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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 $\frac{1}{3}$

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9 | 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 5:42
What is your understanding of the notation, $0.333\dots$? – Gerry Myerson Mar 20 '13 at 5:44
Also, did you know that $0.9999\ldots = 1$? – Kaster Mar 20 '13 at 5:48
This has to be a duplicate. – Thomas Andrews Mar 20 '13 at 6:21
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
up vote 36 down vote accepted

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 $0.3333...=\frac{1}{3}$.

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Cool note: You can extend this to show that 0.9999... = 1 – GraphicsMuncher Mar 20 '13 at 5:50
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 6:38
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 9:23
@MarcvanLeeuwen: any of this without knowledge of limits would be gibberish. This is often where the confusion lies. – robjohn Mar 20 '13 at 9:42
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

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.

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


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

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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 $0.\overline{3}=3/9=1/3$.

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You didn't follow the thread here Does .99999... = 1?.

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

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

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$\dfrac13 \neq 0.33333$ – user17762 Mar 20 '13 at 5:51
@Marvis: Fine, now? :) – Inceptio Mar 20 '13 at 5:53
You may want to use \ldots inside math mode to display $\ldots$ . I have edited your post now. – user17762 Mar 20 '13 at 5:54
Thanks. Sure.:) – Inceptio Mar 20 '13 at 6:02

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.

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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 $N\to\infty;(\frac{1}{10})^{N+1}\to0$ $$=3\cdot \lim_{N\to\infty}\frac{(1-(\frac{1}{10})^{N+1})*10}{9}$$ $$=3\cdot \frac{10}{9}$$ $$=\frac{10}{3}$$ $$\Rightarrow 3+0.33333...=\frac{10}{3}$$ $$\Rightarrow 0.33333...=\frac{10}{3}-3$$ $$\Rightarrow 0.33333...=\frac{10-9}{3}$$ $$\Rightarrow 0.33333...=\frac{1}{3}$$

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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 8:13
Error in your first line: $0.33333\ne\frac{1}{3}$. – Joel Reyes Noche Mar 20 '13 at 9:22

You can say The limit of 0.3333... = 1/3 The difference is infinitesimally small. This leads to problems of infinity and equality. If the difference between two things is infinitesimally small are they equal? "it should offend no sensibilities if we make no distinction".

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I think that you meant $\frac13$ rather than $1$... – Asaf Karagila Mar 20 '13 at 17:25
@AsafKaragila Oh, yes, fixed, thanks. – QuentinUK Mar 20 '13 at 17:27
0 is the only infinitesimal in $\mathbb{R}$. – Charles Mar 20 '13 at 17:29
@Charles: $0$ is not an infinitesimal. Infinitesimal is a positive number which is smaller than $\frac1n$ for every positive integer $n$. $0$ is not positive. – Asaf Karagila Mar 20 '13 at 17:54
@AsafKaragila: I suppose it's a matter of definition, mine explicitly includes 0. Using your definition I would have to say "the difference is infinitesimal or 0, and if it's real then it can't be infinitesimal". – Charles Mar 20 '13 at 17:57

$1/3 = 0.3 + 1/(3*10)= 0.33 + 1/(3*10^2) = 0.333 + 1/(3*10^3) = ... = 0.333... + 1/(3*10^{\infty}$)

They are obviously not equal

There is a strong corelation between the number of threes and the n in 1/(3*10^n). The infinite number of threes does not solve the problem - expressing 1/3 in decimal. It is just impossible.

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"They are obviously not equal" is the cry of those who have just make a mistake. – RghtHndSd Oct 8 '14 at 21:11
expressing 1/3 in decimal is possible. It is just an infinite recurring of 3 from the first decimal place. – Kenny Guy Feb 28 at 0:47

protected by Eric Wofsey Feb 28 at 0:32

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