complete it with respect to the p-adic absolute value.)

    • @Kram1032
      @Kram1032 3 года назад

      @@MuffinsAPlenty ah I see, makes sense

  • @stevendouglas4437
    @stevendouglas4437 2 года назад +3

    0.999 recurring is the sum of terms 0.9, 0.09, 0.009 etc which is a geometric sequence with ratio = 0.1 < 1. So the terms converge to 0 and the series converges. Now the formula for an infinite convergent series can be used and shows that the series converges to 1.

    • @thevalarauka101
      @thevalarauka101 Год назад +3

      time to prove the formula for an infinite convergent series...

  • @wardellcastles
    @wardellcastles 3 года назад +4

    My "proof" to those who say it's not equal is to ask, "If it's not equal, then what is 1 - .999...?" So far no one has been able to give me an answer!

    • @deadpixel7123
      @deadpixel7123 3 года назад

      Wouldn't it be equal to 1*10^-1000....+1?

    • @romilgoel4191
      @romilgoel4191 3 года назад +1

      Perhaps, ask them..
      ok ... What number lies between them ? 🤔

    • @leonardovalente9772
      @leonardovalente9772 3 года назад

      @@deadpixel7123 10000... doesn't converge to a number, it just blows up to infinity. That's why the proof of that ...9999999 = -1 is wrong

    • @crateer
      @crateer 3 года назад

      Well, the answer is a *0.* (What ever many zero's) and a 1 at the end
      So: 0.0...01

    • @kwan3217
      @kwan3217 3 года назад

      It might be the hyperreal number (en.wikipedia.org/wiki/Hyperreal_number ) epsilon. I can't rigorously prove if it is or isn't, but I know that hyperreal numbers are complete and consistent enough for there to be a proof one way or the other.
      Hyperreal epsilon is an example of an infinitesimal number, a number which is smaller than any positive finite real number but greater than zero. As commonly defined (by for instance Dedekind cuts en.wikipedia.org/wiki/Dedekind_cut ), the real number system doesn't have any infinitesimal numbers, but including one allows one to construct a logical system which is consistent if and only if real numbers are consistent.
      Admittedly, hyperreal epsilon is not a *real* number, but it is a perfectly valid number in the hyperreal system, just like sqrt(-1) is not a real number either, but a perfectly valid number in the complex system.

  • @leighlewis5514
    @leighlewis5514 2 года назад +1

    5:26 So... what is the summation for n = infinity ? 1 and a bit?

  • @АнтонРусанов-ф5к
    @АнтонРусанов-ф5к 2 года назад +5

    To me the biggest confusion about the statement 0,999... = 1 was, that for example if we say something approaches 0, I didn't consider it being equal to 0, because it doesn't behave itself as 0.

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 года назад +1

      0.999... does not approach 1.

    • @pluto8404
      @pluto8404 2 года назад +1

      yeah this video was kind of dumb. It through out the idea of moving the decimal right when multiplying by 10 because that would be to complex for people to understand, then goes on to explain limits as if that would be more helpful to the laymen. And I agree with the whole converges, always seemed like some hacky way to just make math work. Like how do I know the sequence doesnt diverge eventual at infinity eventualy.

    • @pluto8404
      @pluto8404 2 года назад +1

      his whole proof is on the assumption we believe 10^-i converges to 0, which is no different than saying 0.333 * 3 = 0.999, 0.3333*3=0.9999 because patterns and stuff.

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 года назад

      @@pluto8404 *It threw out the idea of moving the decimal right when multiplying by 10 because that would be too complex for people to understand,...*
      No, he did not do that. He explained that the assumption that you can move the radix punctuation when multiplying by 10 may not be true when dealing with the notation 0.(9), and he even provided a counterexample that demonstrated that the algebraic proof is invalid. It has nothing to do with it being complex.
      *...then goes on to explain limits as if that would be more helpful to the laymen.*
      No, he did not claim this is more helpful to the "laymen." He explained that in order for the proofs to be valid, 0.(9) must be well-defined, and for it to be well-defined, you must state the fornal definition of 0.(9), and whether you like it or not, 0.(9) is defined as the limit of a sequence of real numbers.
      *And I agree with the whole converges, always seemed like some hacky way to make math work.*
      This is an unintelligible sentence. This to say, that the way in which you arranged the words together is nonsensical, and as such, I have no idea what it is that you are trying to communicate.
      *How do I know the sequence does not diverge to infinity eventually?*
      He proved it. I am starting to think you did not watch the video carefully at all.
      *this video was kind of dumb.*
      You obviously did not understand it, considering that every statement you have made about what was claimed in the video has been false. It seems to me you need to spend more time trying to understand the video, and less time being rude and making unintelligent comments.
      *His whole proof is on the assumption that we believe 10^(-i) converges to 0,...*
      No, this is wrong. He never used the words "we believe" in the video. Also, he did not assume 10^(-i) converges to 0. He proved it converges to 0.
      *...which is no different than saying 0.3·3 = 0.9, 0.33·3 = 0.99, because patterns and stuff.*
      It absolutely is very different. Nowhere in the proof is the fact that you mention utilized at all.
      It is clear you lack even a basic understanding of the topic at hand, so you would save yourself some public embarrassment by not pretending that you are qualified to present criticism to the video. In fact, this is not merely a problem with understanding, but a problem with attitude. It seems clear to me that you are not interested in knowing the truth and developing an understanding of why mathematicians present 0.(9) = 1 as a fact.

