Prev: Small Phase Angle Between Two Functions
Next: grounds for dismissal of Julia F. Knight and Chandler Davis as math journal editors #5.10 Correcting Math
From: Archimedes Plutonium on 10 Aug 2010 12:04


sttscitrans(a)tesco.net wrote:
sttscitr...(a)tesco.net wrote:
> You still have not answered my question.
> If the Key Theorem
> "Every natural >1 has a prime divisor"

state the true **full theorem**, idiot,


*every natural >1 is divisible by itself and has a prime divisor*


When asked to define prime number, most people would say: a prime
number is a Natural Number greater than 1, divisible only by itself
and 1.


Ask Iain Davidson to define prime number and the likelihood from the
above "key theorem omission" answer that Iain would give is that a
"prime number is divisible only by 1. Where Iain forgets to say
"divisible by itself"


You fooled Lwalk, but you will not fool anyone else.

L. Walker do you still think the below is a valid proof and not a
flop?

sttscitr...(a)tesco.net wrote:
> 1) A natural is prime if it has preceisly two distinct divisors

 > 2) Every natural >1 has at least one prime divisor
  > 3) GCD(m,m+1) = 1, for any natural m
  > 3) Assume pn is the last prime
  > 4) w = the product of all primes
  > 5) 3) => gcd(w,w+1) =1 => no prime divides w+1
  >    This contradicts 2)
  > 6) Therefore: Assumption 3 is false
  >   - pn is not last prime
From: sttscitrans on 10 Aug 2010 13:32
On 10 Aug, 17:04, Archimedes Plutonium
<plutonium.archime...(a)gmail.com> wrote:
> sttscitr...(a)tesco.net wrote:
> sttscitr...(a)tesco.net wrote:
> > You still have not answered my question.
> > If the Key Theorem
> > "Every natural >1 has a prime divisor"
>
> state the true **full theorem**, idiot,

Buffoon -
"Every natural >1 has at least one prime divisor"
is the full theorem.

"Every natural >1 has a prime divisor"

This statement is either true or false.
If it is true it is a theorem.

You claim it is false.
So which n> 1 has no prime divisors ?

n = nx1 is trivially true - 1 is the identity element.

The conjunction of any self-evident truth
e.g. "Archie Poo is a slow learner" (truth value =1)
and some other statement A

A and "Archie Poo is a slow learner"

has the same truth value as A

"Paris is the capital of Germany" and "Archie Poo is a slow learner"

is false

"Paris is the capital of France " and "Archie Poo is a slow learner"

is true.

A&1 <=> A

So as usual, Archie Poo has yet again demonstrated that he has no idea
what he is talking about.

If the "full theorem" is true then

"Every natural >1 has a prime divisor"
must also be true.

But AP thinks
"Every natural >1 has a prime divisor"
is false, so AP;s "full-theorem" isn't a theorem after
all.

Poor old Archie Poo - not the slightest clue
about anything at all
From: Barb Knox on 11 Aug 2010 01:24
In article
<4b99bf87-a8a6-43f8-969d-0deeb8cf741d(a)l14g2000yql.googlegroups.com>,
"sttscitrans(a)tesco.net" <sttscitrans(a)tesco.net> wrote:

> On 10 Aug, 17:04, Archimedes Plutonium
> <plutonium.archime...(a)gmail.com> wrote:
> > sttscitr...(a)tesco.net wrote:
> > sttscitr...(a)tesco.net wrote:
> > > You still have not answered my question.
> > > If the Key Theorem
> > > "Every natural >1 has a prime divisor"
> >
> > state the true **full theorem**, idiot,
>
> Buffoon -
> "Every natural >1 has at least one prime divisor"
> is the full theorem.
>
> "Every natural >1 has a prime divisor"
>
> This statement is either true or false.
> If it is true it is a theorem.
>
> You claim it is false.
> So which n> 1 has no prime divisors ?
>
> n = nx1 is trivially true - 1 is the identity element.

Yes, but 1 is not a prime.

The simplest proof that that every n > 1 has at least one prime divisor
uses strong induction on n when n is not itself prime.


