G.H.Hardy gave an invalid proof of Infinitude of Primes in "A Mathematician's Apology"
Archimedes Plutonium
But my business here today is to analyze Hardy's attempt to prove
Euclid's Infinitude of Primes. As in the title of this post, Hardy
failed, but the bright light is that one can see his pretty poetry. I
quote from his book.
--- quoting A MATHEMATICIAN'S APOLOGY, Godfrey H. Hardy, 1940,
pages 92-94 ---
I can hardly do better than go back to the Greeks. I will state
and prove two of the famous theorems of Greek mathematics. They are
'simple' theorems, simple both in idea and in execution, but there is
no doubt at all about their being theorems of the highest class. Each
is as fresh and significant as when it was discovered-- two thousand
years have not written a wrinkle on either of them. Finally, both the
statements and the proof can be mastered in an hour by any intelligent
reader, however slender his mathematical equipment.
1. The first is Euclid's (Elements IX 20. The real origin of many
theorems in the Elements is obscure, but there seems to be no
particular reason for supposing that this one is not Euclid's own)
proof of the existence of an infinity of prime numbers.
The prime numbers or primes are the numbers (A)
2,3,5,7,11,13,17,19,23,29,... which cannot be resolved into smaller
factors. (There are technical reasons for not counting 1 as a prime.)
Thus 37 and 317 are prime. The primes are the material out of which all
numbers are built up by multiplication: thus 666 = 2x3x3x37. Every
number which is not prime itself is divisible by at least one prime
(usually, of course, by several). We have to prove that there are
infinitely many primes, i.e. that the series (A) never comes to an end.
Let us suppose that it does, and that 2,3,5,..., P is the complete
series (so that P is the largest prime); and let us, on this
hypothesis, consider the number Q defined by the formula Q =
(2x3x5x..xP) + 1. It is plain that Q is not divisible by any of
2,3,5,...,P; for it leaves the remainder 1 when divided by any one of
these numbers. But, if not itself prime, it is divisible by some prime,
and therefore there is a prime (which may be Q itself) greater than any
of them. This contradicts our hypothesis, that there is no prime
greater than P; and therefore this hypothesis is false.
The proof is by reductio ad absurdum, and reductio ad absurdum,
which Euclid loved so much, is one of a mathematician's finest weapons.
It is a far finer gambit than any chess gambit: a chess player may
offer the sacrifice of a pawn or even a piece, but a mathematician
offers the game.
--- end quoting A MATHEMATICIAN'S APOLOGY, Godfrey H. Hardy, 1940,
pages 92-94 ---
Hardy's flaw is that Q is NECESSARILY prime and once Q is formed it is
the
end of the proof since it forms the contradiction that simultaneously
not prime and prime; or, that Q is a larger prime than P. To search for
some prime factor as Hardy does is to show that he failed and did not
understand Q, for
it ends the proof and it is the only candidate of a prime not on the
original
finite list. But Hardy made two mistakes, one he did not look at
Euclid's IP
closely for Euclid gave a direct proof of IP by increasing cardinality
of any-
and-thus-all finite sets, boosting the set of primes to infinity , and
his
second mistake was that he could not give a valid indirect proof of IP.
[lines deleted]
---
It is worth repeating, that in the direct proof method of increasing
cardinality a prime factor search must be undertaken. But in the
indirect proof method, the moment P!+1 is formed, is the exact moment
that the contradiction arises and that P!+1 is necessarily a new prime.
So when one does the Indirect IP and starts to make a search for a
"prime factor", well the person has failed and made a invalid proof.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
Nathan
> It is worth repeating, that in the direct proof method of increasing
> cardinality a prime factor search must be undertaken. But in the
> indirect proof method, the moment P!+1 is formed, is the exact moment
> that the contradiction arises and that P!+1 is necessarily a new
> prime. So when one does the Indirect IP and starts to make a search
> for a "prime factor", well the person has failed and made a invalid
> proof.
I have 2 (rhetorical) questions:
1. Why in the world would you think P!+1 must necessarily be a new
prime? In reality, it scarcely ever is.
2. Even if this idea of yours had any validity, why would adding an
unnecessary step render a valid proof invalid?
Torkel Franzen
> I have 2 (rhetorical) questions:
>
> 1. Why in the world would you think P!+1 must necessarily be a new
> prime? In reality, it scarcely ever is.
A foolish question! It is a Plutonium prime, with electrons swirling
about it.
Arthur Fischer
> --- first posted by me in the early to mid 1990s to the Internet ---
>
> But my business here today is to analyze Hardy's attempt to prove
> Euclid's Infinitude of Primes. As in the title of this post, Hardy
> failed, but the bright light is that one can see his pretty poetry. I
> quote from his book.
> --- quoting A MATHEMATICIAN'S APOLOGY, Godfrey H. Hardy, 1940,
> pages 92-94 ---
>
> [snip of quotation]
>
> Let us suppose that it does, and that 2,3,5,..., P is the complete
> series (so that P is the largest prime); and let us, on this
> hypothesis, consider the number Q defined by the formula Q =
> (2x3x5x..xP) + 1. It is plain that Q is not divisible by any of
> 2,3,5,...,P; for it leaves the remainder 1 when divided by any one of
> these numbers. But, if not itself prime, it is divisible by some prime,
> and therefore there is a prime (which may be Q itself) greater than any
> of them. This contradicts our hypothesis, that there is no prime
> greater than P; and therefore this hypothesis is false.
>
> [snip of quotation]
>
> --- end quoting A MATHEMATICIAN'S APOLOGY, Godfrey H. Hardy, 1940,
> pages 92-94 ---
>
> Hardy's flaw is that Q is NECESSARILY prime and once Q is formed it is
> the
> end of the proof since it forms the contradiction that simultaneously
> not prime and prime; or, that Q is a larger prime than P.
Suppose that we thought, wrongly, that the only prime numbers were
2,3,5,7,11,13,and 17 (which is the starting point of Hardy's proof ---
we think that we have some finite list that enumerates all primes; the
fact that I have chosen a ridiculously short initial segment of the
sequence of primes is, of course, immaterial). Then simple calculation
of Hardy's method gives
Q = 2x3x5x7x11x13x17 + 1 = 510511 = 19(26869)
Are you claiming that there are natural numbers which are both prime and
composite?
__
Arthur
sttsc...@tesco.net
Nathan wrote:
> Archimedes Plutonium wrote:
> > It is worth repeating, that in the direct proof method of
increasing
> > cardinality a prime factor search must be undertaken. But in the
> > indirect proof method, the moment P!+1 is formed, is the exact
moment
> > that the contradiction arises and that P!+1 is necessarily a new
> > prime. So when one does the Indirect IP and starts to make a search
> > for a "prime factor", well the person has failed and made a invalid
> > proof.
>
> I have 2 (rhetorical) questions:
>
> 1. Why in the world would you think P!+1 must necessarily be a new
> prime? In reality, it scarcely ever is.
"In reality", P!+1 must be divisible by some prime, but in the
"possible world" AP is considering there are only
n primes. In this "possible world", P!+1 is not divisible by any prime
whatsoever as P!+1 = 1 mod pk. Since P!+1 is not equal to 1, P!+1 can't
be an integer. To say that P!+1 is prime or composite or divisible by
some other prime
greater than pn is irrelevant.
AP is, therefore, being "illogical" himself.
GCD(P!+1,p) = 1, for any p,p a prime => P!+1 = 1 => P!=0
Abraham Buckingham
<snip>
>
> Hardy's flaw is that Q is NECESSARILY prime and once Q is formed it is
> the
> end of the proof since it forms the contradiction that simultaneously
> not prime and prime; or, that Q is a larger prime than P.
Hardy never assumes that Q is prime, only that Q is not divisible by
any prime on his finite list of primes. Q is relatively prime to every
prime on his initial list and is constructed specifically with that in
mind but does not have to be a prime itself. Since Q cannot be divided
by any prime on the list, it's either a prime itself OR divisable by a
prime not on the list (since it can't be divided by any of the listed
primes). This proves that another prime number not on the list exists,
and contradicts the original assumption that only finitely many primes
exist since the choice of primes was arbitrary.
Archimedes Plutonium
>
> Suppose that we thought, wrongly, that the only prime numbers were
> 2,3,5,7,11,13,and 17 (which is the starting point of Hardy's proof ---
> we think that we have some finite list that enumerates all primes; the
> fact that I have chosen a ridiculously short initial segment of the
> sequence of primes is, of course, immaterial). Then simple calculation
> of Hardy's method gives
> Q = 2x3x5x7x11x13x17 + 1 = 510511 = 19(26869)
>
> Are you claiming that there are natural numbers which are both prime and
> composite?
>
> __
> Arthur
You mistake method with specific example.
In the *indirect method* yes of course 510511 is necessarily prime because
your total universe space of primes is your assumption that 2,3,5,7,11,13,17
were all the primes that existed. In that restricted space 510511 is
necessarily prime and larger than 17 and hence the proof.
In the *indirect method* let us assume 3,5 was the total universe of primes
and thus P!+1 yields 16. It is necessarily prime because the assumption was
that 3,5 were the only primes in existence.
Now, shifting to the *direct method* which Euclid's ancient proof was a
direct method of increasing set cardinality. Given the set {2,3,5,7,11,13,17}
then in the proof 510511 is no longer necessarily prime and a *prime factor*
search must be added into the proof yielding two new primes of 19 and
possibly 26869??. The set {3,5} also requires a prime-factor search to yield
the new prime not in the list which is "2".
So in the Indirect Method there is never a search for a prime-factor because
P!+1 is necessarily prime. In the Direct Method where every set of primes is
increased with a new prime not in the set itself requires a *prime factor
search*.
You cannot mix the two.
The two mistakes Hardy made was that he thought Euclid's proof was Indirect
when it was Direct all along and secondly, Hardy does a prime-factor search
in a Indirect method.
If Infinitude of Primes proof were written out in solely the language of
Symbolic Logic then it would be evident that a "Prime Factor Search" arises
only in the Direct set theory method.
sttsc...@tesco.net
Abraham Buckingham wrote:
> Archimedes Plutonium <a_plu...@iw.net> wrote in message
news:<4249637C...@iw.net>...
> <snip>
> >
> > Hardy's flaw is that Q is NECESSARILY prime and once Q is formed
it is
> > the
> > end of the proof since it forms the contradiction that
simultaneously
> > not prime and prime; or, that Q is a larger prime than P.
>
> Hardy never assumes that Q is prime, only that Q is not divisible by
> any prime on his finite list of primes. Q is relatively prime to
every
> prime on his initial list and is constructed specifically with that
in
> mind but does not have to be a prime itself.
By assumption Q can't be prime itself or divisible by a prime
not on the list . At this stage of
the proof all you know is that Q is composite
because Q > pn, the last prime.
Since Q cannot be divided
> by any prime on the list, it's either a prime itself OR divisable by
a
> prime not on the list (since it can't be divided by any of the listed
> primes).
If you are assuming there is "no other prime on the list"
how can you say that Q is divisible by a prime not
on the list. You are after all assuming that all primes that exist are
on the list. If all primes that exist are on the list then there is no
other prime which can divide Q.
You are still thinking as if there were an
inexhaustible supply of primes available.
If 43 is the last prime is Q= 2*3*5*7*.......*43+1 prime or divisible
by a prime >43?
Not divisible by 2
Not divisible by 3
.
.
Not divisible by 43
Therefore prime or divisbile by a prime > 43.
But as your assumption is that 43 is the last
prime you cannot simply state that there exists some other
prime that divides Q
If Q is not divisible by any of the primes on the list
then Q is not an integer. It is a theorem that every Z>1 is divisible
by some prime, but Q is not divisible by any prime at all.
You have got to take your assumption that the list
contains all the primes that exist seriously. If Q is not an integer it
can't be prime, composite or a unit or even exist.
This proves that another prime number not on the list exists,
> and contradicts the original assumption that only finitely many
primes
> exist since the choice of primes was arbitrary.
No, Q does not exist, therefore the original
assumption was wrong. There are an infinitude of
primes and Q is therefore itself a prime not on the list or divisible
by some prime not on the list.
As n is arbitrary, there is an inexhaustible
supply of primes.
You don't have to show the existence of
"another prime" to prove an infinitude of primes exist.
Take Kummer's proof
If pn is the last prime
and Q = 2*3*...*pn +1 then Q is compsoite (Q>pn)
As only n primes exist
2*3*...*pn +1 = (2^a)(3^b)*....((pn)^z)
This implies that some pk divides 1 - a contradiction
So pn is not the last prime.
tomle...@hotmail.com
there are two possibilities.
1. you have no clue what you are talking about
2. you are a troll
i'm erring on the side of 2.
sttsc...@tesco.net
Archimedes Plutonium wrote:
> Tue, 29 Mar 2005 10:44:35 -0500 Arthur Fischer wrote:
> In the *indirect method* yes of course 510511 is necessarily prime
because
> your total universe space of primes is your assumption that
2,3,5,7,11,13,17
> were all the primes that existed. In that restricted space 510511 is
> necessarily prime and larger than 17 and hence the proof.
If 2,3,5,7,11,13,17 are all the primes that exist
and GCD(51051, p) =1, p in PRIME ={2,3,5,7,11,13,17)
then you have a contradiction. 51051 is an integer >
1 that is not divisible by any prime, but it is a theorem
that every integer >1 is divisible by at least one prime.
You are not following through on your own logic.
If there is no point looking for a prime greater
than the last prime on your list, then it does
not matter whether this prime is W+1 or a divisor of W+1
GCD(x, p) =1, for every prime p => x =1
Are you saying this statement is false ?
