Is the constant $e$ infinitely long? The number $e = 2.718281828...$ is the base of the natural logarithm. Its decimal representation is infinitely long.
Why does this mathematical constant contain an infinite number? What is the reason behind this?
added for clearance: it contains infinitely long numbers, which does not repeat itself, how is this proven? it should at some point has some repeated numbers.
can it be represented by a fraction? ex: 1/2?
 A: I'm not sure what level of understanding you're at, or exactly what you're looking for as an answer, but it might help to notice that the decimal expansion of all numbers are effectively infinite. The number '2' can be written as $2.00000000.....$ continued forever (or $1.999999...$ for that matter :) ). Similarly 10/3 can be written as $3.33333......$, while 1/7 is $0.142857 142857 142857.....$ So there's nothing that special about e in that sense. 
Where e does become special is that the decimal expansion never repeats itself, like all the examples above did. Now why is this (the inevitable rhetorical question...)? It's because e is an irrational number (as you mentioned, and as is proved above). It is irrational because it cannot be represented as a fraction of 2 whole numbers.
Now, what does this have to do with repeating decimal expansions (I can't help myself... :D )? Well it actually tells us that there can be no forever-repeating pattern in the decimal expansion of e. This is because, if there was a repeating pattern, we'd be able to show that the number is rational, which would be ridiculous, as we know it's irrational.
But how would be able to show it's rational? It's very simple really, and I'll give an example. Imagine that we discovered that after, say, 5 digits, the decimals started repeating:
$$e=2.718287182871828....$$
Then we'd be able to say 
$$100000\times e=271,828.7182871828....$$
And therefore,
$$100000\times e-e=99999\times e=271,828.7182871828....-2.718287182871828...$$
Giving us (as the infinite decimal expansions cancel)
$$99999\times e=271826$$
$$e=\frac{271826}{99999}$$
Which is rational. As we could apply the same procedure for any infinite cycles in the decimal expansion of e, we can only conclude that the decimal expansion of e has none of these cycles. Of course, whether there aren't any other patterns is very, very tricky. Take a look at this, for instance http://en.wikipedia.org/wiki/Normal_number
EDIT:
After considering some feedback on the answer, I should probably highlight one or two things. Firstly, e is irrational because we can prove it is irrational, (c.f. http://en.wikipedia.org/wiki/Proof_that_e_is_irrational, as has been linked to). Secondly, if you're wondering why the decimal expansion of e never terminates (which you probably are...) then just consider that any terminating decimal is really a decimal that ends up in a repeating cycle forever (in this case, a cycle of 0's, like $\frac{1}{8}=0.1250000000.....$) and so any decimal that terminates is necessarily rational. Thus e, an irrational number, never terminates. 
A: The constant $e$ was originally defined as a limit,  $e=\displaystyle\lim\limits_{n\to\infty}\left(1+\frac{1}{n}\right)^n$. It is pretty straightforward, using the binomial theorem and basic theorems about limits, to show that $e$ can be computed as a series
$$
e=\sum_{n=0}^\infty\frac{1}{n!}\tag{1}
$$
where $n!=1\cdot2\cdot3\cdot\dots\cdot(n{-}1)\cdot n$, and $0!=1$.
Using $(1)$, we will show that $e$ cannot be written as the ratio of two integers. Suppose that $e=m/n$ where both $m$ and $n$ are integers.  Then, $n!\;e=(n{-}1)!\;ne=(n{-}1)!\;m$ would also be an integer. However,
$$
\begin{align}
n!\;e
&=n!\;\sum_{k=0}^\infty\frac{1}{k!}\\
&=\sum_{k=0}^n\frac{n!}{k!}\;+\;\sum_{k=n+1}^\infty\frac{n!}{k!}\tag{2}
\end{align}
$$
Each term in the left sum of $(2)$ is an integer, but the terms of the right sum are
$$
\begin{align}
\sum_{k=n+1}^\infty\frac{n!}{k!}
&=\frac{1}{n{+}1}+\frac{1}{(n{+}1)(n{+}2)}+\frac{1}{(n{+}1)(n{+}2)(n{+}3)}+\dots\\
&<\frac{1}{n{+}1}+\frac{1}{(n{+}1)^2}+\frac{1}{(n{+}1)^3}+\dots\\
&=\frac{1}{n}
\end{align}
$$
Thus, $n!\;e$ is the sum of an integer (the left sum of $(2)$) and a positive number less than $\frac{1}{n}$ (the right sum of $(2)$), so it can't be an integer.
All real numbers which have a finite, or even a repeating, decimal representation, can be expressed as the ratio of two integers. Therefore, $e$ must have an infinite, non-repeating decimal representation.
A: "Actually in order to know that you must know what are something called Rational and Irrational Numbers, see here
more over if you know something advanced,you can read about Transcendental and Algebraic numbers,see here,
so 'e' is infinitely long as the fraction is not terminated so as in the case of $\pi$
the intuition is that they turn out to be  Decimal numbers that do not end up,
like if you have $\frac{22}{7}$ you can go on dividing but you never end up,
thats why they have infinite precision,
thank you,
cordialmente,
iyengar
A: Of course as a particular Cauchy sequence of rational numbers $\{2,\,\frac{27}{10},\,\frac{271}{100},\,\cdots\}$ the number $e$ is indeed "infinitely long" in the sense requested (i.e is $e$ rational or irrational?) by the OP, as shown by robjohn's answer. Another "reason" for " What is the reason behind this?" is simply that "because most real numbers are like this".
There is another sense wherein the number $e$ and indeed $\pi$ and most of even the transcendental numbers mathematicians talk about are all "finitely long": you can give a finite description of what the $n^{th}$ decimal digit is, or, more generally, given any interval length $\epsilon>0$, you can construct by a finite, always terminating algorithm a rational approximation $e_\epsilon$ to $e$ such that $|e_\epsilon-e|<\epsilon$.
Given the last equation in robjohn's answer one could define the number $e$ as the Cauchy sequence whose $n^{th}$ member is the output of a finite, terminating algorithm. So, as a method of definition, we can equate $e$ with a finite algorithm (or its set of outputs). We can also do this for $\pi,\,\gamma$ and just about any number most mathematicians work with. Indeed in most cases, one can even put hard bounds on how many steps these algorithms need to yield the $n^{th}$ digit for example.
By now you may have noticed something. The set of all such numbers, essentially the constructable numbers must be countably infinite, because the set of all algortihms in a finitely generated language is. To understand this in simple terms, imagine concatenating all of the ASCII codes for the characters in each legal code. This yields a unique integer for each code, and so the set of all codes is a subset of $\mathbb{N}$. In particular, the rationals are countable, a fact wontedly proven by other means.
If you accept that the application $\beth_1=2^\aleph_0 > \aleph_0$ to $\mathbb{N}$ of Cantor's theorem (see Wiki page of this name)  implies that there "exist" entities in $2^{\mathbb{N}}\setminus\mathbb{N}$, it follows that most real numbers are ones which, unlike $e$ and $\pi$ are not computable by a finite algorithm. These are the numbers that are "truly" infinitely long, the uncomputable numbers or unconstructable numbers. 
Chaitin's constant (see Wiki page of this name) is a famous example of one such "infinitely long" number.
It gets even worse. One can give an in-principle description of Chaitin's constant and of many uncomputable numbers. But since the set of descriptions in a finitely generated language is countably infinite, again, if you accept the implication of Cantor's theorem above, "most" of the real numbers are undefinable numbers: as soon as you begin to describe one such entity, by definition you can't be describing that entity!
