Like any type of other organic number, N is divisible by at least one element number (it is feasible that N itself is prime).

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Is there a proof because that this? Hint $\$ If $n > 1$ climate its least element $> 1$ is prime. $\$ And similarly for polynomials:

$\qquad\ \$ if f(x) is nonconstant climate its least-degree nonconstant factor is irreducible.

And likewise for Gaussian integers, by taking the least-norm nonunit factor, etc. Yes (modulo Brian"s comment.)

Hint: intend not, and also let $n$ be the least herbal number $\ge 2$ that is no divisible through a prime. In particular, $n$ is not prime, so... Let $P(n)$ it is in the statement, "$n$ is divisible by a element number." since 2 is prime and 2 divides itself, it follows that $P(2)$ is true. This serves together the based instance of the induction.Let $m>2$ and assume that $P(k)$ is true for all $k$ whereby $2\le k2$ was arbitrary, the result follows by complete Induction. Thanks for contributing an answer to aramuseum.orgematics ridge Exchange!

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