Under the assumption that 2,3,5,7,11,13,17 are all the primes that
exist, then 51051 can be anything a unit,
a composite, a prime, a non-integer. It is false to claim that
it is necessarily prime.
Timothy Little
> Suppose that we thought, wrongly, that the only prime numbers were
> 2,3,5,7,11,13,and 17
> Are you claiming that there are natural numbers which are both prime
> and composite?
In a subsystem based on classical logic that has inconsistent
premises, yes. In such a system every proposition is true, including
all contradictions, and a natural number that is both prime and
composite in particular.
We're still in the hypothetical subsystem which has the premise "Let
us suppose that it does, and that 2,3,5,..., P is the complete series
(so that P is the largest prime)".
In that subsystem, Q *is* prime. And composite. And even, and odd,
and equal to 0. And is a zebra. The absurdity of this subsystem is
what motivates us to strike out the premise that started it all.
- Tim
sttsc...@tesco.net
"Hypothetical subsystem" is a useful
expression.
I would have thought that if you are in a particular
hypothetical subsystem you would not be allowed
to use assumptions that are not considered to belong to that
hypothetical subsystem.
Once you have got to the
Q = 2*3*....*pn +1 stage of the proof
You can demonstrate that no pk (k = 1 to n)
divides Q. To simply claim that Q must be a prime not
on the list or divisible by a prime not on the list
seems to be contradictory as you have not yet shown that
the original assumption was false and have simply
"jumped out" of the hypothetical subsystem to a system
where there are more primes than you have assumed.
The fact that Q is an integer >1 and has no
prime divisors is an immediate contradiction and you need not go any
further to show the original assumption was false.
I suppose you could say no prime less than Q divides Q therefore
Q is prime. This, however, does not imply that
Q is divisible by some prime <>Q and greater then pn.
Archimedes Plutonium
> Archimedes Plutonium wrote:
> > It is worth repeating, that in the direct proof method of increasing
> > cardinality a prime factor search must be undertaken. But in the
> > indirect proof method, the moment P!+1 is formed, is the exact moment
> > that the contradiction arises and that P!+1 is necessarily a new
> > prime. So when one does the Indirect IP and starts to make a search
> > for a "prime factor", well the person has failed and made a invalid
> > proof.
>
> I have 2 (rhetorical) questions:
>
> 1. Why in the world would you think P!+1 must necessarily be a new
> prime? In reality, it scarcely ever is.
It is a prime out of necessity because you assumed you had a complete list
of all the primes that ever existed. But the moment you form P!+1 you
create a new one.
>
>
> 2. Even if this idea of yours had any validity, why would adding an
> unnecessary step render a valid proof invalid?
Because adding the Prime Factor Search as Hardy does to P!+1, Hardy never
produces a new prime does he. It is essential to produce a new prime in
order to proof the statement. This is what I called "the only prime
candidate" in the Indirect proof method.
Let me show you by example which is faster.
Direct Method: Increase the cardinality of any finite set {3,5} form P!+1
which is 16. Now apply the prime factor search yielding the prime 2 out of
16. Thus you produced a new prime not on your list. And since *any* finite
set is increased in cardinality implies it is infinite.
Indirect Method: Suppose {3,5} are the only primes. Form P!+1 which is 16.
It is necessarily prime because 3 and 5 are all the primes that ever exist
but when they divide into P!+1 they leave a remainder of 1, hence P!+1 is
prime and larger than 5. Contradiction therefore proof.
Now notice something important Nathan. In both methods we actually
produced a ***new prime*** not on the original list.
Now mix the prime factor search with the Indirect Method just as Hardy
does. You end up with a statement that either P!+1 is prime or there is a
prime factor not on the list. So you never really produce a new prime do
you.
The whole objective, the whole purpose of the proof is to produce a new
prime not on the list.
So when you mix a prime factor search with Indirect Method you end up
without a new prime at the end but a flimsy lousy statement that either
P!+1 or a prime factor is not accounted for. The point of the proof was to
yield the actual new prime.
So you see, when you mix the method of Direct of a prime factor search
with that of the Indirect Method you end up with not a new prime but some
unproven claim that perhaps a new prime exists.
You see, in the Direct Method the prime factor search yields a brand new
prime not on the list and in the example above it is "2". And the Indirect
Method yields a brand new prime not on the list and the example above
gives 16. But Hardy's attempt yields at the end not a new prime but a
statement that either P!+1 is prime or look for a prime-factor. So Hardy
gets nowhere. The end of Hardy's proof is not much different than the
beginning of the proof where we went out to look for a new prime.
That is why Hardy's attempt is invalid. It does not prove Infinitude of
Primes because it fails to pinpoint a new prime. That is why P!+1 is
essential in being a prime because we cannot find a new prime as a
"prime-factor" in the Indirect Method.
Hope that clears it up for you.
Archimedes Plutonium
> Archimedes Plutonium <a_plu...@iw.net> wrote in message news:<4249637C...@iw.net>...
> <snip>
> >
> > Hardy's flaw is that Q is NECESSARILY prime and once Q is formed it is
> > the
> > end of the proof since it forms the contradiction that simultaneously
> > not prime and prime; or, that Q is a larger prime than P.
>
> Hardy never assumes that Q is prime, only that Q is not divisible by
> any prime on his finite list of primes. Q is relatively prime to every
> prime on his initial list and is constructed specifically with that in
> mind but does not have to be a prime itself. Since Q cannot be divided
> by any prime on the list, it's either a prime itself OR divisable by a
And that is the fatal flaw that makes Hardy's attempt invalid and makes MathWorld attempt
invalid.
You see, Hardy and MathWorld never produce a new prime not on the list.
They end up with what they started with-- a claim that a new prime exists.
So when you add a Prime Factor search in the Indirect Method you end up with not producing a new
prime.
Examples:
Valid Direct Method: {3,5} form P!+1 yields 16 and with a Prime-Factor search
yields a new prime of "2". And since *any* finite set of primes can be
increased in cardinality thence infinite.
Valid Indirect Method: Assume {3,5} are all the primes, form P!+1 yields 16.
And 16 is prime because you assumed 3,5 were all the primes ever
existed. Contradiction because you have a new prime-- 16 and larger than
5.
Now let us show what Hardy and MathWorld did: Assume {3,5} are all the primes, form P!+1 yields
16. Now either 16 is prime or a prime factor is buried somewhere in there.
You see, Abraham, a valid Direct Method produces in hand a new prime of "2". The Valid Indirect
Method produces a new prime not on the list of "16" which is prime in a restricted space of
{3,5}
But note Abraham, that the Hardy and MathWorld attempts do not yield at the end a new prime.
Only a claim that a prime-factor exists buried. The point, the objective of the proof is to
actually generate and produce the new prime, not to mouth off another buried claim.
>
> prime not on the list (since it can't be divided by any of the listed
> primes). This proves that another prime number not on the list exists,
> and contradicts the original assumption that only finitely many primes
> exist since the choice of primes was arbitrary.
I hope the above clarifies to you why mixing a prime factor search in the Indirect Method ends
up with an Invalid proof.
Daniel W. Johnson
> The whole objective, the whole purpose of the proof is to produce a new
> prime not on the list.
No, the purpose is to show the inconsistency of a hypothesis that the
original (finite) list was complete. Once you've done that, attempting
to obtain further conclusions from the false hypothesis is pointless.
--
Daniel W. Johnson
pano...@iquest.net
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
Timothy Little
> I suppose you could say no prime less than Q divides Q therefore
> Q is prime.
That's the idea, yes. That path arrives at the contradiction even if
it may not be the most elegant or efficient path. There are many
valid paths to the contradiction (including some truly weird ones).
> This, however, does not imply that Q is divisible by some prime <>Q
> and greater then pn.
In a system that includes a contradiction, everything follows,
including that Q is both prime and divisible by some other prime.
Outside the hypothetical, nothing is known about Q because it is
isn't defined outside.
- Tim
Archimedes Plutonium
> Archimedes Plutonium <a_plu...@iw.net> wrote:
>
> > The whole objective, the whole purpose of the proof is to produce a new
> > prime not on the list.
>
> No, the purpose is to show the inconsistency of a hypothesis that the
> original (finite) list was complete. Once you've done that, attempting
> to obtain further conclusions from the false hypothesis is pointless.
> --
> Daniel W. Johnson
Rare that I encounter exchanges like this where the other person is arguing
in favor of me and in disfavor of his own thoughts.
I remember some verbal exchanges where the one person trys to put down
another but instead it backfires and lifts the other and puts down their
ownself. This is what has happened above.
If the purpose is to show the inconsistency, then, Daniel, can an
inconsistency arise if no new prime is forthcoming? Of course not. First you
have to produce the new prime and then the next step away is to announce the
inconsistency.
So if the Prime Factor search is installed into the Indirect method for
Infinitude of Primes such as what MathWorld and Hardy have done, and it gets
to the stage where "either P!+1 is prime or there exists a prime factor"
does that look like a new prime was delivered? A new prime was generated? Of
course not.
That is why I say there can be not prime-factor-search in the Indirect
Method and those proofs are invalid, because they never generate a new
prime.
The endpoint of Hardy and MathWorld's attempt at Infinitude of Primes is not
much closer to a "new prime" at the end of the proof than it is at the very
beginning.
You see Daniel, to get an inconsistency you first have to produce a new
prime. When MathWorld or Hardy produce "either P!+1 is prime or has a prime
factor" then both MathWorld and Hardy fail to produce, fail to generate a
new prime.
But when I say P!+1 is necessarily prime and never bother to mention prime
factor, can you see Daniel that I have really generated a new prime. I have
coughed up a new prime, delivered it all packaged and brand new.
So I can then go to the next step of declaring an inconsistency. But
MathWorld and Hardy and Daniel W. Johnson still has not generated a new
prime. They are still in the weeds with "either P!+1 is prime or some
unknown prime factor". They cannot move onto the next step of declaring
inconsistency simply because no new prime was ever produced.
You can forgive me for laughing at your complaint, Daniel, because seldom
has anyone complained about my work whilst supporting it so much and
disfavoring your own position.
darren...@netscape.net
statements and the proof can be mastered in an hour by any intelligent
reader, however slender his mathematical equipment. "
Darren
Archimedes Plutonium
darren...@netscape.net wrote:
Very nice observation Darren. I once owned the book to see if that is
what Hardy goes further on to say. My memory is not that bad yet and I
believe that is what Hardy said about Infinitude of Primes proof. I think
he made the comment before he dug into the proof itself.
Obviously that needs to be updated and revised to our modern times into
something like this:
The statements and comments and history written about the Infinitude of
Primes proof never became clear until about 2005. So this proof,
Infinitude of Primes took about 3 thousand years to get its flaws and
blemishes removed. And it still needs to go through a Symbolic Logic mill
and written out in full in Symbolic Logic.
One of the main reasons of the false history written about Euclid's proof
in declaring it an "indirect method" when it never was such, is that the
history of mathematics has never had a dominant one person figure to
clean out the house of mathematics that physics has had. Physics history
shows us that one dominant person in charge of the whole community--
Kepler, Galileo, Newton, Faraday, Maxwell keeps the community focused.
Whereas an unfocused community such as mathematics has hundreds of
sub-communities doing some silly things such as saying that Euclid's
Infinitude of Primes was a Indirect Method when it never was.
Mathematics is the worst of the sciences because it is run by old men
like a clubhouse who care more for their book royalties than they care
about mathematics. And where noone focuses the entire community onto what
they really should be concentrated upon.
sttsc...@tesco.net
Not at all. The amusing thing is that you
accuse others of "searching for a prime" and then
are inconsistant yourself in a similar way.
If you can find any contradiction at all, this
means that the premise "There is a last prime, pn"
is false. So there is no last prime.
GCD(2*3*...*(pn)+1,q)= 1 for every q, q prime implies
that 2*3*...*(pn)+1 = 1 - an inconsistency.
There is no need to look for any "new prime".
Are you saying that this argument is invalid ?
If so why ?
sttsc...@tesco.net
Timothy Little wrote:
> sttsc...@tesco.net wrote:
> > I suppose you could say no prime less than Q divides Q therefore
> > Q is prime.
>
> That's the idea, yes. That path arrives at the contradiction even if
> it may not be the most elegant or efficient path. There are many
> valid paths to the contradiction (including some truly weird ones).
> > This, however, does not imply that Q is divisible by some prime <>Q
> > and greater then pn.
>
> In a system that includes a contradiction, everything follows,
> including that Q is both prime and divisible by some other prime.
Yes, if a train of reasoning implies 1 =2 then this implies anything.
But surely you have to reason correctly
to reach a contradiction in the first place.
For example,in the hyothetical system, it does not directly follow
from the statement "2,3, ,pn do not divide Q= 2*3*...*pn +1" that there
is a prime less than Q and > pn which divides Q.
While on the other hand "2,3, ,pn do not divide Q= 2*3*...*pn +1"
implies Q is the only proper divisor of Q and so Q is prime" does.
Abraham Buckingham
> 29 Mar 2005 12:13:56 -0800 Abraham Buckingham wrote:
>
> > Archimedes Plutonium <a_plu...@iw.net> wrote in message news:<4249637C...@iw.net>...
> > <snip>
> > >
> > > Hardy's flaw is that Q is NECESSARILY prime and once Q is formed it is
> > > the
> > > end of the proof since it forms the contradiction that simultaneously
> > > not prime and prime; or, that Q is a larger prime than P.
> >
> > Hardy never assumes that Q is prime, only that Q is not divisible by
> > any prime on his finite list of primes. Q is relatively prime to every
> > prime on his initial list and is constructed specifically with that in
> > mind but does not have to be a prime itself. Since Q cannot be divided
> > by any prime on the list, it's either a prime itself OR divisable by a
>
> And that is the fatal flaw that makes Hardy's attempt invalid and makes MathWorld attempt
> invalid.
