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!

Books for Gifted Children

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Match Wits With Mensa: The Complete Quiz Book

This is the #1 Top-Selling book recommended on my website! It includes Mathematical Logic Puzzles from Mensa. Highly recommended for gifted children. Parents, if your child is gifted and you want to stretch his or her learning potential, you may want to buy this book as it is the most complete quiz book on the market. It doesn’t matter whether you are in the Gifted Education Programme, as long as you have an interest in logic puzzles this book is for you.

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.

Puzzle fans have bought more than 650,000 copies of the Mensa Genius Quiz series—the only books that let readers “match wits with Mensa,” comparing how well they do against members of the famous high-IQ society. Here, in a giant omnibus edition, are four best-selling titles: The Mensa Genius Quiz Books 1 & 2, The Mensa Genius Quiz-A-Day Book, and The Mensa Genius ABC Book. Here are more than 800 fun mindbenders to exercise every part of your brain—word games, trivia, logic riddles, number challenges, visual puzzles—plus tips on how to improve your thinking skills. All the puzzles have been tested by members of American Mensa, Ltd., and include the percentage of Mensa testers who could solve each one, so that you can score yourself against some of the nation’s fittest mental athletes.

AM-GM inequality

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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Competition Math for Middle School

Written for the gifted math student, the new math coach, the teacher in search of problems and materials to challenge exceptional students, or anyone else interested in advanced mathematical problems. Competition Math contains over 700 examples and problems in the areas of Algebra, Counting, Probability, Number Theory, and Geometry. Examples and full solutions present clear concepts and provide helpful tips and tricks. “I wish I had a book like this when I started my competition career.” Four-Time National Champion MATHCOUNTS coach Jeff Boyd “This book is full of juicy questions and ideas that will enable the reader to excel in MATHCOUNTS and AMC competitions. I recommend it to any students who aspire to be great problem solvers.” Former AHSME Committee Chairman Harold Reiter

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.

nus ranking


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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.