C.B.BOYER,UTA MERZBACH, EXCELLENT BOOK, BUT PITIFUL FAILURE IN IP PROOF
Ludwig Plutonium
Merzbach,1991 page 115
" Book IX, the last of the three books on theory of numbers, contains
several theorems that are of special interest. Of these the most
celebrated is Proposition 20: "Prime numbers are more than any assigned
multitude of prime numbers." That is, Euclid here gives the well-known
elementary proof that the number of primes is infinite. The proof is
indirect, for one shows that the assumption of a finite number of
primes leads to a contradiction. Let P be the product of all the
primes, assumed to be finite in number, and consider the number N = P +
1. Now, N cannot be prime, for this would contradict the
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
assumption that P was the product of all primes. Hence, N is composite
and
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
must be measured by some prime p. But p cannot be any of the prime
factors in P, for then it would have to be a factor of 1. Hence, p must
be a prime different from all of those in the product P; therefore, the
assumption that P was the product of all the primes must be false. "
If Boyer and Merzbach had just a little math logic they would never
have printed the above silliness. Here is what they should have printed
to make a valid proof. Now, N is prime, because all the primes that
exist leave a remainder of 1, contradiction. Proof. QED
Math professors in the midst of a Ramanujan, Ramanujan at will could
turn them into a laughing stock of jello.
Math professors in the midst of Ludwig Plutonium, are all quiet, for
they know by instinct that they are in the prescence of supergenius.
Watch the movie Tombstone. See the interplay between Doc Holliday and
Ringo. Ringo was a big mouthed arrogant show-off, but he, Doc and Wyatt
Earp knew by instinct that Doc was the fastest. Of course that is a
fiction and a fiction movie. But my point is quite clear. When
Ramanujan lived, entered a room full of math professors, they all knew
they were in "his presence". Ramanujan was the fastest math gun early
1900's. VonNeumann was the fastest math gun after Ramanujan. LP is that
now.
Ludwig Plutonium
Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:
> If Boyer and Merzbach had just a little math logic they would never
> have printed the above silliness. Here is what they should have printed
> to make a valid proof. Now, N is prime, because all the primes that
> exist leave a remainder of 1, contradiction. Proof. QED
I need to talk about this excellent book, and by the way, thanks John
Wiley for printing books on acid free paper as the one here. I own this
book and it is top notch. I nowadays never buy a book that is not on
acid free paper. If not on acid free then I borrow the book from the
library and photocopy it onto acid free paper. Those publishers who do
not state "acid free" I think are scalawag moneygrubs.
I hope Merzbach continues to revise this book. Of course, rewrite IP
into a valid proof, giving me the proper credit due. And the chart or
table at the end of chronological list was very helpful to me in naming
the isotopes. Hope you expand on that chronology. The above book is in
the Plutonium Atom Foundation library as all of my books are.
John Greene
Ludwig.P...@dartmouth.edu (Ludwig Plutonium) wrote:
> If Boyer and Merzbach had just a little math logic they would never
> have printed the above silliness. Here is what they should have printed
> to make a valid proof. Now, N is prime, because all the primes that
> exist leave a remainder of 1, contradiction. Proof. QED
>
There is a flaw in your reasoning, Ludwig. Flaw is probably too strong a term
but as has been pointed out, if a number has divisors (prime or not) other than
one, then that number is not prime. For example, to show that 144 is not prime,
it is enough to show that 144 = 12*12, even though 12 is not prime. Now it
is a theorem that if n is not divisible by any prime except (possibly) itself,
then n is prime. You are using this theorem when you state that your N is
prime. I'll ask you this question: Do you believe the theorem I just
mentioned? If so, can you prove it? The only proofs I know of require
induction, which I believe you do not accept.
John
Ludwig Plutonium
jgr...@frodo.d.umn.edu (John Greene) writes:
> There is a flaw in your reasoning, Ludwig.
No there is not. I could explain it all day and night and still not get
through. In math, many times you should say to yourself, John Greene I
am wrong. Then see how it works. Read Karl Heuer's post or Ron Bruck's,
perhaps the light will go on.
David Kastrup
People giving advice they do not adhere to themselves do seem sort
of discreditable.