>
> You see, Hardy and MathWorld never produce a new prime not on the list.
> They end up with what they started with-- a claim that a new prime exists.
>
> So when you add a Prime Factor search in the Indirect Method you end up with not producing a new
> prime.
I agree, they didn't produce the new prime explicitly, but they didn't
have to. This step he left up to the read to fill in the blank as to
why another prime not on the list must exist. After all 'A
Mathamtician's Apology' isn't ment to be a rigorous treatment of the
infinitude of primes.
Lets think for a moment about the fundemental theorem of arithmatic,
and the definition of a prime number. We know that *every* number must
either be composite or prime because of these things. Now consider a
finite list of primes. {p_1, p_2, p_3,..., p_n} with Q constructed to
be the product of these primes + 1 as usual. Now we know for a fact
that Q is a number - that means either Q is prime or Q is composite.
Since Q > p_k for all k less then or equal to n and we assumed that
our list has all the primes we know Q isn't on our list so naturally
can't be prime. Unfortunately, we also know that Q isn't divisable by
any of our list of primes either, which means it's only divisable by Q
and 1, but that would mean that Q is prime - which we've already
figure out it's not. So Q isn't prime, and Q isn't composite, I'm
forced to conclude that Q isn't a number - which of course is absurd
since it's construction assures us that it is. Our assumption that
only finitely many primes exist led us to conclude that a number is
not a number a contradiction - so we reject the assumption. Since we
can conclude that the list of primes is *not* finite the only other
choice is to suggest that the list is infinite since after all
'infinite' is defined as being 'not finite'.
> Examples:
>
>
> Valid Direct Method: {3,5} form P!+1 yields 16 and with a Prime-Factor search
> yields a new prime of "2". And since *any* finite set of primes can be
> increased in cardinality thence infinite.
>
> Valid Indirect Method: Assume {3,5} are all the primes, form P!+1 yields 16.
> And 16 is prime because you assumed 3,5 were all the primes ever
> existed. Contradiction because you have a new prime-- 16 and larger than
> 5.
>
> Now let us show what Hardy and MathWorld did: Assume {3,5} are all the primes, form P!+1 yields
> 16. Now either 16 is prime or a prime factor is buried somewhere in there.
>
> You see, Abraham, a valid Direct Method produces in hand a new prime of "2". The Valid Indirect
> Method produces a new prime not on the list of "16" which is prime in a restricted space of
> {3,5}
>
> But note Abraham, that the Hardy and MathWorld attempts do not yield at the end a new prime.
> Only a claim that a prime-factor exists buried. The point, the objective of the proof is to
> actually generate and produce the new prime, not to mouth off another buried claim.
I hope I've exposed the buried claim - which is that all numbers are
either prime or composite and Q is a number. That's all the
information we need to generate the contradiction and therefore lead
us to reject the initial assumption.
>
> >
> > prime not on the list (since it can't be divided by any of the listed
> > primes). This proves that another prime number not on the list exists,
> > and contradicts the original assumption that only finitely many primes
> > exist since the choice of primes was arbitrary.
>
> I hope the above clarifies to you why mixing a prime factor search in the Indirect Method ends
> up with an Invalid proof.
I'm very unclear as to why you feel it's a proof rather then an brief
outline of a simple mathamatical idea. Hardy certainly wasn't writing
this with rigor in mind, I think you're taking his comments out of
context. It's not ment to be a rigorous proof so we shouldn't be
suprised that he doesn't meet those standards in a meta-commentary on
the subject. Moreso, I don't see why you're still stuck on the idea
you have to explicitly generate a new prime number in order for the
argument to work.
Bill Dubuque
>
> For example,in the hyothetical system, it does not directly follow
> that there is a prime factor p of Q satisfying pn < p < Q
Sure it does:
Since Q > 1 it has a prime factor p.
p < Q since Q is not prime, being greater than pn
p > pn since all of the primes <= pn aren't factors of Q
(2,3,..,pn are all the primes <= pn, being all the primes)
--Bill Dubuque
Bill Dubuque
>
> Under the assumption that 2,3,5,7,11,13,17 are all the primes that
> exist, then 51051 can be anything a unit, > a composite, a prime,
> a non-integer. It is false to claim that it is necessarily prime.
Ni, it's true: an integer > 1 with no smaller prime factors is prime.
--Bill Dubuque
sttsc...@tesco.net
Abraham Buckingham wrote:
> Archimedes Plutonium <a_plu...@iw.net> wrote in message
news:<424A1D09...@iw.net>...
> > 29 Mar 2005 12:13:56 -0800 Abraham Buckingham wrote:
> >
> > > Archimedes Plutonium <a_plu...@iw.net> wrote in message
news:<4249637C...@iw.net>...
> > > <snip>
>
> and the definition of a prime number. We know that *every* number
must
> either be composite or prime because of these things.
Are you saying that 1 is prime or composite ?
If 1 is prime every number > 1 is composite.
If 1 is composite there must some prime that
divides 1 and every number >1 is composite.
Now consider a
> finite list of primes. {p_1, p_2, p_3,..., p_n} with Q constructed to
> be the product of these primes + 1 as usual. Now we know for a fact
> that Q is a number - that means either Q is prime or Q is composite
or a unit ?
> Since Q > p_k for all k less then or equal to n and we assumed that
> our list has all the primes we know Q isn't on our list so naturally
> can't be prime.
Yes, it's composite if every number is either prime
or composite. It's not prime by assumption.
Unfortunately, we also know that Q isn't divisable by
> any of our list of primes either,
If you have a composite number that has no
prime divisors whatsoever you now know that Q can't
exist. Above you say that every number
is prime or composite, this means that every
number must have some prime divisor.
But you conclude that Q has no prime divisors
whatsoever.
which means it's only divisable by Q
> and 1,
Norwithstanding the status of 1, you have
just shown that Q is a composite with no prime divisors.
- a contradiction.
You now know that Q doesn't exist
and so can have any contrdictory property you require.
Composite and prime, integer and irrational etc.
sttsc...@tesco.net
It's true that an integer >1 with no smaller prime factors
is prime but that is not the claim I made.
The only integer > 0 that has no prime factors
is 1, i.e 1 is not divisible by any prime. There is no number >1 with
no prime factors. 51051 <> 0, there is no prime that
divides 51051, so 51051 must equal 1.
Daniel W. Johnson
> So if the Prime Factor search is installed into the Indirect method for
> Infinitude of Primes such as what MathWorld and Hardy have done, and it gets
> to the stage where "either P!+1 is prime or there exists a prime factor"
> does that look like a new prime was delivered? A new prime was generated? Of
> course not.
What would be the point? Hardy identifies a dead end and marks it as
such. You insist on measuring the length of the dead end.
Sevenhundred Elves
Let's see if I remember it right:
Take the list of all prime numbers below and including any prime number
n. Add one to their product. (The resulting number, A, is then greater
than n.)
The resulting number, A, is not divisible by any number on the list
because of the added 1. This is why we added the 1.
Now there are no more than two possibilities for A: It either is prime,
or it isn't.
If A is prime, it is a greater prime than n.
If A is not prime, it is a composite number, and has a prime factor, B.
Since the primes on the original list can't be divisors of A (we made
sure of this by adding the 1), the prime B can not be on the list. B
must therefore be a greater prime than n.
In either case, we end up with a prime greater than any prime on the
list. Since we specified at the outset that n could be any prime, we
must conclude that for any given prime there must exist a larger prime.
This is what I learned in school, and I can't see what's wrong with it.
Is there really a flaw in that proof? Perhaps I'm too naďve.
S.
Archimedes Plutonium
>
> Let's see if I remember it right:
>
> Take the list of all prime numbers below and including any prime number
> n. Add one to their product. (The resulting number, A, is then greater
> than n.)
>
> The resulting number, A, is not divisible by any number on the list
> because of the added 1. This is why we added the 1.
>
> Now there are no more than two possibilities for A: It either is prime,
> or it isn't.
>
> If A is prime, it is a greater prime than n.
>
> If A is not prime, it is a composite number, and has a prime factor, B.
> Since the primes on the original list can't be divisors of A (we made
> sure of this by adding the 1), the prime B can not be on the list. B
> must therefore be a greater prime than n.
>
> In either case, we end up with a prime greater than any prime on the
> list. Since we specified at the outset that n could be any prime, we
> must conclude that for any given prime there must exist a larger prime.
>
> This is what I learned in school, and I can't see what's wrong with it.
> Is there really a flaw in that proof? Perhaps I'm too naīve.
>
> S.
Well a huge crowd is going to overwhelm me with every one of their tiny nuances
and with their flawed and failed attempts. So I have to streamline the reply to
show them where they are flawed, naive and invalid.
Symbolic Logic Write Up of Infinitude of Primes Proof:
Using the examples as model
Valid Direct Method:
{3,5} form P!+1 yields 16
and with a *Prime-Factor-Search*
yields a new prime of "2".
And since *any* finite set of primes can be
increased in cardinality thence infinite.
Valid Indirect Method:
Assume {3,5} are all the primes,
form P!+1 yields 16.
And 16 is necessarily prime because in this
restricted universe-space of prime you assumed 3,5 were all the primes ever
existed.
But a new prime 16 is produced and larger than the largest prime 5
Contradiction thence primes are infinite.
Now let us show what mixing a *Prime Factor Search* into the
Indirect Method causes. This is what Hardy and MathWorld did:
Assume {3,5} are all the primes,
form P!+1 yields 16.
Now maybe 16 is a new prime or maybe the
new prime is somewhere else for we have not
pinpointed where this new prime is hanging out
can we say there is a contradiction even though
we have not pinpointed the new prime?
No, because in a Symbolic Logic write off
there is a line that says
P_k is precisely a new prime that leads to
the next step of the Contradiction step
Contradiction thence primes are infinite.
The trouble with mixing the PrimeFactorSearch into the Indirect Method is that it
fails to bridge the steps in the Symbolic Logic Write Up of Infinitude of Primes
Proof of a step that says K is precisely this new prime and is larger than any
prime in P. By generating and producing this specific new prime allows the
Symbolic Logic write up to go to the next step of Contradiction.
So, sorry to say, yes you are naive by thinking that a Prime Factor Search in the
Indirect Method will work.
Archimedes Plutonium
Abraham Buckingham wrote:
>
>
> I agree, they didn't produce the new prime explicitly, but they didn't
> have to. This step he left up to the read to fill in the blank as to
> why another prime not on the list must exist. After all 'A
> Mathamtician's Apology' isn't ment to be a rigorous treatment of the
> infinitude of primes.
>
Wrong. The ultimate write up of Infinitude of Primes proof is a Symbolic Logic line by line, step by
step write up. And the reason adding the Prime Factor Search into the Indirect Method fails is
because to reach the step of Contradiction, the preceding step is to state precisely and specifically
what the new prime that causes the contradiction.
Hardy and MathWorld renditions leave it as such:
....... either P!+1 or some other number not yet found will contradict the original assumption...
Symbolic Logic works with rigor and that is not rigorous.
Well a huge crowd is going to overwhelm me with every one of their tiny nuances
and with their flawed and failed attempts. So I have to streamline the reply to
show them where they are flawed, naive and invalid.
####
Symbolic Logic Write Up of Infinitude of Primes Proof:
Using these examples as model
Valid Direct Method:
{3,5} form P!+1 yields 16
and with a *Prime-Factor-Search*
yields a new prime of "2".
And since *any* finite set of primes can be
increased in cardinality thence infinite.
Valid Indirect Method:
Assume {3,5} are all the primes,
form P!+1 yields 16.
And 16 is necessarily prime because in this
restricted universe-space of prime you assumed 3,5 were all the primes ever
existed.
But a new prime 16 is produced and larger than the largest prime 5
Contradiction thence primes are infinite.
Now let us show what mixing a *Prime Factor Search* into the
Indirect Method causes. This is what Hardy and MathWorld did:
Assume {3,5} are all the primes,
form P!+1 yields 16.
Now maybe 16 is a new prime or maybe the
new prime is somewhere else for we have not
pinpointed where this new prime is hanging out
can we say there is a contradiction even though
we have not pinpointed the new prime?
No, because in a Symbolic Logic write up
there is a line that says
P_k is precisely a new prime that leads to
the next step of the Contradiction step
Contradiction thence primes are infinite.
####
The trouble with mixing the PrimeFactorSearch into the Indirect Method is that it
fails to bridge the steps in the Symbolic Logic Write Up of Infinitude of Primes
Proof of a step that says K is precisely this new prime and is larger than any
prime in P. By generating and producing this specific new prime allows the
Symbolic Logic write up to go to the next step of Contradiction.
So, sorry to say, yes you are naive by thinking that a Prime Factor Search in the
Indirect Method will work.
Archimedes Plutonium
sttsc...@tesco.net
Archimedes Plutonium wrote:
> Wed, 30 Mar 2005 16:04:46 GMT Sevenhundred Elves wrote:
>
> Valid Indirect Method:
> Assume {3,5} are all the primes,
> form P!+1 yields 16.
Yes, you have got that right
3*5+1 does equal 16.
Is 16 in the the set {3,5} ?
I can't see it anywhere.
So 16 can't be prime.
It must therefore be composite or 1
Does 3 divide 16 ? No
Does 5 divide 16? No.
16 is composite or 1 but has no prime divisors.
16 is therefore equal to 1 as every composite has a prime divisor.
So 16 is necessarily 1.
As a bonus it also follows that
every number that is not a multiple of
3 or 5 is equal to 1.
Abraham Buckingham
You continue to point out the obvious.
