(Sci Am, 2017)
TABLE OF CONTENTS
1. Introduction
2. The Easy Part: Why 1=0.999 …
3. The Hard Part: Why ~ (1=0.999 …) is True as Well
4. Compatibility with the Standard Real Number Axioms
5. The Archimedean Principle Objection
6. A Tangent: Refuting Cantor’s Diagonal Argument
7. Conclusion
The essay below will be published in two installments; this, the first, contains sections 1-4.
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Why 1=.999… and ~ (1=.999…) are Both True! An Argument for the Inconsistency of the Reals, #1
1. Introduction
The mainstream position in mathematics is that 1= 0.999 … with “1” and the non-terminating recurring decimal real number “0.999 …” denoting the same number. But the issue is hotly debated on the internet by enthusiasts and hobbyists, with sites devoted to the issue. In general, those “skeptics” disputing the equality do so based upon fallacies, or limitations such as:
- Finite Thinking: People tend to treat 0.999 … as a finite sequence (e.g., 0.999 … with a fixed number of 9s), which is indeed less than 1. The infinite nature of 0.999 … is counterintuitive.
- Visually Misleading: The notation “0.999 …” looks smaller than “1” due to its decimal form, leading to a gut feeling that it’s less, even among those with some scientific/mathematical training. Slightly more sophisticated, it may be thought, that 1 > .9, 1 > .99. ,1 > .999, and so on, so 1 > .999 … which is fallacious as it commits a “some” to “all” logical jump.
- Lack of Formal Proof Exposure: Science degree holders may not have encountered rigorous proofs about real numbers (e.g., in analysis or number theory), so their intuition aligns with everyday approximations rather than mathematical precision.
This intuition is reinforced by low numeracy, which distorts perceptions of numerical relationships, even in educated populations. For example, studies show that low numeracy affects risk perception and decision-making (Reyna et al., 2009), suggesting that similar cognitive biases could apply to understanding 0.999 … versus 1.
However, it does not follow from this psychological background that there cannot be a sound argument for ~ (1 = 0.999 …), and for 1 = 0.999 … as well, based upon more sophisticated reasonings. The claim that the proof of 1 = 0.999 … precludes a proof of ~ (1= 0.99 …), assumes that the real numbers are consistent, and such a proof is precluded by Gödel’s Second Incompleteness Theorem. In any case, in this essay, we’ll prove the inconsistency.
2. The Easy Part: Why 1=0.999 …
There are many mainstream proofs of 1=0.999 …, but as the constructive mathematician Fred Richman has shown (Richman, 1999), all of these proofs, while classically acceptable, could be challenged by “skeptics,” defined as those who doubt that 1= 0.999 … . Before Richman’s paper, one of us (JWS) gave similar arguments (Smith, 1991).
One simple algebraic argument for 1 = 0.999 …, is to multiple the equation 1/3 = 0.333… by 3 to get 1 = 0.999 … . The multiplication step is not an avenue for faulting, so the skeptic will challenge the equation 1/3 = 0.333 … by the same type of reasons applied to the issue of 1 = 0.999 … (Richman, 1999, 396).
The use of the principle of the convergence of infinite series can be made, observing that 0.999 = 9/10 +9/100 +9/1,000 + … which converges to 1. Hence it may be concluded that 1 = 0.999 … . But a convergent infinite series does not necessarily have to actually “reach” the limit value, only “tend” to it, so it could be maintained by the skeptic that ~ (1 = 0.999 …) (Richman, 1999, 396).
A more difficult argument to reject is as follows. Let x = 0.999 … Multiple both sides of this equation by 10 giving:
(1) 10x = 9.999 … . Then subtracting the RHS from the LHS, and rearrange to get:
(2) 9x = 9, so:
(3) x = 0.999 …= 1.
