Prime Generalization Conjecture
MeAmI.org
1/2 of some other number +1.
Richard Heathfield
> RULE: EVERY PRIME number is exactly
> 1/2 of some other number +1.
2 is a counter-example unless "other number" can mean "same number".
In general, if P = X / 2 + 1, then X = (P - 1) * 2, so your claim is
tantamount to saying that for each prime number there exists some
other number that can be reached by subtracting one from the prime
number and then doubling the result. This leads us to some other
observations of equal interest:
RULE: EVERY PERFECT SQUARE is exactly 1/2 of some other number +1.
RULE: EVERY PERFECT CUBE is exactly 1/2 of some other number +1.
RULE: EVERY FIBONACCI NUMBER is exactly 1/2 of some other number +1.
RULE: EVERY LUCAS NUMBER is exactly 1/2 of some other number +1.
RULE: EVERY INTEGER is exactly 1/2 of some other number +1.
Good solid stuff, but I don't think it's going to win a Fields Medal
any time soon.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Forged article? See
http://www.cpax.org.uk/prg/usenet/comp.lang.c/msgauth.php
"Usenet is a strange place" - dmr 29 July 1999
William Elliot
> RULE: EVERY PRIME number is exactly
> 1/2 of some other number +1.
>
So what? For all x, x = (2x - 2)/2 + 1.
Trivial rule.
Every integer (rational number, real number, complex number) is half
of some integer (resp. rational number, real number, complex number)
plus one.
Exercise. How many primes are half of some prime plus one?
Musatov
Prime Generalization (by Musatov): EVERY PRIME number is exactly
1/2 of a number +1.
Variant: RULE(0): EVERY PRIME number greater than two is exactly
Musatov
1/2 of a number +1.
Variant: RULE(0): EVERY PRIME number greater than two is exactly
Musatov
None. Half of a prime number is not a whole number.
17/2=8.5+1=9.5 (NP).
So we have the result:
RULE: No prime number is 1/2 another prime number plus one.
Musatov
But perhaps this is what you meant.
Inverse/Additive prime property per Musatov: (below)
RULE: EVERY PRIME number is twice
a number +1.
3=1*2+1
5=2*2+1
7=3*2+1
11=5*2+1
13=6*2+1
17=8*2+1
19=9*2+1
23=11*2+1
29=14*2+1
31=15*2+1
37=18*2+1
41=20*2+1
43=21*2+1
47=23*2+1
51=25*2+1
53=26*2+1
And combined Prime Generalization: (Musatov)
RULE: Every prime is 1/2 a number +1 and twice a number plus +1.
Now consider the series again, but this time plot the additive
difference between first and next doubled number.
In the first two terms we write....
3=1*2+1 #
5=2*2+1 1 because the difference between the doubled numbers from
the first to the next was "1".
And we continue....
(here is the full table)
3=1*2+1 #
5=2*2+1 1
7=3*2+1 1
11=5*2+1 2
13=6*2+1 1
17=8*2+1 2
19=9*2+1 1
23=11*2+1 2
29=14*2+1 3
31=15*2+1 1
37=18*2+1 3
41=20*2+1 2
43=21*2+1 1
47=23*2+1 2
51=25*2+1 2
53=26*2+1 1
I would like to see if these reveals more to clarity to series of
primes...
Ben Bacarisse
> On Jun 20, 12:13 am, William Elliot <ma...@rdrop.remove.com> wrote:
>> Exercise. How many primes are half of some prime plus one?
>
> None. Half of a prime number is not a whole number.
Except for 2.
<snip>
> So we have the result:
>
> RULE: No prime number is 1/2 another prime number plus one.
Just one prime number is exactly one plus 1/2 another prime number.
(Wording changed to avoid the ambiguity between p/2 + 1 and (p + 1)/2.)
--
Ben.
Musatov
Musatov wrote:
> On Jun 20, 4:06 am, Musatov <marty.musa...@gmail.com> wrote:
> > On Jun 20, 12:13 am, William Elliot <ma...@rdrop.remove.com> wrote:
> >
> > > On Fri, 19 Jun 2009, MeAmI.org wrote:
> > > > RULE: EVERY PRIME number is exactly
> > > > 1/2 of some other number +1.
