Counterintuitive Math Problems

Looks like I’ve just added Ian McEwan’s new novel to my reading list:

During one of their Brighton rendezvouses, after a round of oysters and a second bottle of champagne, Tom Haley asks Serena Frome the question every mathematician longs for her lover to utter:

I want you to tell me something…something interesting, no, counterintuitive, paradoxical. You owe me a good maths story.

Frome (“rhymes with plume”), a twenty-something blonde blessed with the looks of Scarlett Johansson might be the last person one would expect capable of satisfying Haley’s request. But readers of Booker-winning author Ian McEwan’s latest novel Sweet Tooth, a Cold War-era romance of multiple deception (semi-autobiographical, some have argued), should get use to mistrusting first readings. Even her newest lover Haley, a short story author and scholar of Spenser, doesn’t yet know her full story. When this conversation takes place Haley believes he is dating his liaison to the Cultural Foundation that is paying him to write the next great English novel. McEwan’s heroine is actually an agent for MI5 who is sent to lure Haley into a cultural war against communism (a la Encounter) as part of the project fittingly named Sweet Tooth. Though a fresh recruit, the service has decided that Frome’s beauty and steady diet of two to three paperbacks a week makes her ideal for the job.

What counterintuitive, paradoxical maths problem does Haley come up with? Why, The Monty Hall Problem, of course. You should read the rest of the article I linked to, if only to admire their impeccable choice of authorities.

I do love a good counterintuitive math problem. Tell people that


0.99999 \ldots =1

and just watch the mayhem that ensues! Which is funny, since no one balks at the idea that

0.33333 \ldots =\frac{1}{3}.

Likewise for

0.66666 \ldots =\frac{2}{3}.

But, somehow, accepting the results of adding these equations together is a bridge too far.

Here’s a little teaser for you. Would you rather receive $4000 for your first year of work, with an $800 raise every year, or $2000 for your first six months of work, with a $200 raise every six months? Don’t think too hard! Give me your first, knee-jerk reaction.

The first one sure sounds more appealing, but a quick calculation will show that you will do better with the second.

How about this classic? Imagine that you wrap a belt around the Earth at the equator, so that the belt hugs the earth tightly. You then cut the belt, straighten it out, and then extend its length by one mile. The lengthened belt is then wrapped around the Earth again, so that the distance between the belt and the ground is uniform. The problem is to estimate how high above the ground the belt will be. Will you be able to slip a pencil under it? Your arm? Could you crawl under it? Walk under it? That sort of thing.

Well, let’s see. The radius of the Earth is roughly 4000 miles. Since the circumference of a circle is given by the formula


C=2 \pi r,

we see that the initial length of the belt, which we shall call L, is

L=8000 \pi.

That’s in miles, incidentally. The stretched belt therefore has a length of

8000 \pi+1,

and we need to determine the radius that corresponds to this circumference. Let’s call that new radius R. Then we must solve the equation

2 \pi R=8000 \pi +1,

to get

R=\frac{8000 \pi+1}{2 \pi}.

From this we conclude that

R=4000+ \frac{1}{2 \pi}.

For approximation purposes, let’s take pi to be three. Then the new radius is one sixth of a mile longer than the old radius, which is close to 900 feet. So, yes, I’d say you could walk under it.

Actually, it gets better. If you look at that calculation carefully, I think you’ll find that the radius of the earth was not actually relevant. Repeat the calculation with some variable, r say, to represent the original radius, and the conclusion will be exactly the same. The new radius will be a sixth of a mile longer than the old radius.

I think that’s cool!

Comments

  1. #1 Steven Carr
    April 2, 2013

    Take a perfectly round orange. You may have to reject several before finding a perfectly round orange.

    Take a very sharp knife.

    Cut the orange into 11 slices with vertical, parallel cuts so that each slice has exactly the same thickness.

    Which slice would have the most orange peel on it? The ones at the end which seem to be made of mostly peel? Or the slice in the middle, which has a bigger volume?

  2. #2 Steven Carr
    April 2, 2013

    Another nice one.

    You throw 10 coins , one after the other.

    What is the probability that you get a run of either 3 heads in a row or 3 tails in a row in that succession of 10 coin tosses?

    To the nearest 10 percent is it….

    a 10 percent

    b 20 percent

    c 50 percent

    d 80 percent

    e 90 percent

  3. #3 Steven Carr
    April 2, 2013

    In the last question, prove that as the number of coin tosses (n) tends to infinity, the probability of 3 heads or 3 tails in a row tends to a limit of about 1 -.0.809^n

  4. #4 MNb
    April 2, 2013

    “Tell people that 0,9999…. = 1″
    Maybe it’s because they are too old. My first graders never have any problem with it. Neither can I remember anything more than some light frowning as a teenager.

  5. #5 Richard Wein
    April 2, 2013

    “If you look at that calculation carefully, I think you’ll find that the radius of the earth was not actually relevant.”

    The belt-round-the-Earth question came up on University Challenge (a TV programme in the UK) a while back. It was one of the “starter” questions where it’s a race to answer first. I had to think so quickly I didn’t even think about the size of the Earth’s radius. I just visualised the situation, realized that C and R had to be scaled up in proportion, thought of “C=2.pi.R”, and jumped to the answer 1/(2pi).

    But when I applied the same thinking here, I had time to question my answer, and I lost my confidence. My intuition told me that the answer should be much smaller, because the 1 mile loosening had to be spread over the thousands of miles of the Earth’s circumference.

    It seems that sometimes it helps to have a bit less time to reflect.

  6. #6 Verbose Stoic
    April 2, 2013

    Whenever I come across the 0.9999… = 1 question, it always strikes me as a conversion error.

    So, you start by converting the fraction 1/3 into a real number, and get 0.33333… So far, so good. But then you do fraction addition and say that adding 1/3 + 1/3 + 1/3 (in some order), gets 3/3 which converts to 1. So, then, adding 0.3333… + 0.3333… + 0.3333… should be 1 as well. But in real addition, it ends up being 0.9999… So then you conclude that 0.9999… = 1. But does it?

    In real numbers, it doesn’t. What you have is 0.9999… = 1.000… (essentially) and that’s false in real numbers (clearly). So, then are you converting to the integer 1? Well. 3/3 does indeed convert directly to the integer 1, but 0.9999… doesn’t. You’d have to round, and if you have to round you are clearly saying that the number is not that number, so again 0.9999… != 1 when expressed as an integer. So that’s out.

    So, the only other move you can make is to take 3/3 and say that that’s really 0.9999… in real notiation. Except, of course, it isn’t. It’s 1.0000… So, then, what’s clearly happening here is that the conversion between reals and fractions is not precise, and as you try to add fractions and reals as if you had a precise conversion you run into exactly this counter-intuitive — and incorrect — result.

    The same thing would happen if you converted reals to integers by strict truncation — a lot of computers do this — and then said that you have 1.5 + 1.5 = 1 + 1 = 2 but 1.5 + 1.5 = 3 that therefore 2 = 3. But that’s a clear conversion error.

  7. #7 schnablo
    April 2, 2013

    “What you have is 0.9999… = 1.000… (essentially) and that’s false in real numbers (clearly)”
    Don’t take the representation of a number for the number itself. You might write instead of 0.9999… = 1.0
    sum_(i=1) ^inf 9*10^-i = 1
    or more precise
    lim_n->inf sum_(i=1) ^n 9*10^-i = 1
    which is obviously the same thing and perhaps easier to accept.

  8. #8 Eric Lund
    April 2, 2013

    The proof that 0.999… = 1 is quite simple. Let X = 0.999…. Then 10X = 9.999…. Therefore 9X = 10X – X = 9.999… – 0.999… = 9, or X = 1.

    More generally, there is the sum rule for geometric series: a * sum(n=1,infinity)(1/x^n) = a / (x – 1). Which, IIRC, can be proven by generalizing the above proof that 0.999… = 1.

  9. #9 Poincare
    April 2, 2013

    In the novel, it is Serena who comes up with the Monty Hall problem. Of course, since it’s McEwan’s novel, the author has posed and answered his own question. It is interesting how often we discuss and even argue over abstract characters as if they are real. The parallel to math is too obvious.

  10. #10 Nick Theodorakis
    April 2, 2013

    Eric Lund posted the proof I like the best, as it is easily understandable and unassailable. It’s also the method used for finding what fraction is represented by a decimal rational number. In other online discussions, I noticed that pushing too hard on adding three 0.333… may simply result in someone deciding to believe that 0.333… is not actually equal to 1/3.

    Demonstrating that rational numbers constitute a countable set is a nice counterintuitive problem I’ve liked.

  11. #11 The Science Pundit
    April 2, 2013

    “Which is funny, since no one balks at the idea that
    0.33333… =1/3.”

    You’ve obviously never read through the comment threads of “0.999…=1″ posts.

  12. #12 Lenoxus
    April 2, 2013

    Surely, Jason, you were aware of what you would do to this page by bringing up .9…? ;)

    I wonder if there’s a term for a statement which no one argues with (such as 1/3=.3…) until a seemingly incorrect conclusion is drawn from it. (Sometimes both the statement and conclusion are right and sometimes they’re both wrong).

    On one .9… forum, someone argued that .3_ may or may not equal 1/3 depending on “how the .3_ was calcuated”, or something to that effect. This and numerous other data (such as the unsual acceptance of .3_=1/3) suggest to me that one significant reason peoplle have trouble with .9_ is that they tend to approach demical fractions (or “decimals”) in a very pragmatc way; each decimal has to be “for” something, the end result of a process, the solution to a pressing real-world problem. .3_ is “allowed” because that’s the best you can do to espress 1/3 in decimal, an occasionally necessary task. .9_, on the other hand, doesn’t “come from” 1 in some obvious way, so it’s not “allowed” to equal 1.

    Anyway, my personal favorite quasi-proof of the equality is to consider the difference between .9~ and 1. Some will call it .0~1, which involves an infinite series of zeroes somehow followed by a one. (In standard arithmetic, any digits after an infinite series are safely ignored.) But what happens if we add this to .9~? Shouldn’t the answer be .9~1, surely as valid a number as .0~1?

    So then we have to adjust our number system even more. The net result would be something like “everything statement made regarding infinite decimal expressions must be true of the equivalent finite expressions.” So just as .9, and .99, and .999 are all less than 1, so too would .9~ be less than 1.

    Well, here’s a challenge to this new arithmetic: Is 1/2 * 1/3 equal to 1/6? Well, the decimal form of 1/6 is .1666~ (one followed by infinite sixes). If we try extrapolating the decimal form of .5 * .3~, we get .1666~5. According to standard arithmetic, that has to be the same as .1666~. According to our fancy new .9~ denying arithmentic, those numbers are different! It’s madness.

