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

Поділитися
Вставка
  • Опубліковано 29 лис 2024

КОМЕНТАРІ • 6 тис.

  • @n0ame1u1
    @n0ame1u1 3 роки тому +5457

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

    • @ujjwalbaranwal6350
      @ujjwalbaranwal6350 3 роки тому +75

      Facts

    • @nexovec
      @nexovec 3 роки тому +37

      The proofs are;

    • @rephirairis
      @rephirairis 3 роки тому +383

      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 роки тому

      no

    • @ujjwalbaranwal6350
      @ujjwalbaranwal6350 3 роки тому +4

      yes

  • @johnmeo1532
    @johnmeo1532 3 роки тому +3437

    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 роки тому +72

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

    • @freedom_aint_free
      @freedom_aint_free 2 роки тому +149

      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 роки тому +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 роки тому +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 роки тому

      @@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 роки тому +2457

    "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 роки тому +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 роки тому +35

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

    • @programaths
      @programaths 2 роки тому +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 роки тому +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 роки тому +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 роки тому +754

    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 роки тому +35

      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 роки тому +10

      @@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 роки тому +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 роки тому +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 роки тому +25

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

  • @andyl.5998
    @andyl.5998 3 роки тому +1587

    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 роки тому +278

      An excellent example!

    • @qwfp
      @qwfp 3 роки тому +209

      So it's just irl overflow?

    • @jenaf372
      @jenaf372 3 роки тому +14

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

    • @MrRyanroberson1
      @MrRyanroberson1 3 роки тому +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 роки тому +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 роки тому +565

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

    • @mCoding
      @mCoding  3 роки тому +74

      Like Art Benjamin!

    • @philippg6023
      @philippg6023 3 роки тому +19

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

    • @mr2octavio
      @mr2octavio 3 роки тому +45

      A mathemagician then

    • @raulf.duarte1856
      @raulf.duarte1856 3 роки тому +3

      Is this a compliment or an offense? lol

    • @ahmedaghadi8281
      @ahmedaghadi8281 3 роки тому +5

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

  • @anonymousjohnson8051
    @anonymousjohnson8051 3 роки тому +387

    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 роки тому +56

      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 роки тому +6

      I think 1brown3blue has a fair explanation of limits.

    • @kingthanatos6093
      @kingthanatos6093 3 роки тому +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 роки тому +61

      @@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 роки тому +8

      @@oscarofastora474 converging sequence but yeah

  • @KingBobXVI
    @KingBobXVI 2 роки тому +382

    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 роки тому +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 роки тому +7

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

    • @MrJonnyk000
      @MrJonnyk000 2 роки тому +218

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

    • @hawks3109
      @hawks3109 2 роки тому +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 роки тому +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.

  • @VivekYadav-ds8oz
    @VivekYadav-ds8oz 3 роки тому +148

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

    • @asmodeuszdewa7194
      @asmodeuszdewa7194 3 роки тому +12

      Look at the description.

    • @prasanna2991
      @prasanna2991 3 роки тому +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 роки тому

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

  • @naptastic
    @naptastic 3 роки тому +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 роки тому +42

      Flat Earthers left the chat

    • @ImNotFine44
      @ImNotFine44 3 роки тому +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 роки тому +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 роки тому +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 роки тому +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).

  • @robin888official
    @robin888official 3 роки тому +836

    "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 роки тому +292

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

    • @EvanOfTheDarkness
      @EvanOfTheDarkness 3 роки тому +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 роки тому +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 роки тому +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 роки тому +4

      @@EvanOfTheDarkness Thanks, that makes sense.

  • @UltimateDurzan
    @UltimateDurzan 2 роки тому +62

    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 роки тому +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 роки тому +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 роки тому +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 роки тому +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 роки тому +2

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

  • @yazanalj1975
    @yazanalj1975 3 роки тому +291

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

  • @johnsavard7583
    @johnsavard7583 3 роки тому +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 роки тому +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 роки тому +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 роки тому +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 роки тому +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 роки тому +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.