    • @pluto8404
      @pluto8404 2 года назад

      @@angelmendez-rivera351 as you state, his whole axoim is that the definition of 0.9999 is the limit of the series of 0.9999 which equals 1... so why would it be wrong to say the limit of 0.3333 repeating multipled by 3 is equal 0.9999 which equals 1. Its the same formula but just dividing by 3 before hand. maths

  • @jounik
    @jounik 2 года назад +3

    06:20 I would have stated the claim to be that the sequence 0.9, 0.99, 0.999 ... approaches a real number and that there exist no real numbers between that number and 1. Given that there are infinitely many real numbers between any two distinct real numbers, the real number limit of the sequence and 1 are not distinct.

    • @Chris-5318
      @Chris-5318 2 года назад

      He kinda did that, even though he didn't express himself as clearly as you just did. Of course, both of you have only tacitly assumed that the sequence also approaches 0.999... (or equivalent). I'm not knocking you BTW, your comment is vastly better than 99+% of the one's posted here.

    • @algotkristoffersson15
      @algotkristoffersson15 Год назад

      There does though, we just can’t write them with the decimal system

    • @algotkristoffersson15
      @algotkristoffersson15 Год назад

      @@Chris-5318well by Defenition of the way it is constructed the limit is necessarily 0.9999999 repeating

    • @Chris-5318
      @Chris-5318 Год назад

      @@algotkristoffersson15 The definition is:
      The sum of a series is the limit of the sequence of its partial sums (if the limit exists).
      With that we have that the sum of 0.999... IS the limit of the sequence 0.9, 0.99, 0.999, ... and that limit is easily seen to be 1 using the epsilon-N definition of limit.
      Are you referring to the construction of the reals as the set of equivalence classes of rational Cauchy sequences?
      What can't we write with the decimal system? If you are referring to terminating decimals, then of course we cannot write 0.999... with that. We need to use infinite decimals expansions for that.

    • @algotkristoffersson15
      @algotkristoffersson15 Год назад

      @@Chris-5318 we can’t write the numbers between 0,9999999… and 1, because the “resolution” of numbers isn’t high enough, like how a 480, p screen can’t display a screenshot from overwatch because there aren’t enough pixels

  • @AlrycaAeveaHexendias
    @AlrycaAeveaHexendias Год назад +3

    After learning about 10-adic / p-adic numbers. The part at 1:50 strangely becomes technically true(?)
    Math is weird.

    • @Chris_5318
      @Chris_5318 Год назад

      That's fine if you specify that you are using a 10-adic pseudo metric. But it was the Euclidean simple difference metric that was implicitly assumed by James (and unknowingly by every schoolkid).

  • @bobon123
    @bobon123 2 года назад +2

    The issue is that this kind of proof is not useful to give to someone that does not know calculus: too many times I heard "Yes, of course it _tends_ to 1, but it _is not_ 1". Those algebraic proofs are not real proofs, but they should be intended in a different way. I interpret them as asking the sceptic: "Do you want to have a definition of periodic numbers and their operations such that 10 x 0.9(9) = 9.9(9)? Then, it is proven like this."
    You can go on and clarify that it is not obvious to accept that definition, and that one should be very careful about assuming things when infinite series are involved. But I believe it is much more persuasive to clarify what are the consequences of your assumptions instead of starting from one specific definition of 0.9(9).

  • @keineangabe8993
    @keineangabe8993 3 года назад +7

    Before watching the video: Absolute Titles are almost always wrong. I am a mathematician and think I can judge for myself if the proofs I've seen are wrong.
    All you need to do, is look at the definition of decimal representation of real numbers and there isn't really a lot to prove left (if you know the geometric series).

    • @TheIllerX
      @TheIllerX 3 года назад +2

      Same here.
      I guess titles like that are meant to be a bit like provoking clickbait.

