Re: prove no largest prime

[Graham Cooper]

On Jul 6, 12:57 pm, quasi <qu...@null.set> wrote:
> Aatu Koskensilta wrote:
> >quasi wrote:
> >> Frederick Williams wrote:
>
> >>>As I remarked elsewhere, I think Quine proved that every
> >>>indirect proof could be made direct.  Further, I think
> >>>that L\"owenheim said as much, much earlier; and the
> >>>transformation is recursive.  I've forgotten the details.
>
> >> Moreover, the reverse transformation is even easier
> >> -- it's obvious that any direct proof can be recast as a
> >> proof by contradiction.
>
> > We can make any direct proof of A indirect by prefixing
> > it with a "Assume not-A" and tacking a triumphant
> > "A contradiction!" at the end.
>
> Right. That's exactly what I had in mind when I said it was
> obvious.
>
> >But Frederick said we can squeeze out of any /indirect/
> >proof a direct one. We can do no such thing.
>
> Intuitively that makes sense, but can you rigorously justify
> the above claim?
>
> quasi



INDIRECT PROOFS USE QUANTIFIERS

DIRECT PROOFS USE EXISTENCE OF FUNCTIONS

so a SKOLEM FUNCTION (to remove existential quantifiers)
is used to convert from 1 Proof type to the other.

*** Permit me to demonstrate ***

PROOF BY CONTRADICTION

ALL(prime) EXIST(BIGGERPRIME) BIGGERPRIME>prime

***

PROOF BY WITNESS (direct proof)

SKOLEMIZE BIGGERPRIME into a partially defined function

BIGGERPRIME(prime) > prime

***

Both proofs proceed to calculate the contradictory witness


Herc
--
http://tinyURL.com/BluePrints-PROOF

[Graham Cooper]

> On Jul 8, 12:59 pm, "Peter Webb" <r.peter.webb...@gmail.com> wrote:
>
> > It may be a proof by contradiction, but it is a constructive proof - the
> > proof algorithm generates an infinite number of primes. In my mind, this
> > puts it in a different category to those proofs which show nothing meets the
> > stated requirement.
>
> which are all syntactic knots that sci.math lap up like dogs
>


You mean like this proof?

PROGRAM START: IF NOT ( INF-LOOPS( 'START' )) THEN GOTO START

CASE A: ABOVE PROGRAM HALTS - INF-LOOPS()=TRUE
CASE B: ABOVE PROGRAM LOOPS - INF-LOOPS()=FALSE

^^^^^^

this is the HOLY GRAIL OF MATHEMATICS!

It's a piece of CRAP!

they proved anything that moves is impossible for 100 years since!

Herc
--
LOGIC
E(Y) Y={x|P(x)} <-> DERIVE( E(Y) Y={x|P(x)} )
DERIVE(T) <-> DERIVE(a) ^ DERIVE(b) ^ (a^b)->T

MATHEMATICS
E(Y) Y={x|P(x)} <-> PRVBLE( E(Y) Y={x|P(x)} )
PRVBLE(T) <-> NOT(DERIVE(NOT(T)))

[Graham Cooper]

On Jul 9, 11:44 pm, Frederick Williams <freddywilli...@btinternet.com>
wrote:
> Graham Cooper wrote:
>
> > On Jul 9, 7:28 am, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> > > Graham Cooper wrote:
>
> > > > On Jul 9, 6:42 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> > > > > On Jul 9, 2:19 am, Barry Schwarz <schwa...@dqel.com> wrote:
>
> > > > > > On Sun, 8 Jul 2012 12:59:17 +1000, "Peter Webb"

>
> > > > > > <r.peter.webb...@gmail.com> wrote:
> > > > > > >It may be a proof by contradiction, but it is a constructive proof - the
> > > > > > >proof algorithm generates an infinite number of primes. In my mind, this
> > > > > > >puts it in a different category to those proofs which show nothing meets the
> > > > > > >stated requirement.
>

> > > > > > How do you figure p(1) * p(2) * ...  * p(n) + 1 generates a prime?
>
> > > > > it's not divisible by p(1) or p(2) or .. p(n)
>
> > > _Which_ prime is "generated"?
>
> > BIGGERPRIME(11) = 2*3*5*7*11 +1
>
> > = 2311
>
> It is not the case for all n, that if you multiply the first n primes
> together and add one, that the result is a prime.
>


It's NOT?

