# 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.