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Why is e irrational?

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

e-irrational

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!

(Reference: http://en.wikipedia.org/wiki/Proof_that_e_is_irrational)

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 $0<x<1$ . 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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Euler: The Master of Us All (Dolciani Mathematical Expositions, No 22)

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!

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Maths and Science is essentially about logical thinking, so logic puzzles will directly benefit studies in maths and science. Above all, logic puzzles are meant to be fun and a good and healthy pastime.

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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 $x_1=x_2=\cdots=x_n$ .

am-gm-inequality

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 $\displaystyle \frac{a+b+c}{3}\geq \sqrt[3]{abc}$ .

$\displaystyle 10\geq \sqrt[3]{abc}$

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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NUS Math Ranked among Top in Asia

In the latest Quacquarelli Symonds (QS) World University Rankings by Subject (2014), NUS Math is ranked among the best mathematics departments in Asia.

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Knowing and Teaching Elementary Mathematics: Teachers’ Understanding of Fundamental Mathematics in China and the United States (Studies in Mathematical Thinking and Learning Series)

Chinese students typically outperform U.S. students on international comparisons of mathematics competency. Paradoxically, Chinese teachers receive far less education than U.S. teachers–11 to 12 years of schooling versus 16 to 18 years of schooling.

Studies of U.S. teacher knowledge often document insufficient subject matter knowledge in mathematics. But, they give few examples of the knowledge teachers need to support teaching, particularly the kind of teaching demanded by recent reforms in mathematics education.

This book describes the nature and development of the “profound understanding of fundamental mathematics” that elementary teachers need to become accomplished mathematics teachers, and suggests why such teaching knowledge is much more common in China than the United States, despite the fact that Chinese teachers have less formal education than their U.S. counterparts.