What's a prime factor of p1*p2*p3*p4..pn +1?

*******

Now since the discussion ventured into high level definitions of
fundamental terminology I will add some strawberries to the waffle....

PROOF BY CONSTRUCTION is writing an algorithm!

PROOF BY CONTRADICTION is Resolution.

Resolution is a theorem proving method, i.e. formal logic.

Now for some meta-theory natural language modelling theory...

******

"What's a prime factor of p1*p2*p3*p4..pn +1?"

This is a RHETORIC QUESTION!

It has no answer! Which actually forms the indirect proof!

Here PRIME-FACTOR is like a PRONOUN.

PRONOUNS in English Language are like EXISTENTIAL VARIABLES!

THEY ARE UNBOUND! i.e. FREE VARIABLES.

"HE" might be bound to a previous paragraph.. "Tom entered the room.."

*******

So by stating a proposition (yes or no English sentence) it is an
assertion.

A RULE in PROLOG!

By stating a PREDICATE (yes or no 'proposition' with a (free)
variable)

you are asking a QUESTION.
i.e. you have a FREE VARIABLE in your STATEMENT that needs to be bound/
answered!

A QUERY in PROLOG!

*******

"something" is another pronoun.

The syntax of this sentence is an assertion of fact.

THERE IS SOMETHING FOGGY ABOUT THE WEATHER TODAY.

But it has a PRONOUN/VARIABLE 'something' which forms it into a
question.

it makes you answer... 'the fog!'

i.e. SOMETHING is bound to FOG. it's a type of rhetoric question.

THERE IS SOMETHING=FOGGY ABOUT THE WEATHER TODAY

is in the form of an assertion.

Resolution is finding contradiction with statements in the form of a
negative assertion, in order to prove the positive form of the
assertion.

THERE IS NO PRIME LARGER THAN SOME MAXIMUM NATURAL NUMBER!

E(m) !E(p) prime(p) & p>m

This is like a sentence in English

"there's no more soup?"

i.e the negative sentence with pronoun/free-variable is in an
assertion form, so it becomes a question to be solved/proven.

Free Variable in sentence = Solve for the Free Variable in the answer.

**** PART II ****

What is the difference between a programming language and a set
theory?

In BASIC you might use
LET x = 4

In ZFC you might use
EXIST(x) x = 4

************

But here is a difference..

ZFC
EXIST(x) x < 4

You can't do that in BASIC!
LET x < 4! //ERROR

Variables are specific bit strings essentially.

OK here's a formula of ZFC.

ALL(y) EXIST(x) x=y

What's the equivalent in BASIC or PASCAL?

You could use SKOLEM FUNCTIONS to eliminate the quantifiers get

PASCAL
x(y) = y

What about?
ALL(y) EXIST(x) x>y

PASCAL
x(y) > y // parse error

So Lambda Calculus is not as concise in it's expressiveness as ZFC.

x(y) > y

"I'm thinking of a number that's bigger than yours!"
"What number is bigger than y?"
"Given y, Solve for x?"

By stating an un-bound quantified variable EXIST(something) it asserts
a question, so in a programming language so would partially defined
Skolem Functions!

Herc
</WAFFLE Excuse='you were warned!'>

[Frederick Williams]

Graham Cooper wrote:
>
> On Jul 9, 11:44 pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:

> >
> > It is not the case for all n, that if you multiply the first n primes
> > together and add one, that the result is a prime.
> >
>
> It's NOT?
>
> What's a prime factor of p1*p2*p3*p4..pn +1?

That is unanswerable unless we know what n is.

--
The animated figures stand
Adorning every public street
And seem to breathe in stone, or
Move their marble feet.

