# 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

## 4 Answers

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

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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"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

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thank you sir ,i never knew that recurring refers to the repeating decimals ,i just found it now,my intention was to use a term which means that not to end,or terminate,so i used recurring with the notion that it refers to something non-terminating,but now i saw that it is means repeating decimals which would mean a blunder here – Iyengar Sep 11 '11 at 8:48
sir a small request,could you please add something to your name ,i mean the moniker "George",is used by many people here (about 6 people use your name,and i need to identify by your reputation ),so that to avoid problems arising due to redundancy ,you could modify your name,so that i can identify correctly,its just a suggestion,and its purely left to you,@George – Iyengar Sep 11 '11 at 8:51
22/7 does ends up repeating itself as 22/7 is only an approximation of π – Lie Ryan Sep 11 '11 at 8:54
@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
@George:and did i serve as a cause for misleading your goal???,if so i am extremely sorry sir,all the best and good luck for your future – Iyengar Sep 11 '11 at 11:05

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!

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