# 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?

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What do you mean when you say it contains an infinite number? And, BTW, the expansion for $e$ is not periodic. –  gary Sep 11 '11 at 3:16
You mean, why did anybody define such a number in the first place? Because "it's irrational" is the reason why it has an infinite non-repeating decimal expansion. –  Arturo Magidin Sep 11 '11 at 3:16
The "part to the left of the decimal point" is obviously always only finitely long. You mean to the right of the decimal point. And, by the way, do you know that the decimal expansion of $1/3$ is $0.333\ldots$, with a string of infinitely many 3's following the decimal point? –  Srivatsan Sep 11 '11 at 3:42
It isn't clearer to me. What do you mean when you say that $e$ contains an infinite number? If all you mean is that $e$ is not a terminating decimal, well, that's implied by the irrationality of $e$, and you've been given a reference to a proof of that. What do you mean when you say $e$ should have some repeated numbers? Of course it has some repeated digits, just look at all those 8s, but it isn't periodic, it isn't what's called a repeating decimal; again, that's implied by its irrationality. So I don't see what you are asking that hasn't already been answered. –  Gerry Myerson Sep 11 '11 at 13:27

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.

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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.

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actually, there is nothing special about irrational number, what's special is rational number. There are much, much more irrational numbers than there are rational numbers. –  Lie Ryan Sep 11 '11 at 8:52
You ASSERT that $e$ is irrational. But you don't even mention that that can be fairly easily proved. Instead, you talk at length about the fact that repeating decimals are those of rational numbers. But the question was how it is known that the decimal expansion of $e$ does not terminate. –  Michael Hardy Sep 11 '11 at 11:33
@Michael Hardy I'm sorry, I suppose you interpreted the question different from myself, though to be honest I think I favour your interpretation, that he is looking for a reason why the decimal expansion of e never terminates (though, as I said at the top, 'termination' really just means repeating 0's forever, so I suppose falls under the argument that here's no infintely repeating sequence :) ). And I did mention that a previous comment had given a proof of the irrationality of e (which, in any case, the questioner had already asserted in the first place). Sorry for the lack of clarity though –  tom Sep 11 '11 at 12:15
sorry for the question guys, but i edited it now, so it is much clearer what im trying to say. –  jeo Sep 11 '11 at 13:19
Haha, in that case, you're definitely looking for en.wikipedia.org/wiki/Proof_that_e_is_irrational –  tom Sep 12 '11 at 0:12

"Actually in order to know that you must know what are something called Rational and Irrational 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,
@iyengar: 22/7 = 3. 142857 142857 142857 ... is a repeating/recurring decimal (i.e. 22/7 is rational). However, the real value of $\pi$ is $3.1415926535897932384626433832795028841971693993751058209749...$ which never repeats (i.e. $\pi$ is irrational). –  Lie Ryan Sep 11 '11 at 9:26