    • @enzzz
      @enzzz 3 года назад

      I am not a mathematician and so I obviously don't understand and have studied all of this, but looking at the definition, it does not satisfy me at all. My first thought is that the definition must have got it wrong or it doesn't make sense. Obviously I'm not smarter than millions of people and years of practicing mathematicians put together, so I must be missing something. But my thoughts are following:
      1) Infinity can't exist really, and so the number 0.(9) can't exist. 0.(9) can exist as a concept, but it can't be used in any calculation as this calculation would take infinite time to compute. Even if 0.(9) could equal 1, it couldn't be proven, and as far as we can go, it will never equal 1.
      2) Because 0.(9) does not truly exist it can't equal to 1, because 1 exists.
      Some other notes. I see that it's mentioned that real numbers should always have infinite other numbers in-between. This is another thing I don't get. Why should this be part of the definition? Bringing this up, because Wikipedia does the first proof by showing that it must be equal, because no other numbers can be put between 0.(9) and 1.
      Other note: Also 0.(3) shouldn't equal 1/3. 0.(3) also would exist as a concept, but not as a number, and as a concept it can be thought of as approximation to 1/3, but I don't see how it can in any way equal 1/3.

    • @keineangabe8993
      @keineangabe8993 3 года назад

      @@enzzz There is a philosophy of mathematics called Finitism, where only finite objects are considered to exist. You might want to have a look at that. Most of mathematics however doesn't care if there are real world representations of mathematical objects, because it doesn't lead to very interesting results if you restrict yourself that way. Note, that many of these results still have implications for the real world.
      All of mathematics starts with a set of axioms, that you assume to be true in order to prove other things. Those axioms don't have proofs, they are the basis for all other deduction. The most widely used basic axioms are the Zermelo-Fraenkel set theory.
      One of these axioms in ZF is the axiom of infinity which assumes the existence of an infinite set.
      My point is: if you don't think, infinite sets exist, then you can't use ZF as the basis for your mathematics. Most of mathematics you see however, including everything in this video, does use ZF as basis. It doesn't make sense to argue about the validity of mathematical results if your base assumptions are different.
      A simple analogon: Some says: if I am 4 years old now, then I will be 6 in 2 years.
      You are now like a person saying: I don't think that is true because I don't believe that you are 4 years old.
      Point here is, it doesn't matter if the assumption (the person is 4 years old) is true or not, it only matters if the deduction is true, given that the assumption IS true.
      Mathematics does the same, it doesn't claim it's basic axioms are correct or represent the real world, it only claims that it's theorems are true given those axioms.

    • @enzzz
      @enzzz 3 года назад

      @@keineangabe8993 Yeah, the thing that I seem to be arguing I suppose is that intuitively it's difficult to understand why such decision was made for the definition of real numbers. If I invented mathematics, I wouldn't define it like that. There could very well be a reason down the road, why it makes more sense to define things like that, but I don't understand it therefore I can't be satisfied.
      So the explanations or proofs why 1 = 0.(9) for me, the layman, don't work as a I'm just left with a conclusion that something, somewhere is broken. At least the common and main branch of mathematics should not define things like that. I can understand if you create some other secondary branch of mathematics where you have those axioms, but not based on reality.
      As a layman, what's the use of having interesting results with infinity existing?

    • @enzzz
      @enzzz 3 года назад

      @@keineangabe8993 So in this case my argument is that we should use Finitism as the main branch and not ZF as you said. As you said that without infinity math wouldn't lead to interesting results? I'm not sure what the interesting results are, but I don't think this argument holds water. Main branch of mathematics should be logical, intuitive and practical. Unless there's a good example of there being practical value provided by infinity being able to exist. If you want to say 1 = 0.(9) then you can so so, but you must add an asterisk or similar, to indicate that this equation is only true in a separate, fantastical branch of mathematics.
      Adding elements like this to mathematics, seems just unnecessary and can only cause issues down the road. It could be a separate branch, where you can test out the interesting results, but definitely not the main branch. Math should be deducible purely from intuition and rules like this unnecessarily complicate things without adding any practical value.
      And while it seems I said this with confidence, I'm not saying this with confidence. I may be wrong and there's good practical value for infinity to somehow exist.

  • @pepn
    @pepn 2 года назад +3

    so if 0.99999 has 1 as a limit when n tends towards infinity or something, why is it correct to say that they're strictly equal ? shouldnt we say that the limit of the sum is equal to one ? or is it implied in the notation "9 repeating" ?

    • @psuw
      @psuw Год назад

      This is pretty much my issue with that as well.

  • @egoreremeev9969
    @egoreremeev9969 3 года назад +4

    I think historically these infinite decimals were even before actual definition of the limit, and they were derived from infinite division.

  • @ethanandrews3076
    @ethanandrews3076 2 года назад +2

    I learned this proof. Between every pair of different real numbers, there exists a number in between them. Because .9 repeating and 1 do not, they are not different numbers

    • @mCoding
      @mCoding  2 года назад

      What this proof (if details are flushed out) shows is that *if* .9 repeating exists, then it equals 1. But it would still remain to show that it exists in the first place.

  • @andrewweirny
    @andrewweirny 3 года назад +22

    Funny that we can’t assume that the audience understands calculus but we can assume that every proof the audience has ever seen is wrong.

