# 0.999...=1

Although some people find it counterintuitive, the decimal expansions $$0.999\dotsc$$ and $$1$$ represent the same real number.

# Informal proofs

These “proofs” can help give insight, but be careful; a similar technique can “prove” that $$1+2+4+8+\dotsc=-1$$. They work in this case because the series corresponding to $$0.999\dotsc$$ is absolutely convergent.

• \begin{align} x &= 0.999\dotsc \newline 10x &= 9.999\dotsc \newline 10x-x &= 9.999\dotsc-0.999\dotsc \newline 9x &= 9 \newline x &= 1 \newline \end{align}

• \begin{align} \frac 1 9 &= 0.111\dotsc \newline 1 &= \frac 9 9 \newline &= 9 \times \frac 1 9 \newline &= 9 \times 0.111\dotsc \newline &= 0.999\dotsc \end{align}

• The real numbers are dense, which means that if $$0.999\dots\neq1$$, there must be some number in between. But there’s no decimal expansion that could represent a number in between $$0.999\dots$$ and $$1$$.

# Formal proof

This is a more formal version of the first informal proof, using the definition of decimal notation.

$$0.999\dots$$ is the decimal expansion where every digit after the decimal point is a $$9$$. By definition, it is the value of the series $$\sum_{k=1}^\infty 9 \cdot 10^{-k}$$. This value is in turn defined as the limit of the sequence $$(\sum_{k=1}^n 9 \cdot 10^{-k})_{n\in\mathbb N}$$. Let $$a_n$$ denote the $$n$$th term of this sequence. I claim the limit is $$1$$. To prove this, we have to show that for any $$\varepsilon>0$$, there is some $$N\in\mathbb N$$ such that for every $$n>N$$, $$|1-a_n|<varepsilon$$.

Let’s prove by induction that $$1-a_n=10^{-n}$$. Since $$a_0$$ is the sum of {$0$ terms, $$a_0=0$$, so $$1-a_0=1=10^0$$. If $$1-a_i=10^{-i}$$, then

\begin{align} 1 - a{i+1} &= 1 - (ai + 9 \cdot 10^{-(i+1)}) \newline &= 1-a_i − 9 \cdot 10^{-(i+1)} \newline &= 10^{-i} − 9 \cdot 10^{-(i+1)} \newline &= 10 \cdot 10^{-(i+1)} − 9 \cdot 10^{-(i+1)} \newline &= 10^{-(i+1)} \end{align}

So $$1-a_n=10^{-n}$$ for all $$n$$. What remains to be shown is that $$10^{-n}$$ eventually gets (and stays) arbitrarily small; this is true by the archimedean property and because $$10^{-n}$$ is monotonically decreasing. <div><div>

# Arguments against $$0.999\dotsc=1$$

These arguments are used to try to refute the claim that $$0.999\dotsc=1$$. They’re flawed, since they claim to prove a false conclusion.

• $$0.999\dotsc$$ and $$1$$ have different digits, so they can’t be the same. In particular, $$0.999\dotsc$$ starts “$0.$,” so it must be less than 1.

Decimal expansions and real numbers are different objects. Decimal expansions are a nice way to represent real numbers, but there’s no reason different decimal expansions have to represent different real numbers.

• If two numbers are the same, their difference must be $$0$$. But $$1-0.999\dotsc=0.000\dotsc001\neq0$$.

Decimal expansions go on infinitely, but no farther.$$0.000\dotsc001$$ doesn’t represent a real number because the $$1$$ is supposed to be after infinitely many $$0$$s, but each digit has to be a finite distance from the decimal point. If you have to pick a real number to for $$0.000\dotsc001$$ to represent, it would be $$0$$.

• $$0.999\dotsc$$ is the limit of the sequence $$0.9, 0.99, 0.999, \dotsc$$. Since each term in this sequence is less than $$1$$, the limit must also be less than $$1$$. (Or “the sequence can never reach $$1$$.”)

The sequence gets arbitrarily close to $$1$$, so its limit is $$1$$. It doesn’t matter that all of the terms are less than $$1$$.

• In the first proof, when you subtract $$0.999\dotsc$$ from $$9.999\dotsc$$, you don’t get $$9$$. There’s an extra digit left over; just as $$9.99-0.999=8.991$$, $$9.999\dotsc-0.999\dotsc=8.999\dotsc991$$.

There are infinitely many $$9$$s in $$0.999\dotsc$$, so when you shift it over a digit there are still the same amount. And the “decimal expansion” $$8.999\dotsc991$$ doesn’t make sense, because it has infinitely many digits and then a $$1$$.

Parents:

• If these are included I think it would be good to also include explanations of why each one is wrong.