Polymathematics

Initial 1's

A friend of mine has been thinking about the well-known fact that many lists of seemingly random numbers (addresses, physical constants, populations) contain far disproportionately many initial 1's (compared to other initial digits, which decrease in frequency up to 9).

He noted that many physical constants are ratios—constants of proportionality, for example.  So he calculated the probability that any two random numbers x and y between 0 and 1 (random with a flat distribution) have a ratio y/x that starts with a 1 (meaning its first non-zero digit:  .00013 counts as an initial 1).

Cool answer:  1/3

The proof is left as an exercise (unless there's an outcry for it).  Just a little tidbit to tide you over, since I haven't been posting much.

Triskaidekaphilia: the 13th Carnival of Mathematics

Hello, and welcome to the 13th Carnival of Mathematics.  Since the previous Carnival was posted on Friday the 13th, and since this one is the 13th Carnival, I'll start with some links involving the number 13.  The pdf file you can download by clicking (EDIT: you have to cut and paste this link in two parts (it's too long for one line)...for some reason, I can't get it to work: public.carnet.hr/ccacaa/CCA-PDF/cca2004/v77-n3/
CCA_77_2004_447-456_pogliani) is a paper involving various facts about the number 13, although it digresses far and often.  This site contains a lot of facts about 13, though some stretch pretty far to be labeled seriously interesting;  a few that really are kind of interesting:  the squares of 13 and its digit reversal (31) are digit reversals (169 and 961);  the sum of the primes up to (and including) 13 is 41 (the 13th prime);  "twelve plus one" and "eleven plus two" are anagrams (that's my favorite).

Browsing the On-Line Encyclopedia of Integer Sequences for something interesting, I came up with:

• 13 is a Fibonacci number (the sequence starting with 0,1, and where each subsequent term is the sum of the previous two), a tribonacci number (the sequence starting with 0,0,1, and where each subsequent term is the sum of the previous three), and in the "skipped Fibonacci sequence" (the sequence starting with 0,0,1, and where each subsequent term is the sum of the previous one and the one two before that).  In sequence terms, it's in the sequences defined by a(n)=a(n-1)+a(n-2), and a(n)=a(n-1)+a(n-2)+a(n-3), and a(n)=a(n-1)+a(n-3).
• 13 is also a tetranacci number (each term being the sum of the previous four), but only if the first four terms are 1,1,1,1 (not 0,0,0,1, which generates another sequence that is also called the tetranacci sequence elsewhere in the database—there aren't really standards for these things, I guess).
• 13 is the smallest sum of the squares of distinct primes.
• 13 is the smallest emirp (a term that just learned from the database:  a prime whose reversal is also a prime).

I had always thought that the fear of the number 13 (triskaidekaphobia) derived from the 13 people present at the Last supper (Jesus and 12 disciples).  But Wikipedia reports that Hammurabi's Code (dated 1760 B.C., which would be fully 1790-ish years before the Last Supper) omits 13 in its numbered list much like airplanes omit rows labeled 13 and buildings omit floors labeled 13 (news flash to the occupants of apartment 1403:  you really are on the 13th floor, you know).  A more interesting possibility raised in that article:  a year has almost exactly 13 lunar (28-day) cycles (13 times 28 is 364), and the moon is associated in many cultures with femininity (for the obvious 28-related and therefore 13-related reasons).  Thus a masculine culture might have thought they had something to fear from a mystical feminine connection to nature.

But on to the Carnival proper.

A math Ph.D. student named Michi works on homological algebra.  I didn't know what that was either.  And I admit that my understanding of formal algebra isn't deep enough to really understand his post explaining it.  But that's my failing entirely—it's well-written and I'm sure it will be enlightening to many.

One link that I just found myself is a very nice proof of the Pythagorean theorem via java applet that roughly follows Pythagoras's own reasoning.  But with the animation, it's crystal clear—impressive from a teaching point of view, and it fits in exactly with my own interests (shamelessly plugging the AHSMP in the sidebar).

Another page that describes a more advanced idea from a basic point of view is this one that discusses the poorly-understood-in-lower-math constant e.  Very clear and intuitive.