> Unfortunately, we also know that Q isn't divisable by
> > any of our list of primes either,
>
> If you have a composite number that has no
> prime divisors whatsoever you now know that Q can't
> exist. Above you say that every number
> is prime or composite, this means that every
> number must have some prime divisor.
> But you conclude that Q has no prime divisors
> whatsoever.
>
> which means it's only divisable by Q
> > and 1,
>
> Norwithstanding the status of 1, you have
> just shown that Q is a composite with no prime divisors.
> - a contradiction.
> You now know that Q doesn't exist
> and so can have any contrdictory property you require.
> Composite and prime, integer and irrational etc.
Why do you insist on re-posting what I've already pointed out? You
realize this is an explaination not a proof correct?
sttsc...@tesco.net
Only because you continue to ignore the obvious.
It is not true that evrry number is prime or composite.
1 is not a prime nor is it composite.
At least according to the usual definitions.
If you are defining these concepts in
a non-standard way, you should say so.
> > Unfortunately, we also know that Q isn't divisable by
> > > any of our list of primes either,
> >
> > If you have a composite number that has no
> > prime divisors whatsoever you now know that Q can't
> > exist. Above you say that every number
> > is prime or composite, this means that every
> > number must have some prime divisor.
> > But you conclude that Q has no prime divisors
> > whatsoever.
> >
> > which means it's only divisable by Q
> > > and 1,
> >
> > Norwithstanding the status of 1, you have
> > just shown that Q is a composite with no prime divisors.
> > - a contradiction.
> > You now know that Q doesn't exist
> > and so can have any contrdictory property you require.
> > Composite and prime, integer and irrational etc.
>
>
> Why do you insist on re-posting what I've already pointed out? You
> realize this is an explaination not a proof correct?
It's not a very clear explanation if you are claiming
that 1 is a prime number or a composite number.
A correct proof would be interesting.
Kummer's proof which I gave earlier from memory seems
to be much more concise and free from circuitous or extraneous
deductions than any I've seen so far.
It is "indirect" and does not rely on showing that
"another prime exists" to prove that there are an infinitude of primes
as AP claims.
Nathan
sttsc...@tesco.net wrote:
> Once you have got to the
> Q = 2*3*....*pn +1 stage of the proof
> You can demonstrate that no pk (k = 1 to n)
> divides Q. To simply claim that Q must be a prime not
> on the list or divisible by a prime not on the list
> seems to be contradictory as you have not yet shown that
> the original assumption was false and have simply
> "jumped out" of the hypothetical subsystem to a system
> where there are more primes than you have assumed.
This is enough. If you assumed there are only n
primes, then can prove there exist at least n+1, this is
a contradiction. It means the assumption was false.
Barb Knox
Sevenhundred Elves <sevenh...@elves.invalid> wrote:
[snip Archie's poo]
>Let's see if I remember it right:
>
>Take the list of all prime numbers below and including any prime number
>n. Add one to their product. (The resulting number, A, is then greater
>than n.)
>
>The resulting number, A, is not divisible by any number on the list
>because of the added 1. This is why we added the 1.
>
>Now there are no more than two possibilities for A: It either is prime,
>or it isn't.
>
>If A is prime, it is a greater prime than n.
>
>If A is not prime, it is a composite number, and has a prime factor, B.
>Since the primes on the original list can't be divisors of A (we made
>sure of this by adding the 1), the prime B can not be on the list. B
>must therefore be a greater prime than n.
>
>In either case, we end up with a prime greater than any prime on the
>list. Since we specified at the outset that n could be any prime, we
>must conclude that for any given prime there must exist a larger prime.
>
>This is what I learned in school, and I can't see what's wrong with it.
>Is there really a flaw in that proof? Perhaps I'm too naïve.
There's no flaw, but it is a bit more complex than necessary -- there is no
need to explicitly consider two separate cases if you use the following
Lemma:
Every n > 1 has at least 1 prime factor.
Now, assuming there is a non-empty finite list of primes, the number A by
construction is > 1 and has *no* prime factors. Contradiction.
Proof of lemma (using well-ordering):
Assume there is at least one number > 1 with no prime factors. Let n be the
smallest of these. Then n can not be prime (since it would thereby have
itself as a prime factor), so it is some composite j*k (with both factors >
1). Then j has some prime factor p (since n is the _smallest_ number > 1
with no prime factors, and j < n because k > 1). But if p is a factor of j
then it is also a factor of j*k, i.e. of N. Contradiction.
If you dislike indirect proofs, there is an equivalent direct proof of the
lemma via strong induction, using the hypothesis that every j < n and > 1
has at least 1 prime factor. Details left to the reader...
--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------
abe.buc...@gmail.com
context it's clear that Q can never be 1 so 1's status as prime or
composite is irrelevent. Once again let me be pefectly clear - rigor is
not my goal and I'm *intentionally* overlooking the obvious because
it's just that - obvious. I'm not claiming anything at all about the
number 1 - I'm making claims about the number Q which by construction
is always greater then 1.
Your proof however appealing to you personally is cryptic and
obviously AP is struggling with the idea or he wouldn't have brought
the question up in the first place. Proofs are only useful to those who
can follow them, while wishy-washy rhetoric can often provide enough of
an intuitive understanding so they can fill in the gaps and actually
understand the proof.
Justin Young
news:d2f7jt$1gq$1...@lust.ihug.co.nz...
> In article <y6A2e.20830$d5.1...@newsb.telia.net>,
> Sevenhundred Elves <sevenh...@elves.invalid> wrote:
> [snip Archie's poo]
>
>>Let's see if I remember it right:
>>
>>Take the list of all prime numbers below and including any prime number
>>n. Add one to their product. (The resulting number, A, is then greater
>>than n.)
>>
>>The resulting number, A, is not divisible by any number on the list
>>because of the added 1. This is why we added the 1.
>>
>>Now there are no more than two possibilities for A: It either is prime,
>>or it isn't.
>>
>>If A is prime, it is a greater prime than n.
>>
>>If A is not prime, it is a composite number, and has a prime factor, B.
>>Since the primes on the original list can't be divisors of A (we made
>>sure of this by adding the 1), the prime B can not be on the list. B
>>must therefore be a greater prime than n.
>>
>>In either case, we end up with a prime greater than any prime on the
>>list. Since we specified at the outset that n could be any prime, we
>>must conclude that for any given prime there must exist a larger prime.
>>
>>This is what I learned in school, and I can't see what's wrong with it.
>
> There's no flaw, but it is a bit more complex than necessary -- there is
> no
> need to explicitly consider two separate cases if you use the following
> Lemma:
>
> Every n > 1 has at least 1 prime factor.
If we want to zoom ahead, we can get this from the fact that Z is a UFD.
The existence of non-units in Z implies that there exists one prime in Z (we
need to show the
set of primes is non-empty)
Now, if we assume {p_1,...,p_n} is a set of primes. The new number
p_1*...*p_n + 1 is bigger than
one and thus not a unit and thus has a prime factor, and this factor is not
in the set. Thus, there is a set of
primes of size n+1. This implies that the set of primes must be infinite,
since no finite set contains all the primes.
Proof by contradiction is not required here. I'm sure you know that, I just
wanted to say this.
abe.buc...@gmail.com
line by line and logically flawed you should have no trouble pointing
out the flaw. As it stands you keep using a Straw Man by providing some
mock-up of the original arguement and then attacking it. P!+1 isn't
even mentioned in the proof if infinite primes at the MathWorld site,
it's mentioned as part of a different theorem that's left unproved on
the site.
Amadeus Train-Owwell Zirconium
as the post taht was posted immediately after yours notes,
and is not explicitly stated in Conway's or Hardy's,
the contradiction of finitude is based upon the contradiction
of P+1 being neither prime nor composite (although
it is easy to show that direct & indirect proofs are isomorphs;
see *Mathematics Magazine* of some issue; that book
of Guy and Conway is great, with lots of pictures).
also, although P+1 is usually not prime,
it will always have a prime factor larger than p;
is that correct, folks?
Archimedes Plutonium wrote:
> > Apparently, another error by Hardy is the statement "Finally, both
the
> > statements and the proof can be mastered in an hour by any
intelligent
> > reader, however slender his mathematical equipment. "
> sub-communities doing some silly things such as saying that Euclid's
> Infinitude of Primes was a Indirect Method when it never was.
--Chairman George and Strep Throat, back in print!
http://larouchepub.com
http://tarpley.net/bush12.htm
sttsc...@tesco.net
No the the point is that it is more
than enough, see post by Barb Knox.
An obvious contrdiction was "staring you
in the face".
Bill Dubuque
>Bill Dubuque wrote:
>> sttsc...@tesco.net wrote:
>>>
>>> Under the assumption that 2,3,5,7,11,13,17 are all the primes that
>>> exist, then 51051 can be anything a unit, a composite, a prime,
>>> a non-integer. It is false to claim that it is necessarily prime.
>>
>
> It's true that an integer >1 with no smaller prime factors
> is prime but that is not the claim I made.
>
> The only integer > 0 that has no prime factors is 1,
> i.e 1 is not divisible by any prime. There is no number >1
> with no prime factors. 51051 <> 0, there is no prime that
> divides 51051, so 51051 must equal 1.
I think we have different interpretations of "necessarily prime".
Precisely what did you mean by "necessarily" here?
I interpreted it to mean that 51051 can be proved prime in
the theory of the integers enlarged by adjoining the single
additional axiom that 2,3,..,17 are the only primes.
Perhaps you are interpreting it to mean that, additionally,
the enlarged theory is also consistent (it doesn't imply a
contradiction) so it doesn't imply 51051 is prime & composite.
--Bill Dubuque
Amadeus Train-Owwell Zirconium
Amadeus Train-Owwell Zirconium
to see whether they actually fully proved it (I assume, so .-)
sttsc...@tesco.net
abe.bucking...@gmail.com wrote:
> Once again my post isn't a proof, it's an explaination. From the
> context it's clear that Q can never be 1 so 1's status as prime or
> composite is irrelevent.
Ok. I see what you mean now.
But if you are proving that there is an infintude of primes
by contradiction,you must remember you are going into a logical
"parallel universe" where every definition and theorem except
"there is an infintude of primes" is true. In "our universe" it is
true that Q is greater than 1, but in the
"parallel universe" you may find that "there
is a last prime" can chnage things a lot and
Q might equal 1 or 0 or not really exist at all.
Once again let me be pefectly clear - rigor is
> not my goal and I'm *intentionally* overlooking the obvious because
> it's just that - obvious. I'm not claiming anything at all about the
> number 1 - I'm making claims about the number Q which by construction
> is always greater then 1.
Ok, Let us suppose that Q is greater than 1 in the "parallel
universe" and continue to suppose that there
is a last prime pn. This also means that Q is composite as Q >pn and pn
is the last prime. Obviously we are assuming that in
"our universe" the primes are inexhaustible.
In the "parallel universe", assuming that all the usual
laws of logic, aritmetic etc. apply, no prime whatsoever divides
Q. In "our universe" every composite >1 is divisible by a prime
so we have encountered our first "anomaly" in the
"parallel universe". As the only difference
between "our universe" and the "parallel universe"
is the existance or non-existance of a last prime. You
can conclude that in our universe there is
no last prime. You can go on to deduce
as many contradictions as you want, but that
would be a waste of time.
> Your proof however appealing to you personally is cryptic and
> obviously AP is struggling with the idea or he wouldn't have brought
> the question up in the first place. Proofs are only useful to those
who
> can follow them, while wishy-washy rhetoric can often provide enough
of
> an intuitive understanding so they can fill in the gaps and actually
> understand the proof.
I hope you now think the alternative
proof is less cryptic. It is a counterexanple
to AP's claim that you have to prove the
existance of "another prime" to show that
the list of primes has no final entry.
sttsc...@tesco.net
Bill Dubuque wrote:
> sttsc...@tesco.net wrote:
> >Bill Dubuque wrote:
> >> sttsc...@tesco.net wrote:
> >>>
> >>> Under the assumption that 2,3,5,7,11,13,17 are all the primes
that
> >>> exist, then 51051 can be anything a unit, a composite, a prime,
> >>> a non-integer. It is false to claim that it is necessarily prime.
> >>
> >> No, it's true: an integer > 1 with no smaller prime factors is
prime.
> >
> > It's true that an integer >1 with no smaller prime factors
> > is prime but that is not the claim I made.
> >
> > The only integer > 0 that has no prime factors is 1,
> > i.e 1 is not divisible by any prime. There is no number >1
> > with no prime factors. 51051 <> 0, there is no prime that
> > divides 51051, so 51051 must equal 1.
>
> I think we have different interpretations of "necessarily prime".
> Precisely what did you mean by "necessarily" here?
>
> I interpreted it to mean that 51051 can be proved prime in
> the theory of the integers enlarged by adjoining the single
> additional axiom that 2,3,..,17 are the only primes.
AP claims that the only way of showing that
"there is another prime on the list" is by showing that 2*3*..*17 +1
must be prime. This is what I
was taking "necessarily prime" to mean.
I see now that this could cause some confusion.
You can, of course, prove that 31051 is prime,
"is necessarily prime" in the standard interpretation
of the phrase.
> Perhaps you are interpreting it to mean that, additionally,
> the enlarged theory is also consistent (it doesn't imply a
> contradiction) so it doesn't imply 51051 is prime & composite.
No,absolutely not, all I was saying was that depending on how you
phrase the argument you can prove that 51051 is prime and copmosite or
composite and 1, etc.. You don't have to prove
that 51051 is prime to find a contradiction.
In fact, the proof seems to be more concise if you don't.
Bill Dubuque
>
> all I was saying was that depending on how you phrase
> the argument you can prove that 51051 is prime and copmosite
> or composite and 1, etc.. You don't have to prove
> that 51051 is prime to find a contradiction.
