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Groups > comp.lang.javascript > #29280 > unrolled thread
| Started by | jonas.thornvall@gmail.com |
|---|---|
| First post | 2016-01-17 01:40 -0800 |
| Last post | 2016-01-19 19:43 +0000 |
| Articles | 14 on this page of 74 — 13 participants |
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Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-17 01:40 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-17 02:20 -0800
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-17 16:24 +0000
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-17 09:03 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-17 09:17 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-17 09:22 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-17 11:32 -0800
Re: Primality sieve challenge Stefan Weiss <krewecherl@gmail.com> - 2016-01-18 00:13 +0100
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-18 04:10 -0800
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-18 01:20 +0000
Re: Primality sieve challenge "Evertjan." <exxjxw.hannivoort@inter.nl.net> - 2016-01-18 12:16 +0100
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-18 11:25 +0000
Re: Primality sieve challenge "Evertjan." <exxjxw.hannivoort@inter.nl.net> - 2016-01-18 12:43 +0100
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-18 04:18 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-18 04:25 -0800
Re: Primality sieve challenge Gene Wirchenko <genew@telus.net> - 2016-01-18 09:52 -0800
Re: Primality sieve challenge "Evertjan." <exxjxw.hannivoort@inter.nl.net> - 2016-01-18 19:23 +0100
Re: Primality sieve challenge "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2016-01-18 10:48 -0800
Re: Primality sieve challenge Gene Wirchenko <genew@telus.net> - 2016-01-18 13:09 -0800
Re: Primality sieve challenge Scott Sauyet <scott.sauyet@gmail.com> - 2016-01-18 13:14 -0800
Re: Primality sieve challenge "Evertjan." <exxjxw.hannivoort@inter.nl.net> - 2016-01-18 22:45 +0100
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-18 14:23 -0800
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-18 22:56 +0000
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-18 15:02 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-18 16:04 -0800
Re: Primality sieve challenge Stefan Weiss <krewecherl@gmail.com> - 2016-01-19 02:42 +0100
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-18 23:34 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-19 01:34 -0800
Re: Primality sieve challenge "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2016-01-19 06:23 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-19 07:10 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-19 07:15 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-19 08:34 -0800
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-19 16:54 +0000
Re: Primality sieve challenge "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2016-01-19 08:57 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-19 10:18 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-19 10:59 -0800
Re: Primality sieve challenge Stefan Weiss <krewecherl@gmail.com> - 2016-01-20 01:33 +0100
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-20 02:10 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-20 02:21 -0800
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-20 10:45 +0000
Re: Primality sieve challenge "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2016-01-20 06:43 -0800
Re: Primality sieve challenge Scott Sauyet <scott.sauyet@gmail.com> - 2016-01-20 15:21 -0800
Re: Primality sieve challenge Stefan Weiss <krewecherl@gmail.com> - 2016-01-21 03:18 +0100
Re: Primality sieve challenge Jon Ribbens <jon+usenet@unequivocal.co.uk> - 2016-01-21 12:12 +0000
Re: Primality sieve challenge Dr J R Stockton <reply1600@merlyn.demon.co.uk.invalid> - 2016-01-21 23:20 +0000
Re: Primality sieve challenge Stefan Weiss <krewecherl@gmail.com> - 2016-01-22 19:44 +0100
Re: Primality sieve challenge Dr J R Stockton <reply1600@merlyn.demon.co.uk.invalid> - 2016-01-23 21:50 +0000
Re: Primality sieve challenge Stefan Weiss <krewecherl@gmail.com> - 2016-01-24 01:26 +0100
Re: Primality sieve challenge "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2016-01-23 17:54 -0800
