Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

The Irrationality of e

Date: 09/04/98 at 00:17:10
From: Shawn Brady
Subject: How is it possible to prove a number is transcendental?

Hi, I was wondering if you could help me. 

I'm reading through a book, and it's talking about transcendental 
numbers. I know that a transcendental number is a number such that it 
can't be put in normal algeabraic forms, such as pi. However, I was 
wondering how this is proven. Thanks in advance for any response.

- Shawn Brady

Date: 09/04/98 at 11:57:49
From: Doctor Wilkinson
Subject: Re: How is it possible to prove a number is transcendental?

In spite of the fact that most numbers are transcendental (a fact 
proved by Cantor), proving that any particular number is transcendental 
is usually quite difficult. There is a large class of such numbers 
known as Liouville numbers, after the French mathematician Liouville 
who defined them and showed they were transcendental. A typical 
Liouville number is:


Here the successive groups of zeros are of length 1!, 2!, 3!, 4!, and 
so on.

As you mentioned, we know that pi is transcendental, but this was 
proved only in 1882 after being studied for more than 2000 years. The 
number e is also transcendental, and so is a^b, if a and b are 
algebraic and b is irrational (like 2 to the square root of 2 power, 
for example). But all these theorems are very hard.

The basic idea, however, in all these proofs is the same: you assume 
that the number in question is algebraic, and you arrive at a 
contradiction. The contradiction always takes the form of showing that 
the assumption implies the existence of an integer between 0 and 1. 
I do not know any of these proofs, although I have studied the proof 
for e, but I can give you some of the idea by showing that e is 

e is defined as 1 + 1/1! + 1/2! + 1/3! + 1/4! + ....

Suppose e is rational, say e = a/b, where a and b are integers. Then 
b! e is certainly an integer. So multiply the equation for e by b!.  
This gives us:

   b! e = b! + b!/1! +... + b!/b! + b!/(b+1)! + b!/(b+2)! + ...

Now all the terms up to b!/b! on the right are integers, and the left 
side is an integer, so the sum of the rest of the terms on the right 
must be an integer. This is:

   1/(b+1) + 1/(b+1)(b+2) + 1/(b+1)(b+2)(b+3) + ...

Now if we replace b+2, b+3, ... by b+1 we make the terms on the right
bigger, so the right side is less than:

   1/(b+1) + 1/(b+1)^2 + 1/(b+1)^3 + ...

which is a geomtric series. Its sum is:

   1/(b+1) (1 - 1/(b+1)) = 1/b.  

Note that b is certainly bigger than 1, since e is not an integer, and 
we have arrived at the promised contradiction: an integer between 0 
and 1.
- Doctor Wilkinson, The Math Forum
  Check out our web site! http://mathforum.org/dr.math/   
Associated Topics:
High School Transcendental Numbers

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.