[Graham Cooper]

On Jul 11, 9:16 pm, Frederick Williams <freddywilli...@btinternet.com>

wrote:
> Graham Cooper wrote:
>
> > On Jul 9, 11:44 pm, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
>
> > > It is not the case for all n, that if you multiply the first n primes
> > > together and add one, that the result is a prime.
>
> > It's NOT?
>
> > What's a prime factor of p1*p2*p3*p4..pn +1?
>
> That is unanswerable unless we know what n is.

you've lost me.

p1 is 2
p2 is 3
..

p1*p2*p3*p4*..*pn has EVERY prime factor. sequence thereof.

so adding 1 it will always have a remainder.




Herc

[Alan Smaill]

let's see:

2.3.5.7.11.13.17.19.23.29 + 1 = 6469693231 = 331.19545901

> Herc
>

--
Alan Smaill

[Graham Cooper]

On Jul 11, 10:12 pm, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

Let's rearrange your list of factors.

2 X (3X5X7X11X13X17X19X23X29)
3 X (1X5X7X11X13X17X19X23X29)
7 X (3X5X2X11X13X17X19X23X29)
11 X (3X5X7X2X13X17X19X23X29)
13 X (3X5X7X11X2X17X19X23X29)
17 X (3X5X7X11X13X2X19X23X29)
19 X (3X5X7X11X13X17X2X23X29)
23 X (3X5X7X11X13X17X19X2X29)
29 X (3X5X7X11X13X17X19X23X2)

Divide the product 6469693230 by any of 2,3,7,11,13,17,19,23,29

and you'll get the Right Hand Term in brackets. Add 1 and you'll get
1/2 remainder, 1/3 remainder, 1/7 remainder, 1/13 remainder, .... 1/29
remainder.

Anyway, I'm only interested in the Prime Proof per se because it
started all that "incomplete list" rot!

This finite list of primes is MISSING ONE!
This infinite list of rationals is MISSING SOMETHING ELSE!
This infinite list of REALS is MISSING ONE!
This infinite list of SUBSETS is MISSING A SET!
This infinite list of LOGIC FORMULA is MISSING SOME!
...

You guys think a SET has higher information entropy than a LIST!

Other way round, you couldn't recognise an INDEX to a RELATION (set)
because your maths courses are just compiled notes on computation and
none of you learnt how to program!

Wait! Here's a program that proves uncomputable real numbers make
really really big sets!

PROGRAM START: if not(INF-LOOPS('START')) then goto START

Whoooa Chaitan's Omega! A witness to uncountability!

You guys can't count all right!

Herc

[Frederick Williams]

Graham Cooper wrote:
>
> On Jul 11, 9:16 pm, Frederick Williams <freddywilli...@btinternet.com>
> wrote:
> > Graham Cooper wrote:
> >
> > > On Jul 9, 11:44 pm, Frederick Williams <freddywilli...@btinternet.com>
> > > wrote:
> >
> > > > It is not the case for all n, that if you multiply the first n primes
> > > > together and add one, that the result is a prime.
> >
> > > It's NOT?
> >
> > > What's a prime factor of p1*p2*p3*p4..pn +1?
> >
> > That is unanswerable unless we know what n is.
>
> you've lost me.
>
> p1 is 2
> p2 is 3
> ..
>
> p1*p2*p3*p4*..*pn has EVERY prime factor. sequence thereof.
>
> so adding 1 it will always have a remainder.

That p1*p2*p3*p4...pn + 1 is nor divisible by any of p1, p2, p3, p4,...
pn I don't doubt; but that does not mean that p1*p2*p3*p4...pn + 1 is a
prime. Hence my question upstream: _Which_ prime is "generated"?
Happily Alan Smaill has provided an example to illustrate my point.

