# 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.

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 - (a*i + 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.

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

\(0.999\dotsc\) 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 \(1\).”)

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 \(9.99-0.999=8.991\), \(9.999\dotsc-0.999\dotsc=8.999\dotsc991\).

Parents:

- Decimal notation
The winning architecture for numerals

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