Re: Primality sieve challenge "Chris M. Thomasson" <nospam@no-spam.ws> - 2016-01-24 13:34 -0800
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-25 00:21 +0000
Re: Primality sieve challenge Stefan Weiss <krewecherl@gmail.com> - 2016-01-25 03:08 +0100
Re: Primality sieve challenge "Michael Haufe (TNO)" <tno@thenewobjective.com> - 2016-01-24 19:35 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-25 03:53 -0800
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-18 19:32 +0000
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-17 19:51 +0000
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-17 13:43 -0800
Re: Primality sieve challenge "Chris M. Thomasson" <nospam@nospam.nospam> - 2016-01-17 12:46 -0800
Re: Primality sieve challenge Gene Wirchenko <genew@telus.net> - 2016-01-18 09:50 -0800
Re: Primality sieve challenge Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-01-18 20:09 +0100
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-18 19:28 +0000
Re: Primality sieve challenge Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-01-18 21:37 +0100
Re: Primality sieve challenge Scott Sauyet <scott.sauyet@gmail.com> - 2016-01-18 13:12 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-18 13:46 -0800
Re: Primality sieve challenge jonas.thornvall@gmail.com - 2016-01-18 14:00 -0800
Re: Primality sieve challenge Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-01-19 13:47 +0100
Re: Primality sieve challenge Ben Bacarisse <ben.usenet@bsb.me.uk> - 2016-01-19 14:04 +0000
Re: Primality sieve challenge Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-01-19 19:59 +0100
Re: Primality sieve challenge Scott Sauyet <scott.sauyet@gmail.com> - 2016-01-19 10:55 -0800
Re: Primality sieve challenge Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-01-19 20:04 +0100
Re: Primality sieve challenge Scott Sauyet <scott.sauyet@gmail.com> - 2016-01-19 13:42 -0800
Re: Primality sieve challenge Erwin Moller <erwinmollerusenet@xs4all.nl> - 2016-01-21 17:21 +0100
Re: Primality sieve challenge Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-02-03 19:52 +0100
Re: Primality sieve challenge Dr J R Stockton <reply1600@merlyn.demon.co.uk.invalid> - 2016-01-19 19:43 +0000
Page 4 of 4 — ← Prev page 1 2 3 [4]
| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2016-01-18 19:28 +0000 |
| Message-ID | <87ziw27kln.fsf@bsb.me.uk> |
| In reply to | #29323 |
Thomas 'PointedEars' Lahn <PointedEars@web.de> writes:
> Gene Wirchenko wrote:
>
>> […] All primes >= 5 are of the form 6k +/- 1 where k is a positive
>> integer.
>
> Interesting thesis. Prove it.
It's almost trivial. All integers >= 5 can be written in the form 6k+n
where n is in {-1, 0, 1, ... 4} and k > 0, but all those that have the
form 6k + {0, 2, 3, 4} are clearly composite.
This is a specific case of the more general observation that all
sufficiently large primes must have the form mk + n with m, n relatively
prime.
--
Ben.
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| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2016-01-18 21:37 +0100 |
| Message-ID | <2865358.ifOrN9Bpor@PointedEars.de> |
| In reply to | #29325 |
Ben Bacarisse wrote:
> Thomas 'PointedEars' Lahn <PointedEars@web.de> writes:
>> Gene Wirchenko wrote:
>>> […] All primes >= 5 are of the form 6k +/- 1 where k is a positive
>>> integer.
>> Interesting thesis. Prove it.
>
> It's almost trivial. All integers >= 5 can be written in the form 6k+n
> where n is in {-1, 0, 1, ... 4} and k > 0,
ACK.
> but all those that have the form 6k + {0, 2, 3, 4} are clearly composite.
I can see why it is so for {0, 2, 4} (all multiples of 6 are divisible by 2,
so are the sums when added 2 or 4), but why also for the summand 3?
Also, I do not see how your argument proves Gene’s assertion.
--
PointedEars
FAQ: <http://PointedEars.de/faq> | SVN: <http://PointedEars.de/wsvn/>
Twitter: @PointedEars2 | ES Matrix: <http://PointedEars.de/es-matrix>
Please do not cc me. / Bitte keine Kopien per E-Mail.
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| From | Scott Sauyet <scott.sauyet@gmail.com> |
|---|---|
| Date | 2016-01-18 13:12 -0800 |
| Message-ID | <0f34c52b-ae45-49b0-8363-080d3d11d5fb@googlegroups.com> |
| In reply to | #29328 |
Thomas 'PointedEars' Lahn wrote:
> Ben Bacarisse wrote:
>
>> but all those that have the form 6k + {0, 2, 3, 4} are clearly composite.
