## 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? 🙂

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Watch this video for another proof that e is irrational!

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# AM-GM inequality

A very useful inequality in Mathematics is the AM-GM Inequality.

The arithmetic mean of numbers $x_1, x_2, \cdots, x_n$ is $\displaystyle \boxed{\frac{x_1+ x_2+\cdots+x_n}{n}}$.

The geometric mean of numbers $x_1, x_2, \cdots, x_n$ is $\boxed{\sqrt[n]{x_1\cdot x_2 \cdots x_n}}$.

The AM-GM Inequality states that:

For any nonnegative numbers $x_1, x_2, \cdots, x_n$, $\displaystyle\boxed{\frac{x_1+x_2+\cdots+ x_n}{n}\geq\sqrt[n]{x_1\cdot x_2 \cdots x_n}}$, and equality holds if and only if $x_1=x_2=\cdots=x_n$. ## How to Apply?

Let say we have three (nonnegative) numbers a, b, c that add up to 30, i.e. $a+b+c=30$. Can we know what is the largest possible product $abc$?

Yes! Using the AM-GM inequality we have just learnt above, we know $\displaystyle \frac{a+b+c}{3}\geq \sqrt{abc}$. $\displaystyle 10\geq \sqrt{abc}$

Cubing both sides, we have, $\displaystyle abc\leq 10^3=1000$.

Also, the AM-GM inequality tells us that there is equality only when $a=b=c$, i.e. $a=b=c=10$. Hence, the largest possible product $abc$ is 1000.

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