# for all integer a>1 there can be found at least one prime divisor

• Jun 8th 2009, 10:27 AM
canuhelp
for all integer a>1 there can be found at least one prime divisor
i am starting to study abstract algebra and at this question i missed some points of the proof i wish you can explain:
prove that-->for all integer a>1 there can be found at least 1 prime divisor.
proof:let S={d (element) Z: d>1 and d|a} be this set.we know that S is not empty since a|a and a>1 so a (element) S and Sis not empty set.
positive integer set is well-ordered set so there exist the minimum element of S and say it p.if we show that p is prime number then the proof is over.
from here;
assume contrary that p is not prime then p=q.r and there can be found at least one q,r (element) Z such that 1<q<p.
and q|p and p|a -->q|a and q (element) S and this contradicts witht the assumption of p be the smallest element of S since p is prime.

my question is why do we try to prove that p is smallest element of S.we are searching at least one prime divisor why that must be the smallest one,what happened if the smallest one is not prime?may there is another elemnt is prime of S.
• Jun 8th 2009, 03:12 PM
Suppose to the contrary that there is an integer greater than 1 that is not divisible by any prime. Let $n$ be the smallest such integer. Then $n$ must be composite (if it were prime, then it would be divisible by a prime, namely itself). Thus $n=ab$ where $1 By minimality of $n,$ there is an prime $p$ which divides $a.$ It follows that $p$ divides $n$ so that $n$ is divisible by a prime after all. This contradiction means that no such integer $n$ exists; hence every integer greater than 1 is divisible by some prime.