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I can write $3/3$ as $(1+1+1)/3$ or $1/3+1/3+1/3$.

Now, $1/3$ is a recurring/repeating/non-ending decimal so if we add these three, i.e. $0.3333... + 0.3333... + 0.3333...$ we will get infinitesimally close to $1$ but not $1$.

Is there a way to show that these decimals do end and will eventually become $1$?

Asaf Karagila
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danish
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    0.999.... is 1. – advocateofnone Nov 24 '16 at 08:51
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    To see that, note that $\frac{1}{9} = 0.1111\dots$ so $1 = \frac{9}{9} = 0.9999\dots$. – HTFB Nov 24 '16 at 08:54
  • $0.999\cdots$ only makes sense as a limit, and that limit is $1$, thus $0.999\cdots=1$. – Richard Ambler Nov 24 '16 at 08:55
  • It is an old issue on this site about 0.99999... , I don't mind it but others sometimes do and put it "on hold". Mathematics does does recognize the sum of infinite number of terms, but limits are OK. The limit of 0.99999... is 1. Look for 0.99999.. on the site. – Mikael Jensen Nov 24 '16 at 08:56
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    If $0.333 \ldots$ is $\frac 1 3$, then $3 \times 0.333 \ldots$ is $3 \times \frac 1 3$. – Mauro ALLEGRANZA Nov 24 '16 at 08:58
  • It shouldn't be surprising that infinite decimals don't all represent different numbers: lots of different representations of fractions are equal. – HTFB Nov 24 '16 at 09:09
  • Note: You accept that 0.33333..... is exactly 1/3 and not "get" "infinitely close to 1/3 but not 1/3". But you don't accept the same statement about 0.99999..... and 1. Why not? The subtle answer is that the reals have the "least upper bound property" which means all sequences of increasingly precise decimals. for example 3, 3.1, 3.14, 3.141.... will have limits that are real numbers. Because these limits have real values we can write infinite decimals and have them mean these limits. we can prove limit .9, .99, .999.... is 1. – fleablood Nov 24 '16 at 09:45

4 Answers4

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$0.\bar 3$ is not in the process of becoming $1/3$ or anything else. If it is a meaningful expression then it is already equal to $1/3$ or not equal to $1/3.$

Your Q is not trivial. A logical foundation for $\mathbb R$ was only developed in the 19th century.

DanielWainfleet
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The decimals do not end, but that's not really a problem, since

$$0.999999999\dots = 1$$

There are several proofs of this, which you can look at yourself.

The simplest one is to say that if $x=0.99\dots$, then $10x = 9.99\dots$, and if you subtract the two equations you get $9x=9$ which means $x=1$.

Another way is to see that $$0.99\dots = 0.9 + 0.09 + 0.009 + \cdots =\\=9\cdot (0.1+0.1+0.001 + \cdots) = 9\cdot \sum_{i=1}^\infty 10^{-i} = 9\cdot \frac19 = 1$$

5xum
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In the real numbers, which is the number system we ordinarily work in, and which is the unique number system for which decimal notation (i.e., writing numbers as arbitrary infinite sequences of decimals) works, there is no such thing as "infinitesimally close but not equal". In fact, $0.333...$ is exactly equal to $\frac13$, and $$ 1 = \frac33 = \frac13 + \frac13 + \frac13 = 0.333... + 0.333... + 0.333... = 0.999... = 1. $$

Mees de Vries
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$0,333333.....= 3\sum _{n=1}^{\infty}\frac{1}{10^n}=\frac{1}{3}$

Fred
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