> >
> > > So what? For all x, x = (2x - 2)/2 + 1.
> >
> > > Trivial rule.
> > > Every integer (rational number, real number, complex number) is half
> > > of some integer (resp. rational number, real number, complex number)
> > > plus one.
> >
> > > Exercise. How many primes are half of some prime plus one?
None. Half of a prime number is not a whole number. (Except 2, in
which case the whole number is 1).
17/2=8.5+1=9.5 (NP).
So we have the result:
RULE: No prime number greater than two is 1/2 another prime number
plus one.
But perhaps this is what you meant.
Inverse/Additive prime property per Musatov: (below)
RULE: EVERY PRIME number greater than 2 is twice a number +1.
3=1*2+1
5=2*2+1
7=3*2+1
11=5*2+1
13=6*2+1
17=8*2+1
19=9*2+1
23=11*2+1
29=14*2+1
31=15*2+1
37=18*2+1
41=20*2+1
43=21*2+1
47=23*2+1
51=25*2+1
53=26*2+1
And combined Prime Generalization: (Musatov)
RULE: Every prime greater than two is 1/2 a number +1 and twice a
number +1.
Now consider the series again, but this time plot the additive
difference between first and next doubled number.
In the first two terms we write....
3=1*2+1 #
5=2*2+1 1 because the difference between the doubled numbers from
first to the next was "1".
And we continue....
(here is the full table)
3=1*2+1 #
5=2*2+1 1
7=3*2+1 1
11=5*2+1 2
13=6*2+1 1
17=8*2+1 2
19=9*2+1 1
23=11*2+1 2
29=14*2+1 3
31=15*2+1 1
37=18*2+1 3
41=20*2+1 2
43=21*2+1 1
47=23*2+1 2
51=25*2+1 2
53=26*2+1 1
I would like to see if these reveals more to clarity to the set of
primes.
How might it?
Ben Bacarisse wrote:
> Musatov <marty....@gmail.com> writes:
> > On Jun 20, 12:13 am, William Elliot <ma...@rdrop.remove.com> wrote:
> <snip>
> >> Exercise. How many primes are half of some prime plus one?
> >
> > None. Half of a prime number is not a whole number.
>
> Except for 2.
>
> > So we have the result:
> >
> > RULE: No prime number is 1/2 another prime number plus one. (Except for )
>
> Just one prime number is exactly one plus 1/2 another prime number.
Oh yeah, which one?
> Copout per Ben.: Wording changed to avoid the ambiguity between p/2 + 1 and (p + 1)/2.)
++
Martin
CBFalconer
> MeAmI.org said:
>
>> RULE: EVERY PRIME number is exactly
>> 1/2 of some other number +1.
>
> 2 is a counter-example unless "other number" can mean "same number".
Oh? 3 is 'another number'. 3+1 = 4 (usually). 4/2 = 2 (usually).
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
Richard Heathfield
> Richard Heathfield wrote:
>> MeAmI.org said:
>>
>>> RULE: EVERY PRIME number is exactly
>>> 1/2 of some other number +1.
>>
>> 2 is a counter-example unless "other number" can mean "same
>> number".
>
> Oh? 3 is 'another number'. 3+1 = 4 (usually). 4/2 = 2 (usually).
Division has precedence over addition. Even if it didn't,
associativity would be left to right unless specified otherwise.
Let "some other number" be X.
(X / 2) + 1 = 2
Subtract one from both sides.
X / 2 = 2 - 1
X / 2 = 1
X = 2
QED
MeAmI.org
Thank you.
--
Musatov
Musatov
> On Fri, 26 Jun 2009 03:15:23 -0700 (PDT), "MeAmI.org"
>
> <Me...@vzw.blackberry.net> wrote:
> >Martin Musatov: (as MeAmI.org wrote)
>
> >Richard Heathfield wrote:
> >> CBFalconer said:
>
> >> > Richard Heathfield wrote:
> >> >> MeAmI.org said:
>
> >> >>> RULE: EVERY PRIME number is exactly
> >> >>> 1/2 of some other number +1.
>
>
> 1/2 of 2 = 1
> 1 + 1 = 2
>
> Does this take into account eg. 4.5/2 = 2.25
> So would that make 3.25 a prime number?