[Graham Cooper]

On Jul 11, 10:49 pm, Frederick Williams

<freddywilli...@btinternet.com> wrote:
> Graham Cooper wrote:
>
> > On Jul 11, 9:16 pm, Frederick Williams <freddywilli...@btinternet.com>
> > wrote:
> > > Graham Cooper wrote:
>
> > > > On Jul 9, 11:44 pm, Frederick Williams <freddywilli...@btinternet.com>
> > > > wrote:
>
> > > > > It is not the case for all n, that if you multiply the first n primes
> > > > > together and add one, that the result is a prime.
>
> > > > It's NOT?
>
> > > > What's a prime factor of p1*p2*p3*p4..pn +1?
>
> > > That is unanswerable unless we know what n is.
>
> > you've lost me.
>
> > p1 is 2
> > p2 is 3
> > ..
>
> > p1*p2*p3*p4*..*pn  has EVERY prime factor.  sequence thereof.
>
> > so adding 1 it will always have a remainder.
>
> That p1*p2*p3*p4...pn + 1 is nor divisible by any of p1, p2, p3, p4,...
> pn I don't doubt; but that does not mean that p1*p2*p3*p4...pn + 1 is a
> prime.  Hence my question upstream: _Which_ prime is "generated"?
> Happily Alan Smaill has provided an example to illustrate my point.

if n=10 then why do you claim you don't know what n is.

Herc

[Alan Smaill]

But you won't get 1/331 remainder that way;
and 331 is a prime factor, I think you'll find.

>
> You guys can't count all right!

Hmm ............

Not the best chosen sign-off line, in the circumstances.

>
> Herc

--
Alan Smaill

[Frederick Williams]

If you had said n = 10, then I would have, but I think this is the first
time you've mentioned it in this thread. But look, if we consider

2.3.5.7.11.13.17.19.23.29 + 1 = 6469693231

_that_ as it stands does not "generate" a prime. Rather, we need to
factor 6469693231 to learn that 331 is its smallest prime factor/

[Graham Cooper]

On Jul 11, 11:10 pm, Frederick Williams

> What's a prime factor of p1*p2*p3*p4..pn +1?


"That is unanswerable unless we know what n is. "

THAT IS UNANSWERABLE??

That's not an ANSWER, that's an EXPLANATION, and a wrong one at that.

Swap less for till.

A deceitful misrepresentation of facts is almost lying!

Herc

[Pubkeybreaker]

On Jul 11, 9:10 am, Frederick Williams <freddywilli...@btinternet.com>

Idiot. You are deliberately mis-using the word 'generate'.

The product of the first 10 primes plus 1 does indeed GENERATE a new
prime.

The new in this case prime is LP(10), where LP(n) is the largest (or
one can choose the
smallest) prime factor of p(1)*p(2) * .... p(n) + 1.

LP(10) is a well defined number whose existence is not in doubt.
Saying that the answer is LP(10)
is every bit as explicit as saying that it is 331. They are just
DIFFERENT REPRESENTATIONS
of the same number.

You however, are totally confused. You believe that 'generate' either
means
'represent in decimal form' or that it means 'give an explicit
elementary algebraic function
that expresses the answer in a form I am willing to accept'.


The existence of a number has nothing to do with its REPRESENTATION.
Numbers may be represented in MANY ways.

It is often in mathematics that in order to find the DECIMAL
representation
of a number one must apply a numerical algorithm. Suppose I told you
that
the answer to some (unspecified) mathematical problem was sin( 2 pi/
n).

You would argue that that does not 'generate' an answer to the
problem.
Yet the function LP(n) is as well defined a function as is sin(x).
But in
both instances one must apply a numerical algorithm to get a decimal
representation of the answer.

You are totally confused over the word 'generate' and fail to
understand that
the representation of a number has nothing to do with its existence.

Stop your prattling. You are beginning to sound like a crank.

[Graham Cooper]

OK it only works up to pn+1 < pn^2

My mistake, based on 4 consecutive erroneous posts I'm reading here!

Herc

[Gary Forbis]

On Jul 11, 5:53 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> if n=10 then why do you claim you don't know what n is.

You didn't specific what n was so it could be any number.
It could be 10.

Euclid's proof doesn't say the product of the first n primes
plus one is a prime. It only says that given any set of
primes their product plus one is either a prime or can
be factored by primes not in the set. For instance
2*5+1 = 11. 11 is a prime. 2*7+1 = 15. 15 = 3*5.
I'm not a familiar with Euclid's proof as I should be.
I don't think the set even has to be unique primes,
that is 2*2+1 = 5. 2*2*2*2+1 = 17. But while limit
oneself to one greater than the product of primes
plus 1? One less than the product of an set of primes
will also be a prime or be a product of primes not
in the set. 2*2-1 = 3. 2*2*2-1 = 7. 2*3-1 = 5.
3*5-1 = 14 = 2*7. Heck, I will go even further.
Any set of primes, unique or not, plus any number
not itself a product of any subset of those primes,
including the set itself, will be a prime or factored by
primes not in the set. 2*3 + 11 = 17. 2*5 + 7 = 17.
3*11 + 1 = 34 = 2 * 17.

