Proof That E Is Irrational |
|
The number e also known as Euler's Number, is irrational.
This page could be replaced by a reference to a good proof
elsewhere.
This will be a proof by contradiction.
The number e can be defined as:
- $e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+...$
Let's suppose (to get our proof by contradiction started) that e is a rational number, so we write $e=a/b$
Now we multiply both sides by b!
- $LHS:\quad{\quad}b!e=b!(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+...)$
- $RHS:\quad{\quad}a(b-1)!$
Many of the terms on the LHS simplify to integers, up to term $\frac{1}{(b+1)!}$ from which point
we start to get left over denominators. We can subtract the first b terms (which are all integers)
from both sides and this expression has to be an integer:
- $\frac{1}{(b+1)}+\frac{1}{(b+1)(b+2)}+\frac{1}{(b+1)(b+2)(b+3)}+...\quad\quad(*)$
Now, we can replace all the (b+c_i) values with (b+1) which makes them all slightly bigger (reducing the denominator increases the value) and now we see that the expression is bigger than 0, but less than
- $b^{-1}+b^{-2}+b^{-3}+b^{-4}+...$
which is a geometric series that converges to $\frac{1}{b-1}$ which is greater than 0, and less than 1.
Hence $\frac{1}{(b+1)}+\frac{1}{(b+1)(b+2)}+\frac{1}{(b+1)(b+2)(b+3)}+...$ is strictly between 0 and 1.
This contradicts the requirement that expression (*) must be an integer.
Thus we cannot have $e=a/b$ and so e is not a rational number - It's an irrational number.
Links to this page /
Page history /
Last change to this page
Recent changes /
Edit this page (with sufficient authority)
All pages /
Search /
Change password /
Logout