The assumption made here is that x is cancellable, which is the most likely objection the skeptic would make, holding that there are cases where real numbers cannot be cancelled (Richman, 1999, 396). The mainstream mathematician assumes that 0.999 … is a number subject to multiplication and subtraction, which requires the completeness of the real numbers. Skeptics might claim these operations are invalid for infinite decimals like 0.999 …, questioning the proof’s validity.
However, this skepticism is self-defeating. If skeptics reject multiplication and subtraction for 0.999 …, they undermine the idea that 0.999 … is a number at all. In any mathematical framework, classical or constructive, a number must support basic arithmetic to be meaningful. If 0.999 … cannot be multiplied (e.g., 0.999 … times 10 does not equal 9.999 …) or subtracted, why assume it exists as a number to compare with 1? For example, claiming 1 is greater than 0.999 … requires arithmetic comparison, which skeptics would deny. This contradiction weakens their position: if 0.999 … lacks number properties, the debate over its equality to 1 becomes incoherent.
Unlike the geometric series proof, the algebraic proof avoids explicit reference to limits or infinity. It treats 0.999 … as a fixed number, making it accessible and robust against constructivist critiques. Even in non-standard analysis, where infinitesimals allow numbers like 1 minus a small value, the standard part of 0.999 … remains 1, supporting the proof.
Skeptics, including laypeople who intuit that 1 is greater than 0.999 … (as seen in online debates and numeracy studies), must accept some arithmetic framework to discuss 0.999 …. By rejecting basic operations, they risk rejecting the number’s existence, making the algebraic proof a compelling resolution to the debate, supposedly.
Hence it can be concluded that there is a sound proof that 1 = 0.999 …, given various arguable assumptions.
3. The Hard Part: Why ~ (1=0.999 …) is True as Well
First, let us assume that for the sake of argument that the reals can be constructed vias infinite decimal representations as argued here: (Hua, 2012; Fardin and Li, 2021; Gowers, n.d.). We note this as some ultra-finitists reject the idea that the reals exist (e.g., N. Wildberger) and that 0.999 … exists. Now to show that ~ (1 = 0.999 …), we need to show that there is a difference between the numbers, done by subtracting them:
1.0000000 …. 000…
-0.9999999 …. 999…
————————-
0.0000000….. ????
We seem to be carrying the “1” with a continuous remainder. The proposal now is to define a real number 0.000 … 001 as the difference between 1 and 0.999 …, arguing that real numbers can be represented as infinite strings terminating at “infinity” with a last digit (e.g., 9 for 0.999 …). We know that 0.999 … if it was to have a “last” digit would have to be a 9, for this number by definition is a string of 9s. So, what is so wrong with rethinking how we organize the writing of 0.999 … and suppose that it is a real number 0.999 … 999, which terminates after an infinite number of 9s, with a 9? After all, mainstream mathematics defines many things to get the results which they want, such as Smooth Infinitesimal Analysis, rejecting the law of the excluded middle (Bell, n.d.), even paraconsistent mathematics (Mortensen, 1995), and in projective geometry, parallel lines meet at “infinity” (as defined (Coxeter, 1987), so we should be able to define reals in this way. You know, intellectual/mathematical freedom and all that.
Thus, we propose that 0.000 … 001 (an infinite string of zeros followed by a 1) is a real number representing the difference between 1 and 0.999 … . This challenges the classical view that 1 = 0.999 … because no real number exists between them. Here is our argument:
- Difference Claim: We assert 1 – 0.999 … = 0.000 … 001, a number with “infinitely many zeros” followed by a 1 at the “last” position.
- Real Number Status: We claim this is a real number, not an infinitesimal (as in non-standard analysis), to avoid reliance on non-standard modelling.
- Proof via Addition: We suggest adding 0.999 … and 0.000 … 001 to get 1, supporting its legitimacy:
0.999 … + 0.000 … 001 = 1.000…000 (carrying the 1 to the “last” digit, yielding 1).