[snip]

--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum videtur.
| BBB aa a r bbb |
-----------------------------
From: sttscitrans on 11 Aug 2010 04:19
On 11 Aug, 06:24, Barb Knox <s...(a)sig.below> wrote:
> In article
> <4b99bf87-a8a6-43f8-969d-0deeb8cf7...(a)l14g2000yql.googlegroups.com>,
>
>
>
>
>
>  "sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> wrote:
> > On 10 Aug, 17:04, Archimedes Plutonium
> > <plutonium.archime...(a)gmail.com> wrote:
> > > sttscitr...(a)tesco.net wrote:
> > > sttscitr...(a)tesco.net wrote:
> > > > You still have not answered my question.
> > > > If the Key Theorem
> > > > "Every natural >1 has a prime divisor"
>
> > > state the true **full theorem**, idiot,
>
> > Buffoon -
> > "Every natural >1 has at least one prime divisor"
> > is the full theorem.
>
> > "Every natural >1 has a prime divisor"
>
> > This statement is either true or false.
> > If it is true it is a theorem.
>
> > You claim it is false.
> > So which n> 1 has no prime divisors ?
>
> > n = nx1 is trivially true -  1 is the identity element.
>
> Yes, but 1 is not a prime.

No one, except perhaps AP, is claiming that 1 is prime.

> The simplest proof that that every n > 1 has at least one prime divisor
> uses strong induction on n when n is not itself prime.

Or by simply assuming that m is the smallest natural >1
with no prime divisors.
m cannot be prime so m = ab, 1<a,b <m, a and b
then being two naturals <m with no prime divisors, a contradiction

From: Barb Knox on 12 Aug 2010 22:05
In article
<5736e570-1771-4f9a-9b8a-36e14aec8e0c(a)j8g2000yqd.googlegroups.com>,
"sttscitrans(a)tesco.net" <sttscitrans(a)tesco.net> wrote:

> On 11 Aug, 06:24, Barb Knox <s...(a)sig.below> wrote:
> > In article
> > <4b99bf87-a8a6-43f8-969d-0deeb8cf7...(a)l14g2000yql.googlegroups.com>,
> >
> >
> >
> >
> >
> > �"sttscitr...(a)tesco.net" <sttscitr...(a)tesco.net> wrote:
> > > On 10 Aug, 17:04, Archimedes Plutonium
> > > <plutonium.archime...(a)gmail.com> wrote:
> > > > sttscitr...(a)tesco.net wrote:
> > > > sttscitr...(a)tesco.net wrote:
> > > > > You still have not answered my question.
> > > > > If the Key Theorem
> > > > > "Every natural >1 has a prime divisor"
> >
> > > > state the true **full theorem**, idiot,
> >
> > > Buffoon -
> > > "Every natural >1 has at least one prime divisor"
> > > is the full theorem.
> >
> > > "Every natural >1 has a prime divisor"
> >
> > > This statement is either true or false.
> > > If it is true it is a theorem.
> >
> > > You claim it is false.
> > > So which n> 1 has no prime divisors ?
> >
> > > n = nx1 is trivially true - �1 is the identity element.
> >
> > Yes, but 1 is not a prime.
>
> No one, except perhaps AP, is claiming that 1 is prime.

Except you seem to have implied that "n = nx1 is trivially true" is
proof (or at least evidence) that every n has a prime divisor. N'est-ce
pas?


> > The simplest proof that that every n > 1 has at least one prime divisor
> > uses strong induction on n when n is not itself prime.
>
> Or by simply assuming that m is the smallest natural >1

Well-ordering is equivalent to strong induction.

> with no prime divisors.
> m cannot be prime so m = ab, 1<a,b <m, a and b
> then being two naturals <m with no prime divisors, a contradiction



--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum videtur.
| BBB aa a r bbb |
-----------------------------
 | 
Pages: 1
Prev: Small Phase Angle Between Two Functions
Next: grounds for dismissal of Julia F. Knight and Chandler Davis as math journal editors #5.10 Correcting Math