Is the constant $e$ infinitely long?

Question

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?

Answer 1

"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

Answer 2

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.

Answer 3

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.