A decimal number is not by classical definition prime. Though if you
have an idea of a decimal equivalent of primality, I would love to
hear it.
--
Musatov
"Tear a whole in Cyberspace"
http://MeAmI.org
Search for Truth!
Richard Heathfield
> On Jun 26, 1:20 pm, John H. Guillory <jo...@communicomm.com>
> wrote:
<snip>
>> So would that make 3.25 a prime number?
>
> A decimal number is not by classical definition prime.
Numbers aren't decimal. They're numbers. Decimal is a system for
/representing/ numbers textually.
> Though if
> you have an idea of a decimal equivalent of primality, I would
> love to hear it.
It's a meaningless concept. Primality has nothing to do with
representation.
Musatov
Dear Mr. Heathfield,
I agree with you in the classical proof sense primality is as much
governed by physics as it is counting numbers we choose to represent
quantities.
My intention with this reference is providing a means to interface
between prime numbers to the left of the decimal point and decimal
numbers to the right of the decimal point. Perhaps a system where 3.17
refers to two primes, and this sense I am speaking mostly toward
computation, but again I assert, numbers to the left or right of the
decimal, are still just numbers.
I have always been fascinated by this notion:
Numerically, our representations do not appear uniform instinctually,
to me at least.
Here is an example.
If we say, "What 10 is to 20 is not what 2.2 is to 3.3," is there any
truth in proportion to justify this assertion in physics or
mathematics?
We are simply counting.
10 is to 20
...is...
20 is 2x 10
2.2 is to 3.3
...is...
3.3 is 1.5x 2.2
...or...
1/2 of 2.2+2.2=3.3
...or...
1/2 of 2.2=1.1*3=3.33
1 and 1/2 of 2.2=3.3
...or...
1.5 of 2.2=3.3
So there is a split of
1/2+2/5+3/5=15/10
.5+.4+.6=1.5
...And...
1/2*2/5*3/5=x
x=5/10*4/10*6/10=120/10=1.2
.5*.4*.6=1.2
Theorem: use of a set of given quantities.
Rule: adding the set produces at least the product.
Proof(1a): 1 apple + 2 apples + 3 apples=6 apples.
Proof(1b):1 apple * 2 apples * 3 apples=6 apples.
Proof(2a): 5 apples + 4 apples + 6 apples =16 apples.
Proof(2b): 5 apples * 4 apples * 6 apples=120 apples.
Contradiction: in the above example the sum=1.5 and the product=1.2.
Fallacy: Multiplying quantities of items does not shrink them. This
applies to measurements and transforms.
But as we count from 1 to 2 and then 2 to 3
10/1.1 = 9.0909091
20/2.2 = 9.0909091
30/3.3 = 9.0909091
What 1
...is to...
10
...is...
What 2.2 is 22
Proof: 1.1/10=.11
2.2/20=.1
2.2/22=.11
Martin Musatov
Musatov
Musatov worte:
Musatov
Musatov wrote:
Musatov wrote:
Musatov wrote: Richard Heathfield wrote:
Musatov said:
On Jun 26, 1:20 pm, John H. Guillory <jo...@communicomm.com>
wrote:
<snip>
So would that make 3.25 a prime number? A decimal number is not by
classical definition prime.
Numbers aren't decimal. They're numbers. Decimal is a system for
/representing/ numbers textually.
Though if
you have an idea of a decimal equivalent of primality, I would
love to hear it.
It's a meaningless concept. Primality has nothing to do with
representation.
Peter Nilsson
> CBFalconer said:
> > Richard Heathfield wrote:
> > > MeAmI.org said:
> > > > RULE: EVERY PRIME number is exactly
> > > > 1/2 of some other number +1.
Every prime is exactly half of +1?!
> > >
> > > 2 is a counter-example unless "other number" can mean "same
> > > number".
> >
> > Oh? 3 is 'another number'. 3+1 = 4 (usually). 4/2 = 2 (usually).
>
> Division has precedence over addition. Even if it didn't,
> associativity would be left to right unless specified otherwise.
If I have 4 dollars and 20 cents, what is half the dollars and cents?
a) $1.20
b) $2.10
c) $2.20
d) $2.40
--
Peter
Musatov
Martin Musatov wrote:
Dear Peter,
Thank you for your reply. Before I respond to your question, please
allow me a brief moment to clarify my conjecture:
"Musatov's Prime Generalization Conjecture": Every prime greater than
2 is 2n+1=P.
