Why doesn't it called like that? It seems fair, $1$ called $1$ while $0.999$... being the largest number before $1$, and not called $1$ while not look like it is. Let's say it isn't, how would that number look like?

  • 3
    $\begingroup$ This has been asked many times before (e.g. here). $\endgroup$ Aug 26, 2014 at 15:10
  • $\begingroup$ This isn't duplicate, I show new argument. $\endgroup$
    – KugBuBu
    Aug 26, 2014 at 15:11
  • $\begingroup$ The question is duplicate because it is based on the same fallacy : taking the symbolic representation of a number for the number itself. $2$ is the double of $1$ independently of the fact that in roman numerals $2$ is represented as $II$ while $1$ is represented as $I$; otherwise, we have to conclude that $IV$ is not the double of $II$. $\endgroup$ Aug 26, 2014 at 15:15
  • $\begingroup$ This is like asking what is the last item in an infinite series $\endgroup$
    – Some Guy
    Apr 3, 2021 at 18:28

1 Answer 1


$0.999999\ldots$ is exactly $1$. It is just another way of writing the number $1$, and is not less than $1$. You can prove this by subtracting $ 1 - 0.999999 \ldots $ and seeing that the answer is $0$. Alternatively, you can prove that there is no "largest number less than 1."

Proof. Note that for any fraction $\frac{p}{q}$ less than one, there is a slightly bigger fraction which is still less than one; in particular, $\frac{p+q}{2q}$ is a fraction such that $\frac{p}{q} < \frac{p+q}{2q} < 1$. $\Box$

  • $\begingroup$ Alternatively: where $x=.999\dots$, $10x=9.999\dots=9+x$, so $9x=9$ so $x=1$. $\endgroup$
    – Rosie F
    Dec 17, 2017 at 11:58

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