  13. #13 Lenoxus
    April 2, 2013

    Verbose Stoic:

    3/3 does indeed convert directly to the integer 1, but 0.9999… doesn’t. You’d have to round, and if you have to round you are clearly saying that the number is not that number, so again 0.9999… != 1 when expressed as an integer.

    “Rounding” always involves adding a specific number to produce a new value. What are you adding to .9~ to result in 1?

    Your entire post is devoid of proof, just assuming the conclusion. This is a very common mistake and you probably didn’t even realize what you were saying because of how obvious it looks to you. People often say “if 0.9… were equal to 1, then it would be 1, not 0.9…!” or variations thereof. It may sound reasonable, but it’s like saying “If 10/14 were equal to 5/7, then it would be 5/7, not 10/14!”

    .9~ and 1 are just two notations for the same number, which can also be written as “one” or as “I” or as “five minus four” or as “the sum from K equals one to K equals infinity for the term nine-tenths to the power of K.”

  14. #14 Eric Lund
    April 2, 2013

    So just as .9, and .99, and .999 are all less than 1, so too would .9~ be less than 1.

    One of Zeno’s paradoxes is mathematically equivalent to this. In Zeno’s version, motion is impossible, because before you get to your destination you must travel halfway there, before you get halfway there you must get 1/4 of the way there, and so on ad infinitum. What Zeno overlooked is that time can be similarly subdivided.

    I think the bottom line is that many people do not understand the concept of taking a limit. The repeating decimal representation is a geometric series. If you truncate the series after a finite number of terms, you have a residual, which gets smaller as you increase the number of terms you include in the sum. The key point that many people miss is that the residual gets arbitrarily small (in technical terms, for every delta there is an epsilon), so that you can treat it as being 0.

    I suspect this issue has gotten more acute in the computer age, where you are forced to truncate the partial sum after some finite number of terms. Verbose Stoic, above, illustrates this issue nicely. If, for instance, you are dealing with IEEE standard floating point numbers (whether single or double precision; all that changes is the number of digits you retain), a number like 1/3 does in fact get truncated to 0.3333333 (or something like that). So 1/3 + 1/3 + 1/3 is computed as 0.9999999, which to the computer is a different number from 1.0000000. This is why, when programming floating point numbers, you generally have to allow some tolerance in equality tests: due to the finite precision of floating point, you will get some roundoff error.

  15. #15 Swimmy
    April 2, 2013

    When I discussed .999… = 1 with a friend his objection was that .999 doesn’t equal one (correct) and .9999 doesn’t (correct), so no matter how many decimals we added on we still couldn’t get to one.

    “Why?” I asked.

    “Because there would always be that one number way down the line separating the two.”

    “But if the 9s repeat infinitely, there is no such number anywhere down the line. There’s never a demarcation; that’s practically the definition of infinity.”

    And he got it right away. I think people just have a really poor understanding of infinity in math. And why not, if they haven’t studied it deeply? It has quite a few quirks. You can come up with endless (ha) counterintuitive math problems when you play with infinity.

  16. #16 Lenoxus
    April 2, 2013

    Regarding Carr’s orange (and assuming that the relationship petween peel size and orange size is identical to that between a true sphere’s surface area and its volume): I suspected the answer would be that they’re all equal. I then cheated by looking up the way to calculate this type of surface area.

    It seems my intuition was right; that formula only takes into account the width of the segment (not the diameters of its ends), and here all the widths are equal.

  17. #17 deepak shetty
    April 2, 2013

    VS is as good at math as he is at philosophy :)

  18. #18 Sean T
    April 2, 2013

    I think the whole .999… = 1 problem becomes even clearer in binary arithmetic. Most of us, somewhere in our mathematical education learn that the series 1/2 + 1/4 + 1/8 + 1/16 … = 1. Since the binary representation of that series is 0.1111…, that shows that 0.1111… = 1 in binary arithmetic. It would not be too difficult, armed with that information to show that 2/3 + 2/9 + 2/27 + 2/81 … = 1. In base 3, that’s represented by 0.2222…. By analogy (or possibly by mathematical induction), it’s then apparent that the decimal expression 0.9999… = 1, since it’s just a way to represent the series 9/10 + 9/100 + 9/1000 + 9/10000 ….

  19. #19 AnonymousCoward
    April 2, 2013

    This post intersects nicely with today’s SMBC comic

  20. #20 Michael Kelsey
    SLAC National Accelerator Laboratory
    April 2, 2013

    @Lenoxus — I’m glad I read through all the comments before posting my response to Carr’s orange.

    @Carr — Your orange slices are a great implementation of throwing a uniform spherical distribution (for which I just happened to be writing a tiny bit of code before reading this blog!). The differential area of a spherical surface is R dphi dcos(theta), _not_ (as many of my undergrads assume) R dphi dtheta.

    Cos(theta) steps are what you get when you slice the unit sphere perpendicular to the z axis. So your orange slices all have areas of 2piR*delta-z => 2piR delta-cos(theta), and hence all equal.

  21. #21 GrayGaffer
    April 2, 2013

    Lots of posts. This is a different one, though like the Earth belt; one of my favorites, ever since we had to do the math in HS:

    Given a frictionless, fully evacuated tunnel, and a passenger car that can traverse said tunnel only under the force of gravity, how long does it take to travel through one from New York to London? Hong Kong and Melbourne? Down the street? No push allowed at the start, just let it go.

    Answer nicely close to HHGTTG’s answer, IIRC.

  22. #22 Poincare
    April 2, 2013

    OK, a little education for a newbie please. What is
    IIRC?
    HHGTTG?
    SMBC comic?

  23. #23 Another Matt
    April 2, 2013

    IIRC = “If I recall [or remember] correctly”
    HHGTTG = Hitchhiker’s Guide To The Galaxy
    SMBC = “Saturday Morning Breakfast Cereal” — a webcomic

    Here’s today’s:
    http://www.smbc-comics.com/index.php?db=comics&id=2934#comic

  24. #24 Another Matt
    April 2, 2013

    There’s another argument that pops up, I think due to familiarity with calculators.

    1/3 = 0.33333…3
    2/3 = 0.66666…7

    “1/3 ends in 3 because the residual rounds down. 2/3 ends in 7 because the residual rounds up.”

    0.333333…3 + 0.66666…7 = 1.0 , not 0.99999…

    As others have pointed out, this is a problem with not taking infinite series as seriously infinite.

  25. #25 Richard Wein
    April 3, 2013

    I think what some people overlook is that the notation of recurring numbers is just an alternative way of expressing rational numbers. (VS seems to think it can express non-rational reals.) So it’s futile to question whether 0.3_ is really equal to 1/3. That’s just what it’s defined to mean. But I suspect that in the popular imagination it means something like “what you get if you keep repeating the digits forever”. So we’re tempted to ask, “what would I get if I kept repeating the digits forever?”. But the question is meaningless, because going on forever means never stopping and so never getting anything. It’s similar to thinking that infinity is just the number you get if you keep adding 1 forever.

    The next problem is that following certain rules mechanically leads us to write 3 x 0.3_ = 0.9_, and 0.9_ is a peculiar way of denoting 1. It might have been better to eschew the use of “0.9_” and write 3 x 0.3_ = 1 instead. It seems to me that the system of notation was only designed to express fractions, not whole numbers. So it’s not surprising we get something peculiar when we use it to express a whole number. Rather than saying that 0.9_ = 1, perhaps it would be better to say that “0.9_” is a misuse of the notation. Just a thought.

  26. #26 Lenoxus
    April 3, 2013

    Richard Wein:

    Rather than saying that 0.9_ = 1, perhaps it would be better to say that “0.9_” is a misuse of the notation.

    But how is it any more a misuse than “0.3_”? Repeating decimals don’t have to be “for” any particular purpose, they’re just representations. Even if no one had ever bothered with the decimal form of a fraction like 1/3 or 1/7, eventually some mathematician would wonder, “Hey, what would be the result of extending 0.111 to an infinite number of places?”

    0.9_ is a valid extension of this model. And on top of everything else, unlike dividing by zero or other generally-invalid operations, it doesn’t lead to any contradictions! (Denying the equality, on the other hand, does – not that I’m accusing you of such!)

  27. #27 Sean T
    April 3, 2013

    Richard,

    Think of it this way, infinite decimals are just a shorthand way to represent infinte geometrical series. The decimal 0.3_ is just shorthand for 3/10 + 3/100 + 3/1000 + 3/10000…, or in other words, sum (i=1 to inf) 3*(1/10)^i. This series is equal to 3 * sum(i=1 to inf) (1/10)^i. By the formula s = 1/(1-r), the series sum(i=0 to inf) (1/10)^i = 10/9. Since that series differs from the series starting at i=1 only by having a lead term of one, the series starting at i=1 has a sum of 1/9. Therefore, our decimal representation 0.3_, which is short for sum(i=1 to inf)3*(1/10)^3 is actually equal to 1/3.

    Doing the same calculation for 0.9_, which is short for sum(i=1 to inf) 9*(1/10)^i: As above, sum(i=0 to inf) (1/10)^i = 10/9. Sum(I=1 to inf)(1/10)^i is 1 less, so it’s equal to 1/9. Multiplying by 9 gives the result that sum(i=1 to inf)9*(1/10)^i = 1.

  28. #28 Richard Wein
    April 3, 2013

    Sean: “Think of it this way, infinite decimals are just a shorthand way to represent infinte geometrical series.”

    Thanks. I was blind but now I see. I retract my previous comment!

  29. #29 Rupert Millard
    UK
    April 3, 2013

    “Here’s a little teaser for you. Would you rather receive $4000 for your first year of work, with an $800 raise every year, or $2000 for your first six months of work, with a $200 raise every six months? Don’t think too hard! Give me your first, knee-jerk reaction.

    The first one sure sounds more appealing, but a quick calculation will show that you will do better with the second.”

    Have you made a mistake here? The second person’s salary goes up by $400 a year, so they never earn as much as the first person in a month, never mind catch up and overtake them!

  30. #30 Lenoxus
    April 3, 2013

    I agree with Rupert. Did you make an error, Jason, or is there some subtlety we’re missing?

  31. #31 Sean T
    April 3, 2013

    Another counterintuitive problem I like:

    There’s an old story about how the earth is supported on the back of a turtle. When questioned about what supports the turtle, believers answer “another turtle”. When questioned about what supports that turtle, the believer answers “it’s turtles all the way down.”

    Taking the story seriously (for the benefit of this problem only, of course), there are presumably an infinite number of turtles. Therefore, the mass of all the turtles is likewise infinite. The counterintuitive question then is since there is an infinite mass to be supported, how many turtles actually have to support an infinite mass?

    The correct answer is zero. Even though there is an infinite mass to support, none of the turtles actually supports an infinite mass.

    Proof: Use mathematical induction.