2*1+1=3
2*2+1=5
...
2*20+1=41
...
(1)Is my conjecture provable?
Okay, thanks for your patience.
Peter, the answer to the question is "b)2.10".
But please bear with me and consider the following scenario:
Suppose I have three quantities of loose coins:
1) $0.50
2) $0.40
3) $0.60
The sum of the three quantities is $1.50.
Now suppose we are multiplying those same three quantities of coins:
$0.50*$0.40*$0.60
.50*.40=.20
And...
.20*.60=$0.12.
(2)Why should multiplying our money result in a loss of $1.38 or a 92%
loss?
If the three sums were dollars the case would be different.
$50+$40+$60=$150
$50*$40+$60=$1200
I do understand how to solve the problem literally, but chose to show
this example to prove decimal representations fail basic rationale.
++
Musatov
Alf P. Steinbach
>
> "Musatov's Prime Generalization Conjecture": Every prime greater than
> 2 is 2n+1=P.
>
> 2*1+1=3
> 2*2+1=5
> ...
> 2*20+1=41
> ...
>
> (1)Is my conjecture provable?
Yes but it's completely silly to conjecture that primes greater than 2 are odd.
Cheers & hth.,
- Alf
--
Due to hosting requirements I need visits to <url: http://alfps.izfree.com/>.
No ads, and there is some C++ stuff! :-) Just going there is good. Linking
to it is even better! Thanks in advance!
Richard Heathfield
> Richard Heathfield <r...@see.sig.invalid> wrote:
<snip>
>>
>> Division has precedence over addition. Even if it didn't,
>> associativity would be left to right unless specified otherwise.
>
> If I have 4 dollars and 20 cents, what is half the dollars and
> cents?
>
> a) $1.20
> b) $2.10
> c) $2.20
> d) $2.40
None of the above. Observe:
#include <stdio.h>
#include <iso646.h>
int half(int x)
{
return x / 2;
}
int main(void)
{
int dollars = 4;
int cents = 20;
int answer = half(dollars) and cents;
printf("The answer is %d\n", answer);
return 0;
}
Output:
The answer is 1
Musatov
*Musatov:
Dear Alf,
Thanks for the reply. I figure I have to start somewhere, thanks for
bearing with me.
(1)So have we established every prime greater than 2 is 2N+1?
(2)Also, "2N+1"="odd", correct?
To quote (play on) the movie "Pi":
Musatov: "11:31pm. Restate my assumptions".
1. Every prime is odd.
2. "2N+1" is every prime greater than 2.
3. "2N+1" is every odd.
4. 2N+1 contains every prime greater than 2 plus every odd composite.
5. All prime factors of odd composites are contained in "2N+1".
6. The set of prime numbers and prime factors of of odd composites is
wholly contained in the set "2N+1".
.......Breather........
The next step is then to establish the case when "N" is even, in every
case "2N+1" is odd. If all of "2N+1" is odd then this is a given.
Now we further examine the states:
When "N" ends in "2":
2*2=4+1=5 is prime, but only the first time. Every number ending in 5
greater than 5 is composite.
So to generate primes by this method we can rule out all numbers
ending in "2".
Then,
N of all even numbers we can say, there only remains numbers ending in
"4, 6, and 8".
Consider each case:
For four: (#4)
2*4+1=9 not prime
2*14+1=29 prime
2*24+1=49 not prime
In the above instances "N" ends in "4" and is not prime, the formula
produces a number with square prime factors (i.e. 3*3=9 and 7*7=49).
(i.e. When it does not produce a prime the composite is a prime
squared).
Also, we note:
When "N" ends in "4" and the formula produces a composite number,
adding "two" to the composite produces a prime.
Shown:
2*4+1=9+2=11 prime
2*24+1=49+2=51 prime
Moving along...
For Six: (i.e. When "N" ends in "6" and is applied 2N+1)
2*6=12+1=13 prime
2*16=32+1=33 is not prime.
2*26=52+1=53 prime
As above when "N" ends in 6 and does not produce prime (as above) we
have two prime factors.
