Question

Negate the following sentences. For every positive number \(\varepsilon\), there is a positive number \(M\) for which \(|f(x)-b|<\varepsilon\) whenever \(x>M\).

Short answer

The negation of the given sentence is 'There exists a positive number \(\varepsilon\) such that for all positive numbers \(M\), there exists a number \(x\) with \(x > M\) and \(|f(x) - b| \geq \varepsilon\)'.

Step-by-step solution

Understand the Original Statement

Given the statement, 'For every positive number \(\varepsilon\), there is a positive number \(M\) for which \(|f(x)-b|<\varepsilon\) whenever \(x>M\)', this is saying that for any positive number \(\varepsilon\), you can find another positive number \(M\) such that the absolute difference between \(f(x)\) and \(b\) is less than \(\varepsilon\) when \(x\) is greater than \(M\).

Formulate the Negation

Negating this requires stating what isn't the case. So let's say that there is some positive number \(\varepsilon\) such that no matter what positive number \(M\) you choose, there will always be some \(x>M\) such that \(|f(x)-b|\geq \varepsilon\). Effectively, no matter how large you choose \(M\) to be, there will always be some \(x\) greater than \(M\) where the absolute difference of \(f(x)\) and \(b\) is more or equal to \(\varepsilon\).

Complete the Negation

The completed negation of the statement then reads: 'There exists a positive number \(\varepsilon\) such that for all positive numbers \(M\), there exists a number \(x\) with \(x > M\) and \(|f(x) - b| \geq \varepsilon\)'. This is the correct negation of the given sentence.