    P(x) = the xth turtle down from the earth supports a finite mass. P(1) obviously holds; the first turtle only supports the earth’s finite mass. Now assume P(n). That means that the nth turtle supports a finite mass. The (n+1)th turtle therefore supports the mass of the nth turtle plus the finite mass supported by the nth turtle. The sum of two finite numbers is finite, so that shows that the (n+1)th turtle supports a finite mass, and therefore P(n) –> P(n+1). Thus, by mathematical induction, for all n, the nth turtle down from the earth supports a finite mass. Thus, no turtle supports an infinite mass.

  32. #32 Eric Lund
    April 3, 2013

    Rupert @29: Jason is correct. The first person gets $4000 in his first year, $4800 in his second year, etc. The second person gets $2000 + $2200 = $4200 in his first year, $2400 + $2600 = $5000 in his second year, etc. The key point being that the second person’s salary increases by $400 per half-year per year, or $800 per year per year. Thus the second person’s income remains $200/year ahead of the first’s.

  33. #33 Rupert Millard
    UK
    April 3, 2013

    Thanks, Eric, I understand now.

  34. #34 Roger
    April 3, 2013

    @32, 33
    I think it was the wording. I, too, was a little confused originally, but your explanation makes it clearer. It’s based on getting that $200 head start, and the perceived gap between the raise amounts is negated by the extra payday per year (each payday benefiting from the raise). Kind of like (but not exactly) making an extra mortgage payment.

  35. #35 Verbose Stoic
    April 4, 2013

    Gee, I posted that comment looking for responses … and then forgot I posted it. My bad.

    Eric Lund,

    The proof that 0.999… = 1 is quite simple. Let X = 0.999…. Then 10X = 9.999…. Therefore 9X = 10X – X = 9.999… – 0.999… = 9, or X = 1.

    I know that this probably wasn’t addressed to me, but why isn’t this seen as what I commented the original proof was: an issue with mathematical games that make a conversion error? How, for example, do you subtract out an infinite number from an infinite number? Does it really work that way? Again, I can simply appeal to the definitions of the numerals themselves and point out that this doesn’t just seem counter-intuitive, but false by definition. Why should we accept that the problem is with us thinking that 0.999_ isn’t the same thing as one as opposed to thinking that something has gone wrong with the multiplication/subtraction in your equation?

    More generally, there is the sum rule for geometric series: a * sum(n=1,infinity)(1/x^n) = a / (x – 1). Which, IIRC, can be proven by generalizing the above proof that 0.999… = 1.

    So, is it really 1, or EFFECTIVELY 1? There’s a big difference here. I’ll accept effectively, but not really, and think that the counter-intuitiveness of the example comes from people thinking that it’s really 1 and not effectively 1.

    Lenoxus,

    “Rounding” always involves adding a specific number to produce a new value. What are you adding to .9~ to result in 1?

    That’s not how I round. I round by looking at the relevant number that I’m rounding and seeing if it is the equivalent of 5 or higher. If it is, then I CONVERT the number to the next highest, otherwise I CONVERT it to the next lowest number. My example of a rather stupid rounding mechanism — but one that computers actually use — should make that clear: the computer doesn’t subtract out the decimal portion, but tosses it away. I argue that that is indeed generally what it means to round, although we generally do it more intelligently. And if we round it that way, then the addition argument doesn’t in any way hit the objection.

    Your entire post is devoid of proof, just assuming the conclusion. This is a very common mistake and you probably didn’t even realize what you were saying because of how obvious it looks to you. People often say “if 0.9… were equal to 1, then it would be 1, not 0.9…!” or variations thereof. It may sound reasonable, but it’s like saying “If 10/14 were equal to 5/7, then it would be 5/7, not 10/14!”

    You missed my key arguments for my conclusions, though:

    1) 1/3 = 0.333_ is not definitional, but is a conversion between two different number systems, and things can go wrong when you convert, depending on how you do it.

    2) 0.999_ to the INTEGER 1 is indeed a conversion, that has to happen through rounding, and if you have to round to get there it isn’t the same number.

    3) If you stick to one number system, 0.999_ would have to equal 1.0 (I argued for 1.000_) … but under the definition of the real number system those are, indeed, different numerals and so represent different numbers. Trying to argue that the different numerals represent the same number is allowed, but you need a reason to make that argument beyond “It leads to a nice, counter-intuitive result” and that makes sense of 0.999_ having that property while being consistent with whatever you’re going to say for, say, 0.888_.

    .9~ and 1 are just two notations for the same number, which can also be written as “one” or as “I” or as “five minus four” or as “the sum from K equals one to K equals infinity for the term nine-tenths to the power of K.”

    So, what two numerals map to the same number for the numeral 0.3333_?

    Eric Lund,

    I think the bottom line is that many people do not understand the concept of taking a limit. The repeating decimal representation is a geometric series. If you truncate the series after a finite number of terms, you have a residual, which gets smaller as you increase the number of terms you include in the sum. The key point that many people miss is that the residual gets arbitrarily small (in technical terms, for every delta there is an epsilon), so that you can treat it as being 0.

    As I said above, I don’t think the problem is saying that you can treat it as 0. I certainly accept that. The problem for me is that you seem to be saying that it IS 0, and to me that is the mistaken view of a limit: a limit is the number that the series approaches but never, in fact, actually reaches. So applying the limit in this case seems to me to be saying that the number approaches but never quite reaches 1 … at which point, saying that 0.999_ = 1 or, as Lenoxus said, that they are two notations for the same number seems to be violating the definition of “limit”.

  36. #36 csrster
    April 4, 2013

    I think the root of the problem people have with 0.9_=1 is that they have some vague (but oddly unshakeable) intuition that the decimal numbers are a unique representation of the reals. They also have an intuition that maths should support their intuitions. The latter is infinitely more dangerous.

  37. #37 Verbose Stoic
    April 4, 2013

    csrster,

    I think the root of the problem people have with 0.9_=1 is that they have some vague (but oddly unshakeable) intuition that the decimal numbers are a unique representation of the reals. They also have an intuition that maths should support their intuitions. The latter is infinitely more dangerous.

    While deepak shetty’s comment wasn’t worth replying to, note that I’m doing more philosophy of mathematics here than mathematics. As such, my comment is that that isn’t really just an intuition, but follows from how number systems usually work. In almost all number systems, you define one numeral for each number. It isn’t a requirement, but by default that’s what we do because a number system where that isn’t true becomes far less useful. If you define, for example, one numeral to map to multiple numbers, you can’t tell by looking at the numeral what number that is supposed to represent. On the other hand — and this is case people are arguing for here — if you have multiple numerals mapping to the same number it seems like you’ve introduced an unnecessary redundancy: why have both numerals if they, effectively, mean the same thing?

    If we try to do this for the case here, we see that we should keep 0.3333_ because that seems like a unique number that we need to represent 1/3 when we convert, and we’d need 0.6666_ for the same reason … but when it comes to 0.9999_ we’d be saying that we don’t need it because it’s really 1. But this would seem quite arbitrary, and would suggest that you should simply never use the numeral 0.999_, even when you add 0.3333_ and 0.6666_. And, of course, you can do that in mathematics if you can get everyone to sign on, but that does seem to me to be the consequence of your argument, and not what most people think of when they think about these things.

  38. #38 Lenoxus
    April 4, 2013

    Verbose Stoic:

    How, for example, do you subtract out an infinite number from an infinite number?

    Be careful with terms like “infinite number.” Surely you agree that .9~ is not infinite in value? It is, after all, larger than zero but smaller than, say, five. Do you think that it’s incorrect to say that 10.141592… (where those digits correspond in full to the digits of pi) minus 7 equals pi?

    Do you accept that 9.9~ is exactly the same as “9 plus .9~”? If you do, then obviously we can subtract .9~ from it, reaching an answer of 9. Yours is probably the most common objection to that proof, that .9~ is somehow “unsubtractable”. But it is just as subtractable from other numbers as pi is.

    Another common objection is to say that .9~ times 10 is not simply 9.9~, but rather a number for which is there is no standard notation: “zero followed by a decimal point followed by infinity-minus-one nines.” This suffers from significant problems. To begin with, we have to reject standard set theory whereby countable-infinity-minus-one equals countable-infinity. A full Hilbert’s Hotel would have to reject incoming guests. Cats and dogs would be living in harmony. Chaos!

    Further, if we try to develop an arithmetic of “infinity-minus-one nines” or “infinity minus one zeros” (in the case of the standard wrong answer to “What’s 1 minus .9~?”), we run into serious complications, as I raised earlier with my example of one-half-times-one-third being apparently unequal to one-sixth.

    I am not a mathematician, but in general, it’s my understanding that mathematicians value consistency over intuitiveness. If you end up with P and not-P, you try to reject one of your premises, not matter how weird the implications seem.

    I round by looking at the relevant number that I’m rounding and seeing if it is the equivalent of 5 or higher. If it is, then I CONVERT the number to the next highest, otherwise I CONVERT it to the next lowest number.

    I concede this point. Truncation is a valid operation; it just happens not be what what’s going on here.

    1) 1/3 = 0.333_ is not definitional, but is a conversion between two different number systems, and things can go wrong when you convert, depending on how you do it.

    This I don’t understand. Does 1/3 = .3~ or not? Remember, .3~ is a single number; there isn’t more than one kind of .3~.

    Trying to argue that the different numerals represent the same number is allowed, but you need a reason to make that argument beyond “It leads to a nice, counter-intuitive result” and that makes sense of 0.999_ having that property while being consistent with whatever you’re going to say for, say, 0.888_.

    0.8_ equals eight-ninths. .9_ equals nine-ninths. No inconsistency occurs.

    So, what two numerals map to the same number for the numeral 0.3333_?

    Are you asking “What is another way to write .3_ in decimal?” or to put it another way, “What is .3_’s equivalent to what .9_ is for 1?” The answer is that there isn’t another such string.

    All rational numbers that (in reduced form) have a prime factor not shared with ten will have a unique expression as a decimal string. All other rational numbers have two: one that ends in a infinite series of 9s, and one that terminates. For example, 1/7 is only expressible in decimal as .142857142857~, but 1/2 is both .5 and .4999~.

    So applying the limit in this case seems to me to be saying that the number approaches but never quite reaches 1 … at which point, saying that 0.999_ = 1 or, as Lenoxus said, that they are two notations for the same number seems to be violating the definition of “limit”.

    Ah, at last we reach the primal misunderstanding of 0.999_. It is that the number “approaches but never quite reaches” one.

    Here’s he thing about 0.999_. It’s not growing. It’s not moving. The number of nines in it is always the same. All the nines are already there.

    0.999_ is not a process that correlates with some variable; it’s just a number. Otherwise, you have to explain what the variable is, and you get into weird situations like “My .999_ has a different value than yours, because my N is 150 and yours is 3,800.”

