Any hint for this demonstration?
Show that if $\displaystyle n$ is a composite (positive integer) number it has a divider that is less or equal to $\displaystyle \sqrt {n}$.
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Suppose $p_{1}$ and $p_{2}$ are primes dividing $n$. If both of them are greater than $\sqrt{n}$, then $p_{1} \times p_{2} > \sqrt{n} \times \sqrt{n} = n$. Contradiction. – MathGod Nov 24 '16 at 10:20
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If $n$ is a positive integer, then $1\le \sqrt n$, and $1$ is a divisor of every integer -- no matter if $n$ is composite or not. – hmakholm left over Monica Nov 24 '16 at 11:07
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Many thanks for the answers! – Oualid Nov 24 '16 at 13:57
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Hint: We know that divisors come in pairs (e.g. $3 \mid 12$ since $3\cdot 4 = 12$, this also yields $4 \mid 12$). What happens if such a pair of divisors are both greater than $\sqrt{n}$?
benguin
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1We don't even need to know that divisors come in pairs in general -- the "$n$ is composite" assumption says that we have such a pair $n=ab$. – hmakholm left over Monica Nov 24 '16 at 11:05