>
> I can see why it is so for {0, 2, 4} (all multiples of 6 are divisible by 2,
> so are the sums when added 2 or 4), but why also for the summand 3?
Because 6k + 3 is divisible by 3. And if it's greater than 5 it's clearly
not equal to 3, so it's not prime.
> Also, I do not see how your argument proves Gene's assertion.
His assertion was that
| [...] All primes >= 5 are of the form 6k +/- 1 where k is a positive
| integer.
Since all positive integers > 5 are (trivially) of one of forms `6k + 0`,
`6k + 1`, `6k + 2`, `6k + 3`, `6k + 4`, or `6k + 5`, and we've easily
demonstrated that all those of the form `6k + {0, 2, 3, 4}` are composite,
all primes must be of the form `6k + 1` or `6k + 5`.
But anything expressible as `6n + 5` can be seen, by substituting
`n = k - 1`, as `6(k - 1) + 5` = `6k - 1`, and if n is a positive integer,
k, which is its successor, must be as well.
Q.E.D.
-- Scott
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| From | jonas.thornvall@gmail.com |
|---|---|
| Date | 2016-01-18 13:46 -0800 |
| Message-ID | <84a7f963-14ba-4203-a74c-00dbcbcab403@googlegroups.com> |
| In reply to | #29330 |
Den måndag 18 januari 2016 kl. 22:12:29 UTC+1 skrev Scott Sauyet:
> Thomas 'PointedEars' Lahn wrote:
> > Ben Bacarisse wrote:
> >
> >> but all those that have the form 6k + {0, 2, 3, 4} are clearly composite.
> >
> > I can see why it is so for {0, 2, 4} (all multiples of 6 are divisible by 2,
> > so are the sums when added 2 or 4), but why also for the summand 3?
>
> Because 6k + 3 is divisible by 3. And if it's greater than 5 it's clearly
> not equal to 3, so it's not prime.
>
>
> > Also, I do not see how your argument proves Gene's assertion.
>
> His assertion was that
>
> | [...] All primes >= 5 are of the form 6k +/- 1 where k is a positive
> | integer.
>
> Since all positive integers > 5 are (trivially) of one of forms `6k + 0`,
> `6k + 1`, `6k + 2`, `6k + 3`, `6k + 4`, or `6k + 5`, and we've easily
> demonstrated that all those of the form `6k + {0, 2, 3, 4}` are composite,
> all primes must be of the form `6k + 1` or `6k + 5`.
>
> But anything expressible as `6n + 5` can be seen, by substituting
> `n = k - 1`, as `6(k - 1) + 5` = `6k - 1`, and if n is a positive integer,
> k, which is its successor, must be as well.
>
> Q.E.D.
>
> -- Scott
Nice that will work good to make a primality test, using my search algorithm.
So i just search in on k.
Thanks guys i think i once knew this but it was gone.
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| From | jonas.thornvall@gmail.com |
|---|---|
| Date | 2016-01-18 14:00 -0800 |
| Message-ID | <2444c08a-2f36-4df5-ba2a-4951a5090163@googlegroups.com> |
| In reply to | #29333 |
Den måndag 18 januari 2016 kl. 22:46:28 UTC+1 skrev jonas.t...@gmail.com:
> Den måndag 18 januari 2016 kl. 22:12:29 UTC+1 skrev Scott Sauyet:
> > Thomas 'PointedEars' Lahn wrote:
> > > Ben Bacarisse wrote:
> > >
> > >> but all those that have the form 6k + {0, 2, 3, 4} are clearly composite.
> > >
> > > I can see why it is so for {0, 2, 4} (all multiples of 6 are divisible by 2,
> > > so are the sums when added 2 or 4), but why also for the summand 3?
> >
> > Because 6k + 3 is divisible by 3. And if it's greater than 5 it's clearly
> > not equal to 3, so it's not prime.
> >
> >
> > > Also, I do not see how your argument proves Gene's assertion.