    And this is entirely accord with the definition of limit, not against it. Yes, if you look at a graph of a function with a limit, you will see that the function “never” reaches the limit. But this doesn’t mean that the limit is not a single specific number! And .9_ is in fact equal to the limit of the function “f(x) = sum from 0 to x for 9/10 to the power of x”. And in fact, when x = infinity, the function does reach the limit. That’s just one of those things about infinity.

    If we try to do this for the case here, we see that we should keep 0.3333_ because that seems like a unique number that we need to represent 1/3 when we convert, and we’d need 0.6666_ for the same reason … but when it comes to 0.9999_ we’d be saying that we don’t need it because it’s really 1.

    We don’t have to make any choice about “keeping” or “rejecting” a notation. If it works, it works. “0.9_” is just as “one-ish” as “1″. It’s not like “1″ is a superior model of “0.9_”, so that “0.9_” should be discarded in favor of the improvement. It may well be less useful than “1″ in most contexts, but there are exceptions, such as that one you bring up here: adding 1/3 and 2/3 in decimal.

  39. #39 JimV
    April 4, 2013

    I have to agree with what I take as the spirit of Mr. Shetty’s reaction, VS has assumed that his intuition counts as philosophical reasoning, which makes his thinking suspect. For example,

    “In almost all number systems, you define one numeral for each number. It isn’t a requirement, but by default that’s what we do because a number system where that isn’t true becomes far less useful. If you define, for example, one numeral to map to multiple numbers, you can’t tell by looking at the numeral what number that is supposed to represent.”

    5/7 = 10/14 = 1.4; 0×20 (hexadecimal) = 32 (decimal). It is very, very useful in mathematics (and in engineering) to have different ways of representing the same number, because the different ways work better than other ways in different situations. As I recall, the Egyptians had just one way of representing all numbers as continued fractions, which if you had to learn to do arithmetic with, you would find very cumbersome. Roman numerals are also bad for doing calculations – but good for embellishing documents and the cornerstones of buildings.

    “,,,not what most people think of when they think about these things.”

    Most people are mathematically illiterate. And some think the moon is made of green cheese.

    In short, nothing in the above “philosophy of mathematics” constitutes an argument. If philosophy means anything except “how to attempt to rationalize one’s innate beliefs”, now would be a good time to provide evidence to the contrary.

    To explain Mr. Shetty’s and my reactions, which do verge on rudeness, the .9999… question has become something of a pons assinorum in recent times, distinguishing those who can make mathematically valid arguments from those who cannot. Which is not to say that I never make any mathematical mistakes. In fact, I take great comfort from my favorite Einstein quotation: “All mathematicians make mistakes; good mathematicians find them.”

    I especially like the lack of a time-limit in that statement, so with a bit of research and thinking perhaps VS will come to understand his mistake.

  40. #40 JimV
    April 4, 2013

    Sorry, that should be 1/1.4 in the first counter-example. A mistake – but found!

  41. #41 Another Matt
    April 4, 2013

    There are still other intuitive approaches.

    Like — between two nonequivalent real numbers there are an infinite number of other real numbers. Between two equivalent numbers there are zero. It borders on the nonsensical to even speak of “between two equivalent numbers”, but let’s take it as a notational question:

    All x such that 1/3 < x < 0.3333… or 0.3333… < x < 1/3 is the null set.

    Likewise for 1 < x < 0.9999…. or 0.9999…. < x < 1 — the set of all such x is also null.

    I think I'm puzzled most by this comment:

    So, is it really 1, or EFFECTIVELY 1? There’s a big difference here. I’ll accept effectively, but not really, and think that the counter-intuitiveness of the example comes from people thinking that it’s really 1 and not effectively 1.

    Here’s how I understand this objection:

    “Just because these two notations in base 10 have rationally indistinguishable behaviors in all contexts where one is substituted for the other does not mean they are really equivalent.”

    Why not? What is this essential difference?

  42. #42 Eric Lund
    April 4, 2013

    VS @35: The problem for me is that you seem to be saying that it IS 0, and to me that is the mistaken view of a limit: a limit is the number that the series approaches but never, in fact, actually reaches.

    Several mathematicians in the 19th century devoted some effort to proving rigorously that yes, if the limit exists (whether it’s the computation of a derivative or the summation of an infinite series), you can do this. Details are a bit beyond my understanding (I am a physicist, not a mathematician), but I know that it can be done.

    The trick, as Lenoxus points out, is that you really are dealing with an infinite number of terms, and infinities have a disturbing tendency to behave contrary to your intuition, as they clearly do here.

  43. #43 Verbose Stoic
    April 4, 2013

    One of the big issues here, at least for me, is that most of the responses harp on “You can’t trust your intutions”, when for me I am not, in fact, relying on intuitions but am, in fact, relying on the definitions of the number systems and other mathematical terms involved, and from my perspective while there have been a lot of attempts to patch up the initial “proofs” or insert new ones, there have been few attempts to actually address the definitions that I’m talking about. So let me try to express it clearer.

    When I was in high school — a long, long time ago — one of the math teachers would put up posters with all sorts of these counter-intuitive results. I seem to recall one of them that proved that 1 = 2, or that if you assume that A != B then A = B (or the other way around). But when we looked at those problems, no one thought that this meant that 1 really did equal 2, for example, because we know that in base 10 integers 1 and 2 are distinct numerals that point to two different numbers. So there had to be a problem in the example that led to that problem, and no one changed the definition of the integer numbers to match that example.’

    My biggest objection, which I have stated before, is that that is what is happening here, in that in the real number system 0.999_ and 1.000_ (which is really what 1 is in the reals; we just don’t normally list trailing zeroes) are two different numerals that represent two different numbers. To me, what you’re doing is basically saying that if I take 0.333_ and multiply it by 3, not only COULD I simply call that 1.000_ without having to round, but in fact I really SHOULD do that. But considering only the reals and not looking at any other number systems or arguments, this is clearly not what we do; the proper thing to do is to use 0.999_. To me, this is because the numerals point to different numbers by definition, and so your examples seem to be challenging that, which creates some rather serious problems, as I have already pointed out. So, not intuition, but my understanding of how the real number system is defined. I could be wrong about that understanding, but simply saying that my intuitions are wrong is too dismissive of my actual argument.

    So, onto the specific comments:

    Lenoxus,

    Yours is probably the most common objection to that proof, that .9~ is somehow “unsubtractable”. But it is just as subtractable from other numbers as pi is.

    Well, except that pi isn’t as easily subtractable as you think. If you take 5.00 and subtract 3.14 from it — using that to represent pi — you get 1.86, which is an approximation to what happens if you subtract pi from 5.00. But it works out well enough for most purposes. But if you said that what you get if you subtract pi from 5.00 really is just 1.86, you’d be wrong; there are a lot of other numbers you’d need to subtract as well, and so you can’t actually do it. Given that example, we have reason to think that the subtraction you posit is perhaps more of an approximation than an exact match, making it problematic to prove what you’re trying to prove. Which is less important, though, because if I am right that 0.999_ and 1.000_ point to different numbers in the reals by definition, then we clearly have an issue with the subtraction in the case that demonstrates it, and if you can demonstrate that they really point to the same number, then we won’t care to examine if there are any issues with subtraction or not.

    I am not a mathematician, but in general, it’s my understanding that mathematicians value consistency over intuitiveness. If you end up with P and not-P, you try to reject one of your premises, not matter how weird the implications seem.

    And the whole thrust of my argument is to question why you are rejecting the premise that 0.999_ and 1.000_ point to different numbers based on one argument that is provably invalid and another that seems right but may have issues.

    This I don’t understand. Does 1/3 = .3~ or not? Remember, .3~ is a single number; there isn’t more than one kind of .3~.

    All mathmatical operations are defined wrt the number system they are in, if I remember my abstract algebra properly. This would include “=”. Now, 1/3 is a fractional number, and 0.333_ is a real, so they aren’t in the same number system. Thus, to do this comparison you have to convert them so that they are in the same number system with a definition for “=”. And so what we generally do is convert the fraction to the real by dividing the numerator by the denominator, and then do the conversion. The problem is that in this case the division leaves a remainder. But we can’t have remainders in the real number system (you can toss them away in integers), so we convert it to the repeating decimal, and the infinite repeating decimal. But 1/3 is not any kind of infinite number in the fractional number system, and is pretty much defined as being the number that if you multiply it by 3 it equals 1. This is, of course, NOT true of the real 0.333_; we generally say that if you multiply that number by 3 you get 0.333_. So the conversion leaves an approximate equality, but not an identical equality, but it’s close enough for most purposes. But it doesn’t work so well here.

    Again, my whole claim about that argument is that those who accept it ignore the fact that conversion is actually happening, and so try to treat them as if there is a non-conversion equivalence. That’s not the case.

    More in the next comment …

  44. #44 Verbose Stoic
    April 4, 2013

    Lenoxus (cont):

    0.8_ equals eight-ninths. .9_ equals nine-ninths. No inconsistency occurs.

    I was actually talking about the results completely in the real number system, without reference to the fractions, because as far as I know we don’t define the real numbers based on fractions, but simply note that we can convert one to the other using division. Under the argument I was talking about, if you claim that you have two numerals that map to the same number then you are saying that in the reals 0.999_ and 1.000_ point to the same number. If you argue that based on issues with infinity, then I point out that 0.888_ is an infinity as well, so does it then have another numeral that points to the same number? If not, why not? You cannot argue that it is infinity that causes the issue and then arbitrarily exclude numbers that have the same property.

    Ah, at last we reach the primal misunderstanding of 0.999_. It is that the number “approaches but never quite reaches” one.

    Nope. People tried to prove their case by appealing to limits, but the definition of limit literally is that it approaches and yet never reaches the limit. So, if the limit argument is true, then my case is proven, as 0.999_ never reaches and so never is just 1.000_.

    And .9_ is in fact equal to the limit of the function “f(x) = sum from 0 to x for 9/10 to the power of x”. And in fact, when x = infinity, the function does reach the limit. That’s just one of those things about infinity.

    Hmmm. If you are defining the number system of reals based on that sort of sum, and it has always been thus, you probably should make sure that you TEACH that that’s what it means before castigating those who go through general mathematics for not understanding the implications of the definition that you only bring up when you claim that they aren’t making mathematically valid deductions (okay, that is more JimV, sorry). To me, that being the definition of a real number — particularly calling it a limit — is a bit too convenient, and runs into two problems:

    First, you turn the numeral 0.999_ repeated by definition into another numeral, and the definition of a numeral should not end up with you defining it as another numeral.

