Monday, November 7, 2011

Arguing Infinite Decimals

Recently the RJLipton blog had two interesting and contentious posts about people who dispute Cantor's diagonal argument (that real numbers have different cardinality than natural numbers), which I'm pretty sure generated more comments than anything else to date on the blog. Apparently this is one of the more popular topics for math-cranks to extensively argue that they've proven the other way -- read for yourself here and here.

I wish that I had the opportunity to address issues like this in the classes I teach, but unfortunately at the moment I don't have any such opportunity. It would be nice to have a venue to refine the argument with a fresh audience every so often, and to work to ferret out the criticisms that arise. If we do so, with a disputatious subject like this (namely: the first few times a student deals with infinite sets and their counterintuitive by-products), then I think it's extra-important that we carefully lay out initial definitions at the start, break down the argument into very atomic numbered steps (so that we can refine discussion and disputes as they come up later), and also give explicit justifications for each step.

Here's another issue which I feel has the same flavor to it: the fact that 0.999... = 1 (or more generally, that any terminating decimal has two different, equivalent representations: the normal one, and a second one that ends with an endless sequence of "9"'s). Here's a suggestion on the careful way that I'd want to do it (again -- not having had this battle-plan encounter the enemy yet):

Definition of 0.999...
(a) The number has infinitely repeating digits.
(b) After every "9" digit, there is another "9".
(c) There is no end to the "9"'s.

Proof that 0.999... = 1 (by algebra)
(1) Let x = 0.999...
(2) Then 10x = 9.999... (multiply each side by 10)
(3) So 9x = 9 (subtract step 1 from step 2; note decimals cancel)
(4) Which means x = 1 (divide each side by 9)
(5) Therefore 0.999... = 1 (substitute from step 1)

And then when the arguments arise you can at least ask your interlocutor to focus on one single step or definition in which they think there's a logical gap.

11 comments:

  1. This algebra proof of 0.999...=1 is a pet peeve of mine. I hold that it's circular. In going from (2) to (3) you skipped some steps:

    (2.5) 10x - x = 9.999... - 0.999...
    (2.6) (10-1)x = 9

    We've used the distributive law. But the distributive law does not work when numbers are naively treated as strings of digits. Indeed, (10-1)x, or 9x, if you work it out "naively", is not 9 at all but 8.999..., and we to assume that 8.999...=9 *in the middle of a proof of 0.999...=1* is a blatant example of circular reasoning.

    The correct way to do things is to define numbers as *equivalence classes* of strings of digits, under the equivalence relation where things are identified if one has an infinite tail of 9's and the other is the simplified form thereof. Indeed, this is Stevin's Construction, one of the alternate constructions of R, named after the inventor of decimals (see: http://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Stevin.27s_construction )

    Under Stevin's construction, 0.999...=1 *by definition*. And this is the only sensible way to prove 0.999...=1. All the other ways involve circular reasoning. (The infinite series proof? It assumes limits are unique-- which is NOT a true property of numbers if by "number" we naively mean "string of digits". It can be made rigorous, but only through a detour which is well outside a high school course.)

    ReplyDelete
  2. Glowing Face Man -- Thanks for commenting. It's interesting, because as I was musing about it last night, I was in fact thinking that the decimal subtraction in Step #3 might be weaker part of the demonstration.

    My take is this -- the standard decimal subtraction algorithm starts on the far-right, which does not exist here. So I think we're making use of this idea: If you know in advance that there are no carries (lower digit always less than or equal to any upper digit), then we can alternatively do all place-value subtractions in any arbitrary order; and of course, in all the digits after the point, 9-9=0. So I'd get the result of 9.000..., and if anything, the unstated assumption I could identify is that that's the same as 9.

    So I don't follow this: "if you work it out 'naively', is not 9 at all but 8.999...". Can you explicate what you think the "naive" work is, because I don't see it?

    ReplyDelete
  3. >So I don't follow this: "if you work it out 'naively', is not 9 at all but 8.999...". Can you explicate what you think the "naive" work is, because I don't see it?

    Certainly. We're calculating
    9*(.999...)
    We multiply 9 by .9 to get 8.1. To this we add 9 times .09 which is .81, giving 8.91. And so on and so on-- we get 8.999...

    A paper which I keep meaning to read, though sadly I haven't gotten around to it yet, is: "The real numbers as a wreath product" by F. Faltin, N. Metropolis, B. Ross and G. C. Rota. If I understand right, this paper addresses issues like the "infinite carrying problem" in your comment.

    ReplyDelete
  4. But Step #3 is a subtraction (not a multiplication). You agree that for all real x, 10x-x = 9x?

    ReplyDelete
  5. I'm roleplaying as a high school math prodigy. I agree that all my life, teachers have told me 10x-x=9x, without proof (none have ever mentioned the axioms of a complete ordered field!) Suddenly a teacher wants to prove something to me-- that 0.999...=1-- and that's great, but along the way, he uses a different fact which was never proved, the distributive law. And indeed, according to my calculations-- multiplying 9 by 0.999...-- it seems as if 10x-x is *NOT* 9x. Unless 8.999... happens to equal 9, but how can I establish that?