> >
> > His assertion was that
> >
> > | [...] All primes >= 5 are of the form 6k +/- 1 where k is a positive
> > | integer.
> >
> > Since all positive integers > 5 are (trivially) of one of forms `6k + 0`,
> > `6k + 1`, `6k + 2`, `6k + 3`, `6k + 4`, or `6k + 5`, and we've easily
> > demonstrated that all those of the form `6k + {0, 2, 3, 4}` are composite,
> > all primes must be of the form `6k + 1` or `6k + 5`.
> >
> > But anything expressible as `6n + 5` can be seen, by substituting
> > `n = k - 1`, as `6(k - 1) + 5` = `6k - 1`, and if n is a positive integer,
> > k, which is its successor, must be as well.
> >
> > Q.E.D.
> >
> > -- Scott
>
> Nice that will work good to make a primality test, using my search algorithm.
> So i just search in on k.
>
> Thanks guys i think i once knew this but it was gone.
But that i can find k in linear time and tell if it is prime or not does not mean i can factor in linear time. So i need another approach for that.
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| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2016-01-19 13:47 +0100 |
| Message-ID | <3532030.noGN7VB15C@PointedEars.de> |
| In reply to | #29330 |
Scott Sauyet wrote:
> Thomas 'PointedEars' Lahn wrote:
>> Ben Bacarisse wrote:
>>> but all those that have the form 6k + {0, 2, 3, 4} are clearly
>>> composite.
>> I can see why it is so for {0, 2, 4} (all multiples of 6 are divisible by
>> 2, so are the sums when added 2 or 4), but why also for the summand 3?
>
> Because 6k + 3 is divisible by 3.
Why? I can see that it follows for k = 1 (9), k = 2 (15), k = 3 (21), and
for several greater k, but why for *all* k?
> And if it's greater than 5 it's clearly not equal to 3, so it's not prime.
I find that a specious argument at best.
>> Also, I do not see how your argument proves Gene's assertion.
>
> His assertion was that
>
> | [...] All primes >= 5 are of the form 6k +/- 1 where k is a positive
> | integer.
Yes.
> Since all positive integers > 5 are (trivially) of one of forms `6k + 0`,
> `6k + 1`, `6k + 2`, `6k + 3`, `6k + 4`, or `6k + 5`, and we've easily
> demonstrated that all those of the form `6k + {0, 2, 3, 4}` are composite,
Yes.
> all primes must be of the form `6k + 1` or `6k + 5`.
Again, why? If something is true for A and B, it does not follow that it is
not true for C ∉ {A, B}: p(A) ∧ p(B) ↛ ¬p(C); here p(X) := “X is composite
(not prime)”. What relation I am missing here?
--
PointedEars
FAQ: <http://PointedEars.de/faq> | SVN: <http://PointedEars.de/wsvn/>
Twitter: @PointedEars2 | ES Matrix: <http://PointedEars.de/es-matrix>
Please do not cc me. / Bitte keine Kopien per E-Mail.
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| From | Ben Bacarisse <ben.usenet@bsb.me.uk> |
|---|---|
| Date | 2016-01-19 14:04 +0000 |
| Message-ID | <87bn8h64yr.fsf@bsb.me.uk> |
| In reply to | #29344 |
Thomas 'PointedEars' Lahn <PointedEars@web.de> writes:
> Scott Sauyet wrote:
>
>> Thomas 'PointedEars' Lahn wrote:
>>> Ben Bacarisse wrote:
>>>> but all those that have the form 6k + {0, 2, 3, 4} are clearly
>>>> composite.
>>> I can see why it is so for {0, 2, 4} (all multiples of 6 are divisible by
>>> 2, so are the sums when added 2 or 4), but why also for the summand 3?
>>
>> Because 6k + 3 is divisible by 3.
>
> Why? I can see that it follows for k = 1 (9), k = 2 (15), k = 3 (21), and
> for several greater k, but why for *all* k?
(6k + 3)/3 = 2k + 1 which is an integer. If that is not enough, you'd
better say what part you doubt rather than just ask another "why?".