    Second, what’s the definition of 1.000_ in the reals, then? Surely not the same thing …

    We don’t have to make any choice about “keeping” or “rejecting” a notation. If it works, it works. “0.9_” is just as “one-ish” as “1″. It’s not like “1″ is a superior model of “0.9_”, so that “0.9_” should be discarded in favor of the improvement. It may well be less useful than “1″ in most contexts, but there are exceptions, such as that one you bring up here: adding 1/3 and 2/3 in decimal.

    If the numeral notation of 0.999_ is useful for operations in the reals where the numeral notation of 1.000_ would not be, then they actually function differently in that number system and so we should greatly wonder if they really do point to the same number.

    JimV,

    The first thing to note about rudeness is that it is quite rude to refer to someone in the third person who a) is in the thread and b) you are REPLYING to.

    Already answered about intuitions, so …

    5/7 = 10/14 = 1.4; 0×20 (hexadecimal) = 32 (decimal). It is very, very useful in mathematics (and in engineering) to have different ways of representing the same number, because the different ways work better than other ways in different situations.

    Interestingly, this is one of the cases where people are replying to my words relying on things that I pointed out that you couldn’t do. I’m not talking about different number systems, but about different numerals in the same system. You are correct to point out that in the fractional number system, you do, although you are strongly encouraged to always reduce it to the lowest numeral. This is because of how fractional operations work; it’s easier to just do the math and reduce than it would be to always work it out to the lowest numeral. But there is no reason for, say, base 10 integers to have two numerals that point to the same number, and they clearly don’t … and I argue that that is the case for the reals by definition before you start playing around with conversions and infinities. And I say we should take the definition over the playing.

    Another Matt,

    All x such that 1/3 < x < 0.3333… or 0.3333… < x < 1/3 is the null set.

    Likewise for 1 < x < 0.9999…. or 0.9999…. < x < 1 — the set of all such x is also null.

    For the first case, remember that 1/3 is not a real number, but is a fractional, so that conversion doesn’t work.

    For the second case, this is what seems to me works out to a denial of the definition of those numbers, because it seems to me that a more reasonable claim is that there are an infinite number of reals between any two “whole” numbers, 0.999_ is a number between 0 and 1 and so is in that infinite number of reals, but since it is between 0 and 1 it cannot be 1. You’d have to deny that 0.999_ is between 0 and 1, but that seems to be just what the definition of that numeral IS. Unless we have a better argument than this, we have a clash of premises. How do we settle that?

    “Just because these two notations in base 10 have rationally indistinguishable behaviors in all contexts where one is substituted for the other does not mean they are really equivalent.”

    Why not? What is this essential difference?

    Take my truncation example, which is what a computer does. In some sense, you can say that 1/3 is effectively 0, but it clearly isn’t really 0; it’s an approximation that works well enough in most cases (we are debating if it works in this one). That’s the difference between effectively and really here.

    As for your example, I argued in the first comment about moving between number systems. Changing base is one way to change number systems, but you can have different number systems in the same base. The reals, fractionals and integers can all be base 10, and yet behave quite differently. For example, there are numbers in the reals that are not in the integers, and fractionals have no actual infinite numbers whereas the reals do. Once you move between number systems, it is no longer obvious what the operations will do, and as I said you will have to do conversions.

    Eric Lund,

    Several mathematicians in the 19th century devoted some effort to proving rigorously that yes, if the limit exists (whether it’s the computation of a derivative or the summation of an infinite series), you can do this. Details are a bit beyond my understanding (I am a physicist, not a mathematician), but I know that it can be done.

    Again, this comes down to “effectively” versus “really”. Yes, I accept that you can use it as if it was 1, but does that make it really 1 in a way to justify the strong equivalence?

  45. #45 Xuuths
    April 4, 2013

    Actually, I was taught that 1/3 is approximately 0.333… exactly because if you were to add three of them together you’d get 0.999999… and that was not equal to 1.

  46. #46 Another Matt
    April 4, 2013

    VS –

    For example, there are numbers in the reals that are not in the integers, and fractionals have no actual infinite numbers whereas the reals do.

    All non-complex rationals and irrationals and transcendentals belong to the reals, including fractions.

    Also, you can have infinite fractions. Check this:
    http://en.wikipedia.org/wiki/Continued_fraction#Infinite_continued_fractions

    (I know this isn’t what you were talking about, having “infinite numbers” in fractions — but there’s nothing special about fractional notation in any case).

    Look at some of these for pi:
    http://en.wikipedia.org/wiki/Pi#Continued_fractions

    Also infinite series for pi:
    http://en.wikipedia.org/wiki/Pi#Infinite_series

    Etc.

    Take my truncation example, which is what a computer does. In some sense, you can say that 1/3 is effectively 0, but it clearly isn’t really 0; it’s an approximation that works well enough in most cases (we are debating if it works in this one). That’s the difference between effectively and really here.

    That’s a contextual difference where the math turns on the distinction. The only context where the distinction you’re trying to make between 0.999… and 1.000… makes a difference seems to be internet arguments. :) Can you name a mathematical context where the results would differ depending on which you used?

    Cf.

    Again, this comes down to “effectively” versus “really”. Yes, I accept that you can use it as if it was 1, but does that make it really 1 in a way to justify the strong equivalence?

    You seem to be saying that you can use it as if it were 1 in all mathematical contexts — that’s an extraordinarily strong equivalence. I think maybe I could understand where you’re coming from if your objection comes down to pointing out something analogous to the difference between extension and intension — the classic of which is that “Morning Star” and “Evening Star” have the same extension (namely the planet “Venus,”), but different intensions.

  47. #47 Nick Theodorakis
    April 4, 2013

    So what real number is between 0.999… and 1? If they are not equal then there must be at least one (actually, an infinite number of them, but I’ll settle for seeing one).

  48. #48 Nick Theodorakis
    April 4, 2013

    Here’s a cute youtube video explaining the equality of 0.999… and 1:

    http://www.youtube.com/watch?v=TINfzxSnnIE
    (promise it’s not a rickroll or anything rude)

  49. #49 Lenoxus
    April 4, 2013

    Verbose Stoic:

    To me, what you’re doing is basically saying that if I take 0.333_ and multiply it by 3, not only COULD I simply call that 1.000_ without having to round, but in fact I really SHOULD do that.

    No, no one is saying that. You are perfectly free to call it either 0.999_ or 1, or any variant thereof. Neither is more correct than the other.

    Now, 1/3 is a fractional number, and 0.333_ is a real, so they aren’t in the same number system.

    Huh? Fractions (if the top and bottom are both integers) are representations of rationals, which are a proper subset of the reals. If a number is a represented by the former, it has to be a member of the latter, though not always vice versa. Later you said that 1/3 is “not a real number”, so I think you just have some terminology confused.

    Given that example, we have reason to think that the subtraction you posit is perhaps more of an approximation than an exact match, making it problematic to prove what you’re trying to prove.

    But I didn’t intend to truncate or approximate the digits of pi at all; my “.1415926…” was intended to mean “and so forth for all the digits of pi”. And just because we can’t write all those digits doesn’t make it invalid. I can’t write all the digits of a googolplex in decimal, but that doesn’t prohibit me from doing math with a googolplex in it. For example, 2(insert a googol zeroes here) minus 1(insert a googol zeroes here) equals 1 (insert a googol zeroes here). There’s no “approximation of googl” involved; what I wrote represents the entire number.

    Similarly, 10.(insert the post-decimal-point digits of pi here) minus 7.(insert the same thing again here) equals exactly 3. The subtraction is completely okay despite the fact that you can’t do it “in full” on a finite blackboard.

    Again, it seems to me that a somewhat better counter-attack to that proof would be to dispute the multiplication step and say there’s “one fewer nine” after the multiplication is performed. (This, of course, leads to cats and dogs living in harmony, and we can’t have that.)

    Actually, there’s another counter-argument I’ve been sitting on, and I need anyone on this board with the relevant math skills to knock it down. Consider the sum 1 + 2 + 4 + 8 + 16…, or sum from k=0 to k=infinity for 2^k. Obviously this is an infinitely large number. Let’s call it Z.

    Z = 1 + 2 + 4 + 8 …
    2Z = 2 + 4 + 8 + 16…
    2Z + 1 = 1 + 2 + 4 + 8…
    2Z + 1 = Z
    Z + 1 = 0
    Z = -1
    ?

    So it’s clear that the standard .9_ proof is not actually complete. I believe it is still valid if enough other premises are added, such as “.9_ is not infinite in value (eg, it is less than five billion and greater than negative five billion).”

    A lot of bogus math proofs involve performing an operation that’s normally okay but has key exceptions. For example, proofs that 1 = 2 usually involve saying A = 1, B = 2, then doing a bunch of fancy-looking algebra, then dividing both sides by (B-2A), by which point we’ve forgotten that this term would have to equal zero and hence division by it is invalid.

    In the case of my false proof regarding 1 + 2 + 4…, the error is in assuming that one can safely subtract Z from both sides and assume one is subtracting “the same amount”. But this is only a problem with infinities such as aleph-null (which I think Z is equal to) and not with decimal numbers that have infinitely long expressions. Overall, I’m pretty darn sure that there is no such issue with 0.9_, especially because no mathematician has been able to raise one. And again, a rejection of the equality leads to contradictions, which are very bad things.

    So, if the limit argument is true, then my case is proven, as 0.999_ never reaches and so never is just 1.000_.

    “Never reaches”? So how far did it get yesterday? How far is it today?”Never reaches” implies the passage of time or some other variable. Do you accept that 0.9_ doesn’t “increase”, any more than 12 “increases”? Again, all the nines are “already there”. 0.9_ is another way of saying “What is the sum of this infinite set of terms?” or “What is the limit of this sequence?”

    If the numeral notation of 0.999_ is useful for operations in the reals where the numeral notation of 1.000_ would not be, then they actually function differently in that number system and so we should greatly wonder if they really do point to the same number.

    Why? It is often cumbersome to add and subtract integers in binary or in Roman numerals, but that doesn’t mean you will get actually different results (if you do everything correctly). That’s all I meant by .9_ being less useful than 1 in most situations.

    You’d have to deny that 0.999_ is between 0 and 1, but that seems to be just what the definition of that numeral IS.

    There is no such definition regarding numbers that begin with a zero followed by a decimal point. It’s just been the rule that applies to nearly every number you encounter in the real world.

  50. #50 Another Matt
    April 4, 2013

    For the second case, this is what seems to me works out to a denial of the definition of those numbers, because it seems to me that a more reasonable claim is that there are an infinite number of reals between any two “whole” numbers, 0.999_ is a number between 0 and 1 and so is in that infinite number of reals, but since it is between 0 and 1 it cannot be 1. You’d have to deny that 0.999_ is between 0 and 1, but that seems to be just what the definition of that numeral IS. Unless we have a better argument than this, we have a clash of premises. How do we settle that?

