Letbe a number that is not a prime.
Show that there is a divisorof
, with
, so that
On my paper, I've written down n=rs where n and s are any natural numbers >1. Not sure if that's needed, though.
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Let
Show that there is a divisor
On my paper, I've written down n=rs where n and s are any natural numbers >1. Not sure if that's needed, though.
If n>2 is not prime, then it has a factor p>=2, so there also exists a q>=2 such that pq=n. Now if both n<p^2 and n<q^2, then pq<n a contradiction.Quote:
Originally Posted by Joel24
Let
Show that there is a divisor
On my paper, I've written down n=rs where n and s are any natural numbers >1. Not sure if that's needed, though.
Hence every composite number >2 has a factor whose square is less than or equal to the number.
RonL
Ah, a proof (or sorts) by contradiction. I would've never have thought of that.
Excellent. Thank you so much!