--
David Kastrup d...@pool.informatik.rwth-aachen.de
Tel: +49-241-72419 Fax: +49-241-79502
Goethestr. 20, D-52064 Aachen
John Greene
Ludwig.P...@dartmouth.edu (Ludwig Plutonium) wrote:
Ludwig,
I have read Karl Heuer's post and Ron Bruck's. But if you ask them,
I think that they would say that they implicitly used mathematical induction.
The proof requires the fact that every positive integer can be factored
into primes. (In particular, it does not ned the full force of unique
factorization.) However, I have never seen a proof of this fact (that all
positive integers can be factored into primes) which does not use induction.
Heuer and Bruck probably accept mathematical induction, but it is my impression
that you don't.
Don't get me wrong--I'm not saying that your proof is wrong, only that
it is incomplete. If you don't want to require mathematical induction in
your proof, then you must find a new proof of the following result:
THEOREM If N is an integer greater than 1, and N has no prime divisors
(except possibly for itself) then N is prime.
If I am wrong in my belief that you don't accept mathematical induction,
then I appoligize for bringing up the whole matter. Otherwise, your proof,
for now, has a gap to be filled. I don't know how to fill it.
John
Ludwig Plutonium
jgr...@frodo.d.umn.edu (John Greene) writes:
> I have read Karl Heuer's post and Ron Bruck's. But if you ask them,
> I think that they would say that they implicitly used mathematical induction.
Two questions: (1) were you condemne? The reason I ask is because I
do not have all of my Internet posts collated yet. And my insufficient
memory suggests that you pestered me and were consequently condemned?
Does any reader out there have a complete up to date list of
condemnations?
(2) I think the Internet is a good place to put academics on probation
for excessive foul ups. Two names who immediately should be put on
probation are Merriman and Adams. Re: Merriman, I noticed on his posts
before condemnation that he has neon lights on---- Plasma umpty ump,
Math dept umpty ump. I have the suspicion that Merriman, as most other
insecure academics, self PROPS himself out of proportion. Does Merriman
say have a part time job there where he comes in to shuffle computer
paper and then posts that he is head honcho of PLASMA -- ---. Regards
to Greene. The statements you are making about math I find it hard to
believe you teach math, and if that is the case I feel you should be
put on teaching probation. It is not the ignorance of math that shines
through your posts so much as the ABSOLUTE ARROGANCE coupled with the
ignorance that makes me say "I pity the students if Greene teaches".
P.S. Dick Adams is a redneck academic that needs to be in "honest
labor". Two years of milking cattle, wheat farming, or West Virginia
coal mining will make a honest man out of Dick. I pity very much, any
student that sits in his classes if he teaches. Teaches is too fine of
a word.
John Greene
Ludwig.P...@dartmouth.edu (Ludwig Plutonium) wrote:
> In article <jgreene-0609...@131.212.112.11>
> jgr...@frodo.d.umn.edu (John Greene) writes:
>
> > I have read Karl Heuer's post and Ron Bruck's. But if you ask them,
> > I think that they would say that they implicitly used mathematical
induction.
>
> Two questions: (1) were you condemne? The reason I ask is because I
> do not have all of my Internet posts collated yet. And my insufficient
> memory suggests that you pestered me and were consequently condemned?
> Does any reader out there have a complete up to date list of
> condemnations?
>
No, I don't believe you have ever condemned me. You are correct that I have
responded to your threads before. The last time I did, it was basically about
this same issue. You asked me if Hardy & Wright use induction to prove
unique factorization and I told you that they did. That is the last time
you responded to me in the past. I assume that I've never been condemned
because I've never attacked you personally or your ideas but rather pointed
out some things that I am uncomfortable with.
> to Greene. The statements you are making about math I find it hard to
> believe you teach math, and if that is the case I feel you should be
> put on teaching probation. It is not the ignorance of math that shines
> through your posts so much as the ABSOLUTE ARROGANCE coupled with the
> ignorance that makes me say "I pity the students if Greene teaches".