An undergraduate gives a brief overview of ANOVA, a technique that I think a lot of people use without really understanding very well.

This blog discusses the Babylonian method for estimating the square root of 2 in the context of attractors and fixed points.  I hope I have time read the whole series—I don't understand those ideas very well, and the explanations seem good.

A blog of math and logic puzzles reviews an important counting technique for finding sizes of intersecting sets.

A blog specializing in K-12 math has a post with a simple puzzle, but does a nice job of explaining how extending the problem can create some thought-provoking math.

And finally, a retired teacher with a blog seems from this post to have a philosophy similar to mine:  general concepts and complete mastery of basic skills are important;  the varying philosophies aren't truly in opposition.  In this post, he presents a nice geometry problem—it has several solution methods, and seeing how they're all linked is instructive.

(Due to some difficulty finding time tomorrow, when this carnival is technically supposed to be posted, I am posting it on Thursday night at about 11pm EDT.  If your submission isn't in this carnival, I apologize;  it wasn't left out intentionally.  Please submit your link again to the next carnival.)

Blog Center for Math Pages

This post serves as Grand Central Station for my Advanced High School Math Project.

The comments section of this post should be used only for corrections or clarifications of pages in that project.  Any other comments will be deleted.  But your corrections are welcome:  please be sure that the page title is in the comment, and if your correction is valid, I will correct the page and credit you.  Corrections include valid claims of previous authorship that should be cited on the bibliography page (which doesn't exist yet).

If you are interested in submitting a proof for publication on these pages please follow the guidelines below, and send it to me at the e-mail address listed on the "About" page.  Feel free to contact me with a proposed topic before you write it.

1. The proofs should require no more than a good understanding of well-rounded high school math curriculum.  I realize this won't mean the same thing to everyone, so I will be the judge of whether it's too advanced or not.
2. The diagrams should be sent to me as JPEG files.  So far, I have not yet learned how to insert equations into the blog except as JPEG files.  Until I do learn that (e-mail me if you know how!!), your simple equations (for example: a+b+c=180) should be in the body of the text, and the complicated ones should be JPEGs.
3. The text should be clearly written, and should avoid complicated notations in favor of concise English.  Clarity is judged by me.
4. Notation should enhance the understanding of the proof, not obscure it.  This judgment will be made by me.
5. The topics should be either significant, well-established mathematical results or surprising facts with especially elegant and/or enlightening proofs.  It is not my intention to merely publish nice solutions to cool problems.  Significance and/or elegance will be weighed by (are you sensing a pattern here?) me.
6. I will try to retain your voice and ideas, but I reserve the right to edit.  Your name or pseudonym will, of course, be included.
7. If you found the proof somewhere else (a book, another web page, whatever), that source must be cited, and your proof must be significantly altered or improved in clarity.  If you do not know the source, but you know you the proof is not original (and I understand that this happens often, especially with clear and enlightening proofs), please say that—I would hope that the original writers of the proofs will eventually be found and cited.  I want to avoid any scent of plagiarism.
8. I'm not trying to discourage submissions;  I just want to be sure the pages retain their original purpose, as explained on the index page of the project.  I would be very happy indeed if I had a lot of clear, well-written contributions.

Thanks for visiting...I hope you find these pages useful.

The Last Few Weeks

Well, she's home.  Finally.  My daughter.  It is still just a little strange to use that word, but it's getting more familiar.

Let me answer some of the frequently heard comments right off the bat:

1. "Hello, dad."  Uhhh, hi.  Yes, I know I'm a dad.  That doesn't freak me out or anything.  I'm not frankly sure why people keep saying that with a kind of wink as if they're seeing whether I really remember the existence of my daughter when she isn't around.  "Hi...and congratulations!" would be fine, thanks.
2. "Well, things are going to be so different for you now, you can't even imagine."  Actually, I think I can imagine, thankyouverymuch.  I had already imagined the midnight feedings, the exhaustion, the elation, the wonder of seeing the world through my child's eyes, the drool, and the diapers.  I did miss foreseeing the cluttered kitchen countertops and the boredom (yes, I'll admit it) of repeatedly entertaining a pre-conversational baby in between feeding and naptime, but I don't think that's what those commenters have in mind.
3. "Are you sleeping okay?"  Take a guess.  I just got back from China where the time difference is approximately one-half a day.  As much jetlag as humanly possible on this planet.  And let me tell you something:  Babies?  Don't understand jetlag.  Yes, I'm tired...get back to me when it might be reasonable to expect our kid to have a decent schedule.

