A Proof that 0.999...=1

2023-07-14

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Any real number can be represented as a finite or infinite decimal: for example, π=3.14159... and e=2.71828.... These representations of π and e do not appear everywhere, because other forms are often more useful. But in general, decimal representations of numbers can run into pitfalls. One interesting property, that has several subtle but profound effects in real analysis, is that certain decimals have two different representations that express the same value. The most famous example of this property is the statement:

1=0.999...

Justifications for the statement often consist of an argument that can be condensed into one line:

1=9/9=9*(1/9)=9*0.111...=0.999...

This is often cited as sufficient proof for the truth of the statement. However, I learned yesterday that while this chain of equalities is useful for understanding the nature of the statement, it is not mathematically rigorous. It depends on the assumption that addition and multiplication rules for infinite decimals behave the same as they do for finite decimals--a true assumption, but whose proof demonstrates that 1=0.999..., making the argument circular.

Rigorous proofs often involve limits, sequences, Dedekind cuts, or other advanced topics from analysis. Fortunately, an elementary proof is demonstrated on Wikipedia, and it relies on a simple fact: the number 0 is the only non-negative number less than every positive number.

This proof uses the notation 0.(9)_n to denote n 9s appearing after the decimal point. For example, 0.(9)_5=0.99999. It is important to note that

0.{(9)}_n<1

for any finite n.

Using this notation, 1/10^n=0.(0)_(n-1)1, and following the previous example, 1/10^5=0.00001. Specifically, note that for any finite n:

0.(9)_n+1/10^n=1

To demonstrate that 0.999...=1, it suffices to show that 1 is the smallest number that is not less than 0.(9)_n for any finite n. Suppose a number x has this property; we want to show that if x is also not greater than 1, it must be 1. In symbolic terms, we wish to find the value of x such that for every positive integer n:

0.(9)_n≤x≤1

We examine the differences between each of these values by subtracting them from 1:

1-1≤1-x≤1-0.(9)_n

We can simplify using the previous equation:

0≤1-x≤1/10^n

This inequality implies that the value of 1-x is non-negative but less than the inverse of any positive integer. 0 is the only number to have this property: any possible nonzero value of 1-x could be written as 1/r for some real r, but by choosing n such that r<10^n, the inequality would fail to hold. Thus 1-x must be 0, and x must be 1. Thus 1 is the smallest number no less than 0.(9)_n for all n; extending this indefinitely, 0.999...=1.

The property key to this proof--that 0 is the only nonnegative number less than every positive number--is an example of the Archimedean Property of the real numbers. The Archimedian Property states that given two real numbers a and b, there is an integer m such that ma>b. This property is used in observing that n can be chosen such that 10^n is greater than any real r, forcing 1-x to be 0.

The concepts involved in this proof are very simple, essentially only using the fact that there is no largest number. However, its application in establishing two distinct decimal representations of the same number is a striking one. It never ceases to amaze me how the simplest facts can lead to the most surprising of results.

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