I don't understand this. I certainly don't mean to sound arrogant and I
appoligize if that is what I conveyed. Ludwid, you are not a detail person-
you look at grand scheme of things, rather than get bogged down in little
details. That is good, but sometimes details are important. I am not
a great mathematician, nor will I ever be a great mathematician. But someone
has to look at details. When I look at your proof of IP, I get worried about
one detail: the claim that the product of the primes + 1 is prime. I can
prove this using mathematical induction, but I can not prove it without using
induction. It is my understanding that you reject induction. If that is the
case, I can not prove that there are infinitely many primes because I don't
know how to get over the detail. To my mind, your proof is incomplete until
this detail is filled in.
Let me one last time raise my objection:
Theorem Let n > 2 be an integer. If n has no proper prime divisors, then
n is prime.
(Note that until proved impossible, one might concieve of a case where a
number has divisors, but no prime divisors.)
You use this theorem in your proof of IP so does Karl Heuer (someone
correct me if this is not true). I believe that this theorem is
true, so I believe your proof. HOWEVER, I can only prove this theorem if
I use mathematical induction. (I'm not saying that it can not be proved
without induction, just that in my ignorance, I don't know how to do it.)
So, if you do not allow proofs by induction, you should first prove this
theorem without it, and then you will have a consistant proof of IP. If
I am misunderstanding your position on induction, I appoligize.
One final point: I would find it helpful if you worded your responces
differently. I understand you are a busy person, but posts like:
"You are ignorant of math" don't do me any good. If you said:
"The detail can be filled in by the following argument..." or "You think
I said ... but I really said ..." That would be very helpful.
John
Ludwig Plutonium
jgr...@frodo.d.umn.edu (John Greene) writes:
> When I look at your proof of IP, I get worried about
> one detail: the claim that the product of the primes + 1 is prime.
A math proof is a slab of logic. Not an extraction of a tidbit to
examine. "To any finite set of primes, there is a largest P, multiply
them add 1. Call it Q. Q is necessarily prime because all in the finite
set leave remainder 1 and is larger than P." Math induction never
enters this picture nor UPFAT which is used. It is a pure slab of
logic. The only hardship is that out of 100,000, nay 1,000,000 people,
perhaps only 1 person has logic ingrained in their genetic makeup that
can "see the flow of the pure slab of logic". All the rest of the
people nitpick and get sidetracked, or like a philosopher to fix on
some dumb stupid nonessential.
Ludwig Plutonium
> Theorem Let n > 2 be an integer. If n has no proper prime divisors, then
> n is prime.
This is not a theorem. This is just the definition of prime reworded.
Ludwig Plutonium
> You asked me if Hardy & Wright use induction to prove
> unique factorization and I told you that they did.
I remember in 1993 asking if induction was used in all of the
so-called proofs (actually they are fakes) of FLT for various
exponents. I do not remember the gentleman who responded with a
"definite yes", but I do not believe it was you Mr. Greene, that all
alleged proofs of FLT for exponents used Math Induction, be it regular
Math Induction or Infinite Descent. This was important for me at the
time because I had posted two so-called proofs of Goldbach, both using
Math Induction and Descent. That was my turning away from my proofs,
realizing both were Fakes. That was the turning point of knowing that
counterexamples of Goldbach exist, e.g., ...000010000100010010.
I was never worried about Math Induction with UPFAT. Why you concern
yourself with UPFAT and Math Induction? In fact I can give you a one
paragraph or two sentence proof of UPFAT without math induction ever
entering the picture. It is for this reason that I ask of your math
credentials. Why the fixation on math induction and UPFAT.
David Seal
No, the reworded definition of "prime" talks about "divisors", not
"prime divisors":
Reworded definition: Let n > 2 be an integer. If n has no proper
divisors, then n is prime.
which is a rewording of:
Definition: An integer n > 2 is prime if it has no divisor d such
that d != n and d != 1. (Using "!=" to mean "is not
equal to".)
The theorem that John Greene states is a rewording of:
Theorem: An integer n > 2 is prime if it has no prime divisor d such
that d != n and d != 1.
This isn't a definition of "prime", since it is circular: it won't
tell you what "prime" means until you know what "prime divisor" means.
It is the step of going from the reworded definition to the theorem
John quotes that he doesn't know how to perform without the use of
induction. Nor do I.
So (without using induction) how do you show that if a number has no
proper prime divisors, then it has no proper divisors at all and so is
prime?
