Wikipedia:Reference desk/Archives/Mathematics/2007 May 3

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May 3

1/a = 1/b + 1/c

I have discovered a sequence {0,0}, {640,630}, {1152,1120}, {2016,1920} ... The sets represent integers for which 1/a = 1/b + 1/40320 (a is the first number, b is the second). I derived some terms by testing every number. Is there a faster way? For example, An equation with one variable that gives 0 if it is 0, 640 if it is 1, 1152 if it is 2, etc. Could this be generalized to c≠40320? Thanks *Max* 00:40, 3 May 2007 (UTC)

I must note that {0,0} does not fit - 1/0 is undefined, and the equation is impossible. Also, you can write a program that will check every possible integer b to determine the corresponding a and check if a is an integer. You could also graph the equation and see if it looks familiar, then maybe you can find a relationship. ST47Talk 01:02, 3 May 2007 (UTC)

Also, I think your ordering is reversed: the smaller number should come first, since its reciprocal will be the larger one, by 1/40320 to be precise.

I am not sure one can come up with a formula exactly as you ask, but it is possible to make this easier to find solutions rather than exhaustively searching. Here is a hint: read about the fundamental theorem of arithmetic, then rearrange your equation a bit. Baccyak4H (Yak!) 01:52, 3 May 2007 (UTC)

1/a = 1/b + 1/40320

1/a = (40320 + b)/(40320 * b)

1 = (a * (40320 + b)) / (40320 * b)

Thus

a * (40320 + b) = 40320 * b

a = 40320 * b / (40320 + b)

thus

  (40320 * b) == 0  [mod (40320 + b) ]

Now run through all integer values of b and find a. 202.168.50.40 06:12, 3 May 2007 (UTC)

All triples (a, b, c) of the form (pq(r−q), pr(r−q), pqr), where a, b and c are all non-zero, satisfy 1/a = 1/b + 1/c. If c = 40320, then, conversely, all triples satisfying 1/a = 1/b + 1/c can be expressed in that form. For example, (640, 630, 40320) is obtained by setting (p, q, r) := (10, 63, 64), and (1120, 1152, 40320) by (p, q, r) := (32, 35, 36). In general, all ways of decomposing c into a product of three factors give a solution. Since 40320 is a highly factorable number, it gives many solutions. I think (but have not proved) that in general, for all values of c, all solutions can be found by decomposing c into a product of three factors.  --Lambiam^Talk 10:19, 3 May 2007 (UTC)

Pretty much what I was getting at. I suppose one would have to allow for degenerate factors of 1 for a very sparse factorization (or even being prime), but the formulation would still work (I think). Baccyak4H (Yak!) 13:30, 3 May 2007 (UTC)

Thanks, I thought it would have something to do with the number of factors (I chose 40320 because it is 8!). Has anyone proved whether this produces all solutions? Also ∞ = ∞ + 1/40320 so {0,0} technically works. *Max* 13:40, 3 May 2007 (UTC)

The 1/0 thing is problematic because its value can really only be understood as a limit, and thus two different limiting processes can give two different "values" of 1/0 [!]. But if you deem your limiting convention making them equal, so be it.

The factorization of 8! can be exploited as follows: break it up into prime factors (2 2 2 2 2 2 2 3 3 5 7), then simply enumerate the ways to partition these factors into three sets. Then for each set, multiply the elements together: that gives you p, q and r (note the multiplicities involved—if p and q are both 4, you can select them in many ways, but there is really only one such resulting triple). Then just plug into the relation above. The fact that you are using 8! only means there will be many such partitions, but the same process would work for other integers.

As for all solutions, the only other way to generate solutions would be to let your known number (8!) factor as p r (r-q), that is, the other denominator of the sum. I suspect that the same set of triples will come out of it though... Baccyak4H (Yak!) 14:11, 3 May 2007 (UTC)

I don't know what reasoning leads to the conclusion that this is the only other way, but in any case this way of factoring c will not generate new solutions, as can be seen by applying the substitution q := r−q to the triple (pq(r−q), pr(r−q), pqr).  --Lambiam^Talk 15:43, 3 May 2007 (UTC)

(undent) You're right, I got the abc and pqr triples mixed up in my head [!] However, I think it can be shown that those are the only solutions.

