.999999... = 1?

Not that long ago, a visitor left a note:

Just this weekend, I tried to explain to my father-in-law the problem with the base 10 numbering system that we accept in this world today.

For example:

1 divided by 3 is .3333 repeating
one third times 3 is 1
.3333 repeating times 3 is .9999

Let me say up front, I sincerely believe that the base 10 (decimal) system is very good. I use it all the time, find it convenient and handy, would heartily recommend it to everyone else. I would not hesitate to cast a "No" vote on any referendum suggesting to change our number system to another.

This said, I must admit that the old good decimal system is not much different from a few others. The basic advantage of base 10 is that this is a positional system. The Roman number 12, XII, contains two symbols of units that both add exactly the same quantity (1) to the total. In the decimal system, 11 means 10 + 1, so that the same digit 1 plays different roles depending on its position in the number representation. The Romans had to invent new symbols for bigger numbers, which sometimes they did and sometimes they did not. David Wells has this entry for the number 2,300,000

The earliest inscription in Europe containing a very large number is on the Columna Rostrata, a monument erected in the Roman Forum to commemorate the victory of 260 BC over the Carthaginians. The symbol for 100,000 was repeated 23 times, a total of 2,300,000.

In the decimal system, we only need 7 symbols to express the same quantity. In base 20, this takes just 3 digits: D9A. In base 20 system, all numbers become shorter. The downside is that the number of digits grows to 20. In base 5, there are only 5 digits but numbers become longer: (2300000)10 = (1042100000)5. In base 12, (2300000)10 = (92B028)12 - just two more digits to remember but somewhat shorter numbers.

The reasons our society has settled on base 10 may be explained by our anatomy rather than by any mathematical reasoning: 10 fingers per individual make a good reference point. Also, memorizing 10 digits does not seem to exert an undue pressure on the mental abilities of an average individual.

The problem addressed by the visitor I mentioned at the beginning of this page pops up in all bases. For example, 1/4 = (0.111...)5. And 4·(0.111...)5 = (0.444...)5. Is this 1? In every base, dividing 1 by the largest digit of the system gives 0.111... As another example, (1/F)16 = (0.111...)16. (F)16·(0.111...)16 = (0.FFF...)16. Are we getting 1 back? Yes we do:

0.999... = 1
(0.444...)5 = 1
(0.FFF...)16 = 1

Of course, I might have used other identities: 1/2 = (0.222...)5 and, as before, 2·(0.222...)5 = (0.444...)5. Also, 1/3 = (0.555...)16, and multiplying back by 3 gives 3·(0.555...)16 = (0.FFF...)16. As we see, it is all the same in other bases. The question whether 0.999... = 1 or not can't be resolved by turning to other systems. Then how?

Those who ask this question should also pause to ponder another question, What is a number? For, if 0.999... is just a symbol, a name (like Seven in the Star Trek series or one of its derivatives) or is it a number? Names are individual: each exists in its own right. Numbers exist all together, so we can apply various operations to one or more numbers to get other numbers. My visitor obtained 0.999... by multiplication (3·0.333...). I can then safely assume he thinks of 0.999... as a number subject to other arithmetic operations. For example, 0.999... = 0.9 + 0.099... = 0.99 + 0.0099..., and so on. If we agree on interpretation of ellipses "...", then I would expect that we should also agree that

0.999... + 0.1 = 0.9 + 0.1 + 0.099... = 1 + 0.099... > 1
0.999... + 0.01 = 0.99 + 0.01 + 0.0099... = 1 + 0.0099... > 1
0.999... + 0.001 = 0.999 + 0.001 + 0.00099... = 1 + 0.00099... > 1

However small fraction we add to 0.999..., we always get a number greater than 1. To me, this implies that the difference between 1 and 0.999... is less than any positive number and definitely it is not negative. Thus, it must be zero: 0.999... = 1. Is it surprising that a number can be written in several ways? No, it should not be. 1.5 = 3/2 = 6/4 = (1.1)2. This also true that 1.5 = 1.4999... The question is actually,

What stands behind the ellipses "..." ?

As before, 0.999... = 0.9 + 0.09 + 0.009 + ... = 9/10 + 9/100 + 9/1000 + ... The expression 0.999... implies an infinite sum, i.e., a sum of an infinite number of terms. In mathematics, such infinite sums are called series. Series are introduced and studied rigorously in Calculus, where a distinction is made: some series are convergent, some are divergent. Every convergent series has a unique number associated with it, its sum. Divergent series are not so lucky. All infinite decimal fractions, like 0.999..., are shown to correspond to convergent series (which converge to their respective sums.) 0.999... converges to 1 which is expressed simply as 0.999... = 1. Similarly, 0.333... is a convergent series whose sum is 1/3: 0.333... = 1/3.

