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e is irrational

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e is irrational

From the Taylor series for exsuperscriptexe^{x} we know the following equation:

e=∑k=0∞1k!.esuperscriptsubscriptk01ke=\sum_{{k=0}}^{{\infty}}\frac{1}{k!}. (1)

Now let us assume that eee is rational. This would mean there are two natural numbers aaa and bbb, such that:

e=ab.eabe=\frac{a}{b}.

This yields:

b!⁢e∈ℕ.beℕb!e\in\mathbb{N}.

Now we can write eee using (1):

b!⁢e=b!⁢∑k=0∞1k!.bebsuperscriptsubscriptk01kb!e=b!\sum_{{k=0}}^{{\infty}}\frac{1}{k!}.

This can also be written:

b!⁢e=∑k=0bb!k!+∑k=b+1∞b!k!.besuperscriptsubscriptk0bbksuperscriptsubscriptkb1bkb!e=\sum_{{k=0}}^{{b}}\frac{b!}{k!}+\sum_{{k=b+1}}^{{\infty}}\frac{b!}{k!}.

The first sum is obviously a natural number, and thus

∑k=b+1∞b!k!superscriptsubscriptkb1bk\sum_{{k=b+1}}^{{\infty}}\frac{b!}{k!}

must also be natural. Now we see:

∑k=b+1∞b!k!=1b+1+1(b+1)⁢(b+2)+…<∑k=1∞(1b+1)k=1b.superscriptsubscriptkb1bk1b11b1b2normal-…superscriptsubscriptk1superscript1b1k1b\sum_{{k=b+1}}^{{\infty}}\frac{b!}{k!}=\frac{1}{b+1}+\frac{1}{(b+1)(b+2)}+...<% \sum_{{k=1}}^{{\infty}}\left(\frac{1}{b+1}\right)^{k}=\frac{1}{b}.

Since 1b≤11b1\frac{1}{b}\leq 1 we conclude:

0<∑k=b+1∞b!k!<1.0superscriptsubscriptkb1bk1.0<\sum_{{k=b+1}}^{{\infty}}\frac{b!}{k!}<1.

We have also seen that this is an integer, but there is no integer between 0 and 1. So there cannot exist two natural numbers aaa and bbb such that e=abeabe=\frac{a}{b}, so eee is irrational.

Related: 
ErIsIrrationalForRinmathbbQsetminus0, NaturalLogBase
Major Section: 
Reference
Type of Math Object: 
Theorem
Parent: 
natural log base

Mathematics Subject Classification

11J82 Measures of irrationality and of transcendence
11J72 Irrationality; linear independence over a field

Comments

Submitted by igor on Mon, 05/20/2002 - 07:09

Maybe it would be a good idea to add this
entry as an example under the "irrational"
node.

Submitted by mathwizard on Tue, 05/21/2002 - 16:23

Yes, that seems like a good idea. Have done this now.

Submitted by matte on Sat, 02/04/2006 - 17:32

How do we know that 1/b<1 in this proof? In other words,
how do we know that b cannot be 1, and e a natural number?
Are there some simple way of deducing this? For example,
it is easy to see that

e< 1/0! + 1/1!=2,

but is it easy to see that e<3?

Submitted by mps on Sat, 02/04/2006 - 18:16

One way to show that e<3 is the following. By definition,

log 3 = \int_1^3 1/t dt.

This integral is larger than its trapezoidal approximation

1/2(1/1 + 1/2) + 1/2(1/2+1/3) = 7/6 > 1,

so log 3 > 1. Hence 3 > e.

Submitted by matte on Sat, 02/04/2006 - 19:38

That is a neat proof, but there is one thing I do not
understand. How do you know that the trapezoidal
approximation is larger than the integral? Although
we have
f(1+1/2) = 2/3 < 3/4 = 1/2(f(1)+f(2))
f(2+1/2) = 2/5 < 5/12 = 1/2(f(2)+f(3))
I don't think it is completely trivial.

One can also show that log 3>1 by approximating:

int_1^3 f(t) > 1/n \sum_{k=1}^2n f(1+k/n)

where f(t) = 1/t. Here the inequality holds as
f is strictly decreasing. (In the sum, the function
is sampled at the right side of each interval.)
Setting n=4 yields numerically 1.01 >1. Although
this gives a sum of 8 terms to compute it is not as
neat as your solution, but still
something one can do by hand.

I added a request for a proof that 2<e<3. It is given
here:

http://planetmath.org/?op=getobj;from=requests;id=520

Submitted by mps on Sat, 02/04/2006 - 20:03

> That is a neat proof, but there is one thing I do not
> understand. How do you know that the trapezoidal
> approximation is larger than the integral?

Sorry, my mental picture was slightly off when I wrote that. I had been thinking of the region under y = 1/x as being convex, which of course isn't true.

Submitted by CWoo on Sat, 02/04/2006 - 20:12

You mean e > 1/0! + 1/1! = 2. As for why e < 3, we can look at e's Taylor expansion again, with the first two terms chopped off:

1/2! + 1/3! + ... + 1/(n+1)! + ...

Compare it term by term with

1/2 + 1/4 + ... + 1/2^n + ...

which is 1.

Use induction to show that 1/(n+1)! < 1/2^n for n>1. So e - 1/0! - 1/1! < 1, or e < 3.

Submitted by rmilson on Sat, 02/04/2006 - 20:28

How about this?

It suffices to show that e^(-1)> 1/3

To do so, we use the series representation

e^(-1) = (1-1) + (1/2-1/6) +(1/24-1/120) + ... + (1/(2k)!-1/(2k+1)!)+...

The terms of this series are all positive.

The first two terms add up to 1/3.

Submitted by dwasserm on Fri, 02/10/2006 - 16:43

The simplest solution is to change "1/b < 1" to "1/b <= 1". We don't need < here because there is < in the previous display. This will fix the proof without requiring a separate proof that e is not a natural number.

Submitted by matte on Mon, 02/13/2006 - 20:53

Various pages on the web state that Euler proved this.
Either in 1757

http://www.bath.ac.uk/~ma2mrm/generalinterest.html

or in 1737

http://numbers.computation.free.fr/Constants/Miscellaneous/irrationality...

Currently the whole production of Euler is available on
the web. Here:

http://www.math.dartmouth.edu/~euler/

Does anyone have a reference to the original proof? It would
be nice to have a link to this in this entry. However,
as the works are written in latin, it is not completely trivial
to find a specific result.

Submitted by matte on Mon, 02/13/2006 - 22:56

I do not quite follow this. What do you mean with 'previous
display'?

The proof is based on the counterassumption that e is
rational. If e is a natural number, the proof gives that

sum_(k=b+1)^\infty b!/k! is a natural number,
and
0< sum_(k=b+1)^\infty b!/k! <=1

Now there is no contradiction; there is a natural number k
satisfying 0<k<=1.

Submitted by stevecheng on Tue, 02/14/2006 - 14:16

The 'previous display' refers to the inequality which shows

\sum_{k=b+1}^\infty b!/k! STRICTLY LESS than the geometric series
which is = 1/b.

So it is not needed to know 1/b < 1
to get strict inequality (tail) < 1.

// Steve

Submitted by dwasserm on Tue, 02/14/2006 - 18:47

By 'display' I mean a collection of math symbols that is printed on a line by itself, centered horizontally. Steve correctly identified the display that I was referring to.

Submitted by matte on Tue, 02/14/2006 - 19:23

Silly me. Now I get it. Many thanks.

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Owner: mathwizard
Added: 2002-03-20 - 20:05
Author(s): mathwizard

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