Uhhhh, sorry about that...had to get that griping out of the way.  Some people...sheesh.

But if you're not one of those people...thanks for your good wishes.  Even if you don't know me or comment on this blog, I hope you express your good wishes to other people who have just adopted, and I'll consider this a thanks from them.  It's a pretty amazing experience.  Sometimes excruciating (the wait, especially), sometimes heart-pounding (sitting on the bus waiting to be brought to some bureaucratic office where you're oddly both going to get a baby and have your passport scrutinized), sometimes thrilling (the first time she smiles when you come into the room because she recognizes you).

There's been laughing and crying (happy crying, mostly, on my part and my wife's—hungry crying on the baby's), and my wife and I are very happy to have our daughter and to be done with the whole frustrating adoption process.  Now comes the part where we raise a child, which was much more the whole point of this exercise.

Oh, uhhhh math...right.  Ummm, here's a good one:  pick any cubic polynomial f(x) with 3 real roots;  now choose any two of those roots and call them a and b;  now find their average v, and locate (v,f(v)) on the graph of the function;  draw the tangent line there, and you'll find that it crosses the x-axis at the third root.  The uncreative calculus proof isn't very enlightening, but it's easily doable with basic first-year calc.  Someone explained it to me better a long time ago in terms of the manipulation of polynomials, but I don't remember the explanation.  Feel free to help in the comments.

Proof of Morley's Theorem

Okay, I reconstructed that proof of Morley's theorem.  If you don't know what Morley's theorem says, I urge you not to look it up just now.  I'll tell you what it says at the end of this proof.  This proof is not the shortest one (which you can find in many places, and is due to John Conway—I will post a link at the end to his proof), but it is the one most presentable to people who know high school geometry.  I find other proofs either too advanced for most people (sometimes including myself), or confusing in their notations.  This proof uses only basic, first-year high school geometry.  It was found online by a student of mine, but I've tried to find it myself, and I can't.  I am eager to give credit where it is due, and if anyone finds the link, I will happily cite it here.  My proof is a reconstruction from some notes and from my memory, and the presentation will thus probably differ somewhat from that source anyway.

First, start with an equilateral triangle called XYZ.  It is drawn in red below.  Then replicate that triangle three times, and arrange them as shown in blue below.

Now choose 3 [positive (thanks to commenter for catching that)] numbers called a, b, and c.  But be sure those three numbers add up to 60.  Draw a line from point X that makes an angle of c degrees from the blue triangle, a line from Z that makes an angle of a degrees, then one more from Z making an angle of b degrees, and one from Y making an angle of c degrees again.  Those lines are shown below in black, with the angles labeled to show you what I mean.  Call the intersection of those lines B and A as indicated.  Note that because the angles of the equilateral triangles are each 60°, and no two of a, b, and c add up to more than 60°, the lines I drew can't be parallel, and thus the points B and A are well-defined.

So consider triangle BXZ.  The two angles at X and Z add up to:

(60°+ c) + (60° + a) = 120° + (a + c) = 120° + (60° – b) = 180° – b.

Since the angles in that triangle must add to 180°, angle B must measure b degrees.  Similarly, angle A must measure a degrees.  That is reflected in the next diagram below, as are two extensions of the blue lines out to points S and T, which we will need next.

Because triangles SXZ and TYZ both contain one angle of 60°, one angle of (60 + c)° and a side of the original equilateral triangle, they are congruent by ASA, and thus SZ = TZ.  Furthermore, triangles SBZ and TZA each contain an angle measuring a and b, and are thus similar.  This gives the first ratio below, and the second equals sign is from the congruent segments we just proved.

