by Dylan Hendrickson Aug 3 2016 updated Aug 4 2016

No, it's not "infinitesimally far" from 1 or anything like that. 0.999... and 1 are literally the same number.

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].

Formal proof

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

%%hidden(Show proof): $~$0.999\dots$~$ 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. %%

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.

%%hidden(Why is this wrong?): 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. %%

%%hidden(Why is this wrong?): 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 $~$0$~$. %%

%%hidden(Why is this wrong?): The sequence gets arbitrarily close to $~$1$~$, so its limit is $~$1$~$. It doesn't matter that all of the terms are less than $~$1$~$. %%

%%hidden(Why is this wrong?): 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 $~$1$~$. %%


Eric Rogstad

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\.

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