In the above case of 33 they are "3 and 11". Of those two prime
factors, adding 2 to each produces 2 more primes. Shown:
3+2=5 prime
11+2=13 prime
Now, onto 8:
2*8+1=17 prime
2*18+1=37 prime
2*28+1=57 prime
2*38+1=77 not prime
In the above we have when "N" ends in "8" and does not produce a prime
as applied (2N+1) it has produced a number with two prime factors:
77=11*7
11 and 7 are prime.
Also, 11+2 is prime.
However, 7+4 is prime.
The conjecture contains:
"Musatov Prime Generalization Conjecture": "Every prime number and
prime factors of odd composites, is contained in the set 2N+1."
The natural further question is how and when do prime factors appear
in even composites and what rules apply?
So we have separated the primes and prime factors of odd composites
into a set (2N+1), now the only remaining prime factors exist inside
even composites and outside of this set.
A lot to chew on, but I thank you all in advance and look forward to
the insight gained from your responses.
Signed,
++
Musatov
Musatov
What is "hth"?
++
Musatov
Dik T. Winter
...
When are you going to stop to post these mindless conjectures based on a
very small number of cases?
> Consider each case:
>
> For four: (#4)
>
> 2*4+1=9 not prime
> 2*14+1=29 prime
> 2*24+1=49 not prime
>
> In the above instances "N" ends in "4" and is not prime, the formula
> produces a number with square prime factors (i.e. 3*3=9 and 7*7=49).
> (i.e. When it does not produce a prime the composite is a prime
> squared).
Wrong. 2*34+1 = 69 = 3*23.
2*544+1 = 1089 = 3*3*11*11
2*364+1 = 729 = 3*3*3*3*3*3
> Also, we note:
>
> When "N" ends in "4" and the formula produces a composite number,
> adding "two" to the composite produces a prime.
>
> Shown:
> 2*4+1=9+2=11 prime
> 2*24+1=49+2=51 prime
Since when is 51 prime?
--
dik t. winter, cwi, science park 123, 1098 xg amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Richard Heathfield
> In article
> <115c718c-fe66-4a34...@i6g2000yqj.googlegroups.com>
> Musatov <marty....@gmail.com> writes: ...
<snip>
> > Shown:
> > 2*4+1=9+2=11 prime
> > 2*24+1=49+2=51 prime
>
> Since when is 51 prime?
It's odd, right? And primes (ignoring 2, which is clearly
experimental error) are odd, right? Therefore, 51 is prime, QED.
Musatov
> Dik T. Winter said:
>
> > In article
> > Musatov <marty.musa...@gmail.com> writes: ...
>
> <snip>
>
> > > Shown:
> > > 2*4+1=9+2=11 prime
> > > 2*24+1=49+2=51 prime
>
> > Since when is 51 prime?
>
> It's odd, right? And primes (ignoring 2, which is clearly
> experimental error) are odd, right? Therefore, 51 is prime, QED.
>
> --
> Richard Heathfield <http://www.cpax.org.uk>
> Email: -http://www. +rjh@
> "Usenet is a strange place" - dmr 29 July 1999
No. 3*17 = 51
So 51 has 4 factors: 1, 3, 17, and 51
Richard Heathfield
> On Jun 30, 6:21 am, Richard Heathfield <r...@see.sig.invalid>
> wrote:
>> Dik T. Winter said:
>>
>> > In article
>> >
<115c718c-fe66-4a34-ac7c-02b57c508...@i6g2000yqj.googlegroups.com>
>> > Musatov <marty.musa...@gmail.com> writes: ...
>>
>> <snip>
>>
>> > > Shown:
>> > > 2*4+1=9+2=11 prime
>> > > 2*24+1=49+2=51 prime
>>
>> > Since when is 51 prime?
>>
>> It's odd, right? And primes (ignoring 2, which is clearly
>> experimental error) are odd, right? Therefore, 51 is prime, QED.
>>
>
> No. 3*17 = 51
WHOOOOOOSH!!
mike
r...@see.sig.invalid says...
> Dik T. Winter said:
>
> > In article
> > <115c718c-fe66-4a34...@i6g2000yqj.googlegroups.com>
> > Musatov <marty....@gmail.com> writes: ...