I agree with you on that. In fact I've emphasized this
here in many prior posts, e.g. see my two posts in the
thread [1], and the link there to a long prior thread
on this same topic.
It seems that the moral of these frequent threads is
proof by contradiction is too subtle for a layperson,
or at least for one with a plutonium enriched mind.
This is especially true when the contradiction lies
within such a basic theory as that of the integers.
Since this theory is so familiar, one contradiction
is easily seen to imply many others, and it's not
long before critical mass is reached and sci.math
is bombarded by large numbers of isodope articles.
Authors of mathematics popularizations should
avoid proofs by contradiction at all costs.
So I think Hardy owes us all a big apology
for sparking a permanent place in the world
that hosts plutonium-disfigured mathematics.
Uh-oh, I think I've reached a contradiction
in my theory of Hardy mindspace...nevermind.
--Bill Dubuque
[1] http://groups-beta.google.com/group/sci.math/msg/ed0fb0e88c8465a5
http://google.com/groups?selm=y8zk6p3...@nestle.csail.mit.edu
sttsc...@tesco.net
http://google.com/groups?selm=y8zk6p3...@nestle.csail.mit.edu
I get a "not found" message for this link.
Archimedes Plutonium
>
> There's no flaw, but it is a bit more complex than necessary -- there is no
> need to explicitly consider two separate cases if you use the following
> Lemma:
>
> Every n > 1 has at least 1 prime factor.
>
> Now, assuming there is a non-empty finite list of primes, the number A by
> construction is > 1 and has *no* prime factors. Contradiction.
>
> Proof of lemma (using well-ordering):
> Assume there is at least one number > 1 with no prime factors. Let n be the
> smallest of these. Then n can not be prime (since it would thereby have
> itself as a prime factor), so it is some composite j*k (with both factors >
> 1). Then j has some prime factor p (since n is the _smallest_ number > 1
> with no prime factors, and j < n because k > 1). But if p is a factor of j
> then it is also a factor of j*k, i.e. of N. Contradiction.
>
> If you dislike indirect proofs, there is an equivalent direct proof of the
> lemma via strong induction, using the hypothesis that every j < n and > 1
> has at least 1 prime factor. Details left to the reader...
>
> --
Amazing how I just moments ago finished posting the Courant attempt of IP
and to then see a post by Knox which claims to patch up the invalidity. Amazing
because the Knox patch is what Courant uses.
But does the Knox patch result in a valid proof of Euclid's Infinitude of
Primes? See comments below.
--- quoting WHAT IS MATHEMATICS? Richard Courant and Herbert Robbins
1941 page 22 ---
The proof of the infinitude of the class of primes as given by Euclid
remains a model of mathematical reasoning. It proceeds by the "indirect
method". We start with the tentative assumption that the theorem is
false. This means that there would be only a finite number of primes,
perhaps very many -- a billion or so -- or, expressed in a general and
non-committal way, n. Using the subscript notation we may denote these
primes by p1, p2, ...,pn. Any other number will be composite, and must
be divisible by at least one of the primes p1,p2,...,pn. We now produce
a contradiction by constructing a number A which differs from every one
of the primes p1, p2, ..., pn because it is larger than any of them,
and which nevertheless is not divisible by any of them. This number is
A = (p1xp2x...xpn) +1, i.e. 1 plus the product of what we supposed to
be all the primes. A is larger than any of the p's as a divisor. Since
our initial assumption that there is only a finite number of primes
leads to this contradiction, the assumption is seen to be absurd, and
hence its contrary must be true. This proves the theorem.
--- end quoting WHAT IS MATHEMATICS? Courant and Robbins ---
This is why it is so very important to have a Symbolic Logic proof of the
Infinitude of Primes where every step is crystal clear and leads to the
endresult of a streamlined and valid proof.
Does the Knox patch work? For one thing it creates a new class of number,
instead of prime and composite we thus have a number that has "no factors"
whereas a prime has at least one factor-- the number itself.
So for brevity sake, in the Symbolic Logic write up of IP when we form the new
number P!+1 and recognize that it is prime and a new and larger prime than any
on the assumed list we have the contradiction. The trouble with the Knox patch
is that it is much longer than simply recognizing P!+1 is a new prime, but as I
said above, it creates a new class of number different from composite which has
2 factors and prime which has 1 factor. This new class of number in the Knox
patch has 0 factors.
The Knox patch should be discarded simply because it adds on more than needed.
But discarded as flawed because it creates a new category of number 0-factored
number which has no reality.
This is why I find it so confusing as to why no-one in mathematics has
painstakingly written the Symbolic Logic proof of IP. It must be that all
mathematicians are too lazy, not that they are too dumb but too lazy.
sttsc...@tesco.net
Archimedes Plutonium wrote:
> Thu, 31 Mar 2005 10:01:31 +1200 Barb Knox wrote:
> But does the Knox patch result in a valid proof of Euclid's
Infinitude of
> Primes? See comments below.
The Knox proof is perfectly sound and
eliminates your circuitous attempts at a proof.
>> Does the Knox patch work? For one thing it creates a new class of
number,
> instead of prime and composite we thus have a number that has "no
factors"
> whereas a prime has at least one factor-- the number itself.
Nonsense. The key concept is "no prime factors".
1, a unit, has no prime factors and this is not a new class of number.
Until now you have never claimed it was prime or composite.
Presumably, you aceept that N, the set of naturals
N ={1, 2, 3,....} contains 1, every prime that exists
and every possible product of the primes that exist.
Accepting this leaves open whether there are infinitely
many primes or not
Assume that there are only m primes 2,3,5,.....,pm.
Then N must comprise 1, the m primes that exist and every product of
the m primes that exist, N can contain nothing else under this
assumption.
But obviously N contains Q =2*3*5*...Pm +1
as this is a positive integer.
Q is not equal to 1
Q is not equal to any of the m primes assumed to exist
Q is not divisible by any of the m primes assumed to exist.
So N contains Q and does not contain Q a contradiction.
You don't have to prove Q is prime.
Archimedes Plutonium
I worked it through my mind and have found the flaw of the Knox patch.
When my proof and Knox's which is the same as Courant's version reaches the point
of forming P!+1. Then upon the forming of this number my version would say that it
is prime and thus discharges the assumption.
What puts a fatal flaw into the Knox patch and the Courant attempt is that they
never affix P!+1 as prime and they immediately spill over into a Prime Factor
Search and because they ignore or skip over or deny that P!+1 is necessarily prime
given the initial assumption they commit the fatal flaw.
Both my proof and the Knox patch/ Courant form P!+1. Both proofs work from the
same definitions of what it means to be ** prime**. My definition of prime is the
same as what Euclid and what Knox and what Courant use. So once P!+1 is formed
then Euclid would have seen it as a new prime. I see it as a new prime. But why
would Knox with her patch and Courant not see it as a new prime? Knox and Courant
used the same definition of prime as I. So why does Knox and Courant deny and want
to skip over the fact that in this proof that P!+1 is necessarily prime thus
contradiction and end of proof.
I said the Symbolic Logic write up of Indirect method of IP will show that my
method has far fewer lines than does the Knox patch. But now I am able to pinpoint
the fatal flaw of the Knox patch and its invalidity. If you use the Knox patch
then as soon as you form P!+1 you skip over the fact that P!+1 is necessarily
prime and thus you deny the original definition of what it means to be prime at
the start of the proof. So you deny the definition of primeness which makes this
attempt invalid.
Metaphor analogy to the Knox patch. Suppose instead of wanting to prove infinitude
of primes and Barbara Knox wanted to go to Walmart. So we as a group pack up into
the car and drive to Walmart and I pull into the parking lot of Walmart. So
instead of Barbara accepting that P!+1 is prime, accepting that we are at Walmart,
she denies that we are at Walmart and expects me to get everyone back into the car
and go searching for her Walmart.
So not only is the Knox patch add more lines to a Symbolic Logic write up of IP
but the Knox patch is internally flawed and leads to an invalid proof of IP
because it denies the definition of what it means to be "prime" at the start of
the proof.
Barb Knox
>In article <y6A2e.20830$d5.1...@newsb.telia.net>,
> Sevenhundred Elves <sevenh...@elves.invalid> wrote:
>[snip Archie's poo]
>
>>Let's see if I remember it right:
>>
>>Take the list of all prime numbers below and including any prime number
>>n. Add one to their product. (The resulting number, A, is then greater
>>than n.)
>>
>>The resulting number, A, is not divisible by any number on the list
>>because of the added 1. This is why we added the 1.
>>
>>Now there are no more than two possibilities for A: It either is prime,
>>or it isn't.
>>
>>If A is prime, it is a greater prime than n.
>>
>>If A is not prime, it is a composite number, and has a prime factor, B.
>>Since the primes on the original list can't be divisors of A (we made
>>sure of this by adding the 1), the prime B can not be on the list. B
>>must therefore be a greater prime than n.
>>
>>In either case, we end up with a prime greater than any prime on the
>>list. Since we specified at the outset that n could be any prime, we
>>must conclude that for any given prime there must exist a larger prime.
>>
>>This is what I learned in school, and I can't see what's wrong with it.
>>Is there really a flaw in that proof? Perhaps I'm too naīve.
>
>There's no flaw, but it is a bit more complex than necessary -- there is no
>need to explicitly consider two separate cases if you use the following
>Lemma:
>
> Every n > 1 has at least 1 prime factor.
>
>Now, assuming there is a non-empty finite list of primes, the number A by
>construction is > 1 and has *no* prime factors. Contradiction.
Since many Usenet kooks seem to choke on indirect proofs, here's a direct
version of the above:
Theorem: any non-empty finite list of prime numbers must be incomplete; that
is, there is always another prime that is not in the list.
Proof: Given any list of n primes (p_1, ..., p_n), construct the number A =
(p_1 * ... * p_n) + 1. By the lemma, A has at least one prime factor, call
it q. No p_i is a factor of A, so the prime q must be different from all
the p_i.
sttsc...@tesco.net
Archimedes Plutonium wrote:
> Thu, 31 Mar 2005 13:36:48 -0600 Archimedes Plutonium wrote:
>
> > Thu, 31 Mar 2005 10:01:31 +1200 Barb Knox wrote:
> >
but as I
> > said above, it creates a new class of number different from
composite which has
> > 2 factors and prime which has 1 factor.
Ballocks
But why
> would Knox with her patch and Courant not see it as a new prime?
Probably because they are not severely retarded.
What is this patch that Knox is wearing anyway?
Are you now imagining she's dressed up as a pirate ?
Archimedes Plutonium
(snipped)
> Thu, 31 Mar 2005 13:36:48 -0600 Archimedes Plutonium wrote:
>
> > Thu, 31 Mar 2005 10:01:31 +1200 Barb Knox wrote:
> >
>
>
> I said the Symbolic Logic write up of Indirect method of IP will show that my
> method has far fewer lines than does the Knox patch. But now I am able to pinpoint
> the fatal flaw of the Knox patch and its invalidity. If you use the Knox patch
> then as soon as you form P!+1 you skip over the fact that P!+1 is necessarily
> prime and thus you deny the original definition of what it means to be prime at
> the start of the proof. So you deny the definition of primeness which makes this
> attempt invalid.
>
I am well aware that in Logic lines of irrelevancy can be added to the proof and not
affect the validity. It just makes the proof less elegant and a bit dirty. But I
contend that the Knox patch and the Courant attempt are invalid. Invalid because they
involve ignoring and bypassing the "definition of prime" that was employed at the
start of the proof. By making a prime-factor-search and not calling P!+1 prime as
soon as it was formed belies a misunderstanding of the definition of prime. As soon
as P!+1 is formed, a valid proof then measures P!+1 with the definition of prime and
reacts by saying P!+1 is necessarily prime begetting the contradiction and end of
proof. But in the Knox patch and Courant attempt they dispel or dismiss or deny the
definition of prime by wanting to reach further than P!+1 and look for some other
prime-factor. It shows that they do not understand the definition of prime. So by
hunting and fishing for a number beyond P!+1 they have convoluted the definition of
prime by focusing on a prime factor. This makes their attempt invalid because they
destroyed the definition of prime by not taking P!+1 as prime. And so, as they
destroy the definition of prime and installing a prime factor to yield the
contradiction, one cannot tell for sure whether their contradiction is a result of
their lack of understanding of the definition of being a prime number. So their
contradiction results not from the logic of the argument but rather is generated from
their own confusion of the definition of prime. So they generate a contradiction out
of their own personal misunderstanding.
That is why I say the Knox patch and Courant proof attempt are invalid and fatally
flawed because if they said immediately after forming P!+1 that it must be
necessarily a new prime, then they understood well what the meaning and definition of
prime was that they started from in the beginning of the proof.
Again, I repeat as I oft have repeated, that we need a Symbolic Logic write up of
this proof, broken down into symbols as minute and fine as can be done. It will be
long but the details of it will show that the logic forces a proof wherein most
attempts were invalid and contained fatal flaws.
My proof rendition is not valid just because it is the shortest, but is valid because
in the Symbolic Logic write up follows the framework of my proof rendition.
In the Symbolic Logic write up there will be a point in the proof somewhat like this:
.
.
.
.
form P!+1
P!+1 is necessarily prime
P!+1 is larger than any prime of P
Contradiction to assumption
Primes are infinite.
In Courant and Knox attempts would look like this:
.
.
.
.
form P!+1
P!+1 is maybe prime
or a prime factor q not in P
.
.
.
Contradiction to assumption
Primes are infinite.
Invalid because they destroy the meaning of the definition of being prime when they
glide over the fact that P!+1 is necessarily prime.