Here's another enlightening argument from Burger. I never met anybody who thought 0.999... greater than 1. So, if it's not equal to 1, it is less than 1. Let's think of the average of 0.999... and 1. As an average of any two numbers, it's greater than 0.999... but is less than 1. Can we determine its decimal expansion? Say, what is its integer part. Since it's less than 1 but greater than 0 < 0.999..., its integer part is bound to be 0. What about its first decimal digit. Since 0.9 < 0.999..., that digit must be 9. And the second one? Since 0.99 < 0.999..., the second digit must also be 9. And so on. It appears like the average of 0.999... and 1 is 0.999... If the latter is denoted as X, (X + 1)/2 = X. X + 1 = 2X. X = 1. The conclusion can't be helped.

The is another way to exploit the notion of everage. Even intuitively, the average of two unequal numbers is greater than the smaller of the two is less than the larger one. If the average is equal to one of the numbers then all three are equal. So what is (0.999... + 1)/2? You can see that it's 0.999... because

2×.999... = 1.8 + 0.18 + 0.018 ... = 1.999... = 0.999... + 1.

but this means that all three numbers are equal!

Remarks

  1. One is led to a curious conclusion in the framework of the non-standard analysis. Even in the non-standard analysis, .999... can't be said to be a sum of an infinite number of terms!

  2. Using calculators appears counterintuitive unless one understands their limitations. The calculator program that comes standard with Windows 95 gives 1/9 = 0.1111111111111 (thirteen 1's.) So that successive division and multiplication by 9 produce a wrong identity: 1/9·9 = 0.9999999999999 where ellipses may be assumed by the user but are not actually present. Calculators have fixed space to store each number. If there is not enough room to store a number, a calculator stores its approximation instead. The button "=" on a calculator's pad means "Compute" or "Calculate" which sometimes results in an actual equality and sometimes in an approximate one. Once 1/9 is stored as 0.1111111111111, a calculator has no memory that the number was obtained as an approximation to 1/9. Multiplication by 9 displays now an exact result, 0.1111111111111 · 9 = 0.9999999999999.

  3. A simple example of a divergent series is the sum of all integers 1 + 2 + 3 + ... As more integers are added the sum grows without bound and no number may sensibly be associated with such a sum.

    Another example is the famous 1 - 1 + 1 - 1 + 1 - ... where units are subtracted and added intermittently. Assume a number S is associated with the series: S = 1 - 1 + 1 - 1 + ... Then we should be able to write, on the one hand, S = (1 - 1) + (1 - 1) + ... = 0, and, on the other, S = 1 - (1 - 1) - (1 - 1) - ... = 1. Which does not make a lot of sense. However, the question is not at all trivial. We may also write S = 1 - (1 - 1 + ...) = 1 - S, from which 2S = 1 and S = 1/2. Many a famous mathematician, L. Euler in particular, believed this to be true.

    Euler relied on the formula for the geometric series:

    (*)

    1/(1 - x) = 1 + x + x2 + x3 + ...,

    which he was willing to consider as the definition of the infinite sum on the right for any x, for which the left side was defined. In particular, for x = -1, we obtain

    1/2 = 1 - 1 + 1 - 1 + 1 - ...

    What if we start with S = 1 - 2 + 4 - 8 + ... ? First we can write

    S = 1 - 2(1 - 2 + 4 - ...),

    so that S = 1 - 2S, from where S = 1/3. On the other hand, S = (1 - 2) + (4 - 8) + ... and surely can't be positive. We obtain the same result from (*) with x = -2.

    Euler was certain that the series on the right is uniquely determined by the analytic expression on the left. However, this does not imply that the numeric series obtained by a substitution of a specific value for x could not be obtained from a different formula. For example, Jean-Charles Callet (1744-1799) found that

    (**)

    (1 - xm) / (1 - xn) = 1 - xm + xn - xn + m + x2n...,

    For a finite geometric series,

    1 + x + x2 + ... + xn-1 = (1 - xn)/(1 - x),

    which yields

    (1 - xm) / (1 - xn) = (1 + x + x2 + ... + xm-1)/(1 + x + x2 + ... + xn-1)

    so that, for x = 1, we get from (**)

    m/n = 1 - 1 + 1 - 1 + 1 - ...

Problem

Two cars start simultaneously towards each other from two cities located at the distance of 100 miles. One car goes at the speed of 20 miles per hour, the other at 30 mph. A fly was sitting on one of the cars before the race started. Startled by a revving engine, the fly alights from one car and flies towards the other at the speed of 40 mph. The moment it reaches the other car, it turns back and proceeds at the same speed and turns back again just a moment before it is squashed by the incoming car. Thus it flies back and forth until it meets its sad fate when the cars bump into each other. How big the distance has been covered by the fly before its untimely end?