  • @craigdupree1687
    @craigdupree1687 2 года назад +5

    Probably also worth mentioning that has to be the only solution, since the limit of a sequence is unique. And this is how the irrationals can be defined, namely, as the limit of certain infinite sequences. In fact, this is why you have to bring the irrationals into the real numbers since they make it a complete set of numbers.

    • @Happyduderawr
      @Happyduderawr Год назад

      You have to prove that using reverse mathematics

    • @KEVINDAVIDBALBUENASALVADOR
      @KEVINDAVIDBALBUENASALVADOR Год назад

      i think proving the limit is unique is fairly easy, you just have to assume there exist 2 different limits and the contradiction shows up almost instantly (you can't choose an epsilon smaller than limit 1 - limit 2 mediums)

  • @tpahere8756
    @tpahere8756 3 года назад +12

    I clicked on this video being afraid that it would be some of these terrible "All mathematicians are wrong, I'll tell you why" videos but I was really happily surprised to see it is really good ! Thanks a lot for making this type of videos that kinda teaches people how to think of the domain of mathematics as it is and not as a world of beliefs and magic tricks

    • @mCoding
      @mCoding  3 года назад +3

      Thank you! I try to take simple topics and peel back the curtain just a little bit.

  • @ezhiq
    @ezhiq Год назад +1

    0.(9) is an infinite sum: 9/10 + 9/100 + 9/1000 + ... The easiest way to find this sum is to simply use the school formula for the geometric progression.

    • @Chris-5318
      @Chris-5318 11 месяцев назад

      0.(9) is an infinite series that has a [finite] sum. I know it is a common idiom, but "infinite sum" is an oxymoron (in the context of the real numbers). The school formula for the sum of a geometric series has to be proven using the same sort of limit based definitions that James used for the sum of 0.(9).

  • @pavlopanasiuk7297
    @pavlopanasiuk7297 3 года назад +6

    The most baiting title and the most satisfying video. Thank you for not being "one of those", youtube really needs clarification nowadays

    • @mCoding
      @mCoding  3 года назад +1

      And thank you for keeping an open mind and not clicking away!

  • @ophertheking
    @ophertheking 3 года назад +38

    What I really enjoy about this channel is that it makes me think of the things I didn't think I would think about. Or need. Kudos for the quality content. Well thought out and enriching to all of us laymen.

  • @CraigNull
    @CraigNull 3 года назад +11

    "Did your favorite math creator make one of these mistakes?"
    Which creators do you think actually made a mistake, versus disagreeing with you about the pedagogy of this topic? Being a proof or not is not a binary -- not filling in justifications of each step makes the proof incomplete, at most, not incorrect. It sounds like you'd look at someone's dissertation, see them invoke x^2 >= 0 for real x, and say their proof is *wrong* if this isn't established in an appendix. There is a surrounding context of the state of everyone's understanding that you seem to neglect

    • @haomingli6175
      @haomingli6175 2 года назад +2

      only what is obvious to the reader can be omitted. that is the context that legitimizes any omissions. The rigorous definition of repeating decimals is almost certainly not obvious to the average reader of those proofs on RUclips. Thus, you cannot omit it from your proof. Without it, those proofs are indeed merely "a series of true statements ending in the desired statement", without actually being a proof (a series of statements, starting from established truths, one implying the next, ending in the desired statement". In this sense, they are wrong, if they are considered proofs at all.

  • @alex1stamford779
    @alex1stamford779 2 года назад +1

    There are multiple errors in this video, or more precisely pedantic nitpicks that while leave the proof "incomplete" but it doesn't make the proof wrong.
    - The first criticism is that we are assuming that 0.99.... exists if we write it down. That's true but that assumption is built into the question. When somebody asks "Is 0.99... equal to 1?", it is implied that the existence of 0.99.. as a number is assumed. Otherwise, if we go onto the pedantic route, why even assume that 1 exists? Why assume that "a=b" is a well-defined statement with a well-defined meaning? Pedantry, at worst, makes a statement incomplete, not wrong.
    - The second criticism comes from the fact that we haven't justified why "10 * 0.9... = 9.9...". While this is an incomplete step, it still doesn't render the proof "wrong". Going via the pedantic route, we can again require the person to derive the consistency of algebraic operations, because if students don't know calculus, then they definitely don't know that algebraic operations aren't mutually inconsistent, which would render the proof, not only incomplete, but wrong.
    - Third criticism is that people shouldn't accept 1/3 = 0.3... as a given equality. This is however besides the point. How comfortable students are with the concept of 1/3, or 0.3... is a question different from whether the proof of 1 = 0.9... holds given the assumptions.