Archimedes Plutonium
Namecalling does not speak well for you or the school you go to. Remember it was
only a few posts previous that you yourself failed to give a valid indirect
proof of IP with your injection of the above Lemma of a prime factor search in
the indirect method. So by logic, Barbara Knox is a "kook" as she so claims
above since she failed to deliver a valid indirect proof.
The science communities expect people in academics such as Barbara Knox that
when found in error they admit they were in error and they make some sort of
congratulatory signal to the person or persons who corrected them. Barbara
should not be calling me or others a kook but congratulating me for pointing out
her errors.
And this is a fault of so many in academics, is that when they are found to be
in error, they are too arrogant to admit nor thank those that corrected them. I
think it is partly due to the lack of ethics taught at colleges and universities
that we continue rampant cheating, rampant namecalling, rampant ignoring of new
science that replaces old science.
Yes, Barbara got a correct and valid "direct proof" above but only because of my
guidance. Barbara failed the indirect proof because at the point in the proof
where both she and I arrive at forming P!+1, I call it necessarily prime to
discharge the initial assumption, whereas Barbara makes a further hunt and
search for a prime-factor with her Lemma. She thus makes the error of denying
the original definition of what it means to be prime by conducting this
prime-factor-search. Both Barbara and I start with the same definition of what
it means to be prime at step one and throughout the proof and at step of forming
P!+1. So when I call P!+1 necessarily prime it is because I use the same
"definition of prime" throughout the proof. But Barbara changes or refuses to
use the definition of prime when she forms P!+1 by conducting a
prime-factor-search.
So, please, people in academics, do not namecall and especially namecall those
that correct your errors. And having found your correction, you owe the
corrector at least a word of "thank you".
Tapio
Theorem: Two consecutive integers p and q (p and q >0) does not have a
common factor.
Proof: Their difference equals to 1, i.e. q-p=1.
Assume two consecutive integers p and q have a common factor s.
Thus s divides p and also q. The results are integers - of course.
The rhs 1 divided by s is not the integer - 1/s is never an integer.
A contradiction, QED!
This is true inspite how do you construct p and q, i.e. the two consecutive
integers. You do not have to know primes.
Tapio
"Archimedes Plutonium" <a_plu...@iw.net> wrote in message
news:42502681...@iw.net...
Justin Young
"Tapio" <hurm...@dlc.fi> wrote in message
news:2lY3e.394$fM4...@read3.inet.fi...
> Dear Pluto
>
> Theorem: Two consecutive integers p and q (p and q >0) does not have a
> common factor.
>
> Proof: Their difference equals to 1, i.e. q-p=1.
> Assume two consecutive integers p and q have a common factor s.
> Thus s divides p and also q. The results are integers - of course.
> The rhs 1 divided by s is not the integer - 1/s is never an integer.
I'm not so sure that 1/s is never an integer.
>
> Tapio
Tapio
"Justin Young" <x_sta...@hotmail.com> wrote in message
news:RoY3e.5819$Ms3....@twister.nyroc.rr.com...
As p and q >0 (as stated above) and - of course s >1 that is not a prime.
:-)
Tapio
>
>>
>> Tapio
>
>
Tapio
I mean 1 is not a prime. s can be prime or composite.
The only assumption was that s is a common factor of p and q that are
consecutive integers and s>1. :-)
Tapio
>
> Tapio
>>
>>>
>>> Tapio
>>
>>
>
>
Bill Dubuque
>
> There's no flaw, but it is a bit more complex than necessary -- there is
> no need to explicitly consider two separate cases if you use the following
>
>
> Now, assuming there is a non-empty finite list of primes p1,p2,...,pn
> 1 + p1...pn is > 1 and has *no* prime factors. Contradiction.
>
> Proof of lemma (using well-ordering): ...
SURPRISE One can bypass the lemma, yielding a self-contained 'elementary'
proof, by 'inlining' the lemma into the theorem (in the same way that the
classic proofs of unique factorization by Klappauf, Lindemann, Zermelo
inline Euclid's Lemma on the Prime Divisor Property: P|AB => P|A or P|B).
To do this smoothly we employ an alternative definition of 'prime' that
is trivially seen to be equivalent to the standard definition, namely
DEFINITION P is prime if P is the least factor>1 of an integer>1
THEOREM There are infinite number of primes.
PROOF Suppose not, i.e. suppose there are only a finite number of primes.
We proceed by descent: given any integer N>1 we deduce there is a prime < N.
Let S be the product of all primes >= N (convention: S = 1 if no primes >= N)
Let P be the least factor>1 of S+1. By definition P is prime. But P is not
one of the primes >= N since these leave remainder 1 when divided into S+1.
So P < N. This implies that among the primes there is an infinite descending
sequence of positive integers, a contradiction. QED
NOTE that I have presented the proof in the pure form of infinite descent
discovered by Fermat to emphasize that this proof could easily have been
one of Fermat's (obviously one could shorten the proof by deviating from
this older format). Although I have never seen this particular variant of
the proof presented anywhere, it is so obvious it must surely be well-known
(I found it and variants as a teenager while playing with inlined elementary
proofs of unique factorization).
--Bill Dubuque
Dror Speiser
But, of course the theorem only works for the whole set of primes. So you can't really use it...
Anyway, I don't understand the problem:
i) If P!+1 is a prime, then the set we had is false and we achieved contradiction
ii) If P!+1 is not prime, then it must be divisible by primes in our set, but it isn't, and hence we get a contradiction again.
AP thinks that in (ii) we need to look for a prime factor of P!+1, but we don't, because the argument is not about saying that every number can be decomposed into prime factors, but every number has a prime number dividing it.
The use of the fundamental theorem of algebra is not necessary in the proof of infinitude of primes.
Bill Dubuque
that there are infinitely many primes. Here is the proof in ascent form.
THEOREM There are infinitely many prime integers.
PROOF Given any finite set S of primes, we prove there exists a prime
not in S. Let M = product of all the elements of S ( M = 1 if S = {} )
Let P be the least factor>1 of M+1 > 1. P is prime, else P would have a
smaller factor>1, contra to our choice of P as the least factor>1 of M+1.
P is not in S, since primes in S leave remainder 1 when divided into M+1.
Hence P is a prime not in S. QED
Notice that this form of proof is not as subtle as those that proceed
by way of contradiction, so it should prove more easily comprehensible
to neophytes. Furthermore, it has a vivid constructive interpretation,
--Bill Dubuque
Archimedes Plutonium
> Barb Knox wrote: (*edited*)
> >
> > There's no flaw, but it is a bit more complex than necessary -- there is
> > no need to explicitly consider two separate cases if you use the following
> >
> > LEMMA Every n > 1 has at least 1 prime factor.
> >
> > Now, assuming there is a non-empty finite list of primes p1,p2,...,pn
> > 1 + p1...pn is > 1 and has *no* prime factors. Contradiction.
> >
> > Proof of lemma (using well-ordering): ...
>
> SURPRISE One can bypass the lemma, yielding a self-contained 'elementary'
> proof, by 'inlining' the lemma into the theorem (in the same way that the
> classic proofs of unique factorization by Klappauf, Lindemann, Zermelo
> inline Euclid's Lemma on the Prime Divisor Property: P|AB => P|A or P|B).
> To do this smoothly we employ an alternative definition of 'prime' that
> is trivially seen to be equivalent to the standard definition, namely
>
Bill, the trouble with Knox's attempt is that she uses one definition of prime
for approx 80% of the proof steps until she forms the new number P!+1 and then
she wants to use a different definition of what it means to be prime for the
remainder of the proof. This is invalid because for one reason out of many
reasons, Knox would be hard pressed to say that her contradiction emanated from
her switching of definitions of what it means to be prime in the Indirect
Method.
Apparently Bill Dubuque never really read my posts to full understanding because
Bill falls into the very same mistake that Knox, that is the switching of
definitions of primeness once P!+1 is formed.
>
> DEFINITION P is prime if P is the least factor>1 of an integer>1
Okay, Bill, that is okay but do you adhere to your new definition throughout the
proof argument or do you switch?
>
>
> THEOREM There are infinite number of primes.
>
> PROOF Suppose not, i.e. suppose there are only a finite number of primes.
> We proceed by descent: given any integer N>1 we deduce there is a prime < N.
You do not need descent and I find it annoying, not helpful. In a way, Bill, you
are saying that in the history of mathematics that Infinite Descent should have
been discovered first and not Mathematical-Induction. Keeping the issue aside
that the Natural Numbers are really the Adics or Infinite Integers where both
Infinite Descent and Mathematical-Induction are fakes.
>
> Let S be the product of all primes >= N (convention: S = 1 if no primes >= N)
> Let P be the least factor>1 of S+1. By definition P is prime. But P is not
> one of the primes >= N since these leave remainder 1 when divided into S+1.
> So P < N. This implies that among the primes there is an infinite descending
> sequence of positive integers, a contradiction. QED
>
> NOTE that I have presented the proof in the pure form of infinite descent
> discovered by Fermat to emphasize that this proof could easily have been
> one of Fermat's (obviously one could shorten the proof by deviating from
> this older format). Although I have never seen this particular variant of
> the proof presented anywhere, it is so obvious it must surely be well-known
> (I found it and variants as a teenager while playing with inlined elementary
> proofs of unique factorization).
>
> --Bill Dubuque
There are many mistakes in the above. The biggest is that you would have to
prove that Infinite Descent or Mathematical Induction are valid concepts and not
illusions. The Adics have no Infinite Descent and have no Mathematical
Induction.
The primes are used to build the Adics of the p-adics, but that does not make
the primes a fake set. Natural Numbers as finite integers is a fake set. But to
use the primes to make the p-adics which then makes ALL ADICS is not a fake set.
The other error of Bill's above is that when he forms P!+1, although he calls it
S+1 that given Bill's definition of what it means to be prime is switched or
denied once he forms P!+1, just as Knox switched the definition once she formed
P!+1. Because as soon as Bill forms P!+1 he should have used his definition that
he started off with. Bill should have seen that according to his definition P!+1
is prime because only P!+1 divides P!+1 evenly. Then Bill should have gone to
the next step of declaring a contradiction and thus dispensing of the
unnecessary baggage of Fermat's Infinite Descent.
Math is a tiny subset of Physics that means that a math proof has to have the
characteristics of Maxwell's Laws. One of the Maxwell theory characteristics is
the Principle of Least Action and we know it as Lightening bolts follow the
shortest path and we know it as that gravity follows the shortest path. That
implies that all sound math proofs have a very --- tight logical chain of
deduction --- to duplicate the shortest path. If a body in gravity fall goes off
course just a little bit it would break a law of physics and that never happens.
Likewise, since all of mathematics is just a tiny subset of Physics, that all
math proofs are like that Lightening bolt that always follows the shortest path
or that ball rolling down inclined planes always follows a unique path so that
the law of physics is never violated. So when Dubuque or Knox or Hardy stray
from the unique proof of Infinitude of Primes by just a little bit, they thus
end up with invalid proof attempts. In gravity we do not see falling bodies
magically going up and that is what the Dubuque rendition with Fermat's Last
Descent is like.
Nathan
> Math is a tiny subset of Physics that means that a math proof has
> to have the characteristics of Maxwell's Laws. One of the Maxwell
> theory characteristics is the Principle of Least Action and we
> know it as Lightening bolts follow the shortest path and we know
> it as that gravity follows the shortest path. That implies that
> all sound math proofs have a very --- tight logical chain of
> deduction --- to duplicate the shortest path. If a body in
> gravity fall goes off course just a little bit it would break a
> law of physics and that never happens.
> Likewise, since all of mathematics is just a tiny subset of
> Physics, that all math proofs are like that Lightening bolt that
> always follows the shortest path or that ball rolling down
> inclined planes always follows a unique path so that the law of
> physics is never violated. So when Dubuque or Knox or Hardy stray
> from the unique proof of Infinitude of Primes by just a little bit,
> they thus end up with invalid proof attempts.
This is unbelievably silly. Every theorem has infinitely many valid
proofs. Just because one proof may take a "longer path" than another
doesn't make it any less a proof of the theorem. Indeed, there are
many other features of a proof, other than its length, that are
valuable as well. Often what mathematicians want from a proof is more
than just the result; they want to know WHY that theorem is true.
Clarity of presentation and ease of understanding are frequently much
more important than the actual length of a proof.
Virtually all the proofs you've been considering that there are
infinitely many primes are equally valid proofs. And of course these
are all different versions of the same basic argument, due to Euclid.
There are many other interesting proofs of the same theorem, using
very different arguments; I like the collection of such proofs
contained in Ribenboim's _New Book of Prime Number Records_.
If all proofs were required to be the shortest possible in order to
be valid, you would almost never know for sure whether or not you
had a valid proof.
P.S. Since your post was much longer than it needed to be to convey
its information (that you're an insane troll), it violates the
principle of least action and is invalid.
P.P.S. This post fails the same test.
Archimedes Plutonium
Trouble with both your ascent and descent attempts is that they turn things
180 degrees opposite to direct and indirect proof methods. You use Fermat
Descent for indirect and you use Mathematical-Induction for direct and
instead of contradiction in Descent it appears in Ascent.
I pointed out your flaw in the Descent attempt in that P!+1 was necessarily
prime given your starting assumption and that the Descent was unneeded
excess baggage. The flaw in your above is that you do not actually produce a
new prime for which the Direct Method demands. So someone reading your above
with the given set as {3,5} would then end up with saying that 16 was a new
prime.
So Bill, congratulations are not in order for you. However, your thoughts do
spark a new path of investigation that I have not considered to date.
The idea that you sparked above is that if you believe, and over 99.99
percent of current math professors believe that the Natural Numbers = Finite
Integers is a sound set, then you believe that Mathematical Induction and
Fermat's Infinite Descent are sound devices. Further, you believe that Math
Induction is equivalent to Fermat's Infinite Descent.