In the standard real number system, 0.000 … 001 (with a 1 after infinitely many zeros) is not a well-defined real number, as decimal expansions don’t have a “last” position. The real number 0.000 … (infinite zeros) is 0, so 0.000…0001 would collapse to 0. However, contrary to this, we propose that reals can be represented as strings “terminating at infinity” with a last digit. This requires a new definition of real numbers, where an infinite string like 0.000 … 001 has a 1 at an “infinite position.” This contradicts the classical Archimedean property (no infinitesimals in the reals) and resembles hyperreal numbers, but remains distinct by our insistence on real number status. We consider this objection below.
There is no need for any nonstandard redefinition of the arithmetic operations such as addition and subtraction. As in any sum, the numbers are lined up:
1.000000000 …. 00000
-0.999999999 …. 99999
——————————-
0.000000000 …. 00001
This allows 0.999 … ≠ 1, as 0.000… 001 exists as the difference, with a last digit of 1.
The positive real, greater than zero can thus be constructed in a table like so, with the first number being 000 … 000, the second being 000 …001, right up to what we define to be the last real number which is 999 … 999. Why is this the last number? Because it is possible to subtract any other number such as 100 … 000, from it to get another smaller real in this case 899 … 999. The table looks like this for positive reals (Richman, 1999, 397):
This table lists real numbers as infinite decimal strings terminating at an “infinite” position, starting from 0.000 … 000 to 999 … 999, to demonstrate that all reals are countable. Each number is represented as a decimal string with a last digit at infinity. The table is ordered lexicographically, beginning with all zeros and ending with all nines (equivalent to 1 in standard terms). A finite excerpt is shown below, as the full table is infinite.
| Index | Real Number | Description | |
| 0 | 0.000…000 | All zeros, last digit 0 at infinity | |
| 1 | 0.000…001 | Zeros with last digit 1 at infinity | |
| 2 | 0.000 … 002 | Zeros with last digit 2 at infinity | |
| … | … | Continuing through all possible digits | |
| K | 999 … 999 | All nines, last digit 9 at infinity | |
Explanation:
- Each row represents a real number as an infinite string with a last digit.
- The number 0.000 … 001 is the difference between 1 and 0.999 …, with a last digit of 1.
- The table includes all possible decimal strings, ordered by their digits.
Or more graphically:
0000000000 … 0000000000
0000000000 … 0000000001
0000000000 … 0000000002
…………………………….
……………………………
9999999999 … 9999999998
9999999999 … 9999999999
4. Compatibility with the Standard Real Number Axioms
We consider now to what extent this idea is compatible with the standard axioms of the real numbers (Lay, 2005). The following axioms define the operations of addition and multiplication, along with their identities and inverses:
(A1) Associative law for addition: For all a, b, c, a + (b + c) = (a + b) + c.
(A2) Existence of additive identity: There exists an element 0 such that for all a, a + 0 = 0 + a = a.
(A3) Existence of additive inverse: For every a, there exists -a such that a + (-a) = (-a) + a = 0.
(A4) Commutative law for addition: For all a, b, a + b = b + a.
(A5) Associative law for multiplication: For all a, b, c, a * (b * c) = (a * b) * c.
(A6) Existence of multiplicative identity: There exists an element 1 ≠ 0 such that for all a,
a * 1 = 1 * a = a.
(A7) Existence of multiplicative inverse: For every a ≠ 0, there exists a^(-1) such that
a * a^(-1) = a^(-1) * a = 1.
(A8) Commutative law for multiplication: For all a, b, a * b = b * a.
(A9) Distributive law: For all a, b, c, a * (b + c) = a * b + a * c.
Order Axioms
There exists a subset P of positive numbers satisfying:
(A10) Trichotomy: For every a, exactly one of the following holds: a ∈ P, -a ∈ P, or a = 0.
(A11) Closure under addition: If a, b ∈ P, then a + b ∈ P.
(A12) Closure under multiplication: If a, b ∈ P, then a * b ∈ P.
Completeness Axiom
A least upper bound of a set A is a number x such that x ≥ y for all y ∈ A, and if z is also an upper bound for A, then z ≥ x.