  • @stydras3380
    @stydras3380 3 года назад +15

    I get your point but I disagree over some points: You can take any proof and say its not a real proof, because it hides some steps. But that just what you got to do to not overcomplicate proofs. For example I might as well say "well why is 3 * 1/3 = 1, thats hidden behind the curtain" or even "you did not give a formal definiton of 1/3 or 1, what even are they?". I think the the proofs as given are okay: I don't feel like hiding how multiplication with 10 works for decimal numbers takes away from the idea especially since 3*0.3... = 0.9... is very intuitive and I think very suited "for those who would be interested in proving such a statement". TLDR: I dont consider the proofs wrong but rather they are not detailed to your (or my) liking.

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 года назад

      *For example, I might as well say "well, why is 3·1/3 = 1, that's hidden behind the curtain,"...*
      No, it is not hidden behind the curtain. The equation 3·1/3 = 1 is what defines the symbol 1/3. In other words, this is a definition, not a theorem that needs proving.
      *...or even "you did not give a formal definition of 1/3 or 1, what even are they?"*
      1 is defined by x·1 = 1·x = 1. 3 is defined as (1 + 1) + 1. 1/3 is defined by 3·1/3 = 1. These definitions do not need to be stated. Why not? Because nearly everyone knows what 1, 3, and 1/3 are defined as. On the other hand, most people do not know the definition of 0.9..., as proven by this very comments section. As such, the definition of 0.9... does need to be stated in a proof.

    • @DarinBrownSJDCMath
      @DarinBrownSJDCMath 2 года назад

      @@angelmendez-rivera351 You cannot define a mathematical object by just declaring that's it's the solution of some equation. That applies to 1/3. The fact that "everyone knows" that 3*(1/3) = 1 is an appeal to authority, based on the fact that most everyone has a working intuition about fractions.
      The same justification could be given for the imaginary unit i. What is the definition of i? Well, it's the solution of x^2 = -1. "Everyone" knows that. Except everyone doesn't know that, and they won't accept this statement as a "definition" of i.
      Both 1/3 and i must be defined in the context of their number systems. 1/3 exists because we can define the rational number system as equivalence classes of ordered pairs of integers and define the four operations on them. Similarly, i exists because we can define the complex numbers as ordered pairs of real numbers and the four operations on those.
      No one uses this presentation when first teaching someone fractions or complex numbers, but that doesn't mean there isn't a rigorous foundation.

    • @Jasruler
      @Jasruler 2 года назад +1

      There's a reason for axioms in math folks.

  • @Tentin.Quarantino
    @Tentin.Quarantino 3 года назад +6

    For the x = -1 it breaks because x is not finite. When the recurring 9 is to the right of the decimal, it is finite.

    • @TreeCube
      @TreeCube 3 года назад +1

      So, how do you prove that recurring 9 to the right is finite?

    • @Tentin.Quarantino
      @Tentin.Quarantino 3 года назад +2

      @@TreeCube well 0.99999 recurring is bound between zero and one. Does this need a rigorous mathematical proof?

    • @TreeCube
      @TreeCube 3 года назад +1

      @@Tentin.Quarantino Well, if you want a rigorous proof, yes. You can use real analysis to prove that 0.999.. converges to 1 (or just use the geometric series to show it converges if you're not that rigorous)

  • @phreak1118
    @phreak1118 2 года назад +9

    When I saw a proof of this back in the 90's I was told that by definition... 2 numbers are only different in value if you can put another number inbetween them.

    • @Baruch.Spinoza
      @Baruch.Spinoza 2 года назад +1

      But then how would the number in between be different

    • @ellaser93
      @ellaser93 2 года назад +2

      @@Baruch.Spinoza Between any two real numbers there are a infinite number of numbers that can be put in between them.

  • @Darryl_Frost
    @Darryl_Frost 2 года назад +2

    is 'approaching' the same as 'equals'? as n approaches infinity 0.999... approaches 1.
    1 is a real number, 0.999... is not a real number. 1 is 'computable' 0.999... is not. Therefore in many ways 1 is NOT equal to 0.999...
    0.999... is an 'instruction' or an operator it tells the person that write down 0.999 forever. 1 is a real, computable number not an instruction or operator.
    Like all pseudo numbers you have to round it off for it to be computable. This also applies to Pi, you cannot express Pi as a real number, for it to be computable you need to round it to some precision.
    0.888... Rounds out to 0.88....9.

  • @IRanOutOfPhrases
    @IRanOutOfPhrases 3 года назад +12

    Enjoyed the alternate take here, but I think in most cases, you're trying to convince a non-math person that it equals 1. So maybe its fair to say that multiplying .99_ by 10 becomes 9.9_ is a calculus thing, but man, good luck trying to explain whats going on at 6:59 to someone who doesnt understand you can move the decimal point over when multiplying by 10, haha
    But yes, eventually I always challenge them to do these three things:
    1. If its not 1, then what is the difference between them?
    2. Since they are different numbers, what number comes between them?
    3. Whats the average of those numbers?