    Also this. Perhaps this flows from the idea that numbers notated in fractional form don’t count as among the reals (they do). There are an infinite number of numbers between two reals, not just between two integers. If 0.999_ < 3/3 then 0.333_ < 1/3 for the same reasons you've been proposing above.

  51. #51 GrayGaffer
    April 4, 2013

    Sorry, Nick, your counter-example is making the same number representtion domain confusion VS is talking about.

    I only need refute one claim, in this case the last one, that there are no mbers between 0.9_ and 1.0_:

    In particular, I can show a number that lies between 0.9_ and 1.0_ quite simply.

    0.9_ < 0.f_ < 1.0_

    What's that? you ask. Simple. The video treatment is entirely in the base 10 number representation domain. Quite apart from using a touchy-feely popular non-mathematical concept of the "value" of a written number. But it could as well have been in the base 16 number representation domain – the symbols 0, 1, and 9 are valid in each. Also, and to drive my point home, the difference between hex 0.f and 1.0_ (the same in both reps) is smaller than decimal 0.9_ and 1.0_, since the incremental value of the representational digits is smaller in hex than decimal. One can then take this representation game itself to infinity by specifying larger and larger representation bases, with correspondingly smaller and smaller symbol incrementational values, thus showing there are an infinite number of reals between 0.9_ and 1.0_ based upon number representations chosen.

    (PS: IANAM, I'm a CS, so I do not know the correct jargon for what I called incrementational symbol value, but 1/16 < 1/10 is what I mean)

  52. #52 Another Matt
    April 4, 2013

    GrayGaffer –

    The 0.f_ hex series converges faster than the 0.9_ decimal series, but that doesn’t have anything to do with whether it’s larger.

    An infinite series that converges pi (say) faster than another is not any more equal to pi than the other.

    All this should make those who deny the equality of 0.9_ and 1.0 distrust other things which might depend on infinite series, like Euler’s formula.

  53. #53 Lenoxus
    April 4, 2013

    A question for the disputers, especially GrayGaffer: are there half as many odd integers as there are integers in general?

  54. #54 Ned Rosen
    April 4, 2013

    0.9999… (repeating forever) is “between” 0 and 1 in the same sense that 3 is between 1 and 3.

    Both 1 and 0.999… (repeating) are representations of the same real number. The decimal representation of reals is very useful but it is not perfect because every decimal that terminates represents a real number that is also represented by a decimal that ends in an infinite string of 9′s. The notion that real numbers “are” decimals, is, IMO, the problem here– the real number system does NOT depend on decimal representations for its definition and properties, but it is to cumbersome to keep saying “the real x represented by 0.3333….”, so out of convenience we speak as if reals actually are decimals, which linguistically suggests uniqueness, but that suggestion is false.

    What is sometimes done, in the context of some particular discussion, is to simply ban decimals that end with an infinite string of 9′s, which gives uniqueness. That is, you choose the terminating representation when you have a choice. But it’s still the SAME real number system regardless of whether you (1) use all decimals and accept non-uniqueness or (2) adopt a convention which gives uniqueness but eliminates some decimal representations from the discussion.

  55. #55 GrayGaffer
    April 4, 2013

    Matt: this is not a converging series. This is a notational representation argument, specifically about a positional notation scheme. Number base chosen for the representation is a significant part of it. Decimal is but one such scheme, one of an infinite set of them. Nothing natural about it either – as man-made as all the rest. Hex may have been a poor choice for my example because there is not a simple one to one mapping of the decimal symbol set into ten symbols of the hex symbol set. So I’ll redo it using base 20. With a clearer (I hope) proof that 0.9_ != 1. Or 1.0_. Or whatever else you want the pointer for the first integer expressed as a real to be.

    The ordered set mapping for the symbols is then:

    [10]::= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
    [20]::= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, a, b, c, d, e, f, g, h, i, j}

    Symbols in [10] therefore map to these symbols in [20]:

    {0, 2, 4, 6, 8, a, c, e, g, i}

    thus by simple symbol substitution we can write

    0.9_[10] = 0.i_[20]

    we can also write

    0.8_[10] = 0.g_[20]

    If we can agree that

    0.8_[10] < 0.9_[10]

    then we can agree

    0.g_[20] < 0.i_[20]

    so we have to then agree

    0.i_[20] < 0.j_[20]

    There exists a representation for a number between 0.9_ and 1. Therefore 0.9_ is NOT the same as 1.

    QED

    As to odd numbers: both are infinite sets, because both are uncountable (see Cantor) and hence are the same (infinite) size. Aleph Null, IIRC. Infinities I studied a looong time ago. Representational number bases other than decimal, and in finite representation spaces, I do daily.

  56. #56 GrayGaffer
    April 4, 2013

    I forgot to add: note that the set of bases for representing numbers is itself an infinite set, so my proof extends to showing that there are an infinite number of such numbers between 0.9_ and 1. By simple induction: for any base x, choose base y = 2x. the number represented by 0.(x-1)_[x] is < 0.(y-1)_[y]. Proven for x = 10 and y = 20 above.

  57. #57 Another Matt
    April 4, 2013

    You can’t just map/substitute symbols like that. Take this:

    Binary has {0,1}
    Decimal has {0,1,2,3,4,5,6,7,8,9}

    0.1[2] == 0.5[10] , granted

    BUT

    0.01[2] != 0.05[10]

    rather:
    0.01[2] == 0.25[10]

    This is because the places map to a different quantity in different bases, not just the symbols.

    0.1_[2] != 0.5_[10]

    Both 0.1_[2] and 0.9[10] are infinite series which converge (and converge absolutely) on 1.0

  58. #58 GrayGaffer
    April 4, 2013

    They are not convergent series. Yes it may feel like one as you try and mentally parse what 0.999…. means, but that is an illusion. Drop it. This is number theory, not feelings.

    And yes, you absolutely can perform that symbolic substitution on representational strings. Do you really want me to go that deep?

    Your computer does it millions of times a second. If it did not work we would not be having this conversation.

    Also Godels’ Incompleteness Theorem rest on being able to do symbolic substitution. And the entire domain of formal proofs is based on it. Number theory is but one subset. It’s just a rewrite rule in a more fundamental setting.

  59. #59 Another Matt
    April 5, 2013

    You really can’t do it that way — base conversion is not that trivial.

    Here’s another example which will hopefully drive the point home.
    Let’s compare base 8 and base 16.

    0.2[8] == 0.4[16] == 0.25[10] (yes?)

    But:
    0.22[8] != 0.44[16]

    because:
    0.22[8] == 2/8 + 2/64 == 9/32

    0.44[16] == 4/16 + 4/256 == 17/64

    These differ by 1/64. The real equivalence is:

    0.22[8] == 0.48[16]

    Again, the direct symbolic substitution fails because the places and the symbols have significance.

    I think it follows pretty directly that
    0.2_[8] != 0.4_[16]

    And also 0.7_[8] != 0.e_[16]

    etc.

    I wish there were a way to do base conversions with trivial symbolic substitution — it would make my life easier right now for a number of reasons.

  60. #60 Anton Mates
    April 5, 2013

    Lenoxus,

    In the case of my false proof regarding 1 + 2 + 4…, the error is in assuming that one can safely subtract Z from both sides and assume one is subtracting “the same amount”. But this is only a problem with infinities such as aleph-null (which I think Z is equal to) and not with decimal numbers that have infinitely long expressions.

    It’s not precisely a problem with infinities here; the problem is that your series Z is not convergent, and so operations such as addition and subtraction and constant multiplication are not well-defined.

    For instance, you identify “2Z + 1″ as 1 + 2 + 4 + 8… , by making the “1″ the first term and sliding all the other terms along. But I could just as easily argue that you should instead add the 1 to the first term of “2Z”, which would give you give you 3 + 4 + 8 + 16… Or maybe you should have created “2Z” by combining the terms of Z and Z, so “2Z + 1″ would be something like 1 + 2 + 2 + 4 + 4 + 8 + 8… In either case, “2Z + 1″ doesn’t look like Z any more! And if we then try to subtract Z from “2Z + 1″, should that be:

    (1 + 2 + 4 + 8…) – (1 + 2 + 4 + 8…) = 0 + 0 + 0 + 0… = 0?
    Or (3 + 4 + 8 + 16…) – (1 + 2 + 4 + 8…) = 2 + 2 + 4 + 8… = infinity?
    Or (1 + 2 + 2 + 4 + 4…) – (1 + 2 + 4 + 8…) = 0 + 0 – 2 – 4… = -infinity?

    There’s no way to say which of these is right–really, none of them are.

    (For an example of a series that doesn’t have infinitely large partial sums but still diverges, consider Z’ = 1 – 1 + 1 – 1 + 1 -1… You can produce the same sort of false proofs with that series, “proving” for instance that 1 – Z’ = Z’ and therefore Z’ = 1/2, even though the partial sums of Z’ alternate between 1 and 0 and never get anywhere near 1/2.)

    You don’t run into this problem with decimal numbers, because all decimals represent convergent series. (Basically, their terms shrink toward zero fast enough to be “safe.”)

    By the way, Z here doesn’t really represent any number, finite or infinite. But if you interpret it as a cardinal number in a particular way, you’re right that it would represent aleph null.

  61. #61 Anton Mates
    April 5, 2013

    Verbose Stoic,

    People tried to prove their case by appealing to limits, but the definition of limit literally is that it approaches and yet never reaches the limit.

    That’s not the definition of a limit. A limit is a number L such that, for any positive number epsilon, you can run down your sequence to some point and all of the terms after that point will remain within distance epsilon of L. Whether or not any of those terms actually attain the limit is irrelevant.

    For instance, the limit of (1, 1/2, 1/4, 1/8…) is 0. The limit of (1, 0, 1/2, 0, 1/4, 0, 1/8…) is also 0. The limit of (1, 0, 0, 0, 0…) is also 0. One of those sequences never reaches 0; one of them repeatedly reaches it and then departs again; and one of them reaches 0 and then stays there forever. But they all have the same limit.

    To me, that being the definition of a real number — particularly calling it a limit — is a bit too convenient

    I don’t know what “convenient” means here, but historically, treating the real numbers as limits of rational sequences was one of the first definitions mathematicians came up with. You can also construct them from Dedekind cuts, basically by defining a real number as the upper bound of a set of rational numbers with no largest element. In that case, the real number 1 is the upper bound of the set {9/10, 99/100, 999/1000…}–so, again, 0.999999… = 1.

    If you want to propose a different definition of the reals, feel free, but it needs to lead to the same properties as the traditional definitions.

    Second, what’s the definition of 1.000_ in the reals, then? Surely not the same thing …

    Yep, it’s the same thing. Decimal representations are shorthand for infinite series; the number a decimal represents is the limit of that series; and it happens that two series can have the same limit sometimes.