(A13) Existence of least upper bounds: Every nonempty set A of real numbers that is bounded above has a least upper bound.
Properties (A1)–(A12), and anything that follows from them, are referred to as elementary arithmetic. Adding property (A13) uniquely determines the real numbers.
Consider (A1): a + (b + c) = (a + b) + c
Let a = 0.000 … 001. Then the left hand side, LHS = 0.000 … 001 + (b + c) = (b + c) + 0.000 … 001. The right hand side RHS = (0.000 … 001 + b) + c = 0.000 … 001 + (b + c).
Hence, RHS = (b + c) + 0.000 … 001 = LHS.
Consider (A2): a + 0 = 0 + a = a. For a = 0.000 … 001, LHS = 0 + 0.000 … 001= 0.000 … 001. RHS = 0 + 0.000 … 001= 0.000 … 01. LHS = RHS.
Consider (A3): a + (-a) = (-a) + a = 0. LHS = 0.000 … 001 + (-0.000 …001) = 0.000 … 001- 0.000 … 001 = 0.
RHS = (- 0.000 … 001) + 0.000 … 001 = 0. LHS = RHS.
Consider (A4): a + b = b + a. LHS = 0.000 … 001 + b = b + 0.00 .. 01.
RHS = b + 0.000 … 001. LHS = RHS.
Consider (A5): a*(b * c) = (a*b) * c, for a = 0.000 … 001.
LHS = 0.000 … 001* (b*c) = 0.000 … 0*(b *c).
RHS = 0.000 … 001*b * c =0.000 … 001*(b * c). LHS = RHS.
Consider (A6): There exists an element 1 ≠ 0 such that for all a,
a * 1 = 1 * a = a. For a = 0.000 … 001, LHS = (0.000 … 001).1 = 0.000 … 001.
RHS = 1. (0.000 … 001) = 0.000 … 001 = a = LHS.
Consider (A7): For every a ≠ 0, there exists a^(-1) such that
a * a^(-1) = a^(-1) * a = 1, for a = 0.000 … 001. LHS = (0.000 … 001). (0.000 … 001)-1 = 0.000 … 001/0.000 … 001 = 1.
RHS = (0.000 … 001)-1. (0.000 … 001) = 0.000 … 001/0.000 … 001=1.
LHS = RHS.
Consider (A8): for all a, b, a *b = b* a.
LHS = (0.000 … 001).b = 0.000 … 00b.
RHS = b.(000 … 001) = 0.000 … 00b. LHS = RHS.
Consider (A9): for all a, b, c: a*(b+ c) = a*b + a*c.
LHS = 0.000 … 001.(b+c) = 0.000 … 00(b+ c).
RHS = 0.000 … 001. b + 0.000 … 001.c = 0.000 … 00(b + c).
LHS = RHS.
Consider (A10): for 0.000 … 001, either 0.000 … 001∈ P, – 0.000 … 001∈ P, or 0.000 … 001 = 0. As 0.000 … 001 is not equal to 0, then by classical logic, either 0.000 … 001∈ P, – 0.000 … 001∈ P. But 0.000 … 001 is defined as being greater than 0.
Consider (A11): If a, b ∈ P, then a + b ∈ P. If 0.000 … 001 and b ∈ P, then we also define 0.000 … 001 + b ∈ P. Since 0.000 … 001 + b is a real number, and 0.00 … 001 + b >0, then 0.000 … 001 + b ∈ P.
Consider (A12): If a, b ∈ P, then a * b ∈ P. For 0.000 … 001. b = 0.000 … 00b is also defined to be ∈ P. As 0.000 … 001 and b are reals, and 0.000 … 001.b >0, then 0.000 … 001.b ∈ P.
Consider (A13). Let the set A = {0.000 … 000, 0.000 … 001, 0.000 … 002, … 0.999 …}. The set A is bounded above, and has a least upper bound of 1.
The issues involving the completeness axiom and the Archimedean Principle are complex and will now be considered separately.
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