Let $ n $ be a natural integer strictly greater than $ 1 $.
If $ n $ is not prime, it admits a prime divisor less than or equal to $ \sqrt {n} $.
How can I write an assertion expressing the above proposition. thank you
Let $\mathcal{P} $ be the set of prime numbers.
$$ (\forall n>1)\quad n\notin\mathcal{P} \implies \exists\; p\in\mathcal{P}, \begin{cases}p\leq \sqrt{n} \\ p\mid n \end{cases}$$
Its contraposition
$$(\forall n>1)\quad \forall\; p\in\mathcal{P}, \begin{cases}p> \sqrt{n} \\\rm{Or}\\ p\nmid n \end{cases}\implies n \in\mathcal{P} $$
Doing the contraposition does not lead me to find the same point method:
To show that a natural number $n\geq 2$ is prime, it suffices to check
That it is not divisible by any integer $k$ satisfying $2 \leq k \leq \sqrt{n}$.
You have this: $$(\forall n>1)\quad \forall\; p\in\mathcal{P}, \begin{cases}p> \sqrt{n} \\\rm{Or}\\ p\nmid n \end{cases}\implies n \in\mathcal{P} $$
To check whether $n$ is prime, you have to prove that for every prime $p$, at least one of the two listed conditions holds. For all primes larger than $\sqrt{n}$, the condition $p> \sqrt{n}$ obviously holds. For the remaining primes, i.e. the primes in the range $2 \le p \le \sqrt{n}$ that condition is false, so for those you need the other condition $p\nmid n$ to be true. If that condition is true for all primes in this range, then the whole statement is true for all primes and $n$ is prime.