    • @ThatDevMat
      @ThatDevMat 2 года назад

      To answer your questions you pose to non math people at the end of your comment since they are kinda Irrelevant:
      1. In most practical applications there’s not a real difference, but say it was not .99… and 1 but larger versions and taxes, it could be the difference between two tax brackets.
      2. This question is also dumb. What whole number comes between 1 & 2, no number being between them is not relevant to actual being a unique number
      3. This also has the fatal error of being completely irrelevant to if a number is real (in the non mathematical sense).

    • @romano-britishmedli7407
      @romano-britishmedli7407 2 года назад

      This is very true.
      I at first was confused by the introduction, thinking
      "But wait... We're assuming to still be in the field of real numbers IR, so of course I can assume that 0.9_ exists, because I can approach this value by using real numbers and operations on these infinitely." (Just like π is a value we can only approach by infinity, yet it is a valid real number in IR we can calculate with.)
      Glad he later addressed this objection, because otherwise I would've been severely confused.
      From my perspective, as you write it comes down to the fact that the difference between 0.9_ and 1 is 0,
      therefore 0.9_ = 1.

    • @IRanOutOfPhrases
      @IRanOutOfPhrases 2 года назад

      @@ThatDevMat
      And this is when I reiterate to the person to stick to the questions asked.
      1. You dodged the qiestion entirey
      2. You changed the question entirely. Of course whole numbers behave differently from real numbers
      3. "I said the question is irrelevant, therefore it is" i mean, yeah, those people exist. Cant argue that. But thats a very weak counter when posed with a question

    • @ThatDevMat
      @ThatDevMat 2 года назад

      @@IRanOutOfPhrases I didnt dodge or change the question for 1 or 2, you just didn’t like that the answers I gave are valid and yet don’t work in your favor because they are open to interpretation.
      For question 3 tell me how being able to average two numbers makes them really numbers… you poised a question that fundamentally doesn’t matter so of course I didn’t bother answering since it’s truly irrelevant
      For the record, I’m not arguing if .99.. = 1, just that the challenges you posed don’t actually help anyone come to that conclusion

    • @IRanOutOfPhrases
      @IRanOutOfPhrases 2 года назад

      @@ThatDevMat you didnt answer the questions though. I re-read your response and I still dont see your answers.
      1. No answer. You said something what about larger numbers and taxcodes?" What?? Lets just focus on the first question. Answer please.
      2. Bringing up whole numbers doesnt get you out of this one. However if the person wants to argue that whole numbers work the same as real numbers, I'd point back to question 1 and how much easier it is for me to calculate 2-1=1. So clearly whole numbers behave much differently than real numbers and are not a good comparison. So what number comes between 0.9_ & 1. Or alternatively, what number comes after 1?
      3. I dont understand your criticism. Is the argument here that they're not really numbers? Thats extra confusing now.

  • @georgechauvet2476
    @georgechauvet2476 2 года назад +3

    Nice, but there is an assumption that there is an integer n larger than 1 / epsilon, i.e. that the field we are dealing with is Archimedean. The reason I mention this is that typical definitions of the real numbers tend to leave the impression that there's no room for any more numbers on the number line.

    • @DrJulianNewmansChannel
      @DrJulianNewmansChannel Год назад

      The assumption isn't that "the field we are dealing with is Archimedean" - he never specificied that the reals were a "field" any more than he specified that they are an "Archimedean" field. The assumption is simply that we know what the real numbers are, because construction of the real line is not the topic of this video. (I grant that including construction of the real line would make the picture a bit more "complete" to a mathematically-minded person, but one can hardly blame him for not going into that!!) Under any explicit construction of the reals (e.g. start with the rationals and then take Dedekind cuts), it is easier to prove that the set of real numbers has the Archimedean property (i.e. is an Archimedean monoid under addition) than it is to prove that the set of real numbers is a field.

  • @jkid1134
    @jkid1134 3 года назад +9

    You're selling it short a little bit. Convergent infinite series really are equal to a number, not just in the sense of the limit of partial sums, but in every sense.

    • @rryan916
      @rryan916 3 года назад +1

      As a predetermined definition decided by mathematicians, yes. His explanation is fine, In EVERY sense? Not in the sense that .99999 don't look anything like 1. Not in the real world either. Does physics allow us to travel the speed of light? No. Does physics allow us to travel .99 repeating the speed of light? technically yes. That's one sense.

    • @MuffinsAPlenty
      @MuffinsAPlenty 3 года назад

      "not just in the sense of the limit of partial sums, but in every sense."
      What does this even _mean?_

    • @jkid1134
      @jkid1134 3 года назад

      @@MuffinsAPlenty I meant, by any notion of equality.

    • @MuffinsAPlenty
      @MuffinsAPlenty 3 года назад

      @@jkid1134 What other notion of equality is there, that doesn't use limit of partial sums as a starting point?