Solution 1

The cars race towards each other at the speed of 50 (= 20 + 30) mph. Let, at some moment just before the fly turns towards another car, the distance between the two cars be A. Assume it takes time T for a fly to reach the other car. We thus get the equation

(1)

A - 50T = 40T,

from which T = A/90. In time T, the fly covers the distance 40·A/90 or 4/9 of A. The cars cover the remaining 5/9 of A. The next leg of the fly's journey will be (4/9)A. Taking (4/9)A instead of A we see that the fly will cover the distance (4/9)2A before he turns once more, and so on. With the original A = 100, the fly will cover the distance given by the series:

(4/9)100 + (4/9)2100 + (4/9)3100 + ...

The sum of this geometric series is known to be 100(4/9)/(1 - 4/9) = 80. Answer: 80 miles.

Solution 2

The cars approach each other at the speed of 50 mph. At this speed they will meet in 2 hours (2 = 100/50). Flying at the speed of 40 mph, the fly will cover 80 miles in 2 hours. Answer: 80 miles.

Anecdote

It so happened that a similar problem was once proposed to John von Neumann who immediately gave the right answer. The proposer then remarked that von Neumann probably knew the trick, to which he is said to have replied, "What trick? What's easier than summing the series?"

Sum of the geometric series

With |q| < 1,

(2)

a + aq + aq2 + aq3 + ... = a/(1 - q)

That the series in (2) is convergent is a result from the beginning Calculus. Denote the left-hand side of (2) as S. It is further shown that the common rules of mathematical operations apply to the infinite sums of convergent series. In particular, this is true of the distributive law:

(3)

Sq = aq + aq2 + aq3 + ... = S - a,

from which (2) follows immediately. As an example,

0.999... = 0.9 + 0.09 + 0.009 + ... = (9/10)(1 + (1/10) + (1/10)2 + ...

This is the same as (2) with a = 9/10 and q = 1/10. This gives

0.999... = (9/10)/(1 - 1/10) = 1,

in agreement with a more intuitive derivation above.

As we see, the question whether or not .999... equals 1 is thus reduced to evaluating the same quantity (1, in this case) also as a sum of a geometric series. The situation is quite analogous to the problem we discussed above. Scott Brodie describes a "real world" problem that leads to the sum of a geometric series but also has a natural and simple direct solution. Computing the area of the Sierpinski gasket serves as an additional example. Another convergent series has a nice geometric interpretation.

In his book The Jaguar and the Quark (p 208), The Nobel Price winner Murray Gell-Mann describes a profound situation that leads to the summation of an infinite series:

... [Richard Feynman] found that it could be done. However, the action came out in the form of an infinite series, and summation of that series was virtually impossible in the absence of the geometrical point of view and the invariance principle. The principle, general relativity, yields the answer directly, without any need for the brute force method or the infinite series. ... Perhaps the situation in superstring theory is similar. If theorists understood the invariance principle of superstring theory, they might be able to write down the formula for the action in short order, without resorting to the summation of the infinite series. (While we are waiting for it to be discovered, what should we call that principle? Field marshal relativity? Generalissimo relativity? Certainly it goes far beyond general relativity.)

As a curiosity, you may want to check that, as it was observed by Harold B. Curtis (Mathematics Magazine, Vol. 42, No. 1 (Jan., 1969), pp. 43-52), 0.666... + (0.666...)² = 1.111... .

Remark

  1. It is important to realize that arithmetic manipulations in (3) only worked because the series was convergent. For divergent series such operations make no sense. Consider the series of all the integer powers of 2: 1 + 2 + 4 + 8 + ... This is a divergent series. Assume nonetheless a sum S is ascribed to it:

    S = 1 + 2 + 4 + 8 + ...

    Then 2S = 2 + 4 + 8 + ... = S - 1, from where S = -1 which is surely absurd since all the terms in the series are positive.

    Note:The absurdity of the equality between a "big positive" number and a negative one stems from the arithmetic context. In the theory of p-adic numbers the identity -1 = 1 + 2 + 4 + 8 + ... is perfectly verifiable.

  2. At least two simple geometric series have 1/3 as the limit:

    1/3 = 1/4 + 1/16 + 1/64 + ... (a = 1/4, q = 1/4) and
    1/3 = 1/2 - 1/4 + 1/8 - 1/16 + ... (a = 1/2, q = -1/2)

    Denominators of every term in the series are powers of 2. As we know, for any angle A, we are able to construct (with a straightedge and a compass) A times a sum of any finite number of terms in the series. However, in order to construct A/3 we need an infinite number of steps, which is forbidden by the rules. The problem of trisecting a general angle is, in principle, not solvable.

References

  1. E. B. Burger, M. Starbird, The Heart of Mathematics, Key College Publishing, 2000
  2. M. Gell-Mann, The Jaguar and the Quark, W.H.Freeman and Company, 1994
  3. J. A. Paulos, Beyond Numeracy, Vintage Books, 1992
  4. D. Pedoe, The Gentle Art of Mathematics, Dover, 1973
  5. D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987.

|Contact| |Front page| |Contents| |Algebra| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 62047764

Search by google: