# Proof That E Is Irrational AllPages RecentChanges Links to this page Edit this page Search Entry portal Advice For New Users

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.