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Why 0 .999… is Not Equal to 1!
By Dr A. Crank


Dear friends, this may get a bit complex, so have a few beers first, and I am sure it will go down smoothly.

Abraham Fraenkal, in “Abstract Set Theory,” (North-Holland, Amsterdam, 1976), gives the following auxiliary theorem from the theory of decimals:

Every positive (non-zero) real number A, has one and only one expansion into an infinite decimal A= m.c1c2c3…ck…, where m is a non-negative integer and the digits ck assume the values 0, 1,2, 3, …9, with the proviso that after every ck digits different from 0 will appear. If from a certain place and onwards in the decimal expansion only zeros appear, the decimal expansion is said to be terminating; if otherwise, non-terminating. Therefore, two infinite decimals which are not identical represent different real numbers. (p. 51)

Fraenkel then proposes that a positive real number that can be expanded into a terminating decimal:

(F1) m.c1c2c3…cn, cn not =0

is equal to the infinite decimal:

(F2) m.c1c2c3...(cn -1)99999….

or if it is a positive integer m, to (m-1).9999…. Fraenkel recognized that this claim was surprising, but said that the equality 1=0.999…, can be obtained by multiplying by 9 the equality 1/9 = 0.111… . (p. 4) See also, G. H. Hardy and G. M. Wright, “An Introduction to the Theory of Numbers,” (Clarendon Press, 1960).

The notion that non-terminating decimals with a repetition of digits, such as 0.999…=1, and 1/3= 0.333…, and so on, are generally accepted, outside of non-standard analysis, to be correct, not only because of the definition given above, but because of a number of mathematical arguments, which will be reviewed and refuted here. For the orthodox positions see “Why is 0.999…Equal to 1?” At http://www.quora.com/Why-is-0999-Idots-equal-to-1; M. A. Navarro and P. P. Carreras, “A Socratic Methodological Proposal for the Study of the Equality 0.999…=1,” “The Teaching of Mathematics,” vol. 13, 2010, pp. 17-34; R. Pemantle and C. Schneider, “When is 0.999…Equal to 1?” (April, 2007), at https://www.math.upenn.edu/~pemantle/papers/nearly.pdf.

From the perspective of non-standard analysis, it has been maintained that 0.999…< 1. In general, the number .999…, is thought of as an infinite decimal, and although some reject that idea (including me), many do not: “What is Wrong with Thinking of Real Numbers as Infinite Decimals?” At https://www.dpmms.com.ac.uk/~wtglo/decimals.html. The standard real number 0.999… would be represented by:

.999…; …999… ,

which falls short of 1 by an allegedly infinite amount, 1/10.H See K. U. Katz and M. G. Katz, “When is .999… Less than 1?” “Montana Mathematics Enthusiast,” vol. 7, no. 1, 2010, pp. 3-30, at p. 5.

In Lightstone’s notation:

.999… ; …999… < 1. See: Katz and Katz, above, p. 6; A Lightstone, “Infinitesimals,” “American Mathematical Monthly,” vol. 79, 1972, pp. 242-251. For further literature on this point see K. U. Katz and M. G. Katz, “A Strict Non-Standard Inequality .999…< 1,” at arXiv: 0811.0164v8 [math.HO] February 24, 2009; K. U. Katz and M. G. Katz, “Zooming in on Infinitesimal 1-.9… in a Post-Triumvirate Era,” “Educational Studies in Mathematics,” vol. 74, 2010, pp. 259-273; D. Tall, “Intuitions of Infinity,” “Mathematics in School,” May, 1981, pp. 30-33.

Numbers such as H = 0.000… ; 999…, are not members of the hyper-reals, because if on assumption H was, it would be less than any positive real, as it has only 0s to the left of the “;”, which would make it an infinitesimal. But, if an infinitesimal is added to H, there would be a carry over to the left of the “;” so that H would have a non-zero standard part. Hence, the theorem that the sum of two infinitesimals is an infinitesimal would be violated. See; “Is .999…=1? A Non-Standard view,” at http://www.cut-the-knot.org/WhatIs/Infinity/999.shtml. But, I believe that rather than saying that H is not a hyper-real, I prefer to see this as a counter-example, and see the rejection of H as a hyper-real as ad hoc. This is because I reject non-standard analysis as I have argued in my “Fuck Infinity” series.