But I believe Natural Numbers = Infinite Integers = All Adics. There is no
Mathematical Induction or Fermat's Infinite Descent that is true in p-adics,
is there.
So, what Bill has sparked is the investigation as to whether Fermat's
Infinite Descent is truly equivalent to Mathematical Induction or does there
appear cracks in Number theory. And how badly does Math Induction or
Fermat's Descent react with p-adics.
You see, I can develop the Primes, the set of primes, and I can prove the
Primes are infinite and then go on to using those primes to develop the
P-adics and the All-Adics and still be consistent and logically sound. And
end up with Natural Numbers = Infinite Integers = All-Adics. But I cannot
have Mathematical Induction or Fermat's Infinite Descent in that
composition, can I, and obviously not.
So what Bill has sparked is whether there is a crack between Math Induction
and Fermat's Infinite Descent because to those that believe Natural Numbers
are Finite Integers there cannot be a crack or cracks in the logic. So this
allows us some new testing.
Amadeus Train-Owwell Zirconium
that happens to rely on making a contradiction;
he has another infinite descent,
that actually finds the smallest right trigon
with a side of a certain form,
which is an integer in the millions, as I recall.
--Chairman George and Strep Throat at W'gate!
http://tarpley.net/bush12.htm
http://larouchepub.com
Bill Dubuque
>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>
>> My prior post gave a self-contained 'descent' proof of Euclid's theorem
>> that there are infinitely many primes. Here is the proof in ascent form.
>>
>> THEOREM There are infinitely many prime integers.
>>
>> PROOF Given any finite set S of primes, we prove there exists a prime
>> not in S. Let M = product of all the elements of S ( M = 1 if S = {} )
>> Let P be the least factor>1 of M+1 > 1. P is prime, else P would have a
>> smaller factor>1, contra to our choice of P as the least factor>1 of M+1.
>> P is not in S, since primes in S leave remainder 1 when divided into M+1.
>> Hence P is a prime not in S. QED
>
> for which the Direct Method demands. So someone reading your above with
Not true. The proof deduces that the least factor>1 of 3*5 + 1
is a prime not in S, i.e. P = 2 is a prime not in S = {3,5}.
--Bill Dubuque
Bill Dubuque
>>>
>>> There's no flaw, but it is a bit more complex than necessary -- there is
>>> no need to explicitly consider two separate cases if you use the following
>>>
>>> LEMMA Every n > 1 has at least 1 prime factor.
>>>
>>> Now, assuming there is a non-empty finite list of primes p1,p2,...,pn
>>>
>>> Proof of lemma (using well-ordering): ...
>>
>> SURPRISE One can bypass the lemma, yielding a self-contained 'elementary'
>> proof, by 'inlining' the lemma into the theorem (in the same way that the
>> classic proofs of unique factorization by Klappauf, Lindemann, Zermelo
>> inline Euclid's Lemma on the Prime Divisor Property: P|AB => P|A or P|B).
>> To do this smoothly we employ an alternative definition of 'prime' that
>> is trivially seen to be equivalent to the standard definition, namely
>>
>>
>>
>> PROOF Suppose not, i.e. suppose there are only a finite number of primes.
>> We proceed by descent: given any integer N>1 we deduce there is a prime < N.
>> Let P be the least factor>1 of S+1. By definition P is prime. But P is not
>> one of the primes >= N since these leave remainder 1 when divided into S+1.
>> So P < N. This implies that among the primes there is an infinite descending
>> sequence of positive integers, a contradiction. QED
>>
>> NOTE that I have presented the proof in the pure form of infinite descent
>> discovered by Fermat to emphasize that this proof could easily have been
>
> There are many mistakes in the above. The biggest is that you would have
> to prove that Infinite Descent or Mathematical Induction are valid concepts
> and not illusions.
The proof is correct. If you reject the existence of the natural numbers
then you'll have difficulty accepting almost all of modern mathematics.
> The Adics have no Infinite Descent and have no Mathematical Induction.
That's not relevant. The proof concerns natural numbers, not your 'Adics'.
> The other error of Bill's above is that when he forms P!+1, although he
> calls it S+1 that given Bill's definition of what it means to be prime
> is switched or denied once he forms P!+1, just as Knox switched the
> definition once she formed P!+1. Because as soon as Bill forms P!+1 he
> should have used his definition that he started off with. Bill should
> have seen that according to his definition P!+1 is prime because only
> P!+1 divides P!+1 evenly. Then Bill should have gone to the next step of
> declaring a contradiction and thus dispensing of the unnecessary baggage
> of Fermat's Infinite Descent.
My proof never mentions 1+P! so your remarks do not even apply to it.
Instead of 1+P! it utilizes 1+S := 1 + (product of all primes >= N)
to deduce that the least factor>1 of 1+S is a prime < N; thus descent.
Here N is an arbitrary integer>1 (not necessarily the least prime 2),
so one cannot deduce that 1+S has no proper prime factors, in analogy
with the way one does for 1+P!. Nor does one desire to deduce this
since the aim here is to produce a smaller prime to achieve descent.
> when Dubuque or Knox or Hardy stray from the unique proof of Infinitude
> of Primes by just a little bit, they thus end up with an invalid proof
Your mind seems to be stuck on a local minima -- the proof that you
deem 'best' based upon some subjective extra-mathematical criterion.
If you study mathematics you will quickly learn that one can and one
does often give many proofs of the same theorem because each may shed
a different light on the heart of the matter. The more we know about
a particular theory the more likely we are to spot 'different' ways
of proof. For example, since we know so well very many theorems about
natural integers, there are numerous targets to aim to contradict when
we are searching for a proof by contradiction. That is one reason why
one encounters so many variations on this particular proof.
> You do not need descent and I find it annoying, not helpful.
Such proofs need various applications of induction. Whether one uses the
least number principle, well-ordering, infinite descent, the pigeonhole
principle, or other variants of induction is simply a matter of taste.
Note that induction is used not only in the lemma that every integer>1
has a prime factor but also, implicitly, in other places. For example,
to rigorously define the product of an arbitrary number of integers
requires so-called 'definition by induction' (this phrase is imprecise
for such definitions require not only the axiom of induction but *all*
of the Peano axioms, see Henkin's Monthly paper [2] for more on this).
To better understand precisely what is required for a complete formal
proof of Euclid's theorem it helps to examine presentations of this
proof in various automated proof verification systems, e.g. [3],[4].
Also it should be mentioned that a more specific and technical notion
of descent arises naturally for a large class of Diophantine equations
that occurred already in the work of Diophantus and Fermat, namely
curves of genus 1 (elliptic curves), e.g. see my post [1]. In this
and related contexts a descent-based viewpoint is much more natural.
> you are saying that in the history of mathematics that Infinite Descent
> should have been discovered first and not Mathematical-Induction.
I said nothing of the sort. Rather I said that "this proof could easily
have been one of Fermat's". However, it is in fact true that Fermat
was one of the first to realize the need for a sufficiently rigorous
notion of induction. He achieved this via his method of infinite descent.
As Weil wrote on p.49 of his: Number theory: an approach through history
In those days, and even much later, the word "induction"
was used in an altogether different sense; it indicated no
more than what is now understood by "conjecture". To
verify "by induction" a statement P(n), depending upon an
integer n, meant merely to verify it for some (and sometimes
very few) initial values of n. Of course it could happen that
the procedure used, say, in order to verify P(4) was to derive
it from P(3) in such a manner as to make it obvious that
the same argument would serve to derive P(n) from P(n-1).
In that case, which occurs not seldom in Euclid (e.g. in
Eucl.I.45, VII. 14, VIII.6, to give a few random examples)
a mathematician has the right to regard such a proof as
conclusive, even though it is not couched in the conventional
terms which would be used for it today. For such proofs,
it became customary to use the term of "complete" (or
"mathematical") induction, while "incomplete" induction
came to mean the heuristic argument described above.
Induction of the "incomplete" kind was used quite sys-
tematically by Wallis in his Arithmetica Infinitorum of 1656
and applied to a wide variety of questions, including the
theory of binomial coefficients; eventually his results led to
Newton's binomial series, and in the next century to Euler's
theory of eulerian integrals and the gamma function. What
has to be emphasized here, however, is the criticism offered
by Fermat in 1657 to Wallis's use of "induction":
"One might use this method if the proof of some prop-
osition were deeply concealed and if, before looking for it,
one wished first to convince oneself more or less of its truth;
but one should place only limited confidence in it and apply
proper caution. Indeed, one could propose such a statement,
and seek to verify it in such a way, that it would be valid in
several special cases but nonetheless false and not universally
true, so that one has to be most circumspect in using it; no
doubt it can still be of value if applied prudently, but it
cannot serve to lay the foundations for some branch of science,
as Mr. Wallis seeks to do, since for such a purpose nothing short
of a demonstration is admissible" [Fe.II.351-352]
Predictably, Wallis was displeased, and, in a letter to Digby
(Comm.Epist., letter XVI = Wal.II.782-783), haughtily de-
clined to answer the charge. As his results have largely been
substantiated by later developments, Fermat's criticism may,
in retrospect, strike us as too harsh, even though he sought
to soften it by some high praise for Wallis (cf. also
Fe.II.337,343). What concerns us here is that it seems to
express Fermat's mature views about proofs. Similarly,
writing to Clerselier in 1662, he states that "the essence of
a proof is that it compels belief" ("la qualite essentielle d'une
demonstration est de forcer a croire": Fe.II.483). This is not to
deny, of course, that he was liable to error; he was so, perhaps
more than others, in view of his fatal habit of seldom writing
down his proofs in full and hardly ever keeping copies of
his communications to his friends. Nevertheless, in view of
the above quotations, when Fermat asserts that he has a proof
for some statement, such a claim has to be taken seriously.
Later, on p.75:
In the tantalizingly brief account of his number-theoretical
work sent to Huygens via Carcavi in 1659 (Fe.II.431-436 Huy.II.
458-462) Fermat has the following to say about his early work:
"As ordinary methods, such as are found in the books,
are inadequate to proving such difficult propositions,
I discovered at last a most singular method ... which
I called the infinite descent. At first I used it only
to prove negative assertions, such as: "No number of the
form 3n-1 can be written as x^2 + 3 y^2", "There is no
right-angled triangle in numbers whose area is a square
....To apply it to affirmative questions is much harder,
so that, when I had to prove that "Every prime of the
form 4n+1 is a sum of two squares", I found myself in
a sorry plight ("en belie peine"). But at last such
questions proved amenable to my method ..."
--Bill Dubuque
[1] http://groups-beta.google.com/group/sci.math/msg/f0cfc620a12728c3
http://google.com/groups?selm=y8z8z9r...@nestle.ai.mit.edu
[2] Henkin, L. On Mathematical Induction.
American Mathematical Monthly, 67, No. 4 (Apr., 1960), pp. 323-338
http://links.jstor.org/sici?sici=0002-9890(196004)67:4<323:AOMI>
[3] Wenzel, M; Wiedijk, F. A comparison of the mathematical proof
languages Mizar and Isar
http://www4.in.tum.de/~wenzelm/papers/romantic.pdf
[4] Chapter 4: Euclid's theorem, in: The HOL system tutorial
http://www.cl.cam.ac.uk/users/mn200/hol-tutorial/tutorial006.html
sttsc...@tesco.net
Dror Speiser wrote:
> It seems as if AP is assuming the Fundamental Theorem of Arithmatic
to the initial set of primes, and if that was the assumption, it is
true that P!+1 must be prime.
>
> But, of course the theorem only works for the whole set of primes. So
you can't really use it...
Why would that necessarily be the case ?
Say you defined 2,3 to be the only primes,
1,5,7,11,13,17,19,.... to be units
and the remaining numbers to be composite, i.e divisible
by 2 or 3.
Then the fundamental theorem that every number
except units has a unique factorisation
up to units and order would apply in this case.
Of course if you are assuming that
2,3, are the only primes 2*3 +1 = 2^a*3^b or unit
you can't say that 2*3+1 = pnew
as pnew =2^a*3^b is contrary to the assumption
that 2,3 are the only primes. So 2*3+1 can only be
composite or a unit. 2*3+1 is not composite (1 <> 2n,
1<>3n,) and in fact is a unit (7).
So the fundamental theorem is true even though
{2,3} is a proper subset of the primes in N
Archimedes Plutonium
snipping to save space)
Thanks for the in depth reply Bill.
>
> The proof is correct. If you reject the existence of the natural numbers
> then you'll have difficulty accepting almost all of modern mathematics.
I am concerned that your proof by Descent is not correct. I find your definition
of prime very good and once you formed P!+1, your definition immediately
discharges the initial assumption and thus a contradiction and end of proof. Which
leaves the mechanism of Descent where? In other words you never even have to use
or apply Descent.
Bill, by your insistence that Descent is applied or need be applied is testimony
that you do not believe P!+1 is necessarily prime given your initial assumption
that the set of all-primes if finite. What delivers the contradiction is the fight
between set of all primes is finite forces P!+1 to be prime. So Descent never
enters the picture.
What I was saying about Natural-Numbers is a different issue from being able to
render a clear valid proof of Infinitude of Primes.
I do not want to toss your Descent and Math-Induction completely out the window.
So I see from your Descent a possible future test of cracks in Natural Numbers. If
you believe that Natural-Numbers are Finite Integers then Fermat's Infinite
Descent and Mathematical Induction are equivalent. But if you see that
NaturalNumbers are Infinite Integers or Adics then Fermat's Infinite Descent and
Mathematical Induction are fakes and lead to many cracks in mathematics.