We know there are solutions. Let a, b, and c be a solution. Rearranging the equation gives

$bc=a(b+c)$

Let d be the GCD of b and c: b = d B, and c = d C, for relatively prime B and C. Plug in and cancel the common factor d:

$dBC=a(B+C)$

But since B and C are relatively prime, B+C shares factors with neither B nor C thus not of their product B C either. Thus, by the fundamental theorem of arithmetic, B+C must be a divisor of d. So there is integer D such that d = D (B + C). Plug in:

$D(B+C)BC=a(B+C)$

(see where we're going?)

$DBC=a$.

Now the rest:

$b=dB=(D(B+C))B=BD(B+C)$

$c=dC=(D(B+C))C=CD(B+C)$.

So the general solution is to select positive integer D and relatively prime positive integers B and C. Then a solution is

${\begin{aligned}a=&BCD\\b=&BD(B+C)\\c=&CD(B+C)\end{aligned}}$

Here D is p in Lamb's solution, B + C is r, and either B or C is q, The only loose end here is to see if everything still holds when one or more of B, C, or D is 1. I think it does... Baccyak4H (Yak!) 18:07, 3 May 2007 (UTC)

Magic formula for Magic Squares?

I am a primary school teacher and have always taught my students the following method for creating magic squares. They love it as it's easy to do and works with absolutely any sequence of numbers. Some of the higher ability kids love playing with decimals and are amazed when it works for them too. While the method below works for 3 X 3 squares, are there any (simple?) formulas or methods for creating 4 X 4 squares or larger? My students(age 8 & 9) will love it Thanks and regards. Kirk UK

Start with a three by three grid with an extra square at each middle.

     
      { }
   { }{ }{ }
{ }{ }{ }{ }{ }	   
   { }{ }{ }
      { }

Fill in using diagonals with any sequence of numbers.

      {3}
   {2}{ }{6}
{1}{ }{5}{ }{9}   
   {4}{ }{8}
      {7}

Flip the extra middle squares into their opposite box.

      {3}
   {2}{7}{6}
{1}{9}{5}{1}{9}   
   {4}{3}{8}
      {7}

 Remove the extra, outermost squares. Et Voila, Magic square.
   {2}{7}{6}
   {9}{5}{1}     
   {4}{3}{8}

</nowiki> —The preceding unsigned comment was added by 82.153.13.109 (talk) 10:01, 3 May 2007 (UTC).

Well, if you start at the top row and fill it in with a complete sequence, then continue the sequence on the third row, only starting from the right, then fill in the second row from the right, then do the last row right to left, you should get this:

1  2  3  4
12 11 10 9
8  7  6  5
13 14 15 16

Then, if you flip the middle two columns vertically:

1  14 15 4
12 7  6  9
8  11 10 5
13 2  3 16

You should have a magic square. ST47Talk 10:22, 3 May 2007 (UTC)

Also, from our atrticle:

A method for constructing a magic square of odd order

Starting from the central column of the first row with the number 1, the fundamental movement for filling the squares is diagonally up and right, one step at a time. If a filled square is encountered, one moves vertically down one square instead, then continuing as before. When a move would leave the square, it is wrapped around to the last row or first column, respectively. By way of example:

Order 5
17	24	1	8	15
23	5	7	14	16
4	6	13	20	22
10	12	19	21	3
11	18	25	2	9

—The preceding unsigned comment was added by ST47 (talk • contribs) 10:27, 3 May 2007 (UTC).

The last method is essentially the same as used for the 3×3 square in the original question. Applied to order 5:

             5               
          4    10            
       3     9    15           3 16  9 22 15
    2     8    14    20       20  8 21 14  2
 1     7    13    19    25     7 25 13  1 19
    6    12    18    24       24 12  5 18  6
      11    17    23          11  4 17 10 23
         16    22            
            21               

You can fill out the right-hand square directly by following diagonals + wrapping around, the only differences being that you start at a different position and make a different hop when proceeding to the next diagonal. For other construction methods, including for magic squares of even order, see our article Magic square.  --Lambiam^Talk 10:57, 3 May 2007 (UTC)

Magic square 4+5=9

Slightly off-topic, I recommend that teachers like the original poster visit The Math Forum at Drexel University. There teachers and mathematics experts exchange questions and ideas. Apropos of this question, you will find a wealth of material on magic squares, a perennial favorite. Your students might enjoy the variation shown here; it's based on a description by Harvey Heinz. --KSmrq^T 20:48, 3 May 2007 (UTC)

How do calculus, economics and physics relate?

