I don't consider this proof to be particularly elegant, but it gets the job done, so I thought I'd share.
p is Prime
p and n are integers
p is a factor of n^2
Thus, p is a factor of n
my proof
n^2 is not prime........................(can be factored by n)
p < n^2......................................(p is prime, p is not greater than or equal to n^2)
-----assume p is not a factor of n......(reductio ad absurdum)
-----p is not a factor of n*n...............(there's probably a rule for this)
-----n*n = n^2
-----p is not a factor of n^2
-----p is and is not a factor of n^2....(conjunction creates a contradiction)
p is a factor of n
Elias, there do seem to be some missing justifications in this proof. As you've noted, "p is not a factor of n*n" is missing explanation. I'm not entirely sure what that would be. If I were structuring this proof reductio ad absurdum, then your assumption is correct: "assume p is not a factor of n". But from here I would work to show that from this assumption, p is not a factor of n^2. This would contradict the third premise; therefore p is a factor of n (RA).
ReplyDelete