  • @BurgerFred1
    @BurgerFred1 3 роки тому +322

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

    • @naptastic
      @naptastic 3 роки тому +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 роки тому +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 роки тому +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 роки тому

      @@TheEnmineer kek

    • @user-lb1ib8rz4h
      @user-lb1ib8rz4h 2 роки тому

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

  • @julianfogel5635
    @julianfogel5635 2 роки тому +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 Рік тому +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

  • @michamichalski8633
    @michamichalski8633 3 роки тому +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 роки тому

      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.

  • @bluerojo
    @bluerojo 3 роки тому +14

    engineers: 0.99999 = 0

  • @hxbel
    @hxbel Рік тому +7

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

    • @Chris-5318
      @Chris-5318 Рік тому +1

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

    • @ElusiveEel
      @ElusiveEel 10 місяців тому

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

    • @Chris-5318
      @Chris-5318 10 місяців тому +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.

  • @jhawley031
    @jhawley031 2 роки тому +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

  • @maddo5501
    @maddo5501 3 роки тому +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 роки тому +7

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

    • @sajanramanathan
      @sajanramanathan 2 роки тому +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 роки тому +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 роки тому +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 роки тому +2

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

  • @alvideiectiones5097
    @alvideiectiones5097 2 роки тому +4

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

  • @justsomeboyprobablydressed9579
    @justsomeboyprobablydressed9579 2 роки тому +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 Рік тому +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 Рік тому +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 Рік тому

      ​@@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 Рік тому

      @@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 Рік тому +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.

  • @LaurenceRietdijk
    @LaurenceRietdijk 2 роки тому +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 роки тому +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 роки тому +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 роки тому +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 Рік тому

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

    • @Davs211
      @Davs211 Рік тому +1

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

  • @Daniel_VolumeDown
    @Daniel_VolumeDown 3 роки тому +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)

  • @TrimutiusToo
    @TrimutiusToo 3 роки тому +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 роки тому

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

    • @TrimutiusToo
      @TrimutiusToo 2 роки тому

      @@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 роки тому

      @@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 роки тому

      @@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 роки тому

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

  • @shohamsen8986
    @shohamsen8986 3 роки тому +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 роки тому +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 роки тому +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 роки тому +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 роки тому +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 роки тому +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.

  • @KissyfaceJenkins
    @KissyfaceJenkins 2 роки тому +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.

  • @taufiqutomo
    @taufiqutomo 3 роки тому +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 роки тому +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 роки тому +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 роки тому

      @@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 роки тому +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 роки тому

      @@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.

  • @kiiometric
    @kiiometric 3 роки тому +74

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

    • @naptastic
      @naptastic 3 роки тому +23

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

    • @imoomi4067
      @imoomi4067 3 роки тому +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 роки тому +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 роки тому +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 роки тому

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

  • @andrewweirny
    @andrewweirny 3 роки тому +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.

  • @cola-z9219
    @cola-z9219 2 роки тому +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.

  • @amirh6712
    @amirh6712 3 роки тому +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 роки тому +5

      I appreciate it!

    • @programaths
      @programaths 3 роки тому +1

      You have to be very clean now!

  • @patatopie
    @patatopie 3 роки тому +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 роки тому

      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 роки тому

      @@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 роки тому +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 роки тому +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 роки тому

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

  • @Txominde
    @Txominde 3 роки тому +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 роки тому +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 роки тому +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 роки тому

      ​@@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 роки тому

      @@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 роки тому

      @@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 UA-cam 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.

  • @Biggyweezer69
    @Biggyweezer69 Рік тому +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.