>
> <snip>
>
> > > Shown:
> > > 2*4+1=9+2=11 prime
> > > 2*24+1=49+2=51 prime
> >
> > Since when is 51 prime?
>
> It's odd, right? And primes (ignoring 2, which is clearly
> experimental error) are odd, right? Therefore, 51 is prime, QED.
>
is prime.
Mike
Musatov
> In article <_9qdnbY6rMCzjdfXnZ2dnUVZ8tZi4...@bt.com>,
> r...@see.sig.invalid says...
>
>
>
> > Dik T. Winter said:
>
> > > In article
> > > Musatov <marty.musa...@gmail.com> writes: ...
>
> > <snip>
>
> > > > Shown:
> > > > 2*4+1=9+2=11 prime
> > > > 2*24+1=49+2=51 prime
>
> > > Since when is 51 prime?
>
> > It's odd, right? And primes (ignoring 2, which is clearly
> > experimental error) are odd, right? Therefore, 51 is prime, QED.
>
> Come now Richard, you have to ignore 3 and 17 as well to prove that 51
> is prime.
>
>
> - Show quoted text -
If it is known that if any NP-complete language is sparse (contains no
more than a polynomial number of strings of length n), then P = NP.
[BH08] improved this result, showing that if any language in NP has an
NP-hard set of subexponential density, then coNP is contained in NP/
poly and thus, by [Yap82], PH collapses to the third level.
Conceptually, a decision problem is a problem that takes as input some
string, and outputs "yes" or "no". If there is an algorithm (say a
Turing machine, or a computer program with unbounded memory) which is
able to produce the correct answer for any input string of length
Failed to parse (<math_output_error>): n
in at most c \cdot n^k steps, where k and Failed to parse
(<math_output_error>): c are constants independent of the input
string, then we say that the problem can be solved in polynomial time
and we place it in the class P. Formally, P is defined as the set of
all languages which can be decided by a deterministic polynomial-time
Turing machine. That is,
P = {L:L = L(M) for some deterministic polynomial-time Turing machine
M}
where L(M) = \{ w\in\Sigma^{*}: M \text{ accepts } w \}
and a deterministic polynomial-time Turing machine is a deterministic
Turing machine M which satisfies the following two conditions:
1. M halts on all input w; and
2. there exists k \in N such that T_{M}(n)\in\; O(nk),
where T_{M}(n) = \max\{ t_{M}(w) : w\in\Sigma^{*}, \left|w
\right| = n \}
and tM(w) = number of steps M takes to halt on input w.
NP can be defined similarly using nondeterministic Turing machines
(the traditional way). However, a modern approach to define NP is to
use the concept of certificate and verifier. Formally, NP is defined
as the set of languages over a finite alphabet that have a verifier
that runs in polynomial time, where the notion of "verifier" is
defined as follows.
Let Failed to parse (<math_output_error>): L
be a language over a finite alphabet, Σ.
L\in\mathbf{NP} if, and only if, there exists a binary relation R
\subset\Sigma^{*}\times\Sigma^{*} and a positive integer k such that
the following two conditions are satisfied:
1. For all x\in\Sigma^{*}, x\in L \Leftrightarrow\exists y\in\Sigma^
{*} such that (x,y)\in R\; and \left|y\right|\in\;O(\left|x\right|^
{k}); and
2. the language L_{R} = \{ x\# y:(x,y)\in R\} over \Sigma\cup\{\#\}
is decidable by a Turing machine in polynomial time.
A Turing machine that decides LR is called a verifier for L and a y
such that (x,y)\in R is called a certificate of membership of x in L.
In general, a verifier does not have to be polynomial-time. However,
for L to be in NP, there must be a verifier that runs in polynomial
time. --MartinMichaelMusatov 07:18, 24 February 2009 (UTC)
NPC: NP-Complete
The class of decision problems such that (1) they're in NP and (2)
every problem in NP is reducible to them (under some notion of
reduction). In other words, the hardest problems in NP.
Two notions of reduction from problem A to problem B are usually
considered:
1. Karp or many-one reductions. Here a polynomial-time algorithm is
given as input an instance of problem A, and must produce as output an
instance of problem B.
2. Turing reductions, in this context also called Cook reductions.
Here the algorithm for problem B can make arbitrarily many calls to an
oracle for problem A.