In fact one of those cracks should be a Divergence in proof results such as
Infinitude of Primes. That when Bill Dubuque uses Fermat's Infinite Descent on
Euclid's Infinitude of Primes proof and then uses Mathematical Induction on the
same Euclid's IP proof, that the two should DIVERGE and show some cracks in the
foundation of number theory. Should begin to show that the Natural Numbers as
finite integers is a falsehood.
Maybe Infinitude of Primes is not the best case test but some other mathematical
proof will better show or indicate this DIVERGENCE of what I speak. Perhaps some
geometrical proof that can accommodate both Descent and Math-Induction may reveal
the cracks.
So, Bill, please do not confuse my issue of what the Natural Numbers really are
with my issue of rendering a valid Euclid IP proof.
>
>
> > The Adics have no Infinite Descent and have no Mathematical Induction.
>
> That's not relevant. The proof concerns natural numbers, not your 'Adics'.
Please do not confuse my attempt to deliver a clear valid proof of Euclid's
Infinitude of Primes with a different issue of what the Natural Numbers truly are.
I am trying to help you Bill, with not throwing out completely the Descent and
Math Induction you have raised. So I am trying to salvage your contribution.
>
>
> > The other error of Bill's above is that when he forms P!+1, although he
> > calls it S+1 that given Bill's definition of what it means to be prime
> > is switched or denied once he forms P!+1, just as Knox switched the
> > definition once she formed P!+1. Because as soon as Bill forms P!+1 he
> > should have used his definition that he started off with. Bill should
> > have seen that according to his definition P!+1 is prime because only
> > P!+1 divides P!+1 evenly. Then Bill should have gone to the next step of
> > declaring a contradiction and thus dispensing of the unnecessary baggage
> > of Fermat's Infinite Descent.
>
> My proof never mentions 1+P! so your remarks do not even apply to it.
By honest with yourself, Bill, the moment you formed 1+S you mentioned it.
You seem to be wanting to say that you can form 1+S and never mention 1+S. So why
quibble and why quibble over terminology since your 1+S is P!+1. When people start
losing a math argument they often revert to namecalling, but Bill has not done
that, instead he seems to be quibbling over terminology. So please be honest with
yourself that your S+1 is the same as P!+1. For this quibbling of terminology is
childish.
>
> Instead of 1+P! it utilizes 1+S := 1 + (product of all primes >= N)
> to deduce that the least factor>1 of 1+S is a prime < N; thus descent.
> Here N is an arbitrary integer>1 (not necessarily the least prime 2),
> so one cannot deduce that 1+S has no proper prime factors, in analogy
> with the way one does for 1+P!. Nor does one desire to deduce this
> since the aim here is to produce a smaller prime to achieve descent.
>
> > when Dubuque or Knox or Hardy stray from the unique proof of Infinitude
> > of Primes by just a little bit, they thus end up with an invalid proof
>
> Your mind seems to be stuck on a local minima -- the proof that you
> deem 'best' based upon some subjective extra-mathematical criterion.
> If you study mathematics you will quickly learn that one can and one
> does often give many proofs of the same theorem because each may shed
> a different light on the heart of the matter. The more we know about
> a particular theory the more likely we are to spot 'different' ways
> of proof. For example, since we know so well very many theorems about
> natural integers, there are numerous targets to aim to contradict when
> we are searching for a proof by contradiction. That is one reason why
> one encounters so many variations on this particular proof.
No. Physics creates mathematics and not the other way around. All of mathematics
comes out of physics. And one of the most consistent and logical foundations of
mathematics is the Maxwell theory or Maxwell Equations. Most of the mathematical
foundations such as the Peano Axiom system is flawed, but not the Maxwell theory.
Mathematics has a few axiom systems that are flawless, but Peano's is not one of
them.
And what the Maxwell Axiom system of the Maxwell Equations tells mathematics is
that a proof of mathematics is very tight. Because in Maxwell Equations forces
follow a "shortest path". Lightening bolts follow a shortest path. Force of
gravity follows a shortest path. That means that a mathematics proof to be valid
will have no excess baggage. That means there is one and only one valid Euclid
Infinitude of Primes indirect proof and one and only one valid Euclid Infinitude
of Primes direct proof. All other proof attempts will have excess baggage that
makes the attempt flawed and invalid. Not only Euclid's Infinitude of Primes but
all proofs in mathematics must follow Physics Maxwell Axiom System with its
embedded Principle of Least Action.
>
>
> > You do not need descent and I find it annoying, not helpful.
>
> Such proofs need various applications of induction. Whether one uses the
> least number principle, well-ordering, infinite descent, the pigeonhole
> principle, or other variants of induction is simply a matter of taste.
No, it is not a matter of taste. Your failure of producing a valid IP proof is the
same as Knox's failure. Both of you start the proof with a definition of what it
means to be "prime" and both of you when you formed P!+1 denied your starting off
definition of what it means to be prime and thus denying that P!+1 was a new prime
discharging the finite assumption. By denying your starting definition you deliver
an invalid attempt.
The world of physics is so constrained in its physical laws that gravity must
follow a Least Action and Electricity Magnetism must follow a Least Action. So it
is rather obtuse and ridiculous to think that mathematicians with their science of
mathematics that purports to be the science of precision can have so much leeway
and room for varied proofs. Can you see the ridiculousness of your argument Bill?
Can you see that why should physics be so very tight and narrow because of Least
Action, yet mathematics have ample room or leeway for a huge spectrum of proofs.
Absurd.
>
> Note that induction is used not only in the lemma that every integer>1
> has a prime factor but also, implicitly, in other places. For example,
Again, I repeat, the reason Dubuque's and Knox's Infinitude of Primes proof
attempt fails is because they deny their definition of prime once they form P!+1.
But the good news about Dubuque's attempt is that it can show us a Divergence of
endresults between using Fermat's Infinite Descent and Mathematical Induction. If
NaturalNumbers are Finite Integers then there should never be any divergence. But
if Natural Numbers are Adics then there should be cracks and divergences between
Descent and Math Induction, especially in geometrical proofs where Descent and
Induction are applied.
>
>
> Later, on p.75:
>
> In the tantalizingly brief account of his number-theoretical
> work sent to Huygens via Carcavi in 1659 (Fe.II.431-436 Huy.II.
> 458-462) Fermat has the following to say about his early work:
>
> "As ordinary methods, such as are found in the books,
> are inadequate to proving such difficult propositions,
> I discovered at last a most singular method ... which
> I called the infinite descent. At first I used it only
> to prove negative assertions, such as: "No number of the
> form 3n-1 can be written as x^2 + 3 y^2", "There is no
> right-angled triangle in numbers whose area is a square
> ....To apply it to affirmative questions is much harder,
> so that, when I had to prove that "Every prime of the
> form 4n+1 is a sum of two squares", I found myself in
> a sorry plight ("en belie peine"). But at last such
> questions proved amenable to my method ..."
>
> --Bill Dubuque
Bill, can you tell me, did Fermat discover his principle of Least Time in physics
that is also known as Least Action came before or after he discovered his Infinite
Descent. Just curious as to when Fermat discovered one and the other.
Bill Dubuque
>>
>> PROOF Given any finite set S of primes, we prove there exists a prime
>> not in S. Let M = product of all the elements of S ( M = 1 if S = {} )
>> Let P be the least factor>1 of M+1 > 1. P is prime, else P would have a
>> smaller factor>1, contra to our choice of P as the least factor>1 of M+1.
>> P is not in S, since primes in S leave remainder 1 when divided into M+1.
>> Hence P is a prime not in S. QED
>
> 180 degrees opposite to direct and indirect proof methods. You use Fermat
> Descent for indirect and you use Mathematical-Induction for direct and
> instead of contradiction in Descent it appears in Ascent.
Not true. Proof by contradiction appears in any proof by Infinite Descent.
My Ascent proof is trivially reformulated to avoid proof by contradiction:
replace the 3rd and 4th lines of the above proof with the following lines
Let P be the least factor>1 of M+1 > 1. P is prime since every factor>1
of P must be >= P, being simultaneously a factor>1 of M+1 (since P|M+1).
Above we have essentially inlined a proof that the least factor>1 of an
integer>1 is prime. We implicitly invoke a variant of induction known as
the Least Number Principle (a nonempty subset of N has a least element)
when we define P to be the least factor>1 of M+1. Contrast this with the
classic proof utilizing the Lemma that every integer>1 has a prime factor,
also proved by induction. A novice might find the former more intuitive.
If your prefer not to inline it you could abstract it out into a Lemma:
DEFINITION Ln := least factor>1 of an integer n>1
DEFINITION n>1 is prime if Ln = n (i.e. n has no proper factors)
LEMMA (1) m|n -> Lm >= Ln PROOF k|m -> k|n -> k >= Ln if k>1
LEMMA (2) Ln is prime for all n>1
PROOF Ln|n -> LLn >= Ln via m = Ln in Lemma (1)
But LLn|Ln -> LLn <= Ln
We conclude that LLn = Ln so Ln is prime by definition. QED
It's quite clear that the above doesn't use proof by contradiction.
--Bill Dubuque
Archimedes Plutonium
The objection I have with Infinite Descent and Math-Induction is that there does
not exist a definition of "prime" for which is used at the outset of the proof
and when P!+1 is formed that it does not allow one to say P!+1 is automatically
and immediately and necessarily prime itself discharging the initial assumption
and ending the proof. So that neither Infinite Descent or Math-Induction need be
applied and that their application is seen as some secondary sideshow glitz and
glamour that invalidates the entire proof.
It is the "Definition of Prime" at the outset that forces a unique and valid
proof of IP on forming P!+1. Once P!+1 is formed, the rigors of Logic force all
the steps for a valid proof of Infinitude of Primes. And it shows that adding
Infinite Descent or Math-Induction are window dressing making it invalid.
It is the definition of prime that creates a Unique step pattern of a valid
proof and the sidetracking or adding other bits and pieces to the proof cause it
to be invalid.
And this is generalized to all proofs in mathematics, that all proofs have one
and only one valid proof for direct and indirect.
Bill, I am working on Fermat's Principle of Least Time also known as Principle
of Least Action in physics. There is a claim by physicists and mathematicians
that Least Time or Least Action can derive the Maxwell theory. I believe that is
incorrect. I believe the Maxwell theory (Maxwell Equations) can derive Least
Action and Least Time but not the reverse.
It is well known that the Maxwell Equations can derive the Lorentz
transformation invariance, but that the Lorentz transformation invariance cannot
derive the Maxwell Equations.
So I believe that the example analogy of the Lorentz transformation applies to
Least Action and Least Time.
Maxwell theory can derive both Lorentz transformation and derive Principle of
Least Time and Principle of Least Action. But not the other way around.
I do not know how or why or where the old time physicists made this gross and
silly error. Perhaps their inflated praise of General Relativity caused them to
be dumb and daffy into thinking that Least Time and Least Action can derive
Maxwell theory.
Bill, I realize you have dived and dove into Descent and Math-Induction on
Infinitude of Primes proof, but I wonder if you can lay that aside and help me
work on pinpointing this error of physicists who think that Least Action and
Least Time can derive the Maxwell theory when in truth they cannot.
And help me prove that the Maxwell theory is so vast and broad that it can
easily derive the Principle of Least Time and the Principle of Least Action.
One of the big payoffs of this adventure will be that it reveals in detail why
the theory of General Relativity was all a fake.
I suspect Least Action and Least Time come directly out of the Coulomb law which
is Gauss's Law of the Maxwell Equations. I suspect that you cannot have a
principle of Least Action or a principle of Least Time without a Coulomb Law.
But that those principles of Least Action and Least Time cannot build a Coulomb
Law, or if they can, cannot build the other 3 Maxwell Equations. So like in
mathematics, of subset to subset proving two sets are the same, we have that
Principle of Least Action and Least Time is a subset of one of the 4 Maxwell
Equations but that the Maxwell Equations are never a subset of Principle of
Least Action and Principle of Least Time.
Keith Ramsay
This is correct. I want to note however that in general the
least element principle is nonconstructive. The fact that
there's a smallest factor >1 is constructive but depends on
the fact that there's an effective procedure for determining
whether a given integer >1 is a factor.
Keith Ramsay
Archimedes Plutonium
> It seems as if AP is assuming the Fundamental Theorem of Arithmatic to the initial set of primes, and if that was the assumption, it is true that P!+1 must be prime.
>
> But, of course the theorem only works for the whole set of primes. So you can't really use it...
>
> Anyway, I don't understand the problem:
Well let me make it clear.
>
>
> i) If P!+1 is a prime, then the set we had is false and we achieved contradiction
This is true and what I have said all along. There are 2 methods to the proof. One is Direct and the other is Indirect. In the Indirect P!+1 is necessarily prime and ends the proof.
>
>
> ii) If P!+1 is not prime, then it must be divisible by primes in our set, but it isn't, and hence we get a contradiction again.
In the Direct Method, there has to be a Prime-Factor Search. But not in the Indirect Method because P!+1 is necessarily prime.
The Direct Method is set cardinality increase which generalizes to the infinite. In the Indirect Method is proof by contradiction.
The trouble with the History of Mathematics is that no-one cleaned up this proof. Euclid gave a Direct Method of increasing set cardinality. And it was professors of mathematics who erroneously claimed Euclid gave a Indirect
Method. And it was professors of mathematics who mixed the two methods and jumbled them all up and thus ending with a invalid proof attempt.
>
>
> AP thinks that in (ii) we need to look for a prime factor of P!+1, but we don't, because the argument is not about saying that every number can be decomposed into prime factors, but every number has a prime number dividing it.
>
> The use of the fundamental theorem of algebra is not necessary in the proof of infinitude of primes.
The only thing I have not done for the Infinitude of Primes proof is to write it out fully in terms of Symbolic Logic. If I get time someday, I will do that, but physics and the experimental sciences are more of a priority for