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 \(1\).