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Negation (TIFR 2014 problem 1)

Question: (MCQ)

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:


Before looking at the options, let us understand what rules we have to negate a statement.

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).



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