Some examples of NP-complete problems are discussed under the entry
for NP.
The classic reference on NPC is [GJ79].
Unless P = NP, NPC does not contain any sparse problems: that is,
problems such that the number of 'yes' instances of size n is upper-
bounded by a polynomial in n [Mah82].
A famous conjecture [BH77] asserts that all NP-complete problems are
polynomial-time isomorphic -- i.e. between any two problems, there is
a one-to-one and onto Karp reduction. If that's true, the NP-complete
problems could be interpreted as mere "relabelings" of one another.
NP-complete problems are p-superterse unless P = NP [BKS95]. This
means that, given k Boolean formulas F1,...,Fk, if you can rule out
even one of the 2k possibilities in polynomial time (e.g., "if
F1,...,Fk-1 are all unsatisfiable then Fk is satisfiable"), then P =
NP.
--
Musatov
Ed Prochak
all primes beyond 2 are odd, this formula tells us nothing about
primes. Your conjecture is true but not yours. It is common
knowledge.
Ed
Richard Heathfield
That would have been a different jest. Mine was more along the lines
of "all cats are animals, this dog is an animal, therefore this dog
is a cat".
Constructive Truth
> In article <115c718c-fe66-4a34-ac7c-02b57c508...@i6g2000yqj.googlegroups.com> Musatov <marty.musa...@gmail.com> writes:
> ...
> When are you going to stop to post these mindless conjectures based on a
> very small number of cases?
>
> > Consider each case:
> >
> > For four: (#4)
Dear Mr. Dik T. Winter:
I am afraid, necessarily you misunderstand the direct language of my
claim.
My claim is:
1. every prime is of the form 2N+1.
My claim is:
2. Not every number of the form 2N+1 is prime.
They are two very different things, but may be stated directly with
the statement I made earlier.
"Every prime number is of the form 2N+1"
Do you understand?
--
Martin Musatov
Venn Diagram/Concentric circle inclusions/statements may help me
clarify.
_____________________
[ All Numbers of the ]
[ {Prime Numbers} ]
[ form 2N+1 ]
----------------------
1. All prime numbers are of the form 2N+1
2. Not all numbers of the form 2N+1 are prime.
Alan Morgan
Constructive Truth <scri...@aol.com> wrote:
>I am afraid, necessarily you misunderstand the direct language of my
>claim.
>
>My claim is:
>
>1. every prime is of the form 2N+1.
Not only is your statement trivial and completely uninteresting (every
prime number is odd. Really? Could that be because even numbers are,
by definition, divisible by 2, and thus not prime?), it isn't even
completely correct. 2 is a prime, but is not of the form 2N+1.
So your claim is both trivial and false. Have no fear! One minor
change:
Musatov-Morgan Theorem
Every prime > 2 is of the form 2N+1
and it's now trivial and true. What's next? A variation on the
Goldbach Conjecture where you hypothesize that every number > 2
can be written as the sum of two other numbers?
Alan
--
Defendit numerus
Musatov
> In article <01ecf301-15ba-4bfb-83c3-f4e8e18c9...@o18g2000pra.googlegroups.com>,
> Constructive Truth <scribe...@aol.com> wrote:
>
> >I am afraid, necessarily you misunderstand the direct language of my
> >claim.
>
> >My claim is:
>
> >1. every prime is of the form 2N+1.
>
> Not only is your statement trivial and completely uninteresting (every
> prime number is odd. Really? Could that be because even numbers are,
> by definition, divisible by 2, and thus not prime?), it isn't even
> completely correct. 2 is a prime, but is not of the form 2N+1.
>
> So your claim is both trivial and false. Have no fear! One minor
> change:
>
> Musatov-Morgan Theorem
> Every prime > 2 is of the form 2N+1
>
> and it's now trivial and true. What's next? A variation on the
> Goldbach Conjecture where you hypothesize that every number > 2
> can be written as the sum of two other numbers?
>
> Alan
> --
> Defendit numerus
Hello Alan,
Thank you for your feedback and write up. Very cool!
Can you explain something to me please? By your comments, I
understand, 2 N + 1 = odd. But what else can we say about 2 N + 1 ?
Here are a list of statements, I would like to know if we can properly
decide from 2 N + 1:
(All of the below statements assume P > 2, as you asserted.)
