Proofs in Discrete Math

Question

∀n∈N+, n composite → ∃p∈N+, p is prime and p≤√n and p|n

Am I supposed to prove that p≤√n and p|n when n is composite and p is prime? Could someone fix my translation if I'm wrong? Thanks!

Answer 1

Our argument is based on the number of primes $m$ that divide $n$. Thus if $m = 1$, then $n$ is of the form $n = p^k$ wheareas $p$ is a prime, and $k \ge2$ (for $n$ to be composite). Thus $n \ge p^2 $ and $p \le \sqrt{n}$. if $m \ge 2$, then we can write $n$ as $n = b\cdot p^r\cdot q^s$ whereas $p,q$ are distinct primes, and $b, r, s \ge 1$ and are natural numbers. WLOG, assume $p \le q$. So: $p^2 \le pq \le p^r\cdot p^s \le b\cdot p^r\cdot q^s = n$, and $p^2 \le n$ or $p \le \sqrt{n}$. In both cases, $p \mid n$.

Answer 2

The translation here is that you are supposed to show that if you have a composite number, at least one of the primes that divide it must be less than or equal to the square root of the number

hint for proof: Try contradiction