Why 0.33... is the only expression of 1/3? I am an undergraduate math student who loves mathematics very much, and I am confused by a math problem.
Given $1/3$, we know that $0.33...$ (there are infinite $3$s) is the decimal expression. But how can we prove that it is the only decimal expression of $1/3$?
Sorry for my mistakes for my wrong math typing and not good English. Thank you very much for your answers!
 A: Clearly $1/3<1,$ so a decimal expression for $1/3$ must be of the form 
$x=0.a_1a_2a_3...a_n...=\sum_{i=1}^\infty a_i10^{-i},$ where $0 \le a_i \le 9.$
As suggested by @Yanko, suppose that
for some $n,$ $a_n < 3$ but $a_i=3$ for $i\lt n$ .  Then 
$3x = \sum_{i=1}^{n-1} 9\times 10^{-i} + 3a_n10^{-n} + 3\sum_{i=n+1}^\infty a_i10^{-i}$
$\quad\quad < 1-10^{-n+1}+7\times10^{-n}+3\times 10^{-n}=1,$ 
so $x<1/3$.  
On the other hand, if for some $n, a_n>3$ but $a_i=3$ for $i\lt n,$
then $3x\ge\sum_{i=1}^{n-1} 9\times 10^{-i} + 12 \times 10^{-n} = 1 - 10^{-n+1} + 1.2 \times 10^{-n+1}>1,$ 
so $x > 1/3.$
A: We're considering numbers in the interval $(0,1)$. Suppose we have two alignments
\begin{align}
a &= 0.a_1a_2\dots a_{n-1}a_n\dots \\
b &= 0.b_1b_2\dots b_{n-1}b_n\dots
\end{align}
that coincide up to the digits at place $n-1$ (with $a_0=b_0=0$) and differ in the digit at place $n$. Without loss of generality, we can assume $a_n<b_n$, so $a_n+1\le b_n$. Let
$$
c=\frac{a_1}{10}+\frac{a_2}{10^2}+\dots+\frac{a_{n-1}}{10^{n-1}}=
\frac{b_1}{10}+\frac{b_2}{10^2}+\dots+\frac{b_{n-1}}{10^{n-1}}
$$
The first alignment represents the number
$$
\sum_{k\ge1}\frac{a_k}{10^k}=c+\frac{a_n}{10^n}+\sum_{k>n}\frac{a_k}{10^k}
$$
With standard arguments we can see that
$$
\sum_{k>n}\frac{a_k}{10^k}\le\sum_{k>n}\frac{9}{10^k}=\frac{9}{10^{n+1}}\sum_{k\ge0}\frac{1}{10^k}=\frac{9}{10^{n+1}}\frac{10}{9}=\frac{1}{10^n}
$$
Therefore
$$
a=c+\frac{a_n}{10^n}+\sum_{k>n}\frac{a_k}{10^k}\le c+\frac{a_n}{10^n}+\frac{1}{10^n}
\le c+\frac{b_n}{10^n}
\color{red}{\le} c+\frac{b_n}{10^n}+\sum_{k>n}\frac{b_k}{10^k}=b
$$
where the red $\color{red}{\le}$ can be replaced by the strict inequality $<$ unless $b_k=0$, for every $k>n$.
Therefore the two differing alignments may represent the same number only if the second one is terminating, in the sense that it is eventually the constant $0$. In order this to actually happen, we need also that $a_n=b_n-1$ and $a_k=9$, for every $k>n$ (prove it).
This is not the case for the decimal expansion of $1/3$, so this representation is unique.