—Preceding unsigned comment added by Todayok (talk • contribs)

Do your own homework. This sounds like an open ended question to see what you know on the subject (There's no real "correct" answer, but some certainly are better than others). Or study Calculus, Economics, and Physics and get your own answer. Root⁴(one) 14:22, 3 May 2007 (UTC)

Because some people find doing their own economics and physics homework as unpleasant as having their teeth scraped ? StuRat 19:13, 3 May 2007 (UTC)

A mathematician, physicist and economist were marooned on a desert island. The mathematician proved the logical impossibility of escaping. The physicist discovered that the fundamental particles locally were grains of sand which were of no use in escaping. The economist founded an economy based on the plentiful natural resources and with the wealth created paid for his escape on a passing liner.…Semiable 11:08, 4 May 2007 (UTC)

God Equation

Has someone developed a mathematical equation that represents the existence of God? —The preceding unsigned comment was added by 66.192.59.2 (talk) 15:37, 3 May 2007 (UTC).

File:Godelproof.gif

Gödel's "proof" that God exists.

This may not be quite what you are looking for, but Kurt Gödel has formulated Saint Anselm's ontological argument for God's existence as a formal logical proof. That of course doesn't mean that the argument is valid, but merely that it can be expressed in the language of mathematical logic. --mglg_(talk) 16:38, 3 May 2007 (UTC)

To clarify my last sentence above, you can prove anything with formal logic if you choose appropriate axioms. For example, the axioms {Axiom 1="(The sky is green) or (the sky is purple)", Axiom 2="The sky is not purple"} lead logically to the conclusion that the sky is green. --mglg_(talk) 17:47, 3 May 2007 (UTC)

I once read a paper with a title such as On the structure of Spinoza's God, which went deeply into some abstruse mathematical theory, perhaps something like ordinal numbers of high cardinality. I don't remember any details. An equation proving the existence of God is ascribed to Leonhard Euler. It is remarkably simple:

${\frac {a+b^{n}}{n}}=x\,.$

 --Lambiam^Talk 19:01, 3 May 2007 (UTC)

So how did Euler's proof work? --mglg_(talk) 19:16, 3 May 2007 (UTC)

Sorry, I see that the answer is on the Euler page. --mglg_(talk) 19:25, 3 May 2007 (UTC)

Euler believed in God? If so, that really diminishes my admiration for the man.--Kirby♥time 19:17, 3 May 2007 (UTC)

If that's true (either that Euler believes in a God or that you're disappointed because you believe Euler believes in a God), you'll really be in a shock when you read that Carl Friedrich Gauss was "deeply religious". (of course, our article doesn't delve into that). Root⁴(one) 16:12, 4 May 2007 (UTC)

Ugh... That's really bad... Anyway, I've always thought that it's possible Newton "believed" in God in order to have his scientific theories not prosecuted by the Church and its zealots. --Taraborn 07:58, 6 May 2007 (UTC)

Didn't Newton write a good deal of religious material? —Tamfang 06:36, 7 May 2007 (UTC)

If you define logic as part of mathematics then there is a logical equation that "leads" to the proof of Allah. Curry's paradox

The truth of the Quran is equivalent to the statement if the quran is true then Allah must be true

Let us denote by ALLAH the proposition to prove, in this case "ALLAH exists". Then, let QURAN denote the statement in question, which asserts that ALLAH follows from the truth of itself. Mathematically, this can be written as

QURAN = (QURAN → ALLAH), and we see that QURAN is defined in terms of itself. The proof proceeds:

1. QURAN → QURAN

identity

2. QURAN → (QURAN → ALLAH)

substitute right side of 1, since QURAN = QURAN → ALLAH

3. QURAN → ALLAH

from 2 by contraction

4. QURAN

substitute 3, since QURAN = QURAN → ALLAH

5. ALLAH

from 4 and 3 by modus ponens

A particular case of this paradox is when ALLAH is in fact a contradiction. Then QURAN becomes QURAN = (QURAN → false), or equivalently (QURAN = ¬QURAN), which is exactly the liar paradox. Ohanian 10:00, 6 January 2006 (UTC)

Way to prove .999, or theories against it?