<snip>
>> Since all positive integers > 5 are (trivially) of one of forms `6k + 0`,
>> `6k + 1`, `6k + 2`, `6k + 3`, `6k + 4`, or `6k + 5`, and we've easily
>> demonstrated that all those of the form `6k + {0, 2, 3, 4}` are composite,
>
> Yes.
>
>> all primes must be of the form `6k + 1` or `6k + 5`.
(all suitably large primes...)
> Again, why? If something is true for A and B, it does not follow that it is
> not true for C ∉ {A, B}: p(A) ∧ p(B) ↛ ¬p(C); here p(X) := “X is composite
> (not prime)”. What relation I am missing here?
The key is that every integer > 5 is in one of the six sets Rr = { 6k +
r | k ∈ N } where r is one of 0, 1,... 5. All the primes are there
somewhere in one or more of these sets. But all the numbers in R0, R2,
R3 and R4 are composite. What options are left? There may be no
primes, of course, but if there are any, they must in either R1 or R5.
--
Ben.
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| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2016-01-19 19:59 +0100 |
| Message-ID | <40372295.ejPoVfOnFb@PointedEars.de> |
| In reply to | #29345 |
Ben Bacarisse wrote:
> Thomas 'PointedEars' Lahn <PointedEars@web.de> writes:
>> Scott Sauyet wrote:
>>> Thomas 'PointedEars' Lahn wrote:
>>>> Ben Bacarisse wrote:
>>>>> but all those that have the form 6k + {0, 2, 3, 4} are clearly
>>>>> composite.
>>>> I can see why it is so for {0, 2, 4} (all multiples of 6 are divisible
>>>> by 2, so are the sums when added 2 or 4), but why also for the summand
>>>> 3?
>>> Because 6k + 3 is divisible by 3.
>> Why? I can see that it follows for k = 1 (9), k = 2 (15), k = 3 (21),
>> and for several greater k, but why for *all* k?
>
> (6k + 3)/3 = 2k + 1 which is an integer. […]
It is obvious to me, now that you have put it *this* way :)
> The key is that every integer > 5 is in one of the six sets Rr = { 6k +
> r | k ∈ N } where r is one of 0, 1,... 5. All the primes are there
> somewhere in one or more of these sets. But all the numbers in R0, R2,
> R3 and R4 are composite. What options are left? There may be no
^^^^^^^^^^^^^^^
> primes, of course, but if there are any, they must in either R1 or R5.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
And integers n = 6k − 1 (in your original statement) are members of the same
equivalence class as integers m = 6k + 5. Thank you, I see it now. I
missed the marked part as I misunderstood Gene’s statement so that it would
mean that you can *find* primes that way.
--
PointedEars
FAQ: <http://PointedEars.de/faq> | SVN: <http://PointedEars.de/wsvn/>
Twitter: @PointedEars2 | ES Matrix: <http://PointedEars.de/es-matrix>
Please do not cc me. / Bitte keine Kopien per E-Mail.
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| From | Scott Sauyet <scott.sauyet@gmail.com> |
|---|---|
| Date | 2016-01-19 10:55 -0800 |
| Message-ID | <06c1db3d-5ab3-4d76-bb98-b48adfbaa900@googlegroups.com> |
| In reply to | #29344 |
Thomas 'PointedEars' Lahn wrote:
> Scott Sauyet wrote:
>> Thomas 'PointedEars' Lahn wrote:
>>> Ben Bacarisse wrote:
>>>> but all those that have the form 6k + {0, 2, 3, 4} are clearly
>>>> composite.
>>> I can see why it is so for {0, 2, 4} (all multiples of 6 are divisible by
>>> 2, so are the sums when added 2 or 4), but why also for the summand 3?
>>
>> Because 6k + 3 is divisible by 3.
>
> Why? I can see that it follows for k = 1 (9), k = 2 (15), k = 3 (21), and
> for several greater k, but why for *all* k?
It's extremely similar to the reason you described for {0, 2, 4}:
All multiples of 6 are divisible by 3, so are the sums when added 3. If
this is still surprising, I'd recommend any introductory text on number
theory.
