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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Comments, observations are welcome 🙂

Regards,

Nalin Pithwa

Some number theory (and miscellaneous) coaching for RMO and INMO: tutorial (problem set) III

Continuing this series of slightly vexing questions, we present below:

1. Prove the inequality \frac{A+a+B+b}{A+a+B+b+c+r} + \frac{B+b+C+c}{B+b+C+c+a+r} > \frac{C+c+A+a}{C+c+A+a++b+r}, where all the variables are positive numbers.

2. A sequence of numbers: Find a sequence of numbers x_{0}, x_{1}, x_{2}, \ldots whose elements are positive and such that a_{0}=1 and a_{n} - a_{n+1}=a_{n+2} for n=0, 1, 2, \ldots. Show that there is only one such sequence.

3. Points in a plane: Consider several points lying in a plane. We connect each point to the nearest point by a straight line. Since we assume all distances to be different, there is no doubt as to which point is the nearest one. Prove that the resulting figure does not containing any closed polygons or intersecting segments.

4. Examination of an angle: Let x_{1}, x_{2}, \ldots, x_{n} be positive numbers. We choose in a plane a ray OX, and we lay off it on a segment OP_{1}=x_{1}. Then, we draw a segment P_{1}P_{2}=x_{2} perpendicular to OP_{1} and next a segment P_{2}P_{3}=x_{3} perpendicular to OP_{2}. We continue in this way up to P_{n-1}P_{n}=x_{n}. The right angles are directed in such a way that their left arms pass through O. We can consider the ray OX to rotate around O from the initial point through points P_{1}, P_{2}, \ldots, P_{n} (the final position being P_{n}). In doing so, it sweeps out a certain angle. Prove that for given numbers x_{i}, this angle is smallest when the numbers x_{i}, that is, x_{1} \geq x_{2} \geq \ldots x_{n} decrease; and the angle is largest when these numbers increase.

5. Area of a triangle: Prove, without the help of trigonometry, that in a triangle with one angle A = 60 \deg the area S of the triangle is given by the formula S = \frac{\sqrt{3}}{4}(a^{2}-(b-c)^{2}) and if A = 120\deg, then S = \frac{\sqrt{3}}{12}(a^{2}-(b-c)^{2}).

More later, cheers, hope you all enjoy. Partial attempts also deserve some credit.

Nalin Pithwa.