    . But there is no reason for, say, base 10 integers to have two numerals that point to the same number

    It’s not that anyone particularly wanted two numerals to point to the same number; that’s just how the limits work out. Most numeral systems have similar ambiguities. For instance, 0.1111111… = 1.000000… in binary, and +0 = -0 in integer notation.

    This happens in other coordinate systems, too. The system of latitude and longitude uniquely identifies each point on the Earth’s surface…except for two points, the north and south poles, that can be associated with an infinite number of longitude values. Unique representation is not usually an easy goal to achieve!

    The only numeral system I can think of offhand where no two numerals represent the same real number is the system of continued fractions, mentioned by JimV above. But that’s not a system people want to work in for most purposes!

  62. #62 Anton Mates
    April 5, 2013

    GrayGaffer,

    we can also write

    0.8_[10] = 0.g_[20]

    No, we can’t.

    0.8_[10] = 8/10 + 8/100 + 8/1000 + 8/10000… = 8/9.

    0.g_[20] = 16/20 + 16/400 + 16/8000 + 16/16000… = 16/19.

    Notice that the first terms of the series are equal, but the later terms of 0.g_[20] are much smaller than those of 0.8_[10], because the denominator is in powers of 20 instead of powers of 10.

    Like Another Matt said, symbol substitution doesn’t work for base conversion.

  63. #63 GrayGaffer
    April 5, 2013

    (rolls up shirtsleeves)

    (even rubs hands!)

    (recovers from exploding brain and …)

    P( 0.9_[10] == 1.0_[10]) is false (I maintain)

    but

    P(exist reals between 0.9_ and 1.0_) is also false (I asserted it true)

    That the question even arises is an artefact of how the value 0.9_[10] is derived and how the equivalency is demonstrated (or not).

    that 9 * 1/9 == 0.9_[10] is a result of applying rational decimal division rules first to transform the fraction (1/9) to a fixed point number, a procedure that will never halt since there is always a remainder from which another digit must be derived. Hence the handy shorthand ‘_’ to note this fact. Then 9 * 0.1111…. == 0.9999…. or 0.9_.

    That 9 * 1/9 == 1 is a result of applying integer arithmetic rules involving only addition and subtraction (multiplication is a shorthand for a repeated form of addition) that definitely halts, and produces precisely the integer 1. These rules are not the same, and do not apply the same way, as fixed point rules on 9.0 * 1.0/9.0 – the result is a function of the order of composition.

    To generalize the 0.9_ production to any base x we can find that (x-1) * (1/(x-1))[x] will result in 0.(x-1)_[10] if the division is performed first (there will be a remainder of 1 at every iteration), and 1 if the multiply is performed first and the division is never required (by term cancellation).

    So the big question is: do 0.9_ and 1.0_ and 1 all refer to the same actual real entity whose name is ’1′ or not?

    I was claiming not, that there exist an infinity of numbers of the form 0.x_ between 0.9_ and 1.

    I needed to prove (in your base choice) 0.f_[16] > 0.7_[8]. You effectively proved (i extended it a bit) 0.e_[16] ? 0.7_[8] in binary the answer is obvious. They both convert to

    0.1_[2] == 0.1_[2]

    (face-palms)

    So the first part is disproved. There is no y for which 0.(y-1)_[y] > 0,(x-1)_[x]. They are all equal (in that every Aleph null set is the same size). In any base representation.

    The second part of the question is: is 0.9_ == 1? I now think this is a semantic issue, in that there is no possible comparison function (like ==) since the two sides are expressions in different languages. In terms of infinities, I will maintain that 0.9_ < 1.0_, since both are in the same language but 1.0_ is trivially == 1.0 (without the _) and therefore can not == 0.9_ (which has no such trivial equivalence). The language in which 1 (without the .0) exists is a different space, a different set, distinct from that of the reals and irrationals. It can be mapped into the reals, but (I believe) the reverse map does not exist. 0.9_ and 1.0_ show at least 2 reals that want to map back to 1 in the integers. I think I actually showed there are an infinite set of 0.(x-1)_[x] reps in the reals that want to map back there, but they are not quite identical semantically.

    So thanks for making me stretch my brain back to my college days in the 60's. I had other plans for this evening, but clearly this one grabbed my attention.

  64. #64 GrayGaffer
    April 5, 2013

    markdown lost a bit of my translation. You stuck on 22 and 44 which is a bit of a red herring. We were working in the (x-1)[x] region which is not quite the same, and a slightly different reasoning has to be applied than base conversion (glad my actual code knows the difference:)

  65. #65 GrayGaffer
    April 5, 2013

    So a drama demonstrating counter-intuitive infinities math just played out before our very eyes. And in my very brain.

    tehe.

  66. #66 lordaxil
    April 5, 2013

    Great, now that’s all sorted out, what do people think about this:

    1 + 2 + 3 + 4 + 5 + … = -1/12

    :)

  67. #67 Verbose Stoic
    April 5, 2013

    Okay, I get it now, so let me start with what convinced me, from Anton Mates:

    I don’t know what “convenient” means here, but historically, treating the real numbers as limits of rational sequences was one of the first definitions mathematicians came up with. You can also construct them from Dedekind cuts, basically by defining a real number as the upper bound of a set of rational numbers with no largest element. In that case, the real number 1 is the upper bound of the set {9/10, 99/100, 999/1000…}–so, again, 0.999999… = 1.

    If you want to propose a different definition of the reals, feel free, but it needs to lead to the same properties as the traditional definitions.

    Yep, it’s the same thing. Decimal representations are shorthand for infinite series; the number a decimal represents is the limit of that series; and it happens that two series can have the same limit sometimes.

    So, basically, this argument is that by the definition of real numbers, 0.999_ and 1.000_ are defined by the limits of rational sequences or Dedekind cuts, and when we look at that definition we can see that, yes, they do indeed have the same definition and so are the same number. The problem, then, with “intuitions” is not with intuitions, but is with the loose definition that most people are taught for what it means to be that number. Thus, people who get this wrong are not relying on outdated intuitions, but with a flaw in the rough and ready “folk” definition of reals that covers most cases but goes wrong in this one. Thus, getting this wrong does NOT, in fact, indicate an inability to make valid mathematical deductions or, in fact, say anything about their mathematical ability beyond “Okay, you get this wrong because you haven’t delved into the field enough to get the full technical definition.”

    But note that the examples given here and in the video Nick posted do not ever point out the definition issue. They are purported proofs that, in fact, aren’t proofs if the folk definition was actually right. They are, however, what we would EXPECT to be the case if the definition given here is the right one. So, if that DIDN’T work we’d have an inconsistency somewhere, but they do work so we don’t. So presenting them as proofs simply makes the problem worse, as some of them — the 1/3 conversion — don’t seem to work when examined carefully (as I said, you can easily claim a conversion error) and there’s nothing in the examples to foil the exact argument I made: that doing so contradicts the definitions of the numerals themselves.

    So, then, why don’t mathematicians, when raising this and arguing this, simply point out from the start that, in mathematics, reals are defined in these ways and look, the definitions are the same? At that point, all anyone can do is argue that the reals shouldn’t be defined that way, which then starts to get into other debates and may at worst indicate a disconnect between what a real means in the folk parlance and what it means to mathematicians.

    I will not, however, argue that the definition should be different [grin].

    So, some clean-up:

    Another Matt,

    Perhaps this flows from the idea that numbers notated in fractional form don’t count as among the reals (they do).

    Our big misunderstanding is that you treat the different numbers as different notations, and I treat them as different number systems. I treat them as such because they have their own definitions for their numerals AND for their operations. “+” for fractions does not work the same way as “+” in reals, and so they are different number systems, and so you have to convert between them whenever you use them. Thus, when I said “0″ and “1″, I did NOT mean the integer values. I meant the real values, which might work better as 0.000_ and 1.000_. Just as you deny that you can move blythely between bases, I deny that you can move blythely between number systems.

    Lenoxus,

    Later you said that 1/3 is “not a real number”, so I think you just have some terminology confused.

    The same as above: a fraction is the the fractional number system, and so is not in and of itself in the real number system. You can, however, convert all fractions to reals. I could be wrong about that, but it seems to me that abstract algebra is on my side, and what you’re fighting against is the folk notion where we don’t treat them as separate number systems, even though they are (because they define their own numerals and operations with their own meanings).

    But I didn’t intend to truncate or approximate the digits of pi at all; my “.1415926…” was intended to mean “and so forth for all the digits of pi”. And just because we can’t write all those digits doesn’t make it invalid.

    My comment was more this:

    5 – pi and 5 – 3.14 are only approximately equal, and not identically equal, because there are a large number of things in pi that aren’t being subtracted out. But for most purposes, we really don’t care.

  68. #68 MNb
    April 5, 2013

    “the definition of limit literally is that it approaches and yet never reaches the limit”
    The second part of your definitiona (“yet never reaches”) is wrong. If it weren’t we couldn’t calculate instantaneous velocity and the entire Newtonian Mechanics were wrong.
    But knowing you it won’t lead anywhere to argue about this. It suffices to remark, as somebody did long before in this thread, that 1 not equal 0,9_ leads to inconsistencies and 1 = 0,9_ does not, except if you call this an inconsistency itself.
    Just one question. If the two are not equal, then what is the difference? And with what number do we have to multiply that difference to obtain 1 again?

  69. #69 Lenoxus
    April 5, 2013

    Anton Mates: Thank you for the clarifications about Z and non-convergent sums.

    (For an example of a series that doesn’t have infinitely large partial sums but still diverges, consider Z’ = 1 – 1 + 1 – 1 + 1 -1… You can produce the same sort of false proofs with that series, “proving” for instance that 1 – Z’ = Z’ and therefore Z’ = 1/2, even though the partial sums of Z’ alternate between 1 and 0 and never get anywhere near 1/2.)

    As it happens, though, 1/2 is the Cesàro sum of that series. But that’s not the correct way to calculate a Cesàro sum, so it should be treated as a coincidence.

    Verbose Stoic: I’m glad to see you’ve come around! People rarely do on this issue. It is heartening.

    There’s still something that needs clearing up, though. You wrote:

    A fraction is the the fractional number system, and so is not in and of itself in the real number system.

    The term “real number” has nothing to do with representations, but with a specific set of actual numbers. The “fractional number system” is a matter of representation. Saying 1/3 is “not a real number” because it is a fraction is like saying that roses are not plants because “rose” is a word, not a plant. “1/3″ refers to the same mathematical entity as “one-third” or “0.3_” or “ternary .1″, and this entity is a rational number and therefore a real number.

    None of the representations is “more” rational, or more real, than the others. We can convert any of the representaions to another representation, but that doesn’t involve “converting it to the reals”. I could replace the word “rose” with a a drawing of one, or with the Thai word for it, but that doesn’t mean I converted either the real rose or the word “rose” into a plant.