[Graham Cooper]

Ah OK. Guess I can't blame you guys for doing all the high level
stuff and never checking your calcs!

Herc

[Frederick Williams]

I am quoting Graham Cooper, the word generate is his. I put in scare
quotes because he hadn't made clear what he meant by it.

>
> The product of the first 10 primes plus 1 does indeed GENERATE a new
> prime.

Only if one knows what 'generate' means.

[dilettante]

"Gary Forbis" <forbi...@msn.com> wrote in message
news:592693b1-904b-42e5...@nl1g2000pbc.googlegroups.com...

The last statement is not true. The number added to the product (of course
you meant "any product" where you said "any set") needs to not be a multiple
of any prime in the set - simply requiring that it not be a product of a
subset is not enough. For example, let the set of primes be {2}. Then 2 + 6
= 8 is a counterexample to the claim.

[Graham Cooper]

SUMMING UP on 2 types of Proof!

> What about?
> ALL(y) EXIST(x) x>y
>
> PASCAL
> x(y) > y    // parse error
>
> So Lambda Calculus is not as concise in it's expressiveness as ZFC.
>
> x(y) > y
>
> "I'm thinking of a number that's bigger than yours!"
> "What number is bigger than y?"
> "Given y, Solve for x?"

the main nat-lang reading is of course

"Given any num this can output a bigger num"

BIGGER(num) > num (SKOLEM VERSION)
<=>
ALL(num) EXIST(b) b > num (QUANTIFIER VERSION)

where BIGGER() is a SKOLEM FUNCTION for Existential Var b

*****************************************************
Just put the outer ALL(vars) into a parameter list
since b DEPENDS ON num, num becomes it's parameter
****************************************************

BIGGER() might be

FUNCTION BIGGER(n)
RETURN n+3

A Skolem Function is only PARTIALLY DEFINED THOUGH

SKOLEM-FUNCTION BIGGER(n)
BIGGER(n) > n


************


1 use a NEGATIVE PROPOSITION
2 use a PREDICATE with a PARTIALLY SPECIFIED VARIABLE

to form a THEOREM into QUESTION FORM.

1 IS THIS RIGHT?
2 FILL IN THIS VALUE?

2 THEOREM TYPES <=> 2 QUESTION TYPES <=> 2 PROOF TYPES

1 Proof By Contradiction
2 Proof By Construction / Witness


*************

This is the PROLOG Grammar.

It's a SUPERSET of PREDICATE CALCULUS.
...formula1 -> formula2 is a formula etc..

...by adding the QUERY Command.


VTPROLOG Grammar


sentence ::- rule | query | command
rule ::- head '.' | head ':-' tail '.'
query ::- '?-' tail '.'
command ::- '@' file_name '.'
head ::- goal
tail ::- goal | goal ',' tail
goal ::- constant | variable | structure
constant ::- {quoted string} | {token beginning with 'a' .. 'z'}
variable ::- {identifier beginning with 'A' .. 'Z' or '_' }
structure ::- functor '(' component_list ')'
functor ::- {token beginning with 'a' .. 'z'}
component_list ::- term | term ',' component_list
term ::- goal | list
list ::- '[]' | '[' element_list ']'
element_list ::- term | term ',' element_list | term | term
file_name ::- {legitimate DOS file name, must be surrounded with
single quotes if it contains a '.',':' or '\'}



Herc

[Graham Cooper]

> ************
>
> 1 use a NEGATIVE PROPOSITION
> 2 use a PREDICATE with a PARTIALLY SPECIFIED VARIABLE
>
> to form a THEOREM into QUESTION FORM.
>
> 1 IS THIS RIGHT?
> 2 FILL IN THIS VALUE?
>

English Grammar seems to have metamorphosed over time..