  • @fn3200
    @fn3200 3 роки тому +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 роки тому +2

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

    • @amjadsiddig2085
      @amjadsiddig2085 3 роки тому +2

      Laughs in ordinal numbers

    • @fn3200
      @fn3200 3 роки тому

      @@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 роки тому

      @@amjadsiddig2085 explain the meme pls😂

    • @amjadsiddig2085
      @amjadsiddig2085 3 роки тому +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 😂

  • @Tomyb15
    @Tomyb15 3 роки тому +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 роки тому +3

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

    • @yyeeeyyyey8802
      @yyeeeyyyey8802 3 роки тому +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 роки тому +8

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

    • @leonardovalente9772
      @leonardovalente9772 3 роки тому +14

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

    • @angelmendez-rivera351
      @angelmendez-rivera351 2 роки тому +3

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

  • @mvryan
    @mvryan 2 роки тому +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 😅

  • @meraak1
    @meraak1 2 роки тому +2

    why is jerma in the corner

  • @tricky2014
    @tricky2014 3 роки тому +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 роки тому +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.

  • @SantoshKumar-ku1mq
    @SantoshKumar-ku1mq 3 роки тому +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 роки тому +8

      Thanks for that

    • @roberthorne9597
      @roberthorne9597 2 роки тому +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 роки тому

      @@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 роки тому +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 роки тому

      @@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 роки тому +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 роки тому +2

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

    • @pedrofuster9161
      @pedrofuster9161 3 роки тому

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

  • @thedoublehelix5661
    @thedoublehelix5661 3 роки тому +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 роки тому +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 роки тому +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 роки тому

      @@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 роки тому

      @@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 роки тому

      @@jenaf372 I agree

  • @isodoubIet
    @isodoubIet 3 роки тому +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 роки тому +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 роки тому +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 роки тому +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 роки тому +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 роки тому +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.

  • @SafetySkull
    @SafetySkull 3 роки тому +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 роки тому +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 роки тому +2

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

    • @phremic5867
      @phremic5867 3 роки тому +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 роки тому +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.

  • @Errenium
    @Errenium 2 роки тому +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 роки тому

      Saying "wrong" was wrong.

  • @Krunschy
    @Krunschy 3 роки тому +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 роки тому +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 роки тому +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 роки тому +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 роки тому

      @@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 роки тому +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.

  • @cobbcoding
    @cobbcoding 3 роки тому +46

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

    • @mCoding
      @mCoding  3 роки тому +4

      Any time!

    • @MichaelBrown-gt4qi
      @MichaelBrown-gt4qi 3 роки тому +1

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

  • @chrisjduplessis
    @chrisjduplessis 3 роки тому +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 роки тому +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

  • @Beatsbasteln
    @Beatsbasteln Рік тому +2

    the ghetto proof is: you can't give someone 99.999...% of a pizza without not having any bite left for yourself, so it must've been the whole pizza

  • @stydras3380
    @stydras3380 3 роки тому +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 роки тому

      *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 роки тому

      @@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 роки тому +1

      There's a reason for axioms in math folks.

  • @_tsu_
    @_tsu_ 2 роки тому +8

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

  • @DarinBrownSJDCMath
    @DarinBrownSJDCMath 2 роки тому +3

    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 роки тому

      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 роки тому +1

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

  • @imacds
    @imacds 2 роки тому

    As a computer scientist, I find the idea of ...999.0 = -1 to make perfect sense. Finite versions of this concept are the usual way in which computers represent negative numbers.

    • @Chris-5318
      @Chris-5318 2 роки тому

      But you had to change the interpretation of what ...999 meant from the standard one. You based it on finite length 10s complement representations. I'm pretty sure that you have never worked with a computer than can store infinitely many digits.

    • @imacds
      @imacds 2 роки тому

      ​@@Chris-5318 I don't know if there is a standard definition of ...999.0. Computer scientists work with the concept of infinite computers all the time, the "Turing Machine" has an infinitely long tape (essentially infinite RAM). This example actually makes the point in this video much more clear than in the true case. If you define ...999.0 as the limit as n -> infinity of the sum from 0 -> n of 9*10^n, then yes that 'number' diverges, doesn't exist, etc.
      I do wonder if we can *assume* (like this video said, an active choice you have to make) that ...9999.0 exists and that ...999.0 = -1 (or assume that some usual rules of algebra hold), and then see if we can build a consistent number system out of it the same way that has a ...9999.0 form representation for every negative number that can be multiplied, added, etc (using rules similar to 2s compliment where bits that go off into infinity are lost). I'm almost certain someone has done this already, perhaps it doesn't work (what would ...999.5 represent for example), perhaps it is too unwieldy to work with.