David Seal
ds...@armltd.co.uk
John Greene
Ludwig.P...@dartmouth.edu (Ludwig Plutonium) wrote:
> > Theorem Let n > 2 be an integer. If n has no proper prime divisors, then
> > n is prime.
>
> This is not a theorem. This is just the definition of prime reworded.
Ludwig,
The usual definition of a prime number is a positive integer (>1 actually)
which can not be written as a product of smaller positive integers. Thus,
24 = 6*4 is not prime, even though it has not been shown that 24 has any
prime factors. The theorem I listed is not just a rewording of what it means
to be prime. In other words, to show that a number is not prime, you do not
have to produce a PRIME factor of the number, just a factor of the number.
That is why one must prove things like "every integer > 1 is divisible by
some prime" a statement which is also a theorem and not just a rewording of
the definition.
John
Gareth Rees
> When I look at your proof of IP, I get worried about one detail: the
> claim that the product of the primes + 1 is prime.
Ludwig Plutonium (Ludwig.P...@dartmouth.edu) replied:
> To any finite set of primes, there is a largest P, multiply them add
> 1. Call it Q. Q is necessarily prime because all in the finite set
> leave remainder 1 and is larger than P.
Counterexample:
(2 * 3 * 5 * 7 * 11 * 13) + 1 = 30031 = 59 * 509
[followups redirected accordingly]
--
Gareth Rees
Michael Stemper
>In article <jgreene-0809...@131.212.112.11>
>jgr...@frodo.d.umn.edu (John Greene) writes:
>
>> When I look at your proof of IP, I get worried about
>> one detail: the claim that the product of the primes + 1 is prime.
>
> A math proof is a slab of logic. Not an extraction of a tidbit to
>examine. "To any finite set of primes, there is a largest P, multiply
>them add 1. Call it Q. Q is necessarily prime because all in the finite
>set leave remainder 1 and is larger than P."
I have some questions about your proof. Please help me to understand it,
so that I have the opportunity to learn from a math genius like you.
Is { 2, 3, 5, 7, 11, 13 } a finite set of primes?
Is 13 the largest prime in this set?
Is the product of the members of this set 30030?
Is 30030+1 equal to 30031?
Does this mean that, for this set, Q is 30031?
Does this mean that 30031 is a prime?
When I multiply 59*509 on my calculator, I get 30031. Since 30031
is obviously prime according to your proof, my calculator must be
broken. What is 59*509?
n primes product product-1 product+1
1 2 2 undef prime
2 2,3 6 prime prime
3 2,3,5 30 prime prime
4 2,3,5,7 210 11*19 prime
5 2,3,5,7,11 2310 prime prime
6 2,3,5,7,11,13 30030 prime 59*509
7 2,3,5,7,11,13,17 510510 61*8369 19*97*277
8 2,3,5,7,11,13,17,19 9699690 53*197*929 347*27953
--
Michael F. Stemper | Any true system of justice should not
mste...@empros.com | look simply to averages. True justice
#include <Standard_Disclaimer> | comes from looking and dealing with
| the marginal cases.
Daniel Jimenez
: [deleted]
: I have some questions about your proof. Please help me to understand it,
: so that I have the opportunity to learn from a math genius like you.
:
: Is { 2, 3, 5, 7, 11, 13 } a finite set of primes?
: Is 13 the largest prime in this set?
: Is the product of the members of this set 30030?
: Is 30030+1 equal to 30031?
: Does this mean that, for this set, Q is 30031?
: Does this mean that 30031 is a prime?
:
: When I multiply 59*509 on my calculator, I get 30031. Since 30031
: is obviously prime according to your proof, my calculator must be
: broken. What is 59*509?
A lot of people have observed the same thing in Ludwig's posts. I think
the key is to realize that we are assuming (for example) that
{ 2, 3, 5, 7, 11, 13 }, or, in Ludwig's generalized proof, { a, b, c, d },
are *all* the primes. So if we assume this, then yes, 30031 is prime
because it is not the product of any prime (remember, we just listed all
the primes and 59 and 509 weren't among them).
Of course, if we assume { a, b, c, d } are all the primes, then the moon
is made of green cheese. That's how you prove there are infinitely primes:
by contradiction.
