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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)

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  • $\begingroup$ Your comment should be part of the question. Please edit the question to add that material, and delete the comment. $\endgroup$ Sep 20, 2016 at 14:15
  • $\begingroup$ In your first full sentence, you write $p$ and $q$ are primes such that $p\gt \mathbb N, q \gt \mathbb N and |p-q| < k$ With respect to the "$\gt$" signs, did you mean to write that $p\in \mathbb N$ and $q\in \mathbb N$? $\endgroup$
    – amWhy
    Sep 20, 2016 at 14:25
  • $\begingroup$ $P(x)$ is a predicate; it must be "filled" with the "name" of the number we want to assert the primality, like $P(p)$. The variable $x$ must be used with a quantifier : $\exists x P(x)$ to express: "there exists a prime ...". $\endgroup$ Sep 20, 2016 at 14:27
  • $\begingroup$ @amWhy I wrote it exactly how it appears from my paper. $\endgroup$
    – LizW
    Sep 20, 2016 at 14:28
  • $\begingroup$ @MauroALLEGRANZA So does that answer my 'Mainly, the change of variables from p,q,k to x,y,z has confused me. Are they the same?' question? I think it does. $\endgroup$
    – LizW
    Sep 20, 2016 at 14:30

1 Answer 1

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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 $|p - q| < k$.

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  • $\begingroup$ Thank you! That is along the of what I got. Sorry, that's just how it was written, but thank you for the advice. @mathsarefun $\endgroup$
    – LizW
    Sep 20, 2016 at 14:36

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