# Mathematics: A Very Short Introduction

By Professor Tim Gowers, Fields Medallist. Highly recommended for my young readers and students especially. Perhaps, it might also be recommended for Indian parents who want to know a bit more about Math and related exams like IITJEE Advanced.

Regards,

Nalin Pithwa.

# A cute little math pearl and some tips for studying for IIT-JEE or math Olympiads

Find out and compare roots of unity and roots of negative unity.

You should learn to play with such little gems of math on your own. This is creates ripples in the pond of the intellect, gently giving training to the subconscious mind. In this way we trick the mind rather create a natural uninterrupted flow of thoughts. It should not be that we can think of a math Olympiad or competitive programming contest or IIT-JEE problem only with so many textbooks in front of us… even in the bank queue or while taking shower one should be able to talk math to oneself…Of course, one does require physical solitude and also be introverted ( implies away from social media etc…)

Try it…and check for yourself…

Regards

Nalin Pithwa

# Check your mathematical induction concepts

Discuss the following “proof” of the (false) theorem:

If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:

PROOF BY INDUCTION:

Step 1:

If $n=1$, the result is evident.

Step 2: By the induction hypothesis the result is true when $n=k$; we must prove that it is correct when $n=k+1$. Let S be any set containing exactly $k+1$ real numbers and denote these real numbers by $a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}$. If we omit $a_{k+1}$ from this list, we obtain exactly k numbers $a_{1}, a_{2}, \ldots, a_{k}$; by induction hypothesis these numbers are all equal: $a_{1}=a_{2}= \ldots = a_{k}$.

If we omit $a_{1}$ from the list of numbers in S, we again obtain exactly k numbers $a_{2}, \ldots, a_{k}, a_{k+1}$; by the induction hypothesis these numbers are all equal: $a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}$.

It follows easily that all $k+1$ numbers in S are equal.

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