Really, 0.999999… is equal to 1. Surreally, this is not so!

The value of the recurring decimal 0.999999 … is a popular topic of conversation amongst amateur mathematicians of various levels of knowledge and expertise. Some of the discussions on the web are of little value or interest, but the topic touches on several subtle and deep aspects of number theory.

999999

[Image Wikimedia Commons]

In school we learn that has an infinite recurring decimal expansion, ⅓ = 0.333… . Formally, multiplying both sides by 3, we get the equality 1 = 0.999… . But many people have a sneaking feeling that there is a teeny-weeny gap between 0.999… and 1.0 and that the recurring decimal gets closer and closer but never quite reaches 1.0. This is largely due to the consideration of 0.999… as a process rather than as a number.

Axiom of Completeness

The axiom of completeness of the real number system assumes that every bounded sequence of real numbers has a least upper bound. Intuitively, completeness of the real numbers means that there are no “gaps” in the number line. In the decimal number system, the axiom is equivalent to the statement that any infinite decimal expansion represents some real number. The sequence S = (0.9, 0.99, 0.999, … ) is bounded above by 1.0, so the least upper bound is not greater than 1.0.

We can write the sequence S = (0.9, 0.99, 0.999, … ) as

S = (1 – 10-1, 1 – 10-2, 1 – 10-3, … ).

So, the deviation from 1.0 is 1 – S = (10-1, 10-2, 10-3, … ). Thus, no number less than 1.0 can be an upper bound as, ultimately, there is a term in the sequence that exceeds it. Therefore, since the sequence is increasing, it must converge to 1.0.

Arguments are often advanced about the connection between the real numbers and the points on a physical line. However, we cannot clearly define this relationship; for one reason, we do not know enough about physical space. So, while the arguments may be intuitively appealing, they are not always logically sound.

Simple Proofs

A simple arithmetic demonstration that 0.999… = 1.0 goes as follows:

x = 0.999…

10 x = 9.999… = 9 + x

9 x = 9

x = 1

Another approach is to consider the number as a geometric series

x = 0.999… = (9 / 10) * (1 + 10-1 + 10-2 + 10-3 + … )

= (9 / 10) * [ 1 / ( 1 – 10-1 ) ] = 1

This argument was used by Euler to demonstrate that 0.999 … = 1.

It is not absolutely essential to have 0.999 … = 1. As pointed out be Timothy Gowers (2002, pg. 60) it is largely a matter of convention. However, if we relax this assumption there is a price to pay. After all, if 0.999 … is not equal to 1 then what is it equal to? We are forced either to change some of the arithmetic rules or to introduce new numbers.

Surreal Numbers

The set of real numbers contains no infinitesimals. The Archimedean property ensures that, for any two positive real numbers a and b, there is some number n such that n a > b and n b > a. So, if we assume that x = 1 – 0.999… > 0 there must be a number n such that n x > 1. But

10 x = 10 ( 1 – 0.999… ) = 10 – 9.99… = 10 – 9 – 0.999… = 1 –  0.999… = x

Since there is no n that makes n x > 1 we must have x = 0.

If we consider the surreal number 0.999… (see earlier post), we may assume that there are ω 9s after the decimal. The sum of the geometric series is then

x = 0.999… = (9 /10)*(1 + 10-1 + 10-2 + … )

= (9 /10)*[ (1 – 10ω ) / (1 – 10-1 ) ] = 1 – 10ω.

This is a surreal number that is strictly less than 1 (if only just so).

We finish with an old joke:

Question: How many mathematicians does it take to change a lightbulb?

Answer: 0.9999999999999999999999999999999999999999999999999 .

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