Calculate next the measures of two angles.  First, in triangle ZTA, angle T measures 180° – (a + b).  This angle is marked with a # below.  Second, angle BZA is obtained by subtracting angles measuring a and b from the blue straight angle, making angle BZA also measure 180° – (a + b).  So if we draw in segment BA, the new triangle we formed (BZA) must be similar to triangle ZTA, since they have one congruent angle, and the sides creating that angle are proportional.

This makes the other two angles of triangle BZA also measure b and a.  And if we duplicate that same proof around points X and Y, we create triangle ABC below.  Note that this triangle has angles adding up to 3a + 3b + 3c, which of course is 180°.

But imagine that you were given an arbitrary triangle to begin with.  If you trisect the angles of that triangle, the angles created conform to the initial condition above that a + b + c = 60°.  Thus I could always create a diagram like the one above using the angles of any given triangle—divided by 3—as a, b, and c.  If you scale that diagram appropiately, it will result in a triangle ABC that is the same as the arbitrarily given one.

Morley's theorem (1899):  The three points of intersection of adjacent trisectors of the angles of any triangle form an equliateral triangle.

Morley proved this as a corollary to some other very complicated calculations of conic sections (I think) based on certain properties of triangles.  But the simplicity of the fact (sometimes now known as Morley's Miracle) led to many different proofs, many understandably involving trigonometry.  Conway's proof is the shortest, and doesn't use trigonometry or complex numbers or group theory (like some of the proofs), but it uses a notation that takes getting used to, and I doubt that high school students would be able to follow it as easily as this proof.

I, for one, find Morley's Miracle almost staggeringly unbelievable.

LMAO

Hahahahahahahahahahahaha.

I have to tell you all about hahahahahahahahahahahahahaha....

I can't even get it together enough to tell you.

hahahahahahahahahahahahahahahahaha

(wiping tears.....) hahahahahahahahahaha oh stop...it's too funny.....hahahahahahaha

the latest comment on my post about how .999... = 1.  Yes...people are still posting about that almost 6 months later.  It comes in waves.

It's by a guy named (get this...snort...giggle) Proffesor David S. Schaul:

You are perhaps competant as a teacher, however your skills in mathmatic theory lack common sense as well as what a number represents. The fact behind the truth is that .999... is not really a number. It is merely a way of expressing a number that cannot exsist. There is no possible way to obtain that number through basic mathmatic functions, therefore implying that .999... is not equal to one, but is nonexistant in the rational world of numbers.

I'll wait while you stop laughing.  Hahahahahahahaha.  No, I can't stop either.

• As my wife points out, no actual professors refer to themselves as "Professor".  They use "Dr." if they use anything at all.
• And they certainly don't refer to themselves as "Proffesor".
• If he's trying to convince me that he's actually a professor of mathematics, it's not very convincing that he misspells his own subject, among other words that every mathematician can spell, like competent, exist, nonexistent.
• Like any mathematician would ever argue a mathematical fact by citing "the fact behind the truth..." rather than resorting to proof.
• Like any mathematician would consider his/her field of study to be "the rational world of numbers", like it's the Discovery Channel's "Wild World of Chimps".
• And then, on top of that, of course, are all the logical fallacies that all the other non-believers make.
• He even gives an e-mail address that includes part of his name, which doesn't include the name Schaul, nor the first initial D.
• And...hahahahahahahaha...noooo...I can't go on.  It hurts too much.

UPDATE:  It has been pointed out to me that my reaction to that comment is not among the more mature imaginable reactions.  Nor is my rebuttal among the cleverest imaginable rebuttals.  Perhaps it is just because I have been enduring 6 months of (sometimes inane, sometimes sincere and well-meaning) continuing discussion in the comments section of the original post, and I just needed to vent.  I apologize if anyone is offended or disenchanted or put off by my reaction.  Not taking the high road here might be a character flaw on my part, but I found the pretention (in the literal sense of 'pretending') in that comment almost too extreme to be believable.  I had to say something, and merciless, childish teasing was what came out.  Next time I'll try to be more civil.

Countability

Since there was some discussion of infinity in the whole .999... shebang, I thought I'd post a quick proof related to infinite sets.  The result is well-known:  the set of rational numbers and the set of natural (counting) numbers have the same cardinality (that is, they are the same size).  This is normally shown with Cantor's diagonalization proof.

This result is usually considered counter-intuitive because there are infinitely many rational numbers just between 0 and 1 alone.  Yet, the set of rationals is indeed the same size as the set of natruals.  All I wanted to do in this post was give you my favorite proof of this.  I know there are many.  I know you'll be tempted to post your favorite in the comments (feel free).  But for some reason, this one is so clever, it's by far my favorite.  I learned it from a professor in college.

I will prove that the set of (positive) rational numbers can be associated one-to-one with a subset of the natural numbers.  Since it is easy to associate the natural numbers with a subset of the rationals (the reciprocal function will do), the demonstration below completes the proof.

By definition, each positive rational number can be expressed in a unique manner as p/q, where p and q are natural numbers with no common factors.  Consider the string of symbols composed of (digits of p)/(digits of q), with a slash between the strings of digits.  Now read that string (which is unique for every rational) as an integer in base 11, with the slash being the symbol for 10.  There's your integer associated with that rational number.  QED.

My Vote for 0^0

In the course of the craziness about the whole you know what (see the links in the sidebar), it somehow came up that 0^0 (which is to say, 0 to the 0-eth power) is a tricky calculation.  Several commenters gave their opinions, and so, for fun, here's mine.  Consider the following chart:

Note that negative and zero exponents are pretty much defined to make each column's pattern (successive division by the base for that column) continue.  And when you do define it that way, lo and behold, the standard laws of exponents hold, and we're therefore happy with that definition.  That's how math works, right?  We find patterns and see if our codifications of them continue to make sense after they get more and more abstract.

The problem with 0^0 is that it seems to have to satisfy two patterns at once.  The column pattern is that the right hand side is always 0.  The row pattern is that the right hand side is always 1.  So what goes in the green box?  1 or 0?  Both continue a pattern.

Take a look, though, at the entries in the blue column below the green box.  Since 0 can't appear in a denominator, those entries are missing.  Thus, crucially, the pattern in the blue column will have to end anyway.  So if our goal for this definition is to preserve as many patterns as possible, we don't lose much pattern by saying that 0^0 is 1.  That preserves the horizontal yellow pattern, and breaks the vertical blue pattern, but...hey, it was broken soon anyway.

So that's my vote and my reason.  0^0 should be defined as 1.

You might be a geek if....you devote a whole blog post to this.

Proof of CMFE

My proof of the coolest math fact ever (CMFE) is below (I will explain after the proof why I think it's so cool).  I don't claim that this is the easiest proof, but shorter proofs probably use math that I don't entirely understand.  The CMFE was stated in a previous post, but I will restate it here.  Coincidentally, earlier today, someone commented on that post with a hint for the proof, and I'm going to assume that person knows what he/she is talking about, because that's exactly the right first step.

Unfortunately, there's a little glitch in the proof that I've never quite understood, and maybe someone could clear it up for me.  It doesn't sink the proof, but I'm not quite sure I understand why not.

Okay.  So the CMFE is this:  Put n equally spaced marks around the circumference of a unit circle (radius of 1).  Then from any one of those marks, draw the chords that connect it to all the other (n-1) marks.  The lengths of these chords are then multiplied together, and amazingly, that product is always n.  So:  532 equally spaced marks, 531 chord lengths, and the product is an amazing 532.  Here is the starting diagram for n=9:

The proof hinges (as the commenter suggests) on considering this unit circle to be the very important complex unit circle—that is, the circle of radius 1 centered at the origin of the complex plane.  It is a well-known fact in complex number analysis that the nth roots of 1 (or, as mathematicians say, the roots of unity) are n points, equally spaced around a unit circle, starting at the complex number 1, which appears 1 unit to right of the origin on the real axis of the complex plane.  You'll notice that if you superimpose a set of axes on the diagram above, the number 1 would fall right at the point from which the chords emanate.  That means that collectively, the n marks are indeed the nth roots of 1.

If you call one of the other endpoints r, then the segment connecting it to the number 1 is the vector that represents the complex number 1-r.  Which means the length of that segment is:

Okay.  So with that in mind, note the following polynomial identity, which holds over the complex numbers:

(z is the variable, the r's are the nth roots of 1.)  This is an identity, because it's always true:  note that both polynomials have the same degree (n), the same leading coefficent (1), and the same roots (the r's).  The only way this can happen is if the polynomials are the same.

Now make the following substitutions:  the first root of 1 is clearly just 1, so r_1 turns into 1.  Then also replace all instances of z with Z+1 (for reasons that will become clear).  We now have the first line below:

The second line above comes from using the binomial expansion on the left and the obvious simplification on the right.  Since the 1's cancel on the left, we can divide both sides by Z to get:

Note that this is still an identity, and true for all values of Z (but this is the glitch, so see below before you claim I'm wrong here).  So I pick 0 as my value for Z.  This gives the way simpler equation:

But if you take the absolute value of both sides, you get simply that the lengths of all the chords have a product of n.  Which is JUST ABSURDLY COOL!

Glitch:  when you divide both sides by Z, you have to exclude the case where Z=0, or the division is meaningless.  Thus the equation after the division is only true if Z is not zero.  Which means I ought not be allowed to substitute Z=0 into it in the next step.  The fix, I think, is to take the limit as Z approaches 0 on both sides of the equation, leaving you with the same result, but never requiring Z to actually equal 0.  Note that if Z=0, then z=1, which is the crucial point from which all the chords emanate.  So is this limit process still legal because excluding one endpoint of a segment doesn't change its length?  Or am I confusing a general identity with the application to a specific case?  A clear explanation of this would be greatly appreciated.

So.  Now you know the CMFE.  The main reason it's so cool is this:  the diagonals of a square with a side-length of 1 have lengths expressable with the square root of 2 (notated SQR(2) hereafter).  The diagonals of a pentagon with a side-length of 1 have lengths expressable with SQR(5).  The diagonals of a hexagon with a side-length of 1 have lengths expressable with SQR(3).  Notice that those are the kinds of lengths multiplied in the CMFE.  Thus it is no surprise that products of SQR(5), for example, result in the number 5.  In other words, it isn't a coincidence that SQR(2) shows up in 4-sided figures, SQR(5) shows up in 5-sided (and 10-sided) figures, and SQR(3) shows up in 6-sided (and 12-sided)figures.  If the number k shows up as SQR(k) in the expression for the length of a diagonal of a regular n-gon, then k must be a factor of n.  (Note that not all regular n-gons have diagonals that can be written in that form—some diagonal lengths are transcendental UPDATE: cat lover points out in the comments that those diagonal lengths, while not expressible with roots, are probably not transcendental; I agree completely.)  I don't know why I find that sooooo cool, but I'm a math geek, and I'm not going to try to make excuses for my opinions of coolness.

QED.

And Finally...

(note:  this post has been closed to comments;  comments about it on other pages will be deleted!)

Okay, this is going to be my last post on the topic of how .999...=1, started in this post, and continued here.  I will engage in more refuting, but I have begun to see how useless the refutations are because many of the "non-believers" (and I use that word jokingly because, as I wrote here, I don't really consider it a matter of belief) don't bother to visit the refutations page or don't read it if they do.  This blog has gotten over 70,000 hits since the original post was on the front page of digg.  The discussion there and at numerous other small forums makes it clear that the refutations aren't being read.  There have even been meta-discussions on how this fact can get a warning from digg about containing inaccurate information, even though every knowledgeable source of information agrees with me.  The only reasonable criticism I found on the digg site is that this doesn't belong in the news because the proof has been around for so long that it ought not count as news.  Unfortunately, enough people seem to disbelieve it that even old news needs to be explained just one more time in the hopes that a few people will come to understand the math better.

But on to the refutations.  This time, I will do them in decreasing order of sophistication of the argument.  This puts the most reasonable first.  (And again, I give full credit to some of the comments for their help in refuting.)

Variations on:  1 - .999... = 1/infinity, which is greater than zero.