    • @Chris-5318
      @Chris-5318 2 роки тому

      @@imacds What I meant is the usual/default Euclidean metric, whereby ...999 is a divergent series that is infinitely large.
      You can have the concept of infinity. But no computer can actually store the numeral ...999 as an actual infinite decimal.
      Even if you use a 10-adic pseudo metric or a 10s complement interpretation, then you are effectively defining what ...999 means in doing so. The vast majority of people have never heard of either, so they won't automatically interpret ...999 in any other way than the way they learnt at school from age eight or so.

    • @imacds
      @imacds 2 роки тому

      @@Chris-5318 An "average" person could think of it similarly to how sqrt(-1) doesn't exist, but we call it i and do math with it. So "...999.0" could be represented by the symbol '-', so a value like "...999.5" would be written as "...999.0 * 0.5" or "-0.5". :)

  • @matsim0
    @matsim0 2 роки тому +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 Рік тому

      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 Рік тому

      ​@Ok-se6mkwhat do you mean by exactly?

  • @tpahere8756
    @tpahere8756 3 роки тому +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 роки тому +3

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

  • @pavlopanasiuk7297
    @pavlopanasiuk7297 3 роки тому +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 роки тому +1

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

  • @alecorsmatem4845
    @alecorsmatem4845 2 роки тому +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 роки тому

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

  • @АнтонРусанов-ф5к
    @АнтонРусанов-ф5к 2 роки тому +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 роки тому +1

      0.999... does not approach 1.

    • @pluto8404
      @pluto8404 2 роки тому +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 роки тому +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 роки тому

      @@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 роки тому

      @@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

  • @johnnygodoy8329
    @johnnygodoy8329 3 роки тому +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 роки тому

      *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?

  • @keineangabe8993
    @keineangabe8993 3 роки тому +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 роки тому +2

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

    • @enzzz
      @enzzz 3 роки тому

      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 роки тому

      @@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 роки тому

      @@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 роки тому

      @@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.

  • @_andys
    @_andys 11 місяців тому +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 місяців тому

      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).

  • @baptiste2b31
    @baptiste2b31 3 роки тому +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.

  • @ophertheking
    @ophertheking 3 роки тому +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.

  • @CjqNslXUcM
    @CjqNslXUcM 2 роки тому +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 роки тому +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 роки тому +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 роки тому

      @@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 роки тому +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 Рік тому +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

  • @ethanandrews3076
    @ethanandrews3076 2 роки тому +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 роки тому

      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.

  • @xuepingsong5329
    @xuepingsong5329 2 роки тому +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 Рік тому +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 місяців тому

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

  • @phreak1118
    @phreak1118 2 роки тому +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 роки тому +1

      But then how would the number in between be different

    • @ellaser93
      @ellaser93 2 роки тому +2

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

  • @wardellcastles
    @wardellcastles 3 роки тому +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 роки тому

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

    • @romilgoel4191
      @romilgoel4191 3 роки тому +1

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

    • @leonardovalente9772
      @leonardovalente9772 3 роки тому

      @@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 роки тому

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

    • @kwan3217
      @kwan3217 3 роки тому

      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.

  • @Darryl_Frost
    @Darryl_Frost 2 роки тому +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.

  • @Martykun36
    @Martykun36 3 роки тому +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

  • @Norhther
    @Norhther 3 роки тому +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 роки тому +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 роки тому +1

      well that limit is exactly one.

    • @jenaf372
      @jenaf372 3 роки тому

      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 роки тому

      @@jenaf372 what shorthand of sums?

    • @jenaf372
      @jenaf372 3 роки тому

      @@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.

  • @stevendouglas4437
    @stevendouglas4437 2 роки тому +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 Рік тому +3

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

  • @alexarnold8461
    @alexarnold8461 Рік тому +1

    I feel like the intersection of people who accept the epsilon-delta definition of a limit and also need convincing that 0.999... = 1 is minute .