NEGATIVE PROPOSITION
NOT THIS IS RIGHT!
THIS IS NOT RIGHT!
THIS IS NOT RIGHT?
IS THIS NOT RIGHT?
IS THIS RIGHT?

*try to find a contradiction to negative proposition*

misc.writing added for a reason! ;)

Herc

[Gary Forbis]

On Jul 11, 9:58 am, "dilettante" <n...@nonono.no> wrote:
> "Gary Forbis" <forbisga...@msn.com> wrote in message

6 is a product of 2 it is 2*3.

[dilettante]

"Gary Forbis" <forbi...@msn.com> wrote in message

news:ec7ccb3d-3e77-441e...@y3g2000pbc.googlegroups.com...

You said "product of any subset of those primes". The literal (and natural)
interpretation of that phrase is "a number which can be written as a product
ALL of whose factors are among those primes". Indeed, if you meant simply
"multiple", why invoke "subsets"? Any number that is a multiple of a subset
is a multiple of one of the primes.

[Gary Forbis]

Because I was writing in a hurry before leaving for work and
improvising.
Sorry for any confusion. My mind was clearer than my fingers. I knew
what I intended and had trouble communicating it.

[Michael Stemper]

In article <fwe4npe...@eriboll.inf.ed.ac.uk>, Alan Smaill <sma...@SPAMinf.ed.ac.uk> writes:
>Graham Cooper <graham...@gmail.com> writes:
>> On Jul 11, 9:16 pm, Frederick Williams <freddywilli...@btinternet.com> wrote:
>>> Graham Cooper wrote:

>>> > > It is not the case for all n, that if you multiply the first n primes
>>> > > together and add one, that the result is a prime.
>>>
>>> > It's NOT?
>>>
>>> > What's a prime factor of p1*p2*p3*p4..pn +1?
>>>
>>> That is unanswerable unless we know what n is.
>>
>> you've lost me.
>>
>> p1 is 2
>> p2 is 3
>> ..
>>
>> p1*p2*p3*p4*..*pn has EVERY prime factor. sequence thereof.
>>
>> so adding 1 it will always have a remainder.

When dividing by any prime less than or equal to pn.

>let's see:
>
> 2.3.5.7.11.13.17.19.23.29 + 1 = 6469693231 = 331.19545901

You don't need to go that high:

(2 * 3 * 5 * 7 * 11 * 13) + 1 = 30031 = 59 * 509

--
Michael F. Stemper
#include <Standard_Disclaimer>
Life's too important to take seriously.

[M3]

On Jul 13, 8:14 am, mstem...@walkabout.empros.com (Michael Stemper)
wrote:
> In article <fwe4npeh4ys....@eriboll.inf.ed.ac.uk>, Alan Smaill <sma...@SPAMinf.ed.ac.uk> writes:

> >Graham Cooper <grahamcoop...@gmail.com> writes:
> >> On Jul 11, 9:16 pm, Frederick Williams <freddywilli...@btinternet.com> wrote:
> >>> Graham Cooper wrote:
> >>> > > It is not the case for all n, that if you multiply the first n primes
> >>> > > together and add one, that the result is a prime.
>
> >>> > It's NOT?
>
> >>> > What's a prime factor of p1*p2*p3*p4..pn +1?
>
> >>> That is unanswerable unless we know what n is.
>
> >> you've lost me.
>
> >> p1 is 2
> >> p2 is 3
> >> ..
>
> >> p1*p2*p3*p4*..*pn  has EVERY prime factor.  sequence thereof.
>
> >> so adding 1 it will always have a remainder.
>
> When dividing by any prime less than or equal to pn.
>
> >let's see:
>
> >  2.3.5.7.11.13.17.19.23.29 + 1 = 6469693231 = 331.19545901
>
> You don't need to go that high:
>
>   (2 * 3 * 5 * 7 * 11 * 13) + 1 = 30031 = 59 * 509
>
> --
> Michael F. Stemper
> #include <Standard_Disclaimer>
> Life's too important to take seriously.

let n0=largest prime Q.E.D.
Musatov, Artist