    • @jkid1134
      @jkid1134 3 года назад

      @@MuffinsAPlenty like, I guess what I mean to say is that this limit definition IS equality - it has all the same properties that equality does. You can substitute numbers and sequences which converge to those numbers without any loss of truth or nuance.

  • @alecorsmatem4845
    @alecorsmatem4845 2 года назад +1

    The counterexample you take for the first error is not so acurate, since the number you tried to build tends to infinity it comes with complications that doesn't apply to the case when it tends to 1 (or at least that is what you are trying to proof, but definitely not to infinity). At the end of the day we are trying to calculate a limit, its not wrong asume that that limit exist to try to find it as long as you don't end with a contradicction, in that case, your asumption was mistaken and you proved that the limit didi't exist in first place.

    • @Chris-5318
      @Chris-5318 2 года назад

      You are appealing to a definition when the whole point was to illustrate what happens when you DON'T appeal to a definition.

  • @michaeldamolsen
    @michaeldamolsen 3 года назад +6

    Every time you said "if it exists" I heard it in the voice of Norman Wildberger.

  • @jamscone_
    @jamscone_ 2 года назад +24

    i'm fairly certain this was my issue with the algebraic proofs when i first heard of this in a maths class, although at the time i wasn't familiar with limits so i wasn't comfortable with the idea that an infinitesimal value could indeed equal zero. the formal series definition felt like it solved all of the nagging questions i had at the time. great video, big fan of your explanations

    • @redshift86
      @redshift86 2 года назад +2

      If an infinitesimal is zero, the it is not an infinitesimal by definition; it's the whole point of the concept of infinitesimal.

    • @jamscone_
      @jamscone_ 2 года назад +1

      @@redshift86 i was writing that more from the perspective of me at a young age, but yeah you're totally right in this case. apologies for my poor wording

    • @empathogen75
      @empathogen75 Год назад +1

      I think the algebraic proofs are a good way to explain the intuition for why it’s true because most people don’t question that .3 repeating is 1/3 and the rest are a bunch of other similar calculations that sort of imply that it makes sense that .9 repeating is 1, even if they aren’t formal proofs.

    • @sphaera2520
      @sphaera2520 Год назад

      @@empathogen75that was my impression as well. These were never meant to be absolute foundational proofs where we define everything from axioms and prove the conclusion. It’s just a quick way to lean on a layperson/child’s intuition that it does make sense to say 0.(9) is another way to represent 1 just like 9/9 or 1.(0) also represent 1.

  • @richardcao7390
    @richardcao7390 3 года назад +4

    a/1-r, infinite geometric sequence with a1=0.9 and r = 0.1

  • @doraemon402
    @doraemon402 11 месяцев назад +1

    1:52 That is true in the p-adic sense
    And, in any case, those proofs aren't really wrong because if that number exists, you know it must be equal to 1

    • @Chris-5318
      @Chris-5318 11 месяцев назад

      By the same logic, if ...999 exists then it "must" (according to you) be equal to -1. Small problem for you, ...999 is positive infinitely large. I agree that the word "wrong" is misleading.

  • @martinsanchez-hw4fi
    @martinsanchez-hw4fi 3 года назад +9

    Thank you! It is so necessary for the mathematical discussion to enrich that serius approaches like this one are shared.

  • @jordanweir7187
    @jordanweir7187 3 года назад +4

    Really nice explanation here thanks for clearing this up, was always a bit of a fuzzy topic for me, one thing I would amend slightly is the tolerance bit near the end, I think it might be clearer to say (unless I've misunderstood) that the conclusion of that argument with epsilon and things is that, for any tolerance, there exists a value of n such that the difference 10^(-n) is less than that tolerance, so because you can choose the tolerance to be as large as you want with no limit, this is a suitable definition of approaching

    • @mCoding
      @mCoding  3 года назад +3

      Indeed, in this video I did not go over official definition of what is, perhaps this would be a good topic for a future video! There is a lot of confusion amongst those who are not familiar with calculus about what a limit actually is.

    • @jordanweir7187
      @jordanweir7187 3 года назад

      @@mCoding yeah its a weird concept for most i'd imagine, this particular calculation gets a lot of people because they're taught the watered down version in school, I distinctly remember being taught it and for years not being aware of the subtleties, they didn't even make it fun then anyway lol

  • @MrThezyga
    @MrThezyga 2 года назад +3

    As math students me and my colleague disagree with you. You do not have to prove that 0.(9) exists, it's just a normal number and your contrargument uses (9).0 which is a completely invalid number and doesn't even make sense. 9*10^(-n) series is a converging series and 9*10^n is not, you can't just compare them like that.

