# Number Devil

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If you enjoyed reading Lewis Carroll’s Alice’s Adventures in Wonderland, George Gamow’s Mr Tompkins, Abbott’s Flatland, Malba Tahan’s The Man Who Counted, Imre Lakatos’s Proofs and Refutations or Tarasov’s Calculus, then you will enjoy reading Enzensberger’s The Number Devil. But that is not an if and only if statement.

Originally written in German and published as Der Zahlenteufel, so far it has been translated into 26 languages (as per the back cover).

After reading this book one will have some knowledge of infinity, infinitesimal, zero, decimal number system, prime numbers (sieve of eratothenes, Bertrand’s postulate, Goldbach conjecture), rational numbers (0.999… = 1.0, fractions with 7 in denominator), irrational numbers (√2 = 1.4142…, uncountable), triangular numbers, square numbers, Fibonacci numbers, Pascal’s triangle (glimpse of Sierpinski triangle in it), combinatorics (permutations and combinations, role of Pascal’s triangle), cardinality of sets (countable sets like even numbers, prime numbers,…), infinite series (geometric series, harmonic series), golden ratio (recursive relations, continued fractions..), Euler characteristic (polyhedra and planar graphs), how to prove (11111111111^2 does not give numerical palindrome, Principia Mathematica), travelling salesman problem, Klein bottle, types of infinities (Cantor’s work), Euler product formula, imaginary numbers (Gaussian integer), Pythagoras theorem, lack of women mathematicians  and pi.

Since this is a translation of original work into English, you might not be happy with the language.  Though the author is not a mathematician, he is a well-known and respected European intellectual and author with wide-ranging interests. He gave a speech on mathematics and culture, “Zugbrücke außer Betrieb, oder die Mathematik im Jenseits der Kultur—eine Außenansicht” (“Drawbridge out of order, or mathematics outside of culture—a view from the outside”), in the program for the general public at  the International Congress of Mathematicians in Berlin in 1998. The speech was published under the joint sponsorship of the American Mathematical Society and the Deutsche Mathematiker Vereinigung as a pamphlet in German with facing English translation under the title Drawbridge Up: Mathematics—A Cultural Anathema, with an introduction by David Mumford.

# Education System

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This blog post has nothing to do with mathematics but just wanted to vent out my emotions.

I know that my opinions regarding the education system don’t matter since there always have been smarter people (i.e. people scoring more than me) around me in my home, school and college (and according to this system, only the opinions of top scorers matter). But, since WordPress allows me to express my opinions, here are the few comics which are in sync with my opinions:

couldn’t find the creator of this comic

I don’t think there is any solution to this problem since there are so many human beings on earth (i.e large variety of minds…).

# Listening Maths

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Earlier I told about new mediums now available to enjoy maths. You can find a list of YouTube channels, to begin with here. And a list of dedicated math podcasts here.

In this post, I want to point out some episodes from the  podcast “In Our Time” which discuss mathematics:

Apart from them, another podcast worth listening to is “The Wizard of Mathematics, Srinivasa Ramanujan” by Prof. Srinivas Kotyada on FM Gold:

# A topic I wanted to discuss for long time

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If you are an average maths undergraduate student (like me), you might have ended up in a situation of choosing between “just completing the degree by somehow passing the courses without caring about the grades” and “repeating a course/taking fewer courses so as to pass all courses with nice grades only”. Following is a nice discussion from Reddit:

# A sequence I didn’t like

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Following is a problem I encountered many times in my high school olympiads, but was never able to solve it. Hence didn’t like it.

What is the 100th term in the sequence $1,2,2,3,3,3,4,4,4,4,5,\ldots$?

Following is a quick solution:

In this post, I will discuss the solution given in The Green Book of Mathematical Problems (problem 14).

Determine a function $f(n)$ such that the $n^{th}$ term of the sequence $1,2,2,3,3,3,4,4,4,4,5,\ldots$ is given by $\lfloor f(n)\rfloor$.

