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I have been taught that $\frac{1}{3}$ is 0.333.... However, I believe that this is not true, as 1/3 cannot actually be represented in base ten; even if you had infinite threes, as 0.333... is supposed to represent, it would not be exactly equal to 1/3, as 10 is not divisible by 3.

0.333... = 3/10 + 3/100 + 3/1000...

This occured to me while I discussion on one of Zeno's Paradoxes. We were talking about one potential solution to the race between Achilles and the Tortoise, one of Zeno's Paradoxes. The solution stated that it would take Achilles $11\frac{1}{3}$ seconds to pass the tortoise, as 0.111... = 1/9. However, this isn't that case, as, no matter how many ones you add, 0.111... will never equal precisely $\frac{1}{9}$.

Could you tell me if this is valid, and if not, why not? Thanks!

I'm not arguing that $0.333...$ isn't the closest that we can get in base 10; rather, I am arguing that, in base 10, we cannot accurately represent $\frac{1}{3}$

Mikhail Katz
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elijaheac
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    http://en.wikipedia.org/wiki/Limit_(mathematics) | http://en.wikipedia.org/wiki/Series_(mathematics). Your reasoning is not valid; you do not know what "0.333..." means in the first place. It is a perfectly sensible question that has been asked time and again by those not in the know about limits and analysis, though. – anon Mar 20 '13 at 05:42
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    What is your understanding of the notation, $0.333\dots$? – Gerry Myerson Mar 20 '13 at 05:44
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    The fact that 10 is not divisible by 3 means that 1/3 will have 2 or more digits after dot. – N. S. Mar 20 '13 at 05:44
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    $0.11111\ldots$ is exactly $1/9$ – Kaster Mar 20 '13 at 05:45
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    Also, did you know that $0.9999\ldots = 1$? – Kaster Mar 20 '13 at 05:48
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    This has to be a duplicate. – Thomas Andrews Mar 20 '13 at 06:21
  • Zeno is wrong (as are you), as classical mechanics (and real numbers) exist in a continuum. The erroneous arguments are based on a system with infinitesimals. – OrangeDog Mar 20 '13 at 09:45
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    Can you name a number that's between $\frac{1}{3}$ and $0.333 \ldots$ without being equal to either of them? – yatima2975 Mar 20 '13 at 11:40
  • @yatima2975 good point there. – Soham Chowdhury Mar 20 '13 at 12:24
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    @OrangeDog Why do you say Zeno was wrong? That was not his argument, as he only proposed the paradox. – elijaheac Mar 20 '13 at 17:08
  • @OrangeDog: Careful, the arguments are erroneous but Zeno himself was not wrong. His intention was to show that geometers of the time were being sloppy in their treatment of infinities by showing how their methods resulted in absurdities. – Jim Mar 20 '13 at 17:40
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    This question actually demonstrates extraordinary thought. The reason many people are offended by "$0.999...=1$" is that they confuse (as they were brought up to confuse) a number with the way it is represented, i.e., as a decimal. They therefore assume $0.999...\neq 1$ because, as infinite strings of digits, indeed they are different. Almost nobody doubts $1/3=0.333...$ because what else could $1/3$ be? – Sean Eberhard Mar 20 '13 at 17:45
  • @yatima2975 No, because $0.333...$ is a close as we can get in base 10. In base 3, $\frac{1}{3}$ would be $0.1$, and 3 * 0.1 = 1 (base 3). To my understanding, it is not the same with base 10. – elijaheac Mar 20 '13 at 18:25
  • In base 3, that is trivial, unlike in base 10. – elijaheac Mar 20 '13 at 18:37
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    @XanderHenderson This one is also related. – Arnaud D. Jun 04 '19 at 13:19

6 Answers6

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Here is a simple reasoning that $1/3=0.3333...$.

Lets denote $0.333333......$ by $x$. Then

$$x=0.33333.....$$ $$10x=3.33333...$$

Subtracting we get $9x=3$. Thus $x=\frac{3}{9}=\frac{1}{3}$.

Since $x$ was chosen as $0.3333....$ it means that $0.3333...=\frac{1}{3}$.