--
Daniel Jimenez djim...@ringer.cs.utsa.edu
"I've so much music in my head" -- Maurice Ravel, shortly before his death.
" " -- John Cage
Michael Hardy
> I have some questions about your proof. Please help me to understand it,
> so that I have the opportunity to learn from a math genius like you.
>
>
> Is { 2, 3, 5, 7, 11, 13 } a finite set of primes?
>
> Is 13 the largest prime in this set?
>
> Is the product of the members of this set 30030?
>
> Is 30030+1 equal to 30031?
>
> Does this mean that, for this set, Q is 30031?
>
> Does this mean that 30031 is a prime?
>
> When I multiply 59*509 on my calculator, I get 30031. Since 30031
> is obviously prime according to your proof, my calculator must be
> broken. What is 59*509?
30031 was only proved prime subject to the _assumption_ (later
proved false) that there are only finitely many primes. Since the assumption
was later seen to be false, we cannot take the proof of the conclusion that
Q+1 is prime as valid. It is simply not true that if you multiply the first
several primes and add 1, you always get a prime, as your counterexample shows.
The proof would be less misleading if it read as follows: Start with any
finite set S of primes. Multiply them, getting Q, and add 1, getting Q+1.
Q+1 cannot be divisible by any of the primes in the set S since Q is divisible
by all of them. Factor Q+1, getting some primes. (Maybe Q+1 is prime and
you get only Q+1, or maybe not and you get some others). Whatever primes you
get cannot belong to S since Q+1 is not divisible by anything in S.
Therefore S is not the set of _all_ primes. This works for _any_ finite set
S of primes, so no finite set can contain all primes.
Mike Hardy ha...@stat.umn.edu
Michael Stemper
>Michael F. Stemper (mste...@empros.com) writes:
>
>> I have some questions about your proof. Please help me to understand it,
>> so that I have the opportunity to learn from a math genius like you.
>>
>>
>> Is { 2, 3, 5, 7, 11, 13 } a finite set of primes?
>
> 30031 was only proved prime subject to the _assumption_ (later
>proved false) that there are only finitely many primes. Since the assumption
I would like to refer you to what LvP wrote:
In article <34obrg$2...@dartvax.dartmouth.edu>, Ludwig.P...@dartmouth.edu (Ludwig Plutonium) writes:
> A math proof is a slab of logic. Not an extraction of a tidbit to
>examine. "To any finite set of primes, there is a largest P, multiply
>them add 1. Call it Q. Q is necessarily prime because all in the finite
>set leave remainder 1 and is larger than P."
He did not make the assumption that there are only finitely many primes.
He did not make the assumption that there exists a finite set that contains
all of the primes. He stated "any finite set of primes". He did not attempt
to contradict the assumption that the given set included all primes,
because he didn't make that assumption. What he tried to do was claim that
you can always generate a prime from a set of primes in this way. Of
course, you can't. Actually, a simpler counter-example than I posted
previously would be the set { 2, 7 }.
Any finite set of primes does imply the existence of a prime not in
that set. This is the basis for the proof you outline below. (The
first proof I ever learned, BTW.) However, there does not exist, AFAIK,
any effective procedure to generate new primes. That's why it's such
big news when somebody does discover a new one.
>was later seen to be false, we cannot take the proof of the conclusion that
>Q+1 is prime as valid. It is simply not true that if you multiply the first
>several primes and add 1, you always get a prime, as your counterexample shows.
> The proof would be less misleading if it read as follows:
Well, if he wrote this, it wouldn't just be less misleading, it would
be correct.
> Start with any
>finite set S of primes. Multiply them, getting Q, and add 1, getting Q+1.
>Q+1 cannot be divisible by any of the primes in the set S since Q is divisible
>by all of them. Factor Q+1, getting some primes. (Maybe Q+1 is prime and
>you get only Q+1, or maybe not and you get some others). Whatever primes you
>get cannot belong to S since Q+1 is not divisible by anything in S.
>Therefore S is not the set of _all_ primes. This works for _any_ finite set
>S of primes, so no finite set can contain all primes.
>
> Mike Hardy ha...@stat.umn.edu
>
--