Negating there exists-for all composite

Question

Suppose $P$ is a proposition statement about $p$ and $q$, and $Q$ is a statement about $r$. Consider:

• There is a prime number $p$ such that $P$ is true for all integers $q < 10$.

• $Q$ is false for all real numbers $r \leq 5$.

What are the respective negations for the above sentences?

My attempts:

• For all prime numbers $p$, there is an integer $q \geq 10$ for which $P$ is false.

• $Q$ is true if there exists a real number $r > 5$.

Are these negations correct? Also, is it allowed to move the "there exists" part before "$P$ is false" even though the original statement has the latter before? Any assistance is much appreciated.

Answer 1

I find it a bit ambiguous to say that $P$ states something about $p$ and $q$, yet not denote is as $P(p,q)$. Similarly, I'd find it clearer if $Q$ was written as $Q(r)$. But ok, you use natural language to do first-order logic, so such ambiguities are only natural! :P

To the point:

Assuming that $p$ and $q$ range over integers, for the first sentence, you're almost there! You only do not take into account the case where $p$ may not be prime:

• Either $p$ is not prime or (it is prime and) there is an integer $q \geq 10$ for which $P$ is false.

For the second sentence, assuming that $r$ ranges over reals, I'd say

• There is a real $r > 5$ for which $Q$ is true.

The problem with your formulation is that the $r$ in your sentence

$Q$ is true if there exists a real number $r>5$

does not refer to the $r$ that $Q$ states something about, but refers to some real number for which we happen to use the same letter $r$ to name it.