# Proofs in Discrete Math. forall n in N+, n composite rightarrow exists p in N+, p is prime and p <= sqrtn and p|n.

Proofs in Discrete Math
$\mathrm{\forall }n\in N+$, n composite $\to \mathrm{\exists }p\in N+$, p is prime and $p\le \sqrt{n}$ and p|n.
Am I supposed to prove that $p\le \sqrt{n}$ and p|n when n is composite and p is prime? Could someone fix my translation if I'm wrong?
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Step 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 $k\ge 2$ (for n to be composite).
Step 2
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∣n.