# 0.999…, 0.000…1 and other numbers with infinitely many digits

The `number’ 0.000…1 sometimes comes up in the discussion of the infamous question of whether 0.999… = 1 (in case you didn’t know, 0.999… is equal to 11). For every natural number ${n}$, ${1 - 0.99\dotso9}$ (where there are ${n}$ 9s in the subtracted number) is 0.00…1 (in which there are ${n}$ 0s, including the one before the decimal point), so naturally, people generalise the pattern to the case where there are infinitely many ${n}$s (although you can’t always generalise in this way), and conclude that ${1 - 0.999\dotso}$ (where there are infinitely many 9s in the second number) is 0.000…1 (in which there are infinitely many 0s), which seems to be a positive number, not 0 as it should be if ${0.999\dotso = 1}$.

Even though it’s wrong, this argument is quite an intelligent one. Its problem is that it relies on a lack of understanding of exactly how decimal notation works, which is understandable, given that the subtleties of decimal notation are rarely given a proper explanation in maths education. It’s easy to understand how decimal notation works when there are only finitely many digits, and I think most people understand it on some level (though they may not be able to articulate it in the way I’m about to articulate it). Given a string of digits with a decimal point in between one pair of consecutive digits, one assigns an index to each digit ${d}$ by looking at its position relative to the decimal point. If ${d}$ is in front of the decimal point, the index is the number of digits between ${d}$ and the decimal point, with ${d}$ not being counted among these digits, and if ${d}$ is behind the decimal point, the index is the negation of the number of digits between ${d}$ and the decimal point, with ${d}$ being counted among these digits. In other words, the digit in front of the decimal point has index 0, the digit behind it has index ${-1}$, and the other indices continue the sequence. For example, in 17.25, the digit 1 has index 1, the digit 7 has index 0, the digit 2 has index ${-1}$ and the digit 5 has index ${-2}$. Then, one calculates the place value associated with ${d}$ as ${d \times 10^n}$, where ${n}$ is the index of ${d}$. The number denoted by the string of digits is simply the sum of all the place values. For example, 17.25 denotes the number ${1 \times 10^1 + 7 \times 10^0 + 2 \times 10^{-1} + 5 \times 10^{-2}}$. The number denoted is always non-negative, but we can denote its negation simply by putting a minus sign in front.

Decimal notation with finitely many digits is not very powerful. Every integer can be denoted in this way, but not every rational number (rational numbers are numbers that can be expressed as fractions whose numerators and denominators are both integers, in case you didn’t know). For example, ${\frac 1 3}$ cannot be denoted in this way. In fact, the only rational numbers which can be denoted in this way are rational numbers whose denominator is not divisible by any prime number other than 2 or 5 (the prime factors of 10). So, in order to denote every rational number, you need infinitely many digits.

It’s in this case that the subtleties arise. Going from finite to infinite always brings a lot of complications with it. First of all, in order to be able to assign indices to each digit in the same way, we’re going to have to require that, even though there are infinitely many digits, there are finitely many digits between each digit and the decimal point. After all, if there are infinitely many digits between some digit ${d}$ and the decimal point, then the number of digits between ${d}$ and the decimal point does not exist, since infinity isn’t a number. Therefore, we can see a problem with the argument above right away. There are infinitely many digits between the digit 1 in 0.000…1 and the decimal point, so it is impossible to assign an index to this digit, and for this reason, 0.000…1 is not an example of well-formed decimal notation. There are ways to define alternative forms of decimal notation in which 0.000…1 is well-formed (and usually denotes some kind of infinitesimal number, i.e. one which is positive but smaller than every positive rational number), but I’m not going to go into them here. Certainly, none of these ways can be considered standard: if you just start talking about 0.000…1 to a mathematician without defining it first, they will ask you to clarify.

If there are finitely many digits between each digit and the decimal point, we can associate each of the infinitely many digits with a place value in the same way as before. So then the number denoted by the string of digits is just the sum of all these place values, right? Well, not quite. Remember, we have infinitely many digits, hence infinitely many place values, so this will be a sum of infinitely many terms! The ordinary definition of a sum is that it is the value obtained by letting the value be 0 at first, and then adding each term in the sum to the value, one at a time, until every term has been added. But if there are infinitely many terms, then no matter how long this process continues, terms will remain which still have not been added. So it doesn’t make sense to speak of sums of infinitely many terms, if we use this definition.

But just because this definition doesn’t work, that doesn’t mean we can’t define sums of infinitely many terms in some other way. For example, it certainly seems sensible to say that the infinite sum ${0 + 0 + 0 + \dotsb}$ is 0, because the sum of any finite number of 0s is 0, so we can generalise this to the case where there are infinitely many 0s. In general, if only finitely many terms in an infinite sum are not equal to 0, we can just disregard the 0s, since they should have no effect on the value of the sum, and say that the infinite sum is equal to the finite sum consisting of its terms which are not equal to 0. For example, ${1 + 2 + 3 + 0 + 0 + 0 + \dotsb = 1 + 2 + 3 = 6}$.