>> And if it's greater than 5 it's clearly not equal to 3, so it's not prime.
>
> I find that a specious argument at best.
A number that is divisible by a prime, but is not actually equal to that
prime is, by definition, composite. I can't see why you consider that
specious.
>> Since all positive integers > 5 are (trivially) of one of forms `6k + 0`,
>> `6k + 1`, `6k + 2`, `6k + 3`, `6k + 4`, or `6k + 5`, and we've easily
>> demonstrated that all those of the form `6k + {0, 2, 3, 4}` are composite,
>
> Yes.
>
>> all primes must be of the form `6k + 1` or `6k + 5`.
>
> Again, why? If something is true for A and B, it does not follow that it is
> not true for C ∉ {A, B}: p(A) ∧ p(B) ↛ ¬p(C); here p(X) := “X is composite
> (not prime)”. What relation I am missing here?
I'm not following your objection. All integers greater than 1 are either
prime or composite. All integers are in one of the sets (to use Ben's
notation) Rr = { 6k + r | k ∈ N } for some r ∈ {0, 1, 2, 3, 4, 5}. All
sufficently large integers in R0, R2, R3, R4 have already been shown to be
composite, so any suffiently large primes must be in R1 and R5. I've
already shown that R5 is equivalent to { 6k - 1 | k ∈ N }.
Note that there is nothing magical about the number 6 here. All primes
greater than 2 are also of one of the forms `8k + 1`, `8k + 3`, `8k + 5`,
or `8k + 7`, and in general for all `n`, thereis some `p` such that all
primes greater than `p` are of the form { nk + r | k ∈ N } for some `r`
coprime to `n`. (`r` is coprime to `n` if the greatest common divisor of
`r` and `n` is 1.)
Does that make it any clearer?
-- Scott
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| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2016-01-19 20:04 +0100 |
| Message-ID | <1860806.lZUhBWmd6J@PointedEars.de> |
| In reply to | #29354 |
Scott Sauyet wrote:
> Thomas 'PointedEars' Lahn wrote:
>> Scott Sauyet wrote:
>>> Thomas 'PointedEars' Lahn wrote:
>>>> Ben Bacarisse wrote:
>>>>> but all those that have the form 6k + {0, 2, 3, 4} are clearly
>>>>> composite.
>>>> I can see why it is so for {0, 2, 4} (all multiples of 6 are divisible
>>>> by 2, so are the sums when added 2 or 4), but why also for the summand
>>>> 3?
>>> Because 6k + 3 is divisible by 3.
>> […]
>>> And if it's greater than 5 it's clearly not equal to 3, so it's not
>>> prime.
>> I find that a specious argument at best.
>
> A number that is divisible by a prime, but is not actually equal to that
> prime is, by definition, composite. I can't see why you consider that
> specious.
I find that argument specious *at best* because it does not follow that a
number is not prime when it is greater than 5 and not equal to 3. Perhaps
I am misunderstanding what you mean by “it”.
> [explanation]
Thanks, see my other followup.
--
PointedEars
FAQ: <http://PointedEars.de/faq> | SVN: <http://PointedEars.de/wsvn/>
Twitter: @PointedEars2 | ES Matrix: <http://PointedEars.de/es-matrix>
Please do not cc me. / Bitte keine Kopien per E-Mail.
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| From | Scott Sauyet <scott.sauyet@gmail.com> |
|---|---|
| Date | 2016-01-19 13:42 -0800 |
| Message-ID | <83722eb8-7e98-4983-afd9-8c420ba1299f@googlegroups.com> |
| In reply to | #29357 |
Thomas 'PointedEars' Lahn wrote:
> Scott Sauyet wrote:
>> Thomas 'PointedEars' Lahn wrote:
>>> Scott Sauyet wrote:
>>>> Thomas 'PointedEars' Lahn wrote:
>>>>> Ben Bacarisse wrote:
>>>>>> but all those that have the form 6k + {0, 2, 3, 4} are clearly
>>>>>> composite.
>>>>> I can see why it is so for {0, 2, 4} (all multiples of 6 are divisible
>>>>> by 2, so are the sums when added 2 or 4), but why also for the summand
>>>>> 3?
>>>> Because 6k + 3 is divisible by 3.
>>> [...]
>>>> And if it's greater than 5 it's clearly not equal to 3, so it's not
>>>> prime.
> I find that argument specious *at best* because it does not follow that a
> number is not prime when it is greater than 5 and not equal to 3. Perhaps
> I am misunderstanding what you mean by "it".
Perhaps my comma was ill-conceived. This would certainly have been clearer:
| Because `6k + 3` is divisible by 3. And if it's greater than 5, it's clearly
| not equal to 3. So it's not prime.
The antecedent to my "it" was an arbitrary number of the form `6k + 3`.
In most mathematical proofs, such would be considered plenty, with the
assumption that the reader does not need every detail spelled out explicitly.
But I'm glad that's cleared up.
----------
Much more interesting is the question of whether Jonas is actually one of
Mentiflex's experiments, a long-posited AI, posting things that occasionally
sound as though they're making sense, only to quickly lapse back into babble.
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| From | Erwin Moller <erwinmollerusenet@xs4all.nl> |
|---|---|
| Date | 2016-01-21 17:21 +0100 |
| Message-ID | <56a1058e$0$23762$e4fe514c@news.xs4all.nl> |
| In reply to | #29358 |
On 1/19/2016 10:42 PM, Scott Sauyet wrote: > > Much more interesting is the question of whether Jonas is actually one of > Mentiflex's experiments, a long-posited AI, posting things that occasionally > sound as though they're making sense, only to quickly lapse back into babble. > Ahh, Mentiflex! I remember the first time I fell for his babble. I even tried to run his "AI" code, without much success of course. Then some kind soul on usenet told me to heed elsewhere. But is Mentiflex posting under other names too? He made it to household name, so why change it, I wonder. ;-) Regards, Erwin Moller -- "That which can be asserted without evidence, can be dismissed without evidence." -- Christopher Hitchens
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| From | Thomas 'PointedEars' Lahn <PointedEars@web.de> |
|---|---|
| Date | 2016-02-03 19:52 +0100 |
| Message-ID | <17742798.bhIW9uKqXE@PointedEars.de> |
| In reply to | #29397 |
Erwin Moller wrote: > On 1/19/2016 10:42 PM, Scott Sauyet wrote: >> Much more interesting is the question of whether Jonas is actually one of >> Mentiflex's experiments, a long-posited AI, posting things that >> occasionally sound as though they're making sense, only to quickly lapse >> back into babble. > > Ahh, Mentiflex! > […] _M*ntif*x_, though thou shan’t speake of the devil lest he shall appear! -- PointedEars FAQ: <http://PointedEars.de/faq> | SVN: <http://PointedEars.de/wsvn/> Twitter: @PointedEars2 | ES Matrix: <http://PointedEars.de/es-matrix> Please do not cc me. / Bitte keine Kopien per E-Mail.e
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| From | Dr J R Stockton <reply1600@merlyn.demon.co.uk.invalid> |
|---|---|
| Date | 2016-01-19 19:43 +0000 |
| Message-ID | <H0KYYZwYHpnWFwpQ@invalid.uk.co.demon.merlyn.invalid> |
| In reply to | #29323 |
In comp.lang.javascript message <37264925.IaaGi9xk8D@PointedEars.de>, Mon, 18 Jan 2016 20:09:33, Thomas 'PointedEars' Lahn <PointedEars@web.de> posted: >Gene Wirchenko wrote: > >> […] All primes >= 5 are of the form 6k +/- 1 where k is a positive >> integer. > >Interesting thesis. Prove it. It is obvious on inspection, and well-known among the mathematically literate. In addition, you should have recalled that all primes >= 3 are of the form 2k +/- 1. -- (c) John Stockton, Surrey, UK. ¬@merlyn.demon.co.uk Turnpike v6.05 MIME. Merlyn Web Site < > - FAQish topics, acronyms, & links.
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