Write the negations of these sentences in as “smooth” a way as possible. (In particular, you may not simply append “It is not the case that …”. Also, you should make explicit any “hidden” quantifiers.)
- There is a real number \(x\) such that for every real number \(y\), \(|x-y|>1\).
- A real-valued function that is continuous on a closed interval attains a minimum value on that interval.
- \(3n+1\) is even if and only if \(n^2+4\) is prime.
Solution
The negation is: \[ \text{For every real }x\text{ there exists a real }y\text{ such that }|x-y|\le1. \]
Making the hidden quantifiers explicit, the original assertion says that for every closed interval \([a,b]\) and every continuous \(f:[a,b]\to\mathbb R\), there exists \(x\in[a,b]\) such that \(f(x)\le f(y)\) for every \(y\in[a,b]\). Its negation is therefore: \[ \text{There exist }a\le b\text{ and continuous }f:[a,b]\to\mathbb R\text{ such that for every }x\in[a,b] \] \[ \text{there exists }y\in[a,b]\text{ with }f(y)<f(x). \]
Interpreting \(n\) as an arbitrary integer, the negation of the biconditional is that there exists an integer \(n\) for which exactly one side holds; equivalently, \[ \text{either }3n+1\text{ is even and }n^2+4\text{ is composite,} \] or \[ 3n+1\text{ is odd and }n^2+4\text{ is prime.} \]