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- From: jxh@math.ksu.edu (James C. Hu)
- Newsgroups: sci.math
- Subject: Re: p prime, p divides ab => pdivides a or b
- Date: 3 Sep 1992 00:36:07 -0500
- Organization: Dept. of Mathematics, Kansas State University
- Lines: 32
- Message-ID: <184887INN7jp@hilbert.math.ksu.edu>
- References: <Btyo8q.E63@ux1.cso.uiuc.edu>
- NNTP-Posting-Host: hilbert.math.ksu.edu
-
- Charles Blair (ceblair@ux1.cso.uiuc.edu) wrote:
- > My recollection is that a number theory course I took presented
- >this as a difficult result, only proved after doing some stuff
- >with the Euclidean algorithm. Is there a proof which avoids that?
-
- I don't recall using the Euclidean algorithm, but the proof I had
- learned used a theorem we called Diophantine's Result:
-
- Let a, b be non-zero integers, where gcd(a,b)=d.
- Then, there exists integers x,y such that
- ax + by = d.
-
- The proof to this is not hard, but it depends upon the Division
- Algorithm, which may be what you meant instead. Using this, and the
- definition of a prime, it is quite easy to show that:
-
- p prime, then p | ab => p | a or p | b.
-
- The proof basically assumes p does not divide either. Then,
- gcd(p,a) = 1 and gcd(p,b) = 1. We write down the Diophantine
- equations which are assured to exist, multiply them together, and
- arrive at an contradiction.
-
- There are doubtless many other ways to prove this, but this result is
- required to prove UFT, and so I don't think any proofs of this using
- UFT would be wise. (Of couse, prime decomposition may be enough alone
- to show the result)
- --
- James C. Hu (jxh@math.ksu.edu), 1804 Denholm Dr., Manhattan, KS 66502
- "If it was so, it might be; and if it were so, it would be; but as it
- isn't, it ain't. Thats logic." - Lewis Carroll
- Opinions expressed are mine. Let me know if they are someone else's.
-