N. S.
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    Cool note: You can extend this to show that 0.9999... = 1 – GraphicsMuncher Mar 20 '13 at 05:50
  • @GraphicsMuncher: That's actually plain wrong. Follow the question here : http://math.stackexchange.com/questions/11/does-99999-1/60#60 . – Inceptio Mar 20 '13 at 05:51
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    @Inceptio Do share, I've got the proof sitting in front of me that I'm correct from YOUR link, as well as a number of others. Set x = 0.9999 and repeat the process outlined above. – GraphicsMuncher Mar 20 '13 at 05:55
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    @GraphicsMuncher Actually I adapted the proof of $0.999999...=1$ for this case ;) – N. S. Mar 20 '13 at 05:56
  • @Inceptio All the proofs there explain why what GM said is RIGHT ;) – N. S. Mar 20 '13 at 05:57
  • @GraphicsMuncher: $0.9999...=1$, this isn't the way(Multiply by 3) you prove it though.. – Inceptio Mar 20 '13 at 05:58
  • @Inceptio: they are multiplying by $10$, not $3$. – robjohn Mar 20 '13 at 06:14
  • @Inceptio: In either this proof or the $.999\ldots = 1$ proof they multiply by $10$, not by $3$. – Jim Mar 20 '13 at 06:14
  • @robjohn: $0.33333..=\frac{1}{3}$, but you just can't multiply $3$ on both sides to get $.999..=1$ . – Inceptio Mar 20 '13 at 06:17
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    I don't think that's what GraphicsMuncher meant by "extend". I think they meant that the same argument can be used to show that for any single digit $a$ the decimal $.aaa\ldots$ is $\frac{a}{9}$. – Jim Mar 20 '13 at 06:38
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    All this assumes that we can safely perform certain operations on infinite decimals (not all, since $1-0.9999\ldots$ would probably get us confused) without knowing about limits, and therefore not knowing how infinite decimals are defined. This is an illusion. – Marc van Leeuwen Mar 20 '13 at 09:23
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    @MarcvanLeeuwen: any of this without knowledge of limits would be gibberish. This is often where the confusion lies. – robjohn Mar 20 '13 at 09:42
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    The trick in this is that 10 x 0.3... is not 3.3... it is 3.3... with a 0 on the end ;) – JamesRyan Mar 20 '13 at 17:06
  • PLEASE nobody else upvote this answer! It's perfect right now. – Sean Eberhard Mar 21 '13 at 11:45
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You can find the sum of $\frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots$ using the formula of sum of infinite geometric progression.

$$a_1 = \frac{3}{10}$$

$$r=\frac{1}{10}$$

$$\sum =\frac{a_1}{1-r}=\frac{3}{10}\times\frac{10}{9} =\frac{1}{3}$$

Michael Hardy
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Ashot
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The problematic part of the question is "no matter how many ones you add, 0.111... will never equal precisely 1/9."

In this (imprecise) context $0.111\ldots$ is an infinite sequence of ones; the sequence of ones does not terminate, so there is no place at which to add another one; each one is already followed by another one. Thus, $10\times0.111\ldots=1.111\ldots$ is precise. Therefore, $9\times0.111\ldots=1.000\ldots=1$ is precise, and $0.111\ldots=1/9$.

I say "imprecise" because we also say $\pi=3.14159\dots$ where ... there means an unspecified sequence of digits following. A more precise way of writing what, in the context of this question, we mean by $0.111\dots$ is $0.\overline{1}$ where the group of digits under the bar is to be repeated without end.

In this question, $0.333\ldots=0.\overline{3}$, and just as above, $10\times0.\overline{3}=3.\overline{3}$, and therefore, $9\times0.\overline{3}=3.\overline{0}=3$, which means $0.\overline{3}=3/9=1/3$.

robjohn
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You didn't follow the thread here Is it true that $0.999999999\ldots = 1$?.

Well, $\frac{1}{3}=0.33333\ldots$

You can use $1$.Geometric Progression. Or $2$. The one N.S suggested.

Community
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Inceptio
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This question is similar to show that $0.999\ldots=1$.

I give here a proof and tou can see Does .99999... = 1? for another proofs. We have $$0.999\ldots\leq 1\leq 0.999\ldots+\frac{1}{10^n},\, \forall n\in\mathbb{N},$$ so by passing at limit ($n\to\infty$) we find $$0.999\ldots\leq1\leq 0.999\ldots$$ which allows us to conclude.

Community
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The following uses the limiting concept to prove that $0.33333=\frac{1}{3}$

$$3+0.33333...=\sum_{i=0}^N\frac{3}{10^n}\quad\text{where N}\to\infty$$ $$=3\sum_{n=0}^N\frac{1}{10^n}$$ $$=3\sum_{n=0}^N(\frac{1}{10})^n$$

Known $$1+a^2+a^3....a^n=\frac{a^{n+1}-1}{a-1}$$ So $$=3\cdot \frac{(\frac{1}{10})^{N+1}-1}{\frac{1}{10}-1}$$ $$=3\cdot \frac{(\frac{1}{10})^{N+1}-1}{\frac{1-10}{10}}$$ $$=3\cdot \frac{((\frac{1}{10})^{N+1}-1)*10}{1-10}$$ $$=3\cdot \frac{(1-(\frac{1}{10})^{N+1})*10}{9}$$

Now apply the limiting concept, if $N\to\infty;(\frac{1}{10})^{N+1}\to0$ $$=3\cdot \lim_{N\to\infty}\frac{(1-(\frac{1}{10})^{N+1})*10}{9}$$ $$=3\cdot \frac{10}{9}$$ $$=\frac{10}{3}$$ $$\Rightarrow 3+0.33333...=\frac{10}{3}$$ $$\Rightarrow 0.33333...=\frac{10}{3}-3$$ $$\Rightarrow 0.33333...=\frac{10-9}{3}$$ $$\Rightarrow 0.33333...=\frac{1}{3}$$

Abhijit
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    There is a mistake here, it is 3+0.333.. and not 1+0.333... You managed to correct this with another mistake near the end (tip: 1 is not equal to 9/3). – TonioElGringo Mar 20 '13 at 08:13
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    Error in your first line: $0.33333\ne\frac{1}{3}$. – JRN Mar 20 '13 at 09:22