I pointed this out to a friend, he absolutely denies it's possible. I'm only familiar with the 3/3 method of proving it, his feeling was that 1=3/3 and .999...=3/3, but they're different values of 3/3, the first being 3/3 as a whole number, the second being 3/3 as a fraction. He also kept going on about how you "can't lose a .1" and how .999...=1 only because people are too lazy to write infinite numbers. He also mentioned something about .111... + .999... = 1, which makes absolutely no sense (it would be 1.111...). Anyone have any idea what kind of theory might actually support him? .999's article seems to overwhelmingly prove him wrong =/ I'd really just like a simple way to prove this to him, preferably not involving fractions, as he seems to think that's part of the problem. He's a fairly smart guy, at least as smart as I am as long as I've known him, so I'm not entirely sure he's not just arguing for entertainment, but he's trying to compare this problem to asking for "a third of a dollar = 33 cents, not 33.333... cents", so if he's not just being a goof it should be easy to disprove. I just don't understand math very well, because I've never given much of a care to it -- Phoeba Wright^OBJECTION! 21:56, 3 May 2007 (UTC)

• .999....

.999... is a shorthand way of writing

$.999...=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}{\frac {9}{10^{i}}}$

Please see Wikipedia:Reference_desk/Archives/Mathematics/2007_April_13#.9_repeating_.3D_1.3F.3F.3F.

--Kirby♥time 22:08, 3 May 2007 (UTC)

Try this: If .999... isn't equal to 1, then it must be less than 1, right? If it's less than 1, then it's some amount less than 1, which you could call 'x', so $1-0.999...=x$. So what number is x? It's clearly smaller than 1/10, because .9999.... is bigger than .9, which is 1/10 smaller than 1. Similarly, x is smaller than 1/100, or 1/1000, or 1/10000,.... and so on. Continuing this way, x is smaller than any positive number, no matter how small, so it must be zero. If the difference between 1 and 0.99999.... is zero, then it must be true that $0.999....=1$. -GTBacchus^(talk) 22:09, 3 May 2007 (UTC)

Our article 0.999... discusses this from about all possible viewpoints. But of course your friend is free to maintain that the whole numbers are not "really" real numbers, as long as he agrees that the real numbers contain a subset that is isomorphic to the whole numbers – that is, they are for all purposes equivalent. To consider the whole numbers as forming that subset themselves is a common and often useful, but not indispensable, convention. In fact, category theorists often prefer to see these as separate, and the semantics of several programming languages agrees. In that case there is equally a difference between 1 and 1.000..., which is fine, but hopefully your friend agrees that 0.999... and 1.000... are two ways of denoting the same real number; otherwise we have a real problem – or, rather, he has one. One last remark: your friend would be well advised not to express his sentiment in the form that .999 ≠ 1, since by universal convention (generally agreed upon among all mathematicians) the relation symbols = and ≠, when used between (for example) a whole number and a real number, actually mean comparing two real numbers, where the standard embedding is used of the whole numbers in the set of real numbers.  --Lambiam^Talk 22:23, 3 May 2007 (UTC)

Thanks GT, that's what I needed. I checked the article, as I said rather clearly above, I've just never been much of a math geek so most of these things go in one ear and out the other. I'm good with math that's short and interesting far more than anything that reminds me of DDR two levels too hard when I look at it. That's why I need a simple explanation I can argue from. -- Phoeba Wright^OBJECTION! 22:29, 3 May 2007 (UTC)

If your friend knows too much, the argument will not convince him. In the surreal numbers, for instance, there are numbers that are smaller than 1/10, 1/100, and so on, and yet larger than 0. If he has already conceded, as you wrote, that 0.999... = 3/3 and 1 = 3/3, then the conventional clincher would now be: 0.999 = 3/3 = 1, therefore 0.999 = 1. But he says they are different kinds of numbers. It is not possible to disprove this.  --Lambiam^Talk 23:45, 3 May 2007 (UTC)

We usually defer formal definitions of number systems to a fairly advanced stage of a mathematical education. Without a formal definition, a formal proof is impossible. For example, a surprising amount of sophistication is required to prove 1+1 = 2. Usually we are content to muddle along without the formalities, so long as our "intuition" can keep up. Rarely does anyone doubt that 1+1 = 2, and rarely does anyone challenge the rules for calculating with fractions. (Though in the case of something like ¹⁄₂+¹⁄₃ = ⁵⁄₆ I think it's more a matter of being too intimidated.) When we are told that ²⁄₃ has the same value as ⁴⁄₆, we accept it. After all, we are accustomed to different calculations giving the same answer. And perhaps it is comforting that we can reduce a fraction to lowest terms, as if ⁴⁄₆ is a calculation and ²⁄₃ is the answer. Decimal fractions like 0.866 are convenient tools in the modern world, but they came late to the party. Clearly one of the reasons is because recurring decimals are a nuisance, even without the issue in question.

I claim that we have two alternatives: (1) take it on faith, perhaps with some handwave arguments, or (2) deal with the necessary formalisms. Those who are satisfied with neither option will only frustrate themselves and those around them. Here is a solid formal argument:

---

The step from rationals to reals is huge, and order is an essential part of the construction. In the Dedekind cut approach, each real number z is a partition of the rational numbers into two sets, (B, A), with the numbers in B being all those ordered less than (below) z and the numbers in A being the rest (above or equal). So for any non-empty set of rationals S bounded above, let U(S) be the set of all rationals that are upper bounds of S. (Thus for any x in S and y in U(S), x ≤ y.) With U(S) as A and its complement (in the rationals) as B, a definite real number is selected.

Now let the set S be {⁰⁄₁, ⁹⁄₁₀, ⁹⁹⁄₁₀₀, ⁹⁹⁹⁄₁₀₀₀, …}, the rational numbers obtained as truncations of 0.9999… to 0, 1, 2, 3, or any number of decimal places. In this way, every number in decimal notation determines a Dedekind cut, which is taken to define its meaning as a real number. The task is thus to show that U({1}) is the same set (and thus gives the same Dedekind cut) as U(S), or equivalently, to show that 1 is the least rational greater than or equal to every member of S.

If an upper bound less than 1 exists, it can be written as 1−x for some positive rational x. To bound ⁹⁄₁₀, which is ¹⁄₁₀ less than 1, x can be at most ¹⁄₁₀. Continuing in this fashion through each decimal place in turn shows that x must be less than ¹⁄_10ⁿ for every positive integer n. But the rationals have the Archimedean property (they contain no infinitesimals), so it must be the case that x = 0. Therefore U(S) = U({1}), and 0.9999… = 1.

---

This is, of course, the same idea described informally by GTBacchus; but look at what we had to do formally. We first defined the real numbers in terms of sets of rational numbers. That's work, and unfamiliar; but we can hardly have a serious discussion without such a step. We also formally defined a way to interpret a decimal fraction as a real number. Again, omitting this step would reduce us to handwaving. Finally, we relied on a subtle technical detail about rational numbers, drawing on an observation first made by Archimedes early in the history of mathematics. If rational numbers could be infinitesimal, the proof would fall apart — along with a great deal of other mathematics!

But your friend has a much more serious problem in the distinctions he claims to draw; he would break even more mathematics. --KSmrq^T 03:42, 4 May 2007 (UTC)

In addition to the required formal machinery, something that is usually overlooked is the need to explain how the rationals are (to be considered) embedded in the newly constructed reals. If you construct the real numbers as being partitions of the ordered set of rationals, then of course no rational number is also a real number. Similarly for going from the naturals to the integers, etcetera. Only our article Complex number manages to describe how real numbers may be considered to be complex numbers.  --Lambiam^Talk 06:47, 4 May 2007 (UTC)

The argument above doesn't explicitly use that injection, though it is trivial to state: take the set S to be a singleton set containing the desired rational number. (Usually "1" and "1.0" are non-controversial.) --KSmrq^T 08:53, 4 May 2007 (UTC)

Why do we need to construct the reals to prove that 0.999...=1 ? Surely this is true in Q as well as in R ? Every partial sum in the series $\sum _{i=1}^{n}{\frac {9}{10^{i}}}$ is rational. Working in Q we don't have completeness, so we can't assume that the limit $\lim _{n\rightarrow \infty }\sum _{i=1}^{n}{\frac {9}{10^{i}}}$ exists in Q - we have to show that the limit does indeed exist and that it equals 1. But this is quite straightforward isn't it since all partial sums after the n^th lie within 10^-n of 1. No irrationals required. Gandalf61 12:55, 4 May 2007 (UTC)

Fine with me if it is agreed in advance that the meaning of 0.999... is "the sum of that infinite series in the rational numbers". That is not the usual understanding of the meaning of an infinite decimal expansion.  --Lambiam^Talk 13:49, 4 May 2007 (UTC)

Sorry, but I don't follow what you mean by "usual understanding". 0.999... is the limit of the sequence 0.9, 0.99, 0.999 ..., which is in turn the limit of the series 0.9, 0.9+0.09, 0.9+0.09+0.009 ...., or in other words $\lim _{n\rightarrow \infty }\sum _{i=1}^{n}{\frac {9}{10^{i}}}$. If that isn't the "usual understanding" of 0.999... then what is ? Gandalf61 14:00, 4 May 2007 (UTC)

To avoid the real numbers, how about we start by proving that a decimal expansion which repeats represents a rational number. (Start by doing a couple of examples, then write down the general proof.) Then notice 0.9999... has a repeating decimal expansion so it is a rational number. Now the only question is, which rational number is it? Stefán 17:19, 4 May 2007 (UTC)

I mean that the decimal representation

$a_{0}.a_{1}a_{2}a_{3}\dots \,,$

where $a_{0}\,$ is a nonnegative integer, and $a_{1},a_{2},\dots \,$ are integers satisfying $0\leq a_{i}\leq 9$, is normally understood to denote the real number

$\sum _{i=0}^{\infty }{\frac {a_{i}}{10^{i}}}\,.$

In this particular case that real number also happens to be in the image of the usual injection of the rationals into the reals. If we know that the series has a limit in the domain of rationals, then we also know that that rational limit maps to the same real number. This is not part of the definition; it is a theorem that needs to be proved before you can appeal to it. While that may seem unnecessarily fussy, if people cast aspersions on our proof methods we can either choose to ignore them, or increase the level of meticulousness; it is not the right occasion to start handwaving. In this case it is (in my opinion) easier to prove "directly" that 0.999... equals 1 than to prove this limit-weakly-commutes-with-embedding theorem and then appeal to it.  --Lambiam^Talk 17:21, 4 May 2007 (UTC)

I agree, a direct proof is easier. And pragmatically, lay readers are likely to be more comfortable with sets than with the limit of an infinite series (or sequence). Worse, almost every "proof" I've seen using a limit approach both begs the question and befuddles the questioner. Here is a more careful handling.

---

Another approach to constructing the real numbers uses the ordering of rationals less directly. First, the distance between x and y is defined as the absolute value |x−y|, where |z| is the maximum of z and −z, thus never negative. Then the reals are defined to be the sequences of rationals that are Cauchy using this distance. That is, in the sequence (x₀, x₁, x₂, …), a mapping from natural numbers to rationals, for any positive rational δ there is an N such that |x_m−x_n| ≤ δ for all m, n >N. (The distance between terms becomes arbitrarily small.)

A sequence (x₀, x₁, x₂, …) has a limit x if the distance |x−x_n| becomes arbitrarily small as n increases. Now if (x_n) and (y_n) are two Cauchy sequences, taken to be real numbers, then they are defined to be equal as real numbers if the sequence (x_n−y_n) has the the limit 0. Truncations of the decimal number b₀.b₁b₂b₃… generate a sequence of rationals which is Cauchy; this is taken to define the real value of the number. Thus in this formalism the task is to show that the sequence

(1−0, 1−⁹⁄₁₀, 1−⁹⁹⁄₁₀₀, …) = (1, ¹⁄₁₀, ¹⁄₁₀₀, …)

has the limit 0. But this is clear by inspection, and so again it must be the case that 0.9999… = 1.

---

Although formally sound, the language and concepts are deliberately kept elementary. And again we: define reals in terms of rationals, state the interpretation of a decimal expansion as a real number, and force a zero difference. Sadly, many who think an infinite sum is obvious and natural fail to see the difference between their "proof" and this one.

For those curious about mathematical foundations, it is interesting to compare the two radically different constructions of the reals in the two proofs I presented. Are these truly the same real numbers? The answer has a subtle twist. Ordinarily, it's yes; but if we build our foundations on topos theory, then not necessarily. See, for example, page 414 of Robert Goldblatt's Topoi: The Categorial Analysis of Logic, ISBN 978-0-486-45026-1, available online at the Cornell Library Historical Mathematics Monographs. --KSmrq^T 20:33, 4 May 2007 (UTC)

For didactic purposes, I believe a definition of the reals in terms of (equivalence classes of) shrinking intervals of rationals is superior to both Dedekind cuts and Cauchy sequences.  --Lambiam^Talk 07:43, 5 May 2007 (UTC)

Ohanian saids " I have a wonderfully elegant proof that .999... = 1 but this margin is too short for me to write it down, however I shall not be a Fermat person and WILL WRITE IT IN THIS DAMNED MARGIN anyway! "

Let $U={\mbox{unity}}=1\,$

Let $N={\mbox{nines}}=0.999...\,$

Proof by contradiction.

Assume that $N\neq U$. This means that there are only two cases that can follow.

CASE 1 : $N>U$

$N>U\,$

$\downarrow$

$\sum _{k=1}^{\infty }{\frac {9}{10^{k}}}>1$

$\downarrow$

${\frac {9}{10}}+{\frac {9}{100}}+{\frac {9}{1000}}+...>1\,$

$\downarrow$

Contradiction!!!

CASE 2 : $N<U$

$N<U\,$

$\downarrow$

$U+N<U+U$

$\downarrow$

$U+N<2U$

$\downarrow$

$(U+N)\cdot (U-N)<2U\cdot (U-N)$

$\downarrow$

$U^{2}-N^{2}<2U\cdot (1-\sum _{k=1}^{\infty }{\frac {9}{10^{k}}})$

$\downarrow$

$U^{2}-N^{2}<2U\cdot (\lim _{k\to \infty }{\frac {1}{10^{k}}})$

$\downarrow$

$U^{2}-N^{2}<2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

$U^{2}<N^{2}+2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

$U^{2}<N^{2}+2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}<N+2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

$U^{2}<N+2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

$U<N+2\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

${\mbox{ }}$

${\frac {U}{2}}<{\frac {N}{2}}+{\frac {2}{2}}\cdot \lim _{k\to \infty }{\frac {1}{10^{k}}}$

$\downarrow$

$0.5<0.499...995+0.000...001$

$\downarrow$

Contradiction!!!

Now since both CASE 1 and CASE 2 results in contradiction, the only conclusion we can come up with is that 0.999... = 1

I rest my case. Ohanian 01:01, 5 May 2007 (UTC)

Our article 0.999... already contains many proofs, most of which I find more convincing than this proof. If the sixth and third steps from the bottom are allowed as single-step inferences, then you can also prove in just two steps that U − N = 0.  --Lambiam^Talk 07:27, 5 May 2007 (UTC)

I've tried showing him this, now he's saying "you still can't get me to believe that .0000~00001 doesn't exist". Sounds like now it would be better to explain infinity, but I'm again not sure how to do it in a convincing manner to him. Halp? -- Phoeba Wright^OBJECTION! 20:29, 7 May 2007 (UTC)

What abuot asking him to calcultate (1+0.999...)/2? IF it's not 1 then if sholud be something bigger than 0.999 whith is not possible. —Preceding unsigned comment added by 195.238.64.156 (talk • contribs) 21:12, 2007 May 7

Sorry, no; it would be a misuse of this forum to indulge your friend. As I first said, either one does the work or one accepts the handwaving. We already have incurred one long-winded idiotic non-proof. There are plenty of places on the web where people delight in debating this at length; and we have given pointers to serious discussions as well. It is intellectually dishonest, self-indulgent, and lazy to assert the existence of "0.0000~00001" without formally defining its meaning and how to calculate with it. (See Invisible Pink Unicorn.) --KSmrq^T 23:08, 7 May 2007 (UTC)