I already discussed this a little bit, but there's more to be said.  While there are some mathematicians who have formalized the notion of infinitesimals, those systems require a very careful formal extension of the real numbers to what they call the hyperreal numbers.  I admit that I am not versed in this stuff because it is way beyond my understanding.  However, I understand enough to know that they are not meant to replace the real numbers, but to extend them.  The inventors of hyperreal numbers and the algebra of infinitesimals assuredly still agree that within the real numbers (a caveat that I have been very careful to include ever since this objection arose) .999...=1.

Within the real numbers, 1/infinity (which I will represent in this essay with the variable E, for epsilon, a standard name for very small (but not infinitely small!) numbers) behaves just like the number zero.  As an example of this, I will refute:

Variations on: .333... doesn't really equal 1/3.  It is just slightly less than 1/3 in the same way that .999... is just slightly less than 1.

(Now remember, I have to put some words into objectors' mouths, since they do not present their objections precisely, which is the whole problem.  I apologize if I mischaracterize an objection somehow.)  The only way to make this consistent, I think, is to claim that:

1/3 = .333... + E    (remember that E is 1/infinity)

This would mean that:

1 = 3/3 = 3(.333... + E) = .999... + 3E

So if you object like this, tell me about this 3E number.  Is it greater than E?  If it is, then .999... isn't really as close as you can get to 1, is it?  "Aha, Mr. Polymath!  You yourself said that you can't use normal algebra on infinity, so it's still possible that the 3E number (the residue from tripling 1/3) might be the smallest possible postive number!"  Yes, indeed, I said that.  But if you're going to say that E and 3E are really the same thing, then you are saying exactly what I'm saying.  If E and 3E are the same, then E is behaving exactly like the number zero.  Or, if you don't like using algebra on this, we can just consider it like this:  if .333... lacks something when it tries to be 1/3, and if .999... lacks the same thing when it tries to be 1, then either that lacking quantity is behaving just like the number 0, or .999... is actually not as close to 1 as you can get.

Variations on:  Why do you insist that there be a number between .999... and 1?  Can't .999... simply be the next number down from 1?

The real numbers have been defined in such a way as to guarantee that they are "dense".  That means that between any two real numbers, you can always find another real number (infinitely many, actually, but certainly at least one).  The easiest one to find is typically the average (x+y)/2.  But whether you specifically refer to the density property or not, I would still challenge these objectors to tell me what happens if you add .999... and 1, and then divide by 2.  I assume they would have to answer that 1+.999... = 1.999..., and then 1.999.../2 is what?  .999...5?   I covered in the last refutations essay how .999...5 is an abuse of notation and doesn't have any meaning because you can't tell me the denominator of the place value represented by the 5.  Sooooo.....what is 1.999.../2?  If it's something larger than .999..., then you've contradicted yourself since now .999... is not the "next number down" from 1.  If you say it's less than .999..., then you've just completely ruined the idea of taking an average and ending up between the numbers you started with.  And if you say it's equal to .999... then (using N for .999...) you're saying that:

(1 + N)/2  =  N, which means (if you multiply both sides by 2):
1 + N = 2N, which means (if you subtract N from both sides):
1 = N, which is what I've been saying all along.

The only thing left for you to say is that the average of 1 and the "next number down" from 1 doesn't exist, since that average can't be less than, greater than or equal to .999....  And if the average doesn't exist, you'll have to deny the basic fact (called closure) that if you add two real numbers or divide a real number by 2, you'll end up with another real number.  I suspect you're not really trying to deny the closure of the reals.

Variations on:  .333... is an approximation of 1/3, but it's not actually equal to 1/3.

These objectors seem to think of the number .333... as having some independent existence from the number 1/3, and thus has the option of being the same or different from it.  Actually, .333... is merely a notational description of what happens when you divide 3 into 1, which is the meaning of 1/3 (see below).  If you do long division to determine the decimal equivalent of 1/8 (and I'm not going to demonstrate that here), you'll find that the tenths place contains a 1, the hundredths place contains a 2, and the thousandths place contains a 5, and the ten-thousandths place (and every place after that) contains a 0.  We write that as 1/8=.125 with no problem (although it clearly could also be written as .125000...).  When you try the same thing with the long division for 1/3, you'll find that the tenths place contains a 3, the hundredths place contains a 3, the thousandths place contains a 3, the ten-thousandths place contains a 3, and eventually you'll notice the pattern.  Every place you might ask about contains a 3.  How are we going to write that?  We have decided to write it as .333....  That's all.  It's just a notation.  .333... has to be the same as 1/3 because it's merely a notation for the result of the dividing 1 by 3.