Let’s denote the $n^{th}$ number of the sequence by $a_n$, i.e. $a_n=\lfloor f(n)\rfloor$. The integer $m$ first occurs in the sequence when each of the integers from 1 to $m-1$ have already appeared 1 to $m-1$ times, respectively. Hence, if $a_n=m$ then
$n = [1 + 2 + 3 + \ldots + (m-1)]+1 +\ell = \frac{m(m-1)}{2} + 1 + \ell$
for $\ell = 0,1,2,\ldots, m-1$.

Hence we have:
$\displaystyle{0\leq n - \frac{m(m-1)}{2} - 1 \leq m-1}$

$\displaystyle{\Rightarrow \frac{m^2-m+2}{2}\leq n \leq \frac{m^2+m}{2}}$

$\displaystyle{\Rightarrow m^2-m+2\leq 2n \leq m^2+m}$

$\displaystyle{\Rightarrow \left(m-\frac{1}{2}\right)^2+\frac{7}{4}\leq 2n \leq \left(m+\frac{1}{2}\right)^2-\frac{1}{4}}$

$\displaystyle{\Rightarrow (2m-1)^2+7\leq 8n \leq (2m+1)^2 - 1}$

$\displaystyle{\Rightarrow (2m-1)^2 \leq 8n - 7 \leq (2m+1)^2 - 8}$

$\displaystyle{\Rightarrow (2m-1)^2 \leq 8n - 7 < (2m+1)^2}$

$\displaystyle{\Rightarrow 2m-1 \leq \sqrt{8n-7}<2m+1}$

$\displaystyle{\Rightarrow m \leq \frac{1+\sqrt{8n-7}}{2} < m+1}$

$\displaystyle{\Rightarrow m=\left\lfloor \frac{1+\sqrt{8n-7}}{2} \right\rfloor}$

Hence we have $a_n = \left\lfloor \frac{1+\sqrt{8n-7}}{2} \right\rfloor$. Thus, we have

$\displaystyle{\boxed{f(n) =\frac{1+\sqrt{8n-7}}{2}}}$

Now compared to the earlier solution obtained by observing the pattern, one might ask “Is there is a better formula?”. For that, you might also look at the discussion at Math.SE.

# Prime Number Problem

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Following is a problem about prime factorization of the sum of consecutive odd primes. (source: problem 80 from The Green Book of Mathematical Problems)

Prove that the sum of two consecutive odd primes is the product of at least three (possibly repeated) prime factors.

The first thing to observe is that sum of odd numbers is even, hence the sum of two consecutive odd primes will be divisible by 2. Let’s see factorization of some of the examples:

$3 + 5 = 2\times 2 \times 2$
$5 + 7 = 2 \times 2\times 3$
$7+11 = 2 \times 3\times 3$
$11+13 = 2 \times 2 \times 2 \times 3$
$13+17 = 2 \times 3 \times 5$
$17+19 = 2\times 2\times 3 \times 3$
$19+23 = 42 = 2\times 3\times 7$
$23+29 = 52 = 2\times 2 \times 13$

Now let $p_n$ and $p_{n+1}$ be the consecutive odd primes, then from above observations we can conjecture that either $p_n+p_{n+1}$ is product of at least three distinct primes or $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ such that $k+\ell \geq 3$.

To prove our conjecture, let’s assume that $p_n+p_{n+1}$ is NOT a product of three (or more) distinct primes (otherwise we are done). Now we will have to show that if $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ then $k+\ell \geq 3$.

If $\ell = 0$ then we should have $k\geq 3$. This is true since $3+5=8$.

Now let $\ell > 0$. Since $k\geq 1$ (sum of odd numbers is even), we just need to show that $k=1, \ell=1$ is not possible. On the contrary, let’s assume that $k=1,\ell = 1$. Then $p_n+p_{n+1} = 2p$. By arithmetic mean property, we have

$\displaystyle{p_n < \frac{p_n+p_{n+1}}{2}} = p

But, this contradicts the fact that $p_n,p_{n+1}$ are consecutive primes. Hence completing the proof of our conjecture.

This is a nice problem where we are equating the sum of prime numbers to product of prime numbers. Please let me know the flaws in my solution (if any) in the comments.