Prev: Quantum Gravity 403.6: Another "Negative Probability" Indication: P(A-->B) vs P(B-->A) with the latter as P(A<--B)
Next: where is the factorial value 1/2 of the exponent value in 10^500 #4.13 Correcting Math
From: Archimedes Plutonium on 24 Jul 2010 02:15

Transfer Principle wrote:
> On Jul 23, 1:00 pm, Archimedes Plutonium
(snipped)

This is not quite being fair to me Lwalk. I deliver an entire proof,
one which can be published in a book. I deliver both Direct
and Indirect in long form and in short form:

short form Indirect
(1) definition of prime number
(2) hypothetical assumption, assume the primes are finite and that
the sequence list is 2,3, 5, 7, 11, . . , p_k
(3) multiply the lot and add 1, calling it W+1
(4) W+1 is necessarily a new prime because of definition in (1)
joining with the fact that
division of W+1 by all the primes that exist in (2) leave a remainder
(5) contradiction to (2) that p_k is the largest and last prime, for W
+1 is now the largest prime
(6) reverse supposition step (2) and primes are infinite


Not fair to me to compare mine with someone who never writes out a
proof, never a step by step. Always jumps in, in midair with
hatemongering.

I do not deserve to be compared with someone who is unable to write
out a proof.

LWalk, every Euclid IP Indirect must start off with the definition of
prime, for we
all must know what we are talking about as step 1. And the definition
is critical in
both methods when we inspect W+1. Most authors of Euclid attempts,
most every
auther, even Ore omits the definiton, and technically every proof that
omits the definition
in step one is invalid, but that is a minor invalidity. The major
invalidity in the Indirect is
not realizing that W+1 is necessarily a prime number.

Step 1 ---- definition
Step 2 ---- assume all existing primes with p_k the last and largest
Step 3 ---- form W+1

All Indirect Euclid IP must have those three elements, and forces a
unique proof.

It is because you have this Euclid Number of multiply the lot, which
is divisable by all the primes of that finite set, but because you
added 1, none of the primes of the finite set divides.
So W+1 is necessarily prime.

Now you look back to step 1, and you are forced by the definition to
say W+1 is necessarily
a new prime.

Definition plus p_k the largest prime plus the forming of W+1, makes
Euclid Indirect a unique
proof of its steps. Sure, some hatemonger can tack on extraneous
garbage steps; can waffle about units
or waffle about composites. But all valid Euclid Indirect boil down to
a unique chain of events.

Definition plus p_k plus W+1 forces W+1 as a necessarily new prime and
thus the proof.

And because no-one really had a valid Euclid IP Indirect until the
1990s, that no-one in all of
mathematics was ever going to prove infinitude of twin primes, because
the only route to a proof of twin primes was via the Euclid Indirect.


> And so which side do I believe is right? Answer: _both_ are right!
>

You say this only because you are too nice of a guy that does not want
to
hurt the feelings of a hatemonger. But I do not think you should
compromise
math or a science, just to be nice to another person. I think truth
trumps being
nice.

Question LWalk: is there something in Galois algebra that says the
Twin Primes are
a subset of the Regular Primes, and since we have an infinitude proof
both direct
and indirect, is there something in Galois Algebra that says the
Indirect or Direct must
also be the basis of a proof of twin primes infinitude? I believe
there is, but just do not know
the technical term for this. The idea here is that if a proof of a
larger set is found, then that
proof must be able to wrangle out a proof for the subset.

So in other words, the twin primes proof needed only the fixing of the
invalid proof attempts of the past in order to provide a simple easy
twin primes proof.

Archimedes Plutonium
http://www.iw.net/~a_plutonium/
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
 | 
Pages: 1
Prev: Quantum Gravity 403.6: Another "Negative Probability" Indication: P(A-->B) vs P(B-->A) with the latter as P(A<--B)
Next: where is the factorial value 1/2 of the exponent value in 10^500 #4.13 Correcting Math