Variations on: 1/3 isn't really a number because you can't truly divide anything equally into 3 parts without a residue left over.

(Biiiiig breath here...)  Wow.  1/3 isn't a number.  This represents a pretty basic problem with the understanding of how numbers work, and explaining something this basic to someone who doesn't understand is probably futile, but here goes.

None of these objectors was really claiming that 1/2 isn't a number.  They see that dividing 1 by 2 gives .5, but they claim that 3 doesn't really "fit" into 1 (they might be interested in my theory of remainders).  While I can (sort of) see how you might give a special privilege to dividing something into halves that you don't want to give to dividing into thirds, do you really also want to give that special privilege to 1/5 (which also has a nice, easy decimal representation)?  Or even 1/125?  (Of course the reason some numbers make nice, easy decimals is the accident of our base 10 counting system.  You can't easily take a third of a dollar, but you can take a third of 90 cents.)

The fact is, it's so possible to divide 1 by 3 that the very meaning of the symbol 1/3 is "the number you get when you divide 1 by 3".  We created 1/3 to actually be the result of dividing 1 by 3, and we made up a symbol for it (namely, 1/3) that is supposed to remind you of it.  If you don't think 1/3 is really a number, you'd better stop all those 2nd-grade kids who are learning about it.

(Whew, that's the best I can do at that one.)

(Actual quote from comments) "Is is NOT and WILL not EVER be exactly 1, because, by DEFINITION, it is LESS than 1. If you fail to understand this, then you simply fail to understand the definition of .9 repeating. I was sick in school, and never went on to uni and higher math, but even I know you're waay wrong." (emphasis mine)

Of course, by now you'll know how I'm going to answer this person mathematically.  The definition of .999... actually says that it is equal to 1.  But that's not my point in bringing up this objection, which basically is just disagreement by intuition.

This (and other comments of which this is the most blatant) seem to imply the following:

"You mathematicians are just sooooo pointy-headed and obstinate that you have your little blinders on and you can't think outside the box.  Out here in the real world, we know what's what, and we don't have to study your stupid Cauchy sequences or Dedekind cuts to know stuff about math.  In fact, not having studied formal mathematics actually gives me more credibility, so you should believe what I say."

Along with this seems to come the notion that I am irresponsible for teaching this, and that people like me are what's wrong with education today—namely, a bunch of geeks who don't teach anything about the real world.

No, no, no, no, no.  The thing that's wrong with education today is that so many people (not just kids) just want the shortcutting, life-applicable, "give me the bullet" answer without taking the time to really think about anything.  Do you truly believe that in the thousands of years of brilliant, creative thinking about math, that no one noticed that .999... seems at first to be less than 1?  The sun seems to be revolving around the earth, but we don't question the astronomers.  Matter seems to be infinitely divisible, but we don't question atomic physicists.  Consuming sugar seems to be the right thing to do when I feel that my blood-sugar level is low, but I didn't question my doctor when he (correctly) explained that this would (counterintuitively) just perpetuate the low-blood-sugar cycle.  We trust professional chefs to make our restaurant food taste good, airline mechanics to fix our planes, architects to design our skyscrapers.  All of these people have spent years and years studying their fields and they hold lifetimes of research and intuition in their heads.  But mathematicians are just too dorky and clueless to understand a basic (although slightly counterintuitive) fact about representing numbers in our base 10 notation system?  Give me a break.

I'm not claiming to be right on everything, just this fact about math.  And I'm not claiming to be the greatest teacher in the world.  But I'm not what's wrong with our education system.  Your non-appreciation for rigorous math, and the centuries of thought that went into it, is what's wrong with our education system.  Sticking your fingers in your ears and going La-La-La-La-I-Won't-Listen-To-Math-Geeks-La-La is what's wrong with our education system.  Show some humility.