# Why does $0,\bar{9}$ equal $1$? [duplicate]

I am finding hard to understand why $0,99999..... = 1$

I have the following proof:

Let $x$ be $0,9999...$

then $10x = 9,999...$

So $10x - x = 9,999 - 0,9999$

$9x = 9 \rightarrow x = 1$

From a philosophical respective, it does seem legit to me that if the decimal form of the number is never ending, then, at infinity and beyond it "tries" to reach 1, so it's limit to infinity equals 1.

My objection is that:

Consider the set $S = \left\{ 0, 1, 2, 3, ..., n \right \}$

What is the possiblity that from the set S we obtain the number $2$?

$P = \frac{P(a)}{P(S)}$ = $\frac{1}{n}$

so from there we can see that the possiblity is very low, $0.000...1$ , which again we can consider to be $0$ since :

$\frac{1}{n}$ = $\frac{1}{\infty}$ = $0$

But again, if we accept that $P = 0$, then there is no possibility that we can select the number 2 from the set $S$, which is false.

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## marked as duplicate by response, Tomas, Marc van Leeuwen, egreg, T. Bongers Nov 14 '13 at 18:08

Hey, you have a proof (and a real, valid and sound one, btw): who cares about philosophy now?! –  DonAntonio Nov 14 '13 at 17:23
If your proof doesn't seem entirely satisfying, then the next step is to delve into what exactly a decimal expansion means and what a real number really is. Your probability argument can be clarified slightly to show that there does not exist a uniform probability distribution on the natural numbers, which means any argument which starts by assuming you have such a distribution is immediately dead in the water. –  Aaron Nov 14 '13 at 17:27
The actual proof that $0.(9)=1$ will depend on your exact definition of numbers and of decimal fractions. Some people define real numbers using decimal notation. For them $0.(9)=1$ by definition. –  Dan Shved Nov 14 '13 at 17:28
@Nick Yes, although a lot of people try to argue that it is $0.\overline 0 1,$ which is slightly nonsensical, although understanding why it is nonsensical requires a certain understanding of what a real number is that not everybody has. –  Aaron Nov 14 '13 at 17:36
I also think that any person who asks himself this question should acquaint himself with at least one definition of real numbers. If one relies only on the intuitive understanding, one runs the risk of ending up talking about meaningless things, like the "number" $0.(0)1$. –  Dan Shved Nov 14 '13 at 17:45

Even though your proof is perfectly valid, here is another one; maybe this convinces you more.

The sum of the terms of an infinite geometric series $a_n$ with first term $a$ and ratio $|r|<1$ is $$S_{\infty}=\frac{a}{1-r}$$

Hence, $$0.999\cdots=\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\cdots=\frac{\frac{9}{10}}{1-\frac{1}{10}}=1$$

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That looks interesting, thanks! –  Nickolas Nov 14 '13 at 17:39
The proof in the question is not valid, because it implicitly assumes that $0.\overline{9}$ represents a number and that multiplication by $10$ produces $9.\overline{9}$ (which is equivalent to assuming that $0.\overline{9}=1$): $10x=9+x$ if and only if $x=1$. –  egreg Nov 14 '13 at 18:03

It's important to note that $0.\overline{9}$ doesn't have a limit: it's just a single number.

However, we like to define decimals so that they can be used to represent real numbers. And, in particular, we would like $0.\overline{9}$ to be equal to the limit of the sequence

$$0.9, 0.99, 0.999, 0.9999, \cdots$$

It is this sequence that "tries" to reach 1, and it is this sequence that we talk about having a limit equal to 1.

But $0.\overline{9}$ is not a sequence: it's just a number.

You make a related mistake in your objection: you confuse the single set

$$\{ 0, 1, 2, \cdots, n \}$$

for various $n$ with the "limiting" set of all integers

$$\{ 0, 1, 2, \cdots \}$$

It is worth noting, however, that there are serious technical issues with the notion of a uniformly randomly chosen integer.

(Also "probability zero" doesn't mean "impossible". There are significant dangers in trying to use one's intuition about "selecting" randomly from finitely many objects when studying more general probability spaces)

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So, practical reasons is the short answer then I guess..? –  Nickolas Nov 14 '13 at 17:42

The problem people tend to have with this concept is that we sometimes assume our number system is the exact representation of the numbers themselves. We are, in fact, dealing with a system similar to $\mathbb Z_{10}$, where we roll over to 0 again in a particular position after reaching 9. There isn't a representation for 10 in this system, but instead a 1 in the 10's place followed by a 0 in the 1's place. That said, representing thirds in this system is difficult, so if we change the base from 10 to 3, we can get an exact representation of $0.\overline{9}$ like so:

Given $1_{3} = 1_{10}$ and $0.1_{3} = 0.\overline{3}_{10}$,

\begin{aligned} & 0.1_{3}\\ + & 0.1_{3}\\ \hline & 0.2_{3}\\ and \\ & 0.\overline{3}_{10} \\ + & 0.\overline{3}_{10} \\ \hline & 0.\overline{6}_{10} \\ thus \\ & 0.2_{3}\\ + & 0.1_{3}\\ \hline & 1.0_{3}\\ and \\ & 0.\overline{6}_{10} \\ + & 0.\overline{3}_{10} \\ \hline & 0.\overline{9}_{10} = 1_{3} = 1_{10}\\ \end{aligned}

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Writing "0.3" would normally be forbidden in base $3$... –  Dan Shved Nov 14 '13 at 17:50
Yes, it is, I wrote that to help explain the concept. –  Jonathan Landrum Nov 14 '13 at 17:51
I suppose I should expand this out to a system of additions rather than these aligned equalities. –  Jonathan Landrum Nov 14 '13 at 17:52