    And to bring all that back around: “7/7″ is a fraction corresponding to an integer, usually called “1″. Because 1 is an integer, it’s a rational number, and therefore also a real number. As a word, it can be written “one”, and as a decimal, it can be written as “0.9_” or “1″ or “1.0″ or “1.00_”, and so forth.

  70. #70 Anton Mates
    April 5, 2013

    So, then, why don’t mathematicians, when raising this and arguing this, simply point out from the start that, in mathematics, reals are defined in these ways and look, the definitions are the same?

    Because the definitions are not arbitrarily produced by the whim of mathematicians, but are constructed so as to give real numbers certain properties generally agreed to be “useful.” It’s important to exercise students’ intuitions about what those properties imply. Sure, formal definitions are what you need for ironclad proofs. But, whenever possible, our intuitions should be trained so that we can get to the same result without such proofs.

    The real numbers weren’t just invented as part of a mathematical game; they were developed to represent the real line. Some of the critical properties of the real line are that it’s well-ordered, continuous, connected and unbounded. E.g., between any two distinct points, there’s another point. Any bounded set of points has unique upper and lower bounds which are also points on the line. Any Cauchy sequence (a sequence where the later points become arbitrarily close to each other) has a limit which is also a point on the line. If the real line did not have these properties, it wouldn’t work like a line–geometry, algebra and calculus problems would not have the same answers.

    If .9999… and 1 were treated as different numbers, then there would be two distinct points on the real line without another point in between them. That would be a problem, intuitively. The line would have a “break” in it. Averaging would no longer work–it wouldn’t always be the case that the average of two numbers lies between them. Calculus wouldn’t work–curves wouldn’t have well-defined slopes as they passed 1. All kinds of weird stuff would start to happen, even before you started worrying about whether any formal definitions were violated. That wouldn’t prove that your approach was wrong–sometimes weird stuff happens in math–but it would be a big red flag.

    And no, if it doesn’t look like a big red flag to you, that doesn’t mean you’re inherently bad at math.

    “+” for fractions does not work the same way as “+” in reals, and so they are different number systems,

    What makes you think it doesn’t work the same way? Given any two rational numbers, their sum has the same value whether they’re treated as rational or real, and whether they’re expressed as decimals or fractions.

  71. #71 Another Matt
    April 5, 2013

    I treat them as such because they have their own definitions for their numerals AND for their operations. “+” for fractions does not work the same way as “+” in reals, and so they are different number systems, and so you have to convert between them whenever you use them.

    It absolutely does work the same way for fractions as it does for reals. You just have to remember the order of operations — fraction notation indicates division, and you have to keep in mind that this division has to happen before any addition occurs. The stuff we teach kids in 4th grade about finding common denominators and so forth is about keeping the implications of the order of operations intact. The fractional notation works for anything — you can put 3.14159265… in the numerator and 7 in the denominator and that just means “pi divided by 7.”

    You’re right though — you have to be careful about moving back and forth between number systems — in most math and arithmetic situations “real” is assumed unless otherwise indicated.

  72. #72 Dabbe
    April 6, 2013

    ”The proof that 0.999… = 1 is quite simple. Let X = 0.999…. Then 10X = 9.999…. Therefore 9X = 10X – X = 9.999… – 0.999… = 9, or X = 1.”

    One day late!

  73. #73 AnswersInGenitals
    April 6, 2013

    Does 0.999… = 1 ?
    Some say yes. Others say no.
    In the spirit of world peace and harmony I will argue that both are right. (Or in the interest of world discord and conflict I will show that both are wrong.)

    First some notation: let 0.9(n) represent a decimal followed by n 9s; let S9 be the set of all 0.9(n)s, i. e. S9 = {0.9(n); n = 1 to infinity} (I hope the curly brackets defining a set show up on your font.)

    Consider two more sets: [0,1] is the set of all real numbers between 0 and 1 including the real numbers 0 and 1. (0,1) is the set of all real numbers between 0 and 1 but excludes 0 and 1. If you prefer a geometric picture, [0,1] and (0,1) are the closed and open line segments from 0 to 1 including and excluding the end points, respectively. It should be noted that open sets like (0,1) are very fundamental in mathematics, forming the foundation of topology which forms the foundation of much of mathematics. (The concept of open sets was invented by a mathematician who, as a child, insisted that his mother remove the crust from his peanut butter and jelly sandwiches.)

    The key point here is that the open set (0,1) contains S9 as a subset, i. e., every member of S9, including our mischievous 0.999… is contained in (0,1), but this set does not contain 1! So, if there is a set that contains 0.999… but does’t contain 1, how can 0.999… be equal to 1? How can it be the same as 1 in any sense? On the other hand, any discrete set can be considered as either an open or a closed set (a closed set contains all its limit points) and S9 is a discrete set. So we can consider S9 to be a closed set and it contains 1 as a limit point which is the limit of 0.9(n) as n goes to infinity, i. e., 0.999… So, what gives?

    The problem is that asking if 0.999… = 1 is like asking: if x + y = 12, does x = 7? The question is underdetermined – there is not enough information given, i. e., the value of y. To answer whether 0.999… = 1 we need to state whether the set S9 is being considered open or closed. The various attempts to prove that 0.999… = 1 all assume the conclusion by surreptitiously assuming that a closed set is being considered. E. g., 0.333… = 1/3 iff S(3) is a closed set.

    The concept of mathematical equality can have some very counterintuitive aspects, particularly where infinities are involved. It is best to avoid saying that 0.999… equals 1 and to say that 0.999… converges to 1.

    Since this blog already has over 70 comments, probably no one is still reading it. But if anyone is still reading it and expresses interest, I’ll show that the distance between 0.9(n+1) and 0.9(n) actually increases as n increases; another very counterintuitive result.

  74. #74 Keith M Ellis
    Kansas City, MO
    April 6, 2013

    I wrote and maintain the first site on the web concerning the Monty Hall Problem — I first wrote it and placed on my personal site in 1996, later I moved it to its own domain but have for the most part never re-written it or updated it, so it has that distinctive 90s look.

    Nevertheless, it’s usually one of Google’s top two or three results for the MHP and it’s been widely linked and referenced.

    And so I’ve been getting email from MHP skeptics for sixteen years.

    My main interest in the MHP has always been related to mathematical intuition, what’s counter-intuitive to many people, and how to break through that misleading intuition. It’s been equal parts fascinating and frustrating corresponding with skeptics (and, to be fair, at least an equal number of people writing to thank me for helping them have their eureka moment).

    Many of the exchanges above remind me, in a bad way, of a lot of these discussions with skeptics.

    The one thing that has continued to most astonish and frustrate me is that even when there are countless others and authorities and websites and articles and TV shows and films that argue that it’s better to switch, and especially given that it’s so trivial to empirically test the whole thing to discover the truth, an amazingly large number of people are absolutely certain that their intuition about the MHP is correct, that switching is not better, and that all these experts are wrong and that they have no need to verify their intuition empirically.

    It’s been sixteen years and probably at this point many dozens if not hundreds of exchanges, and the hubris of this still takes my breath away.

    It’s no sin to let one’s mathematical intuition lead one astray. Essentially, that’s pretty much the entire history of mathematics. It’s one long progression of discover that intuition is misleading.

    But there is something wrong with holding to that intuition regardless of better-informed argument; to not, for example, take a course or read a book on number theory (when one has heterodox opinions on matters such as are being discussed in this thread). Someone who writes that 1/3 and 0.333_ are not in the same number system has no business arguing at length here.

    People who privilege their intuition (which, yes, is what happening when you argue that .999… is not equal to 1 because you’re privileging what you think that .999… must mean over what people say who actually know) ought to take a few steps back and look at, say, Euclid and his careful treatment of irrational numbers. People are taught what irrationals are by rote, by their “shape” as, say, decimals, but not what incommensurability really is — and this matters because right there at the very beginning of western mathematics, right there embedded in junior high math, is something that is profoundly counter-intuitive and yet, absolutely true. Better acquaintance with this sort of thing at the foundations would help people better learn to mistrust their innate mathematical intuition.

  75. #75 bad Jim
    April 7, 2013

    AIG, you can only show that every finite sequence 0.999… belongs to (0,1).

    If 0.999… is different than 1, there is a remainder when you subtract the two expressions. If it isn’t zero, what is it?

  76. #76 Anton Mates
    April 7, 2013

    AIG,

    In the spirit of world peace and harmony I will argue that both are right. (Or in the interest of world discord and conflict I will show that both are wrong.)

    And in the latter interest I will argue with you. :)

    The key point here is that the open set (0,1) contains S9 as a subset, i. e., every member of S9, including our mischievous 0.999… is contained in (0,1), but this set does not contain 1!

    But S9 does not contain 0.999…, at least not as you define the set here. Every number in S9 is a decimal followed by n 9s, where n is a positive integer. n may become arbitrarily large, but there is no actual value n = infinity, and hence 0.999…. does not correspond to any value of n.

    You can of course add 0.999… to S9 if you wish, but in that case it is no longer a subset of (0,1). Nor is it a discrete set, because 0.999… is not an isolated point with respect to S9.

    On the other hand, any discrete set can be considered as either an open or a closed set (a closed set contains all its limit points) and S9 is a discrete set.

    Not so. No discrete set of real numbers is open (except, trivially, the empty set). Every point in a discrete set is an isolated point; by definition, there’s a small neighborhood around it which contains no other point in the set. Every point in an open set is an interior point; by definition, there’s a small neighborhood around it where all the points are in the set! Clearly, no member of a set of reals can be both isolated and interior.

    Nor is S9 a closed set, unless, again, you explicitly add 0.999…, in which case it’s no longer discrete. Finite discrete sets are always closed on the reals, but infinite discrete sets need not be. The infinite set {1/1, 1/2, 1/3, 1/4…} is discrete but not closed.

    What you’re probably thinking of is the fact that any subset of a discrete space is both open and closed with respect to that space. But here we’re asking about whether a set is open or closed with respect to the space of real numbers (that is, the real line.) Different parent spaces, with different topologies. The key difference is that, on a discrete space, there are small open neighborhoods around each point which are otherwise totally empty, making that point both interior and isolated. Of course, that’s never true for the real line.

    To answer whether 0.999… = 1 we need to state whether the set S9 is being considered open or closed.

    Well, we don’t really have the freedom to consider it as one or the other. The definitions of “open” and “closed” are unambiguous for any set on the real line. There are sets of real numbers that are both open and closed–namely, the empty set and the complete real line–but S9 isn’t one of them.

    But if anyone is still reading it and expresses interest, I’ll show that the distance between 0.9(n+1) and 0.9(n) actually increases as n increases; another very counterintuitive result.

    This will probably depend on you using a very weird distance metric. :) But let’s hear the argument!

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