[Math] Contrapositive: $\forall\; n > 1, n:$ composite $\implies\exists\; p$ (prime) s.t. $p \leq \sqrt n$ and $p\mid n$

Question

Usually, I find it a cakewalk to write the contrapositive, but the following statement is quite complex for the task:

For all integers $n > 1$, if $n$ is not prime, then there exists a prime number $p$ such that $p \leq \sqrt n$ and $n$ is divisible by $p$.

Is it "There exists no prime number $p$ such that $p \leq \sqrt n$ and $p \, | \, n$ given that $(n > 1)$ for a prime integer $n$"? The weird thing here is that the first universal statement (for all integers $n$) was not converted into a conditional, which makes me uncomfortable.

I'd appreciate some guidance.

Answer 1

You have the statement:

For all integers $n > 1$, if $n$ is not prime, then there exists a prime number $p$ such that $p \leq \sqrt n$ and $n$ is divisible by $p$.

This is a statement of the form $$\text{Let }\;n\in \mathbb Z, n > 1: \quad \forall n\left [\lnot P(n) \implies \exists p (Q(n, p) \land R(n, p))\right]$$

It's contrapositive is:

For all integers $n\gt 1$, if there does not exist a prime number $p$ such that $p \leq \sqrt n$ and $n$ is divisible by $p$, then $n$ is prime.

Which is a statement of the form $$\text{Let}\;n\in \mathbb Z, n > 1:\quad \forall n [\lnot \exists p(Q(n,p) \land R(n,p)) \implies P(n))$$