  • @michaeldamolsen
    @michaeldamolsen 3 роки тому +6

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

  • @pepn
    @pepn 2 роки тому +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 11 місяців тому

      This is pretty much my issue with that as well.

  • @egoreremeev9969
    @egoreremeev9969 3 роки тому +4

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

  • @alexismiller2349
    @alexismiller2349 2 роки тому +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...

  • @alvarozsh6566
    @alvarozsh6566 3 роки тому +8

    I think for a general public that isn't familiarized with Geometric Series and limits it's easier to use the "wrong proofs" pointed out at the start of the video

  • @williambaillargeon5266
    @williambaillargeon5266 3 роки тому +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 роки тому

      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 роки тому

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

    • @vertihippo1274
      @vertihippo1274 3 роки тому

      @@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 роки тому +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.

  • @lunaaurora3994
    @lunaaurora3994 2 роки тому +24

    Man, this video is so good, thank you for making it! I always felt that there was something missing or incomplete about those other videos I've seen about this, but I'm not highly skilled at maths so I could never figure out what. It's cool to see that that hunch was right, and your proof involving the limits of the sequence is really satisfying to see!

  • @bobon123
    @bobon123 2 роки тому +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).

  • @BrianStDenis-pj1tq
    @BrianStDenis-pj1tq 3 роки тому +8

    I really like your videos (I started with programming in Python videos). In this video, you made it sound like there is some uncertainty if 1/3 is .333(repeating). To me, this is just an issue of fractional notation versus decimal notation. There is no need to worry about basic algebra when dealing with a number that can't be represented exactly (like .3 repeating). Its just a representation, so the math of 1/3 * 3 should work like .3(repeating) * 3. Do you feel the same way with root(2) for example? If you had a representation of root(2) in decimal, and squared it, would you doubt the answer is 2?

    • @Chris-5318
      @Chris-5318 3 роки тому +1

      I think you'll find that mCoding is several steps ahead of you. He does not have any uncertainty about the fact that 0.333 . . . = 1/3. The main point of the video is that the only actual proof use the definition that *the sum of an infinite series is the limit of the sequence of its partial sums* (if the limit exists). Almost all of the popular proofs completely ignore that definition. I expect that many of the authors aren't even aware of it.

  • @craigdupree1687
    @craigdupree1687 2 роки тому +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 Рік тому

      You have to prove that using reverse mathematics

    • @KEVINDAVIDBALBUENASALVADOR
      @KEVINDAVIDBALBUENASALVADOR Рік тому

      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)

  • @Tentin.Quarantino
    @Tentin.Quarantino 3 роки тому +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 роки тому +1

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

    • @Tentin.Quarantino
      @Tentin.Quarantino 3 роки тому +2

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

    • @TreeCube
      @TreeCube 3 роки тому +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)

  • @inspireupliftflow
    @inspireupliftflow 2 роки тому +1

    We know
    1/3 = 0.333...
    Therefore, 1/3 + 1/3 + 1/3 must be 0.333... + 0.333... + 0.333...
    or, 3/3 = 0.999...
    which means, 1 = 0.999...
    Am I wrong somewhere?

    • @Chris-5318
      @Chris-5318 2 роки тому

      You haven't proven that 0.333... = 1/3. I doubt that a proof is given at most schools. Before you reply, be aware that the division algorithm only, at best, suggests it.

    • @mCoding
      @mCoding  2 роки тому

      This is one of the proofs specifically debunked in the video. All your statements are true but this is still not a proof of .9 repeating = 1.

  • @cmilkau
    @cmilkau 2 роки тому +3

    This title is quite arrogant.

    • @Chris_5318
      @Chris_5318 2 роки тому +1

      The title is wrong.

    • @sicko5821
      @sicko5821 Рік тому

      do you even know what arrogant means or do you have a reason that you're so set to use it?

  • @CraigNull
    @CraigNull 3 роки тому +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 роки тому +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 UA-cam. 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.