These aren’t very interesting examples of infinite sums, since they are basically finite sums in disguise. It is possible to assign values to more interesting infinite sums. To see how this is done, first let me note that one way of stating the reasoning why an infinite sum of the form ${x_1 + x_2 + \dotsb + x_n + 0 + 0 + 0 + \dotsb}$ is equal to the finite sum ${x_1 + x_2 + \dotsb + x_n}$ is this: if we start with the value of 0, and add each term of the infinite sum in order, then, after we add ${x_n}$, the remaining additions do not change the value at all (since we are just adding 0). Therefore, the value stays at ${x_1 + x_2 + \dotsb + x_n}$, even if infinitely many additional additions are made, because these additional additions have no effect.

Now, if there are infinitely many terms which are not 0 it is impossible for the value of the incomplete sum to stay equal to a particular number after a finite number of terms are added. But it may be possible for the value to stay within a certain distance of a particular number after a finite number of terms are added, and if it does, then we can conclude that the distance of the sum of all the terms from that particular number must be less than or equal to this distance (by generalising to the infinite case). Remember, the distance between two numbers ${x}$ and ${y}$ is either ${x - y}$ or ${y - x}$, whichever one is positive (or, if ${x - y}$ and ${y - x}$ are both 0, so that neither is positive, the distance is 0).

An example will help to make this clear. Consider the number denoted by 0.999\textellipsis: ${\frac 9 {10} + \frac 9 {100} + \frac 9 {1000} + \dotsb}$. For every positive integer ${n}$, the sum of the first ${n}$ terms in this sum is ${\frac 9 {10} + \frac 9 {100} + \dotsb + \frac 9 {10^n}}$. What is the distance of this finite sum from 1? The sum is clearly less than 1, because it can be written (using the rule for addition of fractions) as ${\frac {10^n - 1} {10^n}}$, i.e. ${1 - \frac 1 {10^n}}$. So the distance is the difference ${1 - (1 - \frac 1 {10^n})}$, which is just ${\frac 1 {10^n}}$. Now, for every positive integer ${n'}$ such that ${n < n'}$, it can be seen in the same way that the distance of the sum of the first ${n'}$ terms from 1 is ${\frac 1 {10^{n'}}}$, which is less than ${\frac 1 {10^n}}$ since ${n < n'}$. That is, no matter how many terms are added to the incomplete sum after the ${n}$th term is added, the distance stays less than ${\frac 1 {10^n}}$. Therefore, we can conclude that the distance of 0.999… from 1 is less than ${\frac 1 {10^n}}$ for every positive integer ${n}$. In fact, it is less than every positive rational number, since every positive rational number can be expressed as a fraction ${\frac a b}$, where ${a}$ and ${b}$ are positive integers, and ${\frac 1 {10^b} < \frac a b}$ (since ${b < 10^b}$).

Another example is ${\frac 1 2 + \frac 1 4 + \frac 1 8 + \dotsb}$, the number denoted by 0.111… in binary. By similar reasoning as above, it can be shown that its distance from 1 is less than every positive rational number. Or, you can just look at this neat diagram.

In cases like this, where we can show that for some rational number ${L}$, the distance of the infinite sum from ${L}$ is less than ${\varepsilon}$ for every positive rational number ${\varepsilon}$, it seems natural to assign the infinite sum the value of ${L}$, because, if the value is a rational number, its distance from ${L}$ must also a rational number, and this distance cannot be positive, otherwise it would have to be less than itself. In other words, ${L}$ is the only rational number which the sum could be, and therefore, ${L}$ is a very natural value to assign to the sum. In the case of 0.999…, ${L}$ is 1, and that’s why we assign 0.999… the value 1. Of course, we could assign values to infinite sums in a different way, but this is the standard way we do it in mathematics. I’m not familiar with any other way of assigning values to infinite sums that is more natural and convenient than this one.

There are a few more remarks I’d like to say about decimal notation. First of all, there are strings of digits such that the sum of the place values cannot be assigned a rational number as a value in this way, since there is provably no rational number ${L}$ such that for every positive rational number ${\varepsilon}$, the distance of the incomplete sum from ${L}$ stays less than ${\varepsilon}$ once a certain finite number of terms have been added. In fact, the only strings which can be assigned a rational number as a value in this way are those such that the digit at every index greater than some integer ${n}$ is 0 and the other digits consist of a finite sequence repeated indefinitely (for example, 0.333… ($\frac 1 3$) or 0.142857142857142857… ($\frac 1 7$)). However, as long as the digit at every index greater than some integer ${n}$ is 0, a similar but slightly weaker condition is satisfied: for every positive rational number ${\varepsilon}$, once a certain finite number of terms have been added, if one picks any pair of incomplete sums which can be formed by adding on more terms, the distance between these two sums is less than ${\varepsilon}$. It’s as if the incomplete sums are approaching a number, but that number is not rational. We can therefore define a whole set of new numbers which are called the irrational numbers, with each irrational number being identified with one of these infinite sums which cannot be defined as a rational number. The numbers in the class that is obtained by adding these irrational numbers to the rational numbers are called real numbers. Although I won’t go into the details, it is possible to define how to add and multiply real numbers, by referring to the sequences that define them, in such a way that all the expected properties of addition and multiplication hold, so real numbers do deserve to be called numbers.

#### Footnotes

1. ^ But some people just won’t accept it…