1994-04-12 - Re: Classic Math gone wrong…Re: (n!+1)^(1/2)

Header Data

From: Frank Vernaillen <Frank.Vernaillen@rug.ac.be>
To: Peter Wayner <pcw@access.digex.net>
Message Hash: 9047e1fe0aefc0af47763614f987d03f53af0048d8b1226d43c95dd59c4b755f
Message ID: <Pine.3.89.9404120216.A18697-0100000@eduserv>
Reply To: <199404120007.AA13053@access3.digex.net>
UTC Datetime: 1994-04-12 01:01:41 UTC
Raw Date: Mon, 11 Apr 94 18:01:41 PDT

Raw message

From: Frank Vernaillen <Frank.Vernaillen@rug.ac.be>
Date: Mon, 11 Apr 94 18:01:41 PDT
To: Peter Wayner <pcw@access.digex.net>
Subject: Re: Classic Math gone wrong...Re: (n!+1)^(1/2)
In-Reply-To: <199404120007.AA13053@access3.digex.net>
Message-ID: <Pine.3.89.9404120216.A18697-0100000@eduserv>
MIME-Version: 1.0
Content-Type: text/plain


> Scott Collins:
> (...) 
> The classic proof goes:
> 
> Is there a largest prime number? 
> If there is then collect all primes, p1...pn and multiply them
> together p=p1*p2*...*pn. p+1 is not divisible by p1...pn. Therefore
> p+1 is a prime. 

This last step (therefore p+1 is a prime) is not totally 
correct. You forgot the posibility p+1 NOT prime, but some prime 
number <p+1 but >pn divides p+1. This number is prime and >pn.
So in any case there would exist a prime >pn, which contradicts the
hypothesis, and the conclusion is indeed:

> Therefore there is no largest prime 
number.

Frank.Vernaillen@rug.ac.be




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