    There are 3 ways out of this dilemma:

    1. explicitly assume the axioms of a complete ordered field and generalize our discussion to any isomorphic copy thereof. The problem: the definition in this blog post of "0.999..." doesn't make sense. If reals are taken to be Dedekind cuts or Cauchy sequence equivalence classes, what's a "digit"? It requires a lot more work.

    2. Work in an explicit model of the reals (e.g., Dedekind cuts) and explicitly prove the distributive law as well as the other axioms. Again, "0.999..." becomes very difficult to define, UNLESS the model we are using is Stevin's Construction.

    3. More realistically for high school: Don't try to prove 0.999...=1. Instead, say something like: "Numbers are a tool, and as a tool, we would like them to have the property that distinct numbers have nonzero difference. The difference between 1 and 0.999... seems to be 0, so in order to force numbers to have the distinct-numbers-have-nonzero-distance property, we simply DEFINE 0.999... to be 1. Because otherwise, as you see by this example about 9 times 0.999..., very basic facts like the distributive law would break." Finally the math prodigy is able to understand, and it makes beautiful perfect sense at last, without having to go into graduate level math.

    ReplyDelete
    Replies
    1. To add some later clarity to my comment below: note that every basic algebra course and textbook takes for granted the commutative, associative, and distributive properties over real numbers, usually in the first section of the book, without proof (you might say that they're taken axiomatically in this context). So to say in #2 that one must "explicitly prove the distributive law" flies in the face of practice everywhere, and is certainly not something that a beginning algebra student has the tools to grasp. (And combining like terms such as 10x-x = 9x is proven by that distributive property.)

      Delete
  6. To me, that seems like a fairly roundabout way of saying that for consistency's sake, we accept the distributive law for real numbers (and thus combining like terms, such as that for all x, 10x-x = 9x). :-)

    ReplyDelete
  7. I prefer a different line of reasoning because I expect it to better leverage one's math intuition:

    (1) x = 1/3 = 0.333... (review if necessary with a manual long division)
    (2) 3x = 3 * 1/3 = 3 * 0.333...
    (3) 3x = 1 = 0.999...

    This format may not seem proper to a proof purist, but the math is correct, and I suspect it to be equally or more persuasive to most students. Why?
    a) the algebra is minimal; fewer steps; no substitutions
    b) someone who's done a few long divisions by hand, including 1 divided by 3, can understand deeply that 1/3 = 0.333...
    c) someone who's grasped the simplest fraction illustrations will readily believe 3 * 1/3 = 1
    d) multiplying a sequence by 10 (as in Delta's proof) requires the whole sequence to "shift one place left", which can be a distraction when contemplating what that means for an infinite sequence; by contrast, multiplying 0.333... by 3 allows each digit to neatly "stay in place" in one's mental image, and so is easier to swallow

    Of course, there will always be resistance due to the counter-intuitive results. Perhaps there's a way to address this (bad) intuition directly, rather than prove it wrong?

    ReplyDelete
    Replies
    1. Imho such transformative steps are worthless. They instill entirely incorrect approach that has little foundation in any sort of mathematics I know. You cannot begin addressing this "problem" without explaining what a set is, what numbers are (members of a set), what functions are, and how one defines positional representations. Once those are addressed, you can show a proper proof.

      Anything less muddies the water and uses analogies that are fundamentally broken. Yes, it leads to correct results, but is useless because you can't reuse the analogy for any other purpose, whereas once you know the real mathematical underpinnings, you'll (hopfeully) reuse them in the rest of your education at least.

      A single-purpose analogy is, IMHO, worthless and destructive. Most people will incorrectly reuse such analogies where they don't apply at all, and there's nothing to help them understand what the limits of a certain analogy are. Just because the analogy comes from the same discipline (mathematics), doesn't make it any good. It's all smoke and mirrors. Feynman would wince in pain.

      Delete
  8. I think the root of the issue is that people forget that a decimal positional system is merely a means of representing a certain set of numbers, the latter being abstract objects with certain properties. It so happens that an infinite (but I think countable) set of real numbers has more than one representation in positional decimal system. Of said representations, at least one happens to be infinitely long, too. Once you acknowledge that, there's no controversy anymore -- at least in my mind. I believe I have heard it explained in 6th grade maths in elementary school. I swear.

    ReplyDelete
  9. Further to that discussion, one has to simply point to a number that would be represented by some positional representation. As in giving an expression that would produce the number represented by, for example, 0.(9). By definition of infinite positional representation, you get that such a number is a sum of an infinite series, and then basic calculus tells you the number is the integer 1.

    The arguments where positional numeric representations are transformed multiple times are at best circular arguments, and completely useless because they ignore the key concepts involved: that of the number as an abstract member of a set, the mapping between a number and a positional representation, and the definition of said mapping (the sum of a series).

    Somehow I was taught that *in school* by the end of grade 12, and all that in spite of being usually a C+ maths student. I shiver to think what sort of "education" one has to have to come up with the silliness that usually surrounds the 0.(9) ?= 1 discussions.

    ReplyDelete