INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

September 26, 2017

TIFR 2014 Problem 1 Solution - Negation


TIFR 2014 Problem 1 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta


Problem:


Let A,B,C be three subsets of (\mathbb{R}) The negation of the following statement: For every (\epsilon >1), there exists (a\in A) and (b\in B) such that for all (c\in C), (|a-c|< \epsilon) and (|b-c|>\epsilon ) is:

(A) There exists ( \epsilon \leq 1), such that for all (a \in A) and (b \in B) there exists (c \in C) such that (|a − c| \geq \epsilon) or (|b − c| \leq \epsilon) 

(B) There exists ( \epsilon \leq 1), such that for all (a \in A) and (b \in B) there exists (c \in C) such that (|a − c| \geq \epsilon) and (|b − c| \leq \epsilon) 

(C) There exists ( \epsilon > 1), such that for all (a \in A) and (b \in B) there exists (c \in C) such that (|a − c| \geq \epsilon) and (|b − c| \leq \epsilon) 

(D) There exists ( \epsilon > 1), such that for all (a \in A) and (b \in B) there exists (c \in C) such that (|a − c| \geq \epsilon) and (|b − c| \leq \epsilon) 


Discussion:


Before looking at the options, let us understand what rules we have to negate a statement.
(Please note this topic have discussed in Appendix portion of mentioned book in Helpdesk section.)

When we negate a for-all statement we get a there-exists statement. For example, the negation of (\forall x P(x)) is true is (\exists x) for which (P(x)) is not true.

A negation of a there-exists statement is for-all. Example: the negation of (\exists x) such that ( P(x)) is true is (\forall x) (P(x)) is not true.

To negate a slightly complicated such as that given in question we follow the following simple procedure:

  1. replace the \(\forall\)s with \(\exists\) and replace \(\exists\) by \(\forall\).
  2. and in the end, negate the conclusion.

Following this procedure we obtain:

(\exists \epsilon>1),such that (\forall a\in A),and (b\in B) (\exists c\in C), (|a-c|\ge \epsilon) or (|b-c|\le \epsilon).

Note that

(\exists \epsilon>1),such that (\forall a\in A),and (b\in B) (\exists c\in C) is the part that comes from by 1. And the part (|a-c|\ge \epsilon) or (|b-c|\le \epsilon) comes from the part 2.

We look at the question paper now, and indeed, this answer is the statement of option (d).


Helpdesk

  • What is this topic: Mathematical Logic
  • What are some of the associated concept: Negation of a Statement
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com