Is 0.999 Repeating Equal to 1? Three Simple Proofs

Short answer: yes. $0.999\ldots$ , with the 9s going on forever, is exactly equal to 1, not just very close. It is the same number written in two different ways, the way $\tfrac12$ and $0.5$ are.

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Three short proofs

1. With one third. Everyone accepts $\tfrac13=0.333\ldots$ . Multiply both sides by 3:

3\times\tfrac13=1,

and

3\times 0.333\ldots=0.999\ldots

2. With algebra. Let $x=0.999\ldots$ . Then $10x=9.999\ldots$ , and subtracting removes the endless tail:

\begin{aligned}10x-x&=9.999\ldots-0.999\ldots\\ 9x&=9\\ x&=1.\end{aligned}

3. With a geometric series (the most rigorous of the three). $0.999\ldots$ means the infinite sum

\begin{aligned}0.999\ldots&=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\dots\\ &=\sum_{n=1}^{\infty}\frac{9}{10^n}=\frac{9/10}{1-1/10}=1,\end{aligned}

using the geometric series formula $\sum_{n\ge 1} ar^{n-1}=\frac{a}{1-r}$ for $|r|<1$ .

The “no gap” argument

If two real numbers are different, there is always another number between them (their average, for example). So try to find a number between $0.999\ldots$ and 1. Any candidate would have to be bigger than $0.9$ , bigger than $0.99$ , bigger than $0.999$ , and so on for every number of 9s, while staying below 1. No such number exists: the gap between $0.\underbrace{99\ldots9}_{n}$ and 1 is $10^{-n}$ , which becomes smaller than any positive amount. A distance smaller than every positive number is 0.

Common objections

“It gets closer and closer to 1 but never reaches it.”

That describes the sequence $0.9,\ 0.99,\ 0.999,\ \ldots$ , and it is true of every term. But $0.999\ldots$ is not any of the terms; it is the limit of the sequence, the value the terms approach. That limit is 1.

“There must be a last 9, and then the difference is $0.000\ldots 1$ .”

There is no last 9. “ $0.000\ldots 1$ ” would need a 1 after infinitely many zeros, which is not a position a decimal digit can occupy. Every digit of a decimal sits at a finite place.

“Then is $0.999\ldots 8$ also equal to $0.999\ldots 9$ ?”

The question uses the same impossible “last digit”. A decimal with infinitely many 9s has no final digit where an 8 could go, so $0.999\ldots 8$ does not name any number.

Try these

Puzzle 1

Use the algebra trick to write $0.272727\ldots$ as a fraction.

Show Answer

The repeating block has 2 digits, so multiply by 100: $100x=27.2727\ldots$ . Subtract: $99x=27$ , so $x=\tfrac{27}{99}=\mathbf{\tfrac{3}{11}}$ .

Puzzle 2

Which other numbers have two decimal expansions?

Show Answer

Every terminating decimal other than 0. For example $0.25=0.24999\ldots$ and $3=2.999\ldots$ . Numbers whose decimals never terminate, like $\tfrac13$ or $\pi$ , have only one.

Puzzle 3

What is $0.0999\ldots$ equal to?

Show Answer

It is $\tfrac{1}{10}\times 0.999\ldots=\tfrac{1}{10}\times 1=\mathbf{0.1}$ .

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