J Benardete in “Infinity,” (Clarendon Press, 1964), p. 279, said: “The intelligibility of the continuum has been found—many times over—to require that the domain of real numbers be enlarged to include infinitesimals. This enlarged domain may be styled the domain of continuum numbers. It will now be evident that .999… does not equal 1 but falls infinitesimally short of it. I think that .999… should indeed be admitted as a number … though not as a real number.” He also argued against the proposition that the rational number 2/3= .666…, in his paper, “Continuity and the Theory of Measurement,” “Journal of Philosophy,” vol. LXV, 1968, pp. 411-430, at p. 429:

“I think that it should be fairly evident that in the theory of absolute continuity the decimal number .6666 ... is smaller than the rational number 2/3. We may establish that result as follows. Imagine that there is a road that is exactly 2/3 miles long from its initial to its terminal point. Imagine further that the road terminates in a wall. Then, if one is to walk the full distance of 2/3 miles starting from the initial point of the road, one must come smack up against the wall. Now posit a sheet of paper so thin that, for any arbitrary rational number r, the sheet can be shown by measurement to be less than r inches thick. If one is tempted to regard this sheet as being 0 inches thick, i.e., as having no thickness at all, we have only to stack seven such sheets of paper one on top of the other. The whole stack must be thicker than any one sheet taken separately. Each sheet of paper, and the whole stock taken collectively, must then be ε inches thick, where ε is some actual infinitesimal. Let us now paper the wall with these seven sheets, again laid one upon the other. In walking the full distance of 2/3 miles from the initial point of the road it will not be sufficient merely to come into contact with the outermost sheet of paper. One must penetrate through all seven sheets so as actually to come smack up against the surface of the wall. But suppose one walks only .6666 . . . miles from the initial point. That is to say, one walks 6/10 miles + 6/100 miles + 6/1,000 miles etc. ad infinitum – no more and no less. In this case one will not even penetrate the outermost sheet of paper, much less any of the others. We must then conclude that 2/3 is greater than .6666.... .” This could be taken as one counter-argument to the approach which argues that if 1/3=0.333…, then 1=0.999…, because this result is obtained by multiplying the first equation by 3. The mathematical skeptic would need to deny that 1/3= 0.333… .

Another elementary argument for .999…=1, is to let K=.999…, multiply K by 10, giving 9.999…, so 9K=9, and K=1. The problem, as will be seen with all standard proofs, is that this argument begs the question: R. Rucker, “Infinity and the Mind,” (Harvester Press, 1982), p. 79: “this argument overlooks the fact that the difference between 10K and 10 is ten times as great as the difference between K and 1” as there is a “residual infinitesimal quantity below that does not get cancelled out.” See also, A. Pappula, at http://www.quora.com/why-is-0-999-Idots-equal-to-1.

The mathematical skeptic could also argue, as Fred Richman, “Is 0.999…=1?” “Mathematics Magazine,” vol. 72, 1999, pp. 396-400, points out as devil’s advocate, that 9K is 8.999…, not 9, and 8.999…, is not equal to 9. The skeptic could say, that K cannot be cancelled, so subtraction of real numbers is not always possible. (p. 396)

John Gabriel, “The 0.999…=1 Fallacy,” rightly rejects, on finitist grounds, multiplying the “number” .999…, as it has an ill-defined infinite sequence of digits, 999… . Moreover, the multiplication argument is circular, as one can just as easily argue: let K=0.999…, so 10K=10 x 0.999… . Then 9K=9 x 0.999…, so K=0.999… . The “proof” is equally useless, assuming what has to be proved. If you don’t like John Gabriel, then you can find others who recognize this skeptical argument. Dr. Peter Cotton, who plugs for the Cauchy sequence argument (refuted below), and is thus not a skeptic, says that the mathematical skeptic rejects proofs that 0.999…=1, because they are circular, “they assume a calculus on infinite decimal expansions (without first constructing it… which will involve defining 0.999…and 1 to be equal and hence be circular.)” See https://www.quora.com?Why-is-0-999-Idots-equal-to-1. He says that the difference between 0.999… and 1 is arbitrarily small, and that the only standard arbitrarily small number is zero. I disagree, and refute this below.