  • @yogeshchandrasekharuni4054
    @yogeshchandrasekharuni4054 3 роки тому +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 роки тому +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 роки тому +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 роки тому +1

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

    • @rsa5991
      @rsa5991 3 роки тому +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 роки тому

      @@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.

  • @astroknight5
    @astroknight5 3 роки тому +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 роки тому

      That's the point

    • @user-lb1ib8rz4h
      @user-lb1ib8rz4h 2 роки тому +3

      okay, prove that the first number is finite.

    • @astroknight5
      @astroknight5 2 роки тому

      ​@@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 роки тому +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.

  • @richardcao7390
    @richardcao7390 3 роки тому +4

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

  • @atthesunrise
    @atthesunrise 2 роки тому +46

    Great video!! The algebraic proof has always bugged me for some reason all my life, and this has filled in the holes (for now at least, lol). Awesome channel too, so happy I found this.

    • @MrPlaneCrashers
      @MrPlaneCrashers 2 роки тому

      If you liked this video, there is a chance you will really like Vi hart's video on the subject (not the april fools one obv), it goes into a lot more depth about the subject. Other good math channels include Mathologer, Stand-up maths, numberphile. The math youtube community is pretty vibrant and there is a lot of really great channels. 3blue1brown and primer also make math related content, both are really good as well.
      (Note: Although I do strongly disagree with this video's view about how the algebraic proof is wrong, this comment was not intended as a way to critisize the channel in general. It is always difficult for me to follow educational creators who made content I disagree about because of trust issues, but that doesn't mean mCoding doesn't make good content. I just thought that if you liked this math video, you might enjoy other math channels, since mCoding is mostly focused on coding, believe it or not)

    • @leonardofacchin1452
      @leonardofacchin1452 2 роки тому

      The funny thing is that I have never actually seen this presented as a proof. I'm not saying that it didn't happen. I'm just saying that I never met such proofs.
      The fact is that decimal repeating numbers are just a representation of rational numbers, that is fractions. So, it's implied that the limit exists and that you can multiply by 10^n for any given n just by shfting the period to the left. This algorithm is usually used to derive the rule that is taught to high school students in order to convert a decimal repeating number into a fraction.
      So, it's not so much a proof, but rather a 'recipe' so that people can turn an unwieldy representation into a much more sensible one.

  • @shivanshu6204
    @shivanshu6204 11 місяців тому +1

    0.999.. existing isn't a wrong assumption from the get go, and ..999 = -1 works in the context of 10-adics but in the context of naturals you literally created a number that's undefined because its divergent. 0.999... can exist because every new 9 added is just really an addition to a convergent series in the first case but divergent in the second.
    How are people this wrong with this much confidence, god damn.

    • @Chris-5318
      @Chris-5318 11 місяців тому

      As you say, ...999.0 = -1 is fine if you use the 10-adic metric and the definition of sum that James gave (for it). But the key point is that you have to use definitions to make the proof meaningful and rigorous. James deliberately only used the kind of understanding that a schoolkid would have.

  • @AlrycaAeveaHexendias
    @AlrycaAeveaHexendias Рік тому +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 Рік тому

      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).

  • @jamscone_
    @jamscone_ 2 роки тому +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 роки тому +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 роки тому +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 Рік тому +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 11 місяців тому

      @@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.

  • @jkid1134
    @jkid1134 3 роки тому +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 роки тому +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 роки тому

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

    • @jkid1134
      @jkid1134 3 роки тому

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

    • @MuffinsAPlenty
      @MuffinsAPlenty 3 роки тому

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

    • @jkid1134
      @jkid1134 3 роки тому

      @@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.

  • @Elektrolite111
    @Elektrolite111 2 роки тому +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.

  • @qwertz12345654321
    @qwertz12345654321 3 роки тому +14

    I always hated the algebraic proofs because they felt like "circular reasoning" thanks you for validation those feelings in a rigorous way

  • @Kram1032
    @Kram1032 3 роки тому +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 роки тому +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 роки тому

      @@MuffinsAPlenty ah I see, makes sense