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Random Math Fact:

Did you know:

Euler’s “lucky” numbers are positive integers n such that m2 − m + n is a prime number for m = 0, …, n − 1.

Leonhard Euler published the polynomial x2 − x + 41 which produces prime numbers for all integer values of x from 0 to 40. Obviously, when x is equal to 41, the value cannot be prime anymore since it is divisible by 41. Only 6 numbers have this property, namely 2, 3, 5, 11, 17 and 41.

Why is e irrational?

Anyone who has taken high school math is familiar with the constant $\boxed{e=2.718281828\cdots}$.

Today we are going to prove that e is in fact irrational! We will go through Joseph Fourier‘s famous proof by contradiction. The maths background we need is to know the power series expansion: $\displaystyle \boxed{e=\sum_{n=0}^{\infty}\frac{1}{n!}}$. The proof is slightly tricky so stay focussed!

Suppose to the contrary that e is a rational number, so $\displaystyle e=\frac{a}{b}$.

Using the power series formula mentioned above, we have $\displaystyle\sum_{n=0}^\infty \frac{1}{n!}=\frac{a}{b}$

Multiply both sides by $b!$, $\displaystyle \sum_{n=0}^{\infty}\frac{b!}{n!}=\frac{ab!}{b}=a(b-1)!$

Now, we split the sum into two parts:

$\displaystyle \sum_{n=0}^b \frac{b!}{n!}+\sum_{n=b+1}^\infty \frac{b!}{n!}=a(b-1)!$

Rearranging,

$\displaystyle \sum_{n=b+1}^\infty \frac{b!}{n!}=a(b-1)!-\sum_{n=0}^b \frac{b!}{n!}$

Now, denote $\displaystyle x=\sum_{n=b+1}^\infty \frac{b!}{n!}>0$. $x$ is an integer since both $\displaystyle a(b-1)!$ and $\displaystyle\sum_{n=0}^b \frac{b!}{n!}$ are integers and their difference (which is x) will be an integer.

We now prove that $x<1$. For all terms with $n\geq b+1$ we have the upper estimate

$\displaystyle\begin{array}{rcl} \frac{b!}{n!}&=&\frac{1\times 2\times \cdots \times b}{1\times 2\times \cdots \times b \times (b+1) \times \cdots \times n}\\ &=&\frac{1}{(b+1)(b+2)\cdots (b+(n-b))}\\ &\leq& \frac{1}{(b+1)^{n-b}} \end{array}$

This inequality is strict for every $n\geq b+2$. Changing the index of summation to $k=n-b$ and using the formula for the infinite geometric progression $S_\infty = \frac{a}{1-r}$, we obtain:

$\displaystyle x=\sum_{n=b+1}^\infty \frac{b!}{n!} < \sum_{n=b+1}^\infty \frac{1}{(b+1)^{n-b}}=\sum_{k=1}^\infty \frac{1}{(b+1)^k}=\frac{\frac{1}{b+1}}{1-\frac{1}{b+1}}=\frac{1}{b}\leq 1$

We have that $x$ is an integer but $0. This is a contradiction (since there is no integer strictly between 0 and 1), and so $e$ must be irrational. (QED)

Interesting? 🙂

Did you know the constant e is sometimes called Euler’s number?

Learn more about Euler in this wonderful book. Rated 4.9/5 stars, it is one of the highest rated books on the whole of Amazon.

Leonhard Euler was one of the most prolific mathematicians that have ever lived. This book examines the huge scope of mathematical areas explored and developed by Euler, which includes number theory, combinatorics, geometry, complex variables and many more. The information known to Euler over 300 years ago is discussed, and many of his advances are reconstructed. Readers will be left in no doubt about the brilliance and pervasive influence of Euler’s work.

Watch this video for another proof that e is irrational!