Re: Wolfram's MathWorld fails to give a valid proof of Euclid's

sttscitrans_at_tesco.net
Date: 03/29/05 

Date: 29 Mar 2005 04:05:26 -0800

bertran wrote:

> If p1, p2,..pn are the only primes, then we consider N=p1·p2·..pn +
1. Now you say that we shouldn't search for the prime factors of N, but
why?

As you say, p1, p2,..pn are the only primes that exist,
and no prime divides N. If no prime divides
N you have already found a contradiction as
it is a theorem that every Z>1 has a prime factorization.

We are entitled to do so if we reach a contradiction this way.

Yes, you could say that N is the only proper divisor of N
and so N must be a prime <> p1,p2,...pn.
But if you take the supposition that p1, p2,...pn are
all the primes that exist "seriously" the fact that no prime divides N
>1 is the "first" contradiction you encounter.
At some point, you show that no pk divides N.
Once you have done that, you have discovered a contradiction.
Whether N is another prime or divisible by another prime
is irrelevant.

If I understand you correctly, what you say is that we need not
consider the case that N is prime, since it's larger than all p1,..pn,
which are the only primes by assumption. This is true, but there is
nothing wrong in considering this case as well and conclude the same
thing (that this case contradicts the definition of p1,..pn). This
might be a bit silly (given A implies B and A we can conclude B
directly without supposing the negation of B and then A and reaching a
contradiction), but it's still correct.

Yes, but "psychologically" inconsistent.
Given a natural M, you "know" that you can trial
divide by primes and either some prime < M
divides M or M is itself prime. M must be divisible by some prime.
After some point every number could, of course, be composite.
If, however, you have tried all the primes that exist
and found none that divide M you might be puzzled unlees, of
course, you had been trying to find a prime that divides 1
But, the supposition is that p1, p2..pn are the
only primes that exist. None of these divide N.
As this is a contradiction (every Z>1 is divisible by some prime), this
means that
p1, p2...,pn are not the only primes that exist and so N is either a
prime >pn or divisible by a prime > pn

By the way, none of the proofs require the Uniqueness Prime
Factorisation theorem as you seem to think.

You might not need uniqueness, but unless
it is a theorem that every Z>1 has a prime factorisation,
you could not conclude that N is divisible by a prime or is itself
prime.

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