Can we say....? (If every prime > 2 is 2 N + 1 = odd)
1. 2 N + 1 = P an (odd) prime 1/2 (of an even) 2P?
For example:
2*8+1=17
1/2*34=17
2. Can we then write for every prime > 2:
2 N + 1 = P (odd) and 2P (Even), N is also always even?
3. Can the above statement equivocally be stated:
2 (even N) + 1 = Every Prime = 1/2 2P or 1/2 * 4
(even N + 2)?
2 * 8 + 1 = 17 (a prime) = 1/2 (17*2) or 1/2 * (4 * 8
+ 2)?
Thank you for your time an patience.
> Musatov-Morgan Theorem
> Every prime > 2 is of the form 2N+1
So cool, thanks again!
Martin Musatov
Musatov
For example:
2*8+1=17
1/2*34=17
2 (even N) + 1 = Every Prime > 2 = 1/2 2P or 1/2 * 4 (even
N) + 2?
2 * 8 + 1 = 17 (a prime) = 1/2 (17*2) or 1/2 * (4 * 8) +
2 ?
Thank you for your time and patience.
Musatov-Morgan Theorem
Every prime > 2 is of the form 2N+1
So cool, thanks again! Good save.
Martin Musatov
I just caught a little mistake, fixed ity.
mike
@z4g2000prh.googlegroups.com>, marty....@gmail.com says...
>
> Here are a list of statements, I would like to know if we can
> properly
> decide from 2 N + 1:
>
> (All of the below statements assume P > 2, as you asserted.)
>
> Can we say....? (If every prime > 2 is 2 N + 1 = odd)
>
> 1. 2 N + 1 = P an (odd) prime 1/2 (of an even) 2P?
I assume that you trying to state:
For all primes P of the form P = 2*N + 1, the number 2*P is even.
If so then it is trivially true by definition.
>
> 2. Can we then write for every prime > 2:
>
> 2 N + 1 = P (odd) and 2P (Even), N is also always
> even?
Here you appear to be stating:
For all P > 2, P = 4*N + 1, for some natural number N.
If so then this is trivially false as it doesn't hold true for 7.
>
> 3. Can the above statement equivocally be stated:
>
> 2 (even N) + 1 = Every Prime > 2 = 1/2 2P or 1/2 * 4 (even
> N) + 2?
>
> 2 * 8 + 1 = 17 (a prime) = 1/2 (17*2) or 1/2 * (4 * 8) +
> 2 ?
This seems teh same as 2 above. If so then it is false.
Marty, it may interest you to note that as well as:
For all P > 2, P = 2*N + 1
being true, it is also the case that:
For all P > 3, P = 6*N +/- 1
-- Mike
I.N.R.I. Logic
> In article <88f7a7fc-702c-47ac-8f3e-66088faf4487
This is very interesting. Thanks Mike. So we have all primes > 2 may
be written as 2 N + 1 or 6 N + / - 1. Can these equations be combined
to deduce something further?
Can we write every prime not equal to 3 may be written as 3 N + / -
1 ?
Does
2 (even N) + 1 = Every Prime =/= 7 > 2 = 1/2 2P or 1/2 * 4
(even N) + 2?
Or does the single 7 present only an iceberg head in fatally flawed
logic?
I.N.R.I. Logic (aka, Martin Musatov -- for now)
I.N.R.I. Logic
>
> - Show quoted text -
I want you to pay close attention to the logic. I referred to "all
primes" meaning "all" not each. Therefore the statement is valid, do
you agree?
Though three cannot be written as 6n + or - 1 it can be written as 2N
+1 so the statement is true.
The statement would be false if I said "each prime" but I did not.
mike
@d4g2000yqa.googlegroups.com>, scri...@aol.com says...
For all P > 5, P = 30*N +/- {1,7,11,13}
...with the obvious interpretation of the {} brackets.
Mike
Musatov
> In article <00e96645-4c61-491b-8a79-2a1ae4d3cf60
> @d4g2000yqa.googlegroups.com>, scribe...@aol.com says...
>
> - Show quoted text -
Neat, huh?
Musatov
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Vindicator2009
M.M.M.
Vindicator2009 wrote:
> drop dead spammer
2 + 3 = 5 is prime.
QmE.D.