Why 0.999... Equals 1

Definition The equality $0.999\ldots = 1$ says that the repeating decimal with the digit $9$ in every place after the decimal point names exactly the real number $1$. A decimal with infinitely many digits has no last digit. It is defined as the limit of its finite truncations, which here are

Every truncation is less than $1$, and each added nine makes the gap $10^{-n}$ ten times smaller. The limit is $1$ for two reasons. No truncation is above $1$, so the limit is not above $1$. And no number $1 - d$ with $d > 0$ can be the limit, because once $10^{-m} < d$, the truncation $a_m$ and all the later ones stay above $1 - d$ by at least the fixed positive amount $d - 10^{-m}$, so they never come close to $1 - d$. So $0.999\ldots = 1$ exactly, in the real numbers. The surprise comes from treating $0.999\ldots$ as a very long finite decimal, from picturing a last nine, or from imagining an infinitely small positive gap. None of these exists among the real numbers, and the equality has nothing to do with rounding.

Definition A decimal expansion $0.d_1 d_2 d_3 \ldots$, in which each digit $d_k$ is one of $0, 1, \ldots, 9$, names the real number

which is the limit of its finite truncations. A whole number part in front, as in $9.999\ldots$, is added to this sum. A repeating decimal is one whose digits, from some place on, consist of one fixed block repeated forever. It is written with a bar over the block, as in $0.\overline{3}$ or $0.1\overline{6}$, or with dots. The repeating decimals, which include the terminating ones that end in repeating zeros, name exactly the rational numbers, the fractions $p/q$ of whole numbers with $q \ne 0$. A real number has two decimal expansions exactly when it has a terminating one with a last nonzero digit, and then the other one ends in repeating nines, as in $0.25 = 0.24999\ldots$ and $1 = 0.999\ldots$. Every other real number has only one decimal expansion.

Definition A sequence $a_1, a_2, a_3, \ldots$ of real numbers has the limit $L$, written $\lim_{n \to \infty} a_n = L$, if for every positive number $\varepsilon$ there is a whole number $N$ such that $\lvert a_n - L \rvert < \varepsilon$ for every $n \ge N$. A sequence has at most one limit. If every term is at most a number $M$ and the sequence has a limit, then the limit is also at most $M$. If the sequence is increasing and has a limit, then the limit is at least as large as every term.

Definition The real numbers are the numbers that measure positions on a number line. They include the whole numbers, the fractions and irrational numbers such as $\sqrt{2}$ and $\pi$, with the usual arithmetic and order. Two of their properties matter here. First, every increasing sequence of real numbers whose terms all stay below some fixed number has a limit, and that limit is a real number. This is the completeness of the real numbers, and it is why every infinite decimal names a real number. Second, there is no infinitely small positive real number. For every real $d > 0$ there is a whole number $n$ with $10^{-n} < d$. It follows that a real number $d \ge 0$ that is at most $10^{-n}$ for every $n$ must be $0$, and that two different real numbers $p < q$ always have other real numbers between them, such as $(p + q)/2$. Some larger number systems, such as the hyperreal numbers, do contain infinitely small positive numbers, but they are not the real numbers, and an expression like $0.999\ldots$ would need its own definition there. On this page, as in school mathematics and calculus, decimals name real numbers.

Definition The truncation after $n$ places of a decimal keeps its whole number part and its first $n$ digits after the decimal point and drops all the later digits. For a decimal $0.d_1 d_2 d_3 \ldots$ it is the finite decimal

Each truncation is a rational number that can be written as a fraction with denominator $10^n$. For a decimal without a minus sign that names the number $x$, the truncations satisfy $t_n \le x \le t_n + 10^{-n}$, because the dropped digits add at most

and the upper bound is reached only when every dropped digit is $9$. For every $n$ there is a next truncation $t_{n+1}$, so there is no last truncation, just as there is no last digit.