## Has anyone noticed this?

Expand Messages
• Has anyone noticed this? Finding primes, X to N when (3*X) N, you do not need all of the primes up to the SQRT of N. Instead all you need is the primes up
Message 1 of 17 , Jan 3, 2001
Has anyone noticed this?

Finding primes, X to N when (3*X) > N, you do not need all of the primes up
to the SQRT of N. Instead all you need is the primes up to the SQRT of X
rounded up to the next odd integer.

For instance:
If X=35, then we can set N all the way to ((X-2)*3) or 99 which lets N<X
Now Instead of having to use SQRT of 99=9.9 and rounding down to the next odd
integer to get 9. This is what was on the page you need all the primes from
3 to 9 to eliminate up to 99.

But if you SQRT of 35 =5.9 round up to next odd integer 7 all you need is
primes 3 to 7 to eliminate from 35 to 99.

7 and 9 isn't that much difference but if you use 335 and 999 for X and N
respectively then you only need the primes up to 19 [SQRT 335=18.3 round up
to next whole odd integer 19] instead of all primes to 31
• ... What about testing 667? Is it sufficient to test divisors only up to 19? +-----------------------------------------------------------+ ...
Message 2 of 17 , Jan 3, 2001
At 05:17 PM 1/3/2001 -0500, Stphn828@... wrote:

>Has anyone noticed this?

...

>7 and 9 isn't that much difference but if you use 335 and 999 for X and N
>respectively then you only need the primes up to 19 [SQRT 335=18.3 round up
>to next whole odd integer 19] instead of all primes to 31

What about testing 667? Is it sufficient to test divisors only up to 19?

+-----------------------------------------------------------+
| Jud McCranie |
| |
| Think recursively( Think recursively( Think recursively)) |
+-----------------------------------------------------------+
• ... How do you throw out 23*23, 23*29 and 29*29? Primes up to P(x) can detect composites up to but not including P(x+1)^2, where P(x) is the xth prime. In
Message 3 of 17 , Jan 3, 2001
On Wed, 03 January 2001, Stphn828@... wrote:
> Has anyone noticed this?
>
>
>
> Finding primes, X to N when (3*X) > N, you do not need all of the primes up
> to the SQRT of N. Instead all you need is the primes up to the SQRT of X
> rounded up to the next odd integer.
>
>
> For instance:
> If X=35, then we can set N all the way to ((X-2)*3) or 99 which lets N<X
> Now Instead of having to use SQRT of 99=9.9 and rounding down to the next odd
> integer to get 9. This is what was on the page you need all the primes from
> 3 to 9 to eliminate up to 99.
>
> But if you SQRT of 35 =5.9 round up to next odd integer 7 all you need is
> primes 3 to 7 to eliminate from 35 to 99.
>
> 7 and 9 isn't that much difference but if you use 335 and 999 for X and N
> respectively then you only need the primes up to 19 [SQRT 335=18.3 round up
> to next whole odd integer 19] instead of all primes to 31

How do you throw out 23*23, 23*29 and 29*29?

Primes up to P(x) can detect composites up to but not including P(x+1)^2, where P(x) is the xth prime.
In general P(x+1) cannot be guaranteed to be more than P(x)+2.

Therefore most generous blanket simplification in terms of P(x) is
(P(x)+2)^2 = P(x)^2 + 4P(x) + 4.

Phil

Mathematics should not have to involve martyrdom;
Support Eric Weisstein, see http://mathworld.wolfram.com
Find the best deals on the web at AltaVista Shopping!
http://www.shopping.altavista.com
• I noticed this (quite some time ago). I even submitted it to Eric W. MathWorld, but it never got published. You can prove it by: If no prime factor has been
Message 4 of 17 , Jan 4, 2001
I noticed this (quite some time ago). I even submitted it to Eric W.
MathWorld, but it never got published.

You can prove it by:

If no prime factor has been found by the square root of x, then the product
of any factors that x possesses will be greater than x, or x is prime.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
Brainbench 'Most Valuable Professional' for HTML
Brainbench 'Most Valuable Professional' for JavaScript
http://www.brainbench.com

-----Original Message-----
From: Stphn828@... [mailto:Stphn828@...]
Sent: 03 January 2001 22:17
Subject: [PrimeNumbers] Has anyone noticed this?

Has anyone noticed this?

Finding primes, X to N when (3*X) > N, you do not need all of the primes up
to the SQRT of N. Instead all you need is the primes up to the SQRT of X
rounded up to the next odd integer.

For instance:
If X=35, then we can set N all the way to ((X-2)*3) or 99 which lets N<X
Now Instead of having to use SQRT of 99=9.9 and rounding down to the next
odd
integer to get 9. This is what was on the page you need all the primes from
3 to 9 to eliminate up to 99.

But if you SQRT of 35 =5.9 round up to next odd integer 7 all you need is
primes 3 to 7 to eliminate from 35 to 99.

7 and 9 isn't that much difference but if you use 335 and 999 for X and N
respectively then you only need the primes up to 19 [SQRT 335=18.3 round up
to next whole odd integer 19] instead of all primes to 31

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org
• ... Isn t this trivially obvious? If x is composite it has a factor p, 1
Message 5 of 17 , Jan 4, 2001
At 06:28 PM 1/4/2001 +0000, Jon Perry wrote:

>If no prime factor has been found by the square root of x, then the product
>of any factors that x possesses will be greater than x, or x is prime.

Isn't this trivially obvious? If x is composite it has a factor p, 1 < p
<= sqrt(x).

+--------------------------------------------------------+
| Jud McCranie |
| |
| 137*2^261147+1 is prime! (78,616 digits, 5/2/00) |
+--------------------------------------------------------+
• ... product ... Having no factor below the square root and some over the square root is impossible. If p is a factor of x and sqrt(x)
Message 6 of 17 , Jan 4, 2001
>>If no prime factor has been found by the square root of x, then the
product
>>of any factors that x possesses will be greater than x, or x is prime.

>Isn't this trivially obvious? If x is composite it has a factor p, 1 < p
><= sqrt(x).

Having no factor below the square root and some over the square root is
impossible.

If p is a factor of x and sqrt(x) <= p < x, then q when p * q = x is 1 < q
<= sqrt(x)
because p - sqrt(x) >= 0 and q - sqrt(x) <= 0.

So having p * q = x when p - sqrt(x) >= 0 and q - sqrt(x) >= 0 is not
possible.

The sentence from Jon Perry is not trivially obvious but obviously false.

Jean
• If x is composite it has a factor p, 1
Message 7 of 17 , Jan 4, 2001
If x is composite it has a factor p, 1 < p <= sqrt(x).

Because?

That's exactly what I just said, just re-worked for glamour and mass-appeal.

While I'm here, I thought of a joke.

'I've solved Goldbach Theorem, and discovered the Prime Distribution
Theorem. Here it is:

oghjpoijvmuno mohg
gdljdlfj;df
5097947649867496
gojgo45434656
g436t
grlj5 45j4o5j545

The only problem is, this is written in Alien, and I haven't yet discovered
the decryption code.'

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
Brainbench 'Most Valuable Professional' for HTML
Brainbench 'Most Valuable Professional' for JavaScript
http://www.brainbench.com

-----Original Message-----
From: Gauthier, Jean [mailto:Jean.Gauthier@...]
Sent: 04 January 2001 19:01
To: 'Jud McCranie'; Jon Perry
Subject: RE: [PrimeNumbers] Has anyone noticed this?

>>If no prime factor has been found by the square root of x, then the
product
>>of any factors that x possesses will be greater than x, or x is prime.

>Isn't this trivially obvious? If x is composite it has a factor p, 1 < p
><= sqrt(x).

Having no factor below the square root and some over the square root is
impossible.

If p is a factor of x and sqrt(x) <= p < x, then q when p * q = x is 1 < q
<= sqrt(x)
because p - sqrt(x) >= 0 and q - sqrt(x) <= 0.

So having p * q = x when p - sqrt(x) >= 0 and q - sqrt(x) >= 0 is not
possible.

The sentence from Jon Perry is not trivially obvious but obviously false.

Jean

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org
• ... Because if it has at least 2 factors other than 1 and itself, one must be
Message 8 of 17 , Jan 4, 2001
At 08:02 PM 1/4/2001 +0000, Jon Perry wrote:
>If x is composite it has a factor p, 1 < p <= sqrt(x).
>
>Because?

Because if it has at least 2 factors other than 1 and itself, one must be
<= sqrt(x).

>That's exactly what I just said, just re-worked for glamour and mass-appeal.

Not exactly.

"If no prime factor has been found by the square root of x, then the
product
of any factors that x possesses will be greater than x, or x is prime. "

The "or x is prime" is OK, but no factors of x or their product will be
greater than x.

+-----------------------------------------------------------+
| Jud McCranie |
| |
| Think recursively( Think recursively( Think recursively)) |
+-----------------------------------------------------------+
• It s a proof. A number is prime is the only factors it has are 1 and itself. So, you must prove that if all primes
Message 9 of 17 , Jan 5, 2001
It's a proof.

A number is prime is the only factors it has are 1 and itself.

So, you must prove that if all primes <= sqroot(n) do not divide n, then n
is prime.

And you do this by proving that if there are no prime factors <=sqroot(n),
then if any factors do remain, this forms a contradiction.

It can be re-worded, or even written in Mathematics.

It's not a difficult theorem/proof.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
Brainbench 'Most Valuable Professional' for HTML
Brainbench 'Most Valuable Professional' for JavaScript
http://www.brainbench.com

-----Original Message-----
From: Gauthier, Jean [mailto:Jean.Gauthier@...]
Sent: 05 January 2001 18:11
To: 'Rob Collins'
Cc: Jon Perry; 'Jud McCranie'
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Well in every case, this sentence is very confusing.

-----Original Message-----
From: Rob Collins [mailto:rtpc@...]
Sent: Friday, January 05, 2001 12:20 PM
To: Jean.Gauthier@...
Cc: Jon Perry; 'Jud McCranie'
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Bcc:

From: "Gauthier, Jean" <Jean.Gauthier@...>
Subject: RE: [PrimeNumbers] Has anyone noticed this?
Date: Thu, 4 Jan 2001 14:01:10 -0500

>>>If no prime factor has been found by the square root of x, then the
>product
>>>of any factors that x possesses will be greater than x, or x is prime.
[snip]
>The sentence from Jon Perry is not trivially obvious but obviously false.

Jon's sentence is logically consistent, but I don't think it was meant to
be. To me, it reads like this:

given x, if no candidate U{a->sqrt(x);a} is a factor of x, x is prime (since
if x is not prime then the product of any non-trivial factors of x will
exceed x).

The last part of his statement always holds true; it is a well known
property. The way it is written you might think that it is possible that x
can have no factors less than sqrt(x) and still have factors, with the
incredible property that the product of any (see he said any) of these
factors exceed x... but that is certainly, obviously, not meant.

--------------------------------------------------------------
Get "Your Linux Desktop on the Net" at http://www.workspot.com
• Ok, but sometimes a proof is more readable with mathematical symbols than words because a sentence may be sometimes interpreted another way you wrote it.
Message 10 of 17 , Jan 5, 2001
Ok,

but sometimes a proof is more readable with mathematical symbols
than words because a sentence may be sometimes interpreted another way you
wrote it.

That's exactly what happens here.

-----Original Message-----
From: Jon Perry [mailto:perry@...]
Sent: Friday, January 05, 2001 1:33 PM
Subject: RE: [PrimeNumbers] Has anyone noticed this?

It's a proof.

A number is prime is the only factors it has are 1 and itself.

So, you must prove that if all primes <= sqroot(n) do not divide n, then n
is prime.

And you do this by proving that if there are no prime factors <=sqroot(n),
then if any factors do remain, this forms a contradiction.

It can be re-worded, or even written in Mathematics.

It's not a difficult theorem/proof.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
Brainbench 'Most Valuable Professional' for HTML
Brainbench 'Most Valuable Professional' for JavaScript
http://www.brainbench.com

-----Original Message-----
From: Gauthier, Jean [mailto:Jean.Gauthier@...]
Sent: 05 January 2001 18:11
To: 'Rob Collins'
Cc: Jon Perry; 'Jud McCranie'
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Well in every case, this sentence is very confusing.

-----Original Message-----
From: Rob Collins [mailto:rtpc@...]
Sent: Friday, January 05, 2001 12:20 PM
To: Jean.Gauthier@...
Cc: Jon Perry; 'Jud McCranie'
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Bcc:

From: "Gauthier, Jean" <Jean.Gauthier@...>
Subject: RE: [PrimeNumbers] Has anyone noticed this?
Date: Thu, 4 Jan 2001 14:01:10 -0500

>>>If no prime factor has been found by the square root of x, then the
>product
>>>of any factors that x possesses will be greater than x, or x is prime.
[snip]
>The sentence from Jon Perry is not trivially obvious but obviously false.

Jon's sentence is logically consistent, but I don't think it was meant to
be. To me, it reads like this:

given x, if no candidate U{a->sqrt(x);a} is a factor of x, x is prime (since
if x is not prime then the product of any non-trivial factors of x will
exceed x).

The last part of his statement always holds true; it is a well known
property. The way it is written you might think that it is possible that x
can have no factors less than sqrt(x) and still have factors, with the
incredible property that the product of any (see he said any) of these
factors exceed x... but that is certainly, obviously, not meant.

--------------------------------------------------------------
Get "Your Linux Desktop on the Net" at http://www.workspot.com

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org
• I disagree. I think it was a well worded sentence. Not everyone understands maths - and it is useful to be able to put theorems/proofs into words. BTW I didn t
Message 11 of 17 , Jan 5, 2001
I disagree. I think it was a well worded sentence. Not everyone understands
maths - and it is useful to be able to put theorems/proofs into words.

BTW I didn't understand which part of the sentence you found confusing.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
Brainbench 'Most Valuable Professional' for HTML
Brainbench 'Most Valuable Professional' for JavaScript
http://www.brainbench.com

-----Original Message-----
From: Gauthier, Jean [mailto:Jean.Gauthier@...]
Sent: 05 January 2001 18:59
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Ok,

but sometimes a proof is more readable with mathematical symbols
than words because a sentence may be sometimes interpreted another way you
wrote it.

That's exactly what happens here.

-----Original Message-----
From: Jon Perry [mailto:perry@...]
Sent: Friday, January 05, 2001 1:33 PM
Subject: RE: [PrimeNumbers] Has anyone noticed this?

It's a proof.

A number is prime is the only factors it has are 1 and itself.

So, you must prove that if all primes <= sqroot(n) do not divide n, then n
is prime.

And you do this by proving that if there are no prime factors <=sqroot(n),
then if any factors do remain, this forms a contradiction.

It can be re-worded, or even written in Mathematics.

It's not a difficult theorem/proof.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
Brainbench 'Most Valuable Professional' for HTML
Brainbench 'Most Valuable Professional' for JavaScript
http://www.brainbench.com

-----Original Message-----
From: Gauthier, Jean [mailto:Jean.Gauthier@...]
Sent: 05 January 2001 18:11
To: 'Rob Collins'
Cc: Jon Perry; 'Jud McCranie'
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Well in every case, this sentence is very confusing.

-----Original Message-----
From: Rob Collins [mailto:rtpc@...]
Sent: Friday, January 05, 2001 12:20 PM
To: Jean.Gauthier@...
Cc: Jon Perry; 'Jud McCranie'
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Bcc:

From: "Gauthier, Jean" <Jean.Gauthier@...>
Subject: RE: [PrimeNumbers] Has anyone noticed this?
Date: Thu, 4 Jan 2001 14:01:10 -0500

>>>If no prime factor has been found by the square root of x, then the
>product
>>>of any factors that x possesses will be greater than x, or x is prime.
[snip]
>The sentence from Jon Perry is not trivially obvious but obviously false.

Jon's sentence is logically consistent, but I don't think it was meant to
be. To me, it reads like this:

given x, if no candidate U{a->sqrt(x);a} is a factor of x, x is prime (since
if x is not prime then the product of any non-trivial factors of x will
exceed x).

The last part of his statement always holds true; it is a well known
property. The way it is written you might think that it is possible that x
can have no factors less than sqrt(x) and still have factors, with the
incredible property that the product of any (see he said any) of these
factors exceed x... but that is certainly, obviously, not meant.

--------------------------------------------------------------
Get "Your Linux Desktop on the Net" at http://www.workspot.com

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org
• ... This is confusing: If no prime factor has been found by the square root of x, then the product of any factors that x possesses will be greater than x, or
Message 12 of 17 , Jan 5, 2001
At 07:24 PM 1/5/2001 +0000, Jon Perry wrote:
>I disagree. I think it was a well worded sentence. Not everyone understands
>maths - and it is useful to be able to put theorems/proofs into words.
>
>BTW I didn't understand which part of the sentence you found confusing.

This is confusing:

"If no prime factor has been found by the square root of x, then the
product of any factors that x possesses will be greater than x, or x is
prime. "

How can the product of the prime factors of x be greater than x?

+-----------------------------------------------------------+
| Jud McCranie |
| |
| Think recursively( Think recursively( Think recursively)) |
+-----------------------------------------------------------+
• ... The or x is prime to me is confusing. ... The or puts there make the whole thing very confusing. For example : I will shoot OR i will kill you. There
Message 13 of 17 , Jan 5, 2001
>>>If no prime factor has been found by the square root of x, then the
>product
>>>of any factors that x possesses will be greater than x, or x is prime.

The "or x is prime" to me is confusing.

It would have been better to wrote something like:

>>>If no prime factor has been found by the square root of x, then the
>product
>>>of any factors that x possesses will be greater than x, "so" x is prime.

The "or" puts there make the whole thing very confusing.

For example : "I will shoot OR i will kill you."
There is more than one way to understand this sentence.
People may be found this sentence funny.

It was better to write: "I will shoot AND i will kill you."
This is sentence have stronger effect on people than the preceding :)

The same thing if you say a girl: "I will marry you OR i will love you."

Jean

-----Original Message-----
From: Jon Perry [mailto:perry@...]
Sent: Friday, January 05, 2001 2:25 PM
Subject: RE: [PrimeNumbers] Has anyone noticed this?

I disagree. I think it was a well worded sentence. Not everyone understands
maths - and it is useful to be able to put theorems/proofs into words.

BTW I didn't understand which part of the sentence you found confusing.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
Brainbench 'Most Valuable Professional' for HTML
Brainbench 'Most Valuable Professional' for JavaScript
http://www.brainbench.com

-----Original Message-----
From: Gauthier, Jean [mailto:Jean.Gauthier@...]
Sent: 05 January 2001 18:59
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Ok,

but sometimes a proof is more readable with mathematical symbols
than words because a sentence may be sometimes interpreted another way you
wrote it.

That's exactly what happens here.

-----Original Message-----
From: Jon Perry [mailto:perry@...]
Sent: Friday, January 05, 2001 1:33 PM
Subject: RE: [PrimeNumbers] Has anyone noticed this?

It's a proof.

A number is prime is the only factors it has are 1 and itself.

So, you must prove that if all primes <= sqroot(n) do not divide n, then n
is prime.

And you do this by proving that if there are no prime factors <=sqroot(n),
then if any factors do remain, this forms a contradiction.

It can be re-worded, or even written in Mathematics.

It's not a difficult theorem/proof.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
Brainbench 'Most Valuable Professional' for HTML
Brainbench 'Most Valuable Professional' for JavaScript
http://www.brainbench.com

-----Original Message-----
From: Gauthier, Jean [mailto:Jean.Gauthier@...]
Sent: 05 January 2001 18:11
To: 'Rob Collins'
Cc: Jon Perry; 'Jud McCranie'
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Well in every case, this sentence is very confusing.

-----Original Message-----
From: Rob Collins [mailto:rtpc@...]
Sent: Friday, January 05, 2001 12:20 PM
To: Jean.Gauthier@...
Cc: Jon Perry; 'Jud McCranie'
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Bcc:

From: "Gauthier, Jean" <Jean.Gauthier@...>
Subject: RE: [PrimeNumbers] Has anyone noticed this?
Date: Thu, 4 Jan 2001 14:01:10 -0500

>>>If no prime factor has been found by the square root of x, then the
>product
>>>of any factors that x possesses will be greater than x, or x is prime.
[snip]
>The sentence from Jon Perry is not trivially obvious but obviously false.

Jon's sentence is logically consistent, but I don't think it was meant to
be. To me, it reads like this:

given x, if no candidate U{a->sqrt(x);a} is a factor of x, x is prime (since
if x is not prime then the product of any non-trivial factors of x will
exceed x).

The last part of his statement always holds true; it is a well known
property. The way it is written you might think that it is possible that x
can have no factors less than sqrt(x) and still have factors, with the
incredible property that the product of any (see he said any) of these
factors exceed x... but that is certainly, obviously, not meant.

--------------------------------------------------------------
Get "Your Linux Desktop on the Net" at http://www.workspot.com

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org
• I think the basic prob here is a lack of comprehension. x is a factor of x, so you will find x as a factor of itself,and x sqrt(x). If x is not prime, then x
Message 14 of 17 , Jan 5, 2001
I think the basic prob here is a lack of comprehension.

x is a factor of x, so you will find x as a factor of itself,and x>sqrt(x).

If x is not prime, then x will have at least two factors, both greater than

The sentence carries the definition of prime with it, it uses the word, and
as such, implies a knowldege of what prime means.

Hope this is clear.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
Brainbench 'Most Valuable Professional' for HTML
Brainbench 'Most Valuable Professional' for JavaScript
http://www.brainbench.com

-----Original Message-----
From: Jud McCranie [mailto:jud.mccranie@...]
Sent: 05 January 2001 20:13
To: Jon Perry
Subject: RE: [PrimeNumbers] Has anyone noticed this?

At 07:24 PM 1/5/2001 +0000, Jon Perry wrote:
>I disagree. I think it was a well worded sentence. Not everyone understands
>maths - and it is useful to be able to put theorems/proofs into words.
>
>BTW I didn't understand which part of the sentence you found confusing.

This is confusing:

"If no prime factor has been found by the square root of x, then the
product of any factors that x possesses will be greater than x, or x is
prime. "

How can the product of the prime factors of x be greater than x?

+-----------------------------------------------------------+
| Jud McCranie |
| |
| Think recursively( Think recursively( Think recursively)) |
+-----------------------------------------------------------+

Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
The Prime Pages : http://www.primepages.org
• Hi, I don t want to do a english lesson here but if we want to use word, let s use it a good way. ... x sqrt(x). This one great. ... than ... In that one,
Message 15 of 17 , Jan 5, 2001
Hi,

I don't want to do a english lesson here but if we want to use word, let's
use it a good way.

> x is a factor of x, so you will find x as a factor of itself,and
x>sqrt(x).

This one great.

> If x is not prime, then x will have at least two factors, both greater
than
> sqrt(x), which is a contradiction.

In that one, something is missing.

You say:
" If x is not prime, then x will have at least two factors, ... "

That's good until that point.

" ... , both greater than sqrt(x), ..."

Here we must stop, because at that point we don't know where you want to go
with
your idea. So it is confusing. Better will be : "..., and if we evaluate the
possibility that
only two factors are greater than x, ..."

Finally:
" ..., which is a contradiction."

According to your sentence, what is the exact contradiction? The fact that
if x is not prime
it can't have at least two factors? Or it is that both cannot be greater
than sqrt(x)?

Sure we all now know the answer, but it is important to express ourself
correctly if
we want to use words. If not, then we do not have to write correctly because
we all know
the answer of every thing, so we can be as confusing as we want and every
body will always
understand us.

If you let me extrapolate again on the subject, maths symbols are use
exactly to prevent
errors of comprehension. Words may have more signification than maths
symbols, so errors
of comprehension may happens during scientifics discussion.

Ex.:
Maths: 1 + 1 = 2; This will never confuse anybody.

Words: Take a number, which is the unit number of the integer class, then
make a copie of that number also,
execute an operation that consist of adding the two numbers you have, then
the result equals two.

Tells me the one you prefer.

Jean
• ... I believe this has been known to be true for some time. In fact, it was discussed in my computer science I course for majors this past semester. Nathan,
Message 16 of 17 , Jan 5, 2001
Jon Perry wrote:
>
> It's a proof.
>
> A number is prime is the only factors it has are 1 and itself.
>
> So, you must prove that if all primes <= sqroot(n) do not divide n, then n
> is prime.
>
> And you do this by proving that if there are no prime factors <=sqroot(n),
> then if any factors do remain, this forms a contradiction.
>
> It can be re-worded, or even written in Mathematics.
>
> It's not a difficult theorem/proof.
>
> Jon Perry

I believe this has been known to be true for some time. In fact, it was
discussed in my computer science I course for majors this past
semester.

Nathan, not sure how to reacy
• I m sorry - I thought it was perfectly understandable. We have two facts . Fact A: We have a number n. Fact B: We know all primes to sqrt(n) do not divide n.
Message 17 of 17 , Jan 5, 2001
I'm sorry - I thought it was perfectly understandable.

We have two 'facts'.

Fact A: We have a number n.
Fact B: We know all primes to sqrt(n) do not divide n.

Theorem:

Fact B --> n is prime.

Proof: Fact B --> n is prime OR Fact B --> n is not prime.

If n is not prime, then it's factors must be greater than sqrt(n). But this
means that the product of the factors is greater than n. Contradiction,
hence Reductio Ad Absurdum. (Reduction to the Absurb - a recoginsed
'proof').

Therefore Fact B --> n is prime.

Which is all I said, despite numerous attempts otherwise to say that I
didn't. I find the Unique Factorization Theorem a confusing proof, but I
don't blab this fact around the known universe.

Sure, it looks good in numbers