Yet another possibility is that even if this proof is accepted, there could still be independent arguments for 0.999…< 1, even given arguments for 0.999…=1, in a kind of neo-Kantian antinomy of reason. So, real number theory is negation inconsistent. Such situations of proof “over-determination” are generally ignored by mathematicians, who simply assume that the whole game is consistent, and sound, thus begging the question of the soundness of mathematics. The logical paradoxes indicate that this is not so, and that there can be arguments both for and against a proposition. Thus, even a proof that 0.999…=1, is not necessarily a refutation of 0.999…<1, because consistency is presupposed.

Another popular argument is based on the limit argument, which is found right across the internet. Here it is argued that “0.999…” is “shorthand” for the limit of the sequence of real numbers, 9/10 +9/100 + 9/1,000 +…., which converges to 1. Therefore, 0.999… and 1 are equal. See for example many pages on quora.com; P. Eisenmann, “Why is it Not True that 0.999…<1?” “The Teaching of Mathematics,” vol. 11, 2008, pp. 35-40. We have already seen in the previous article, that there is a need for great caution in identifying the limit of a sequence with the sequence itself. Richman (as above, p. 397), says: “some distinction between convergence and equality in the present case might be appropriate.” D. Tall, cited from Katz and Katz (“Educ. Stud. Math,” paper, as above, pp. 260-261), said: “the infinite decimal 0.999…is intended to signal the limit of the sequence 0.9, 0.99, 0.999… which is 1, but in practice it is often imagined as a limiting process which never quite reaches 1.”

Carl B. Boyer, “The History of the Calculus and its Conceptual Development,” (Dover Publications, New York, 1949), said; “Cauchy had stated in his “Cours d’ analyse,” that irrational numbers are to be regarded as the limits of sequences of rational numbers. Since the limit is defined as a number to which the terms of the sequence approach in such a way that ultimately the difference between this number and the terms of the sequence can be made less than any given number, the existence of the irrational number depends, in the definition of limit, upon the known existence, and hence the prior definition, of the very quantity whose definition is being attempted. That is, one cannot define the number √2 as the limit of the sequence 1, 1.4, 1.41, 1.414, …, because to prove this sequence has a limit one must assume, in view of the definitions of limit and convergence, the existence of this number as previously demonstrated or defined.

Cauchy appears not to have noticed the circularity of the reasoning in this connection, but tacitly assumed that every sequence converging within itself has a limit.” (pp. 281-282) The same circularity objection can be made against the argument using an infinite series to allegedly show that 0.999…=1.

Real numbers may be defined as the limit of Cauchy sequences of rational numbers. Thus, if xn and yn are two Cauchy sequences, then they are equal if the sequence xn – yn has limit 0. Here (1-0, 1-9/10, 1-99/100, …) = (1, 1/10, 1/100, …) would have limit 0.

Another argument is the Dedekind cut argument. A real number R is defined as the infinite set of all rational numbers less than R. Hence, the real number 1 is the set of all rational numbers less than 1. The number 0.999… is the set of rational numbers r such that: r<0 or r<0.9, or r<0.99… and r is less than a number 1 – (1/10).n Therefore, arguing along these lines, every element of 0.999… is less than 1. Hence, it is an element of the real number 1. An element of 1 is a rational number a/b<1, so a/b<1 – (1/10).b As 0.999… and 1 contain the same rational number, they are allegedly the same set, so 0.999…=1.

Richman rightly observes that the mathematical skeptic will reject both of these arguments. A definition is set up to rule out the existence of distinct numbers 0.999… and 1, as 0.999… is the cut {x ε D: x <1} and 1 is the cut {x ε D: x ≤ 1}. Dedekind said that these cuts were “only unessentially different” (p. 398, Richman, as above), which begs the question, because, the signs “<” and “≤” have different meanings, and contrary to Dedekind, the cuts are essentially different. Richman says, “in the traditional definition of the real numbers, the equation 0.9…=1 is built in from the beginning,” and anyone “who challenges that equation is, in fact, challenging the traditional formal view of the real numbers.” (p. 399) As the traditional view of numbers is what is being challenged by the mathematical skeptic, it is obviously question begging to use that account, undefended, in defense.

