Writing and negating statements with quantifiers

Question

There exists a natural number $k$ such that for all natural numbers n, there exists primes $p$ and $q$ such that $p > n$ , $q > n$, and $\lvert p - q\rvert < k$.

First, by using the open statements P(x): x is prime, Q(x,y): x > y, and R(x,y,z):$\lvert$x - y$\rvert$ < z, write the statement in symbols.

Then, write the negation in symbols.

Lastly, write the negation in words.

Mainly, the change of variables from p,q,k to x,y,z has confused me. Are they the same? I just need help writing in symbols and negation, no proofing (for those that are interested in looking at it, it’s from number theory and proved recently)

Answer 1

It doesn't matter how you name the variables. The statement would be $$\exists k\in\mathbb{N}(\forall n\in\mathbb{N}(\exists q,p\in\mathbb{N}:(P(p)\wedge P(q)\wedge Q(p,n)\wedge Q(q,n)\wedge R(p,q,k))))$$ and the negation would be $$\forall k\in\mathbb{N}(\exists n\in\mathbb{N}(\forall q,p\in\mathbb{N}:\neg(P(p)\wedge P(q)\wedge Q(p,n)\wedge Q(q,n)\wedge R(p,q,k)))).$$

The negation in words: For every natural number $k$ there exists a natural number $n$ such that for all primes $p,q>n$ we get $|p-q|\geq k$.

Also: You should not use $\mathbb{N}$ as a symbol for a natural number. Your statement should therefore look along lines of this:

There exists a natural number $k$ such that for all natural numbers $n$, there exists primes p and q such that $p > n$ , $q > n$, and $|p - q| < k$.