    • @Chris-5318
      @Chris-5318 2 года назад

      NB I wrote my responses before reading your entire comment. I think that it is appropriate that I don't edit it in view of how it progressed.
      For someone claiming to be a mathematician I'm shocked. Perhaps you meant that because the proofs were made a long time ago, that we can just accept them i.e. defer to authority. It is very strange that many math textbooks such as Terence Tao's "Analysis I" 3rd edition, do bother with such proofs.
      How did you decide that 0.(9) makes sense and (9).0 doesn't make sense. Did you use some sort of mathematical divining rod?
      Now I see, you said "9*10^(-n) series is a converging series and 9*10^n is not". So you have already completely U-turned on your original statement. You now are saying that you do use the definition of sum (and limit) to make the decision. OK, perhaps you don't have to evaluate the sum, just establish the convergence of the sequence of partial sums. That's fair.
      Really I think you were just a bit too hasty in your response, but you got it right in the end. As I said, I won't edit this. It can serve as an object lesson to you.

    • @MrThezyga
      @MrThezyga 2 года назад +2

      @@Chris-5318 Listen, 0.(9) is literally by definition a limit of a converging series of 10^(-n)*9. Saying that you need to prove it first is like saying that you can't just say that 2+(2+4)=4+4=8 because you have to first prove that addition in (R, +, *) is associative. It's looking for a problem in a place where it's not present.

    • @Chris-5318
      @Chris-5318 2 года назад

      @@MrThezyga Almost right. 0.(9) is a series that has a sum. That sum is the limit of the sequence 0.9, 0.99, 0.999, ...
      I wonder if you are aware of what a sum is? I know that Rudin and Spivak are badly confused by the term.
      NB (9).0 is a series that doesn't have a sum. (Obviously I am assuming the Euclidean metric).
      Oh, it's you again. You are very confused/confusing. You are also being deliberately awkward. Most schoolkids are familiar with finding the sum of a finite series. Many do NOT know as a formal fact what the sum of an infinite series is. You keep using the definition yourself to prove what you are saying, yet seen to be saying that you didn't need to do that. You really need to clear your head.
      As every digit in 0.999... represents a finite quantity, and there are infinitely many digits, the natural conclusion would be that they add up to an infinite quantity. It might even be one of Zeno's paradoxes. Despite that you claim that you don't need to bother with the formalities. Without appealing to convergence and divergence etc., how do you decide that 0.(9) is a real number and (9).0 isn't? Are you going to get out you divining rod?

    • @MrThezyga
      @MrThezyga 2 года назад +2

      @@Chris-5318 0.(9) is a just a repeating decimal with a period equal to 9. You're saying that you can't use numbers like idk 0.(45) or 0.(1) in algebraic equations without first proving that they exist in the first place? Besides, if we're not sure if 0.(9) is a real non-contradictory number that we can use then what's the point of even trying to prove that 0.(9)=1? It's like saying that a proof that there is an infinite number of primes first needs to prove that primes even exist by finding a prime and then proving somehow analytically (because you appearantly can't use algebra for anything) that you in fact can't divide it by anything other than itself and 1. Or maybe you need to first prove that you even can divide anything because what if the operation doesn't work now in real numbers field for some reason? You see the rabbit hole.

    • @Chris-5318
      @Chris-5318 2 года назад

      @@MrThezyga Nope. I said no such thing. Try to use context. This video is about PROVING that 0.999... = 1. My understanding is that an actual proof is a formal/rigorous demonstration. You obviously don't agree. I'm not claiming that this video achieves its goal, but it does vastly better than most of the eighth grader versions that litter YT.
      Of course I wouldn't normally be pedantic about the difference between a series and its sum. I am quite happy to say 1 + 2 = 3, and don't bother to say the sum of 1 + 2 = 3 or the sum of 1 + 2 is 3. And I certainly recite all the Peano axioms every time I calculate 1 + 1.
      For some people, this may be the first (and only) time that they even knew there was a definition for the sum of a series (or limit).
      The really special thing that James wanted to emphasise was the definition of sum (and to a lesser extent the definition of limit) . But he made the notional concession that many people probably have an adequate grasp of limit.
      You don't seem to be getting the point or making any concession for the sort of audience that this video may have been targeting. I think James got it about right. Your rabbit hole is just being childish.
      As this clearly is not going anywhere, I think I'm going to bow out.

  • @Random-User555
    @Random-User555 Год назад

    2:15 Is it divergent? comparing to 0.99…. which is convergent?

    • @mCoding
      @mCoding  Год назад

      Correct! Because it diverges all the algebra to solve it is not valid.

    • @Chris-5318
      @Chris-5318 Год назад

      You only know it's divergent because you are using some definitions (possibly not consciously). James was deliberately ignoring those definitions in order to get a result that everybody(?) "knew" was crazy.

  • @Martykun36
    @Martykun36 3 года назад +4

    nope, slightly unrigurous is not the same as wrong, published papers also do these kinds of jumps in reasoning, it is expected from the reader to fill in the gaps