Consider numbers to be formal strings of symbols. Thus, 1 is denoted by “1.” A number such as .999…, consists of the concatenation of signs “9” followed by “9,” and so on. Further, there is no good reason why one cannot start at the right hand-side of the number, and write .999…, as .999…999. Mathematics is, after all, just a formal game and one can do what one likes, bar absolute inconsistency. In a sense, one could “terminate” a conventional non-terminating decimal because we know, by definition, that .999… is only going to have 9s in its sequence. This will not be so for say irrational numbers, such as √2, or transcendental numbers such as e and π. But, for all practical purposes, it would not matter what number, if any, was put as a “last’ number, or even a special disjunctive last place could be defined; “0 v1 v 2 v 3 v 4 v 5 v 6 v 7 v 8 v 9.” Or, in the alternative, there may be no last digit characterizable at all. It matters not in a game with symbols.

Numbers, then, are constructed using the concatenation of signs “0,” “1,” “2,” “3,” “4,’ “5,’ “6,” “7,’ “8,” and “9.” The first positive non-zero real number can be defined as:

0.000…001, and the next: 0.000…002, and so on. The number 0.999… is thus, 0.999…999. This number is 0.000…001 less than 1.000…000. One can add 0.000…001 to 0.999…999, to get 1.000…000. Whether one calls 0.000…001 an infinitesimal or not, is moot. It is obtained from an elementary rethinking of the way we view real numbers. Given all of the bs that we have looked at in this series of articles, this approach makes as much sense as anything else.

Interestingly enough we can define the largest real number. It is:

…999.9999…999…, an infinite string of 9s going left and right. The reason this is the largest number, is that if one attempts to add to it, one gets a smaller number. There is no way of symbolically representing a larger number.

As well, there will be an infinity of numbers, with no number between them. Thus, it is not true for all real numbers that if x, y, if x < y, then (x + y)/2 < y, as that average number may not be defined, if they are next to each other, so to speak.

It has also been objected by Ted Alper, at https://quora.com/why-is-0-999-Idots-equal-to-1,” that there will be difficulties for this theory, for the usual arithmetic operations. To take addition, for example, he asks: what is: x = 0.4999… + 0.4999…? His suggestion is 0.999…998. He says that for arithmetic whenever a < b, then a + c < b + c. So we have: 0.4999… 999 < 0.5. Then:

x = 0.4999… + 0.4999…( = 0.999…998) < 0.5 + 0.4999… = 0.999…999.

But, he says, x will need to be some number less than 0.999…999, but greater than 0.4999… + 0.4999… = 0.999…998. There is no such number capable of representation, because nothing falls between 0.999…998 and 0.999…999. However, the above inequality is correct, for x = 0.999…998, which is less than 0.5 + 0.499…999 = 0.999…999. Hence, the objection collapses. But, even if it did not, one could maintain that the usual rules of arithmetic do not mechanically apply to infinite numbers, which is what mathematicians say in other areas. Thus, the rule above would fail for some numbers, such as the last and second to last numbers.

Philosophically this approach should be a joy to mathematical skeptics, iconoclasts and cranks, as it gives a way of refuting Cantor’s diagonal argument (the diagonal number, by definition will be in the infinite sequence of concatenated numbers). Further, it gives an example of a number speculated by Graham Priest, to be the largest possible number. See G. Priest, “Inconsistent Models of Arithmetic II: The General Case,” “Journal of Symbolic Logic,” vol. 65, 200, pp. 1519-1529. However, where I disagree with Priest is that he seems to think that this last number is inconsistent, while I see no reason for doing that. In fact, adding to the largest number Ω= 9999999… produces a smaller one, so ~( Ω = Ω + 1).

It is concluded that 0.999… is not equal to 1. All of the standard proofs that 0.999…=1, are circular. Mathematics is in flames, but looking at the dreamy universities, with the masters of the symbolic arts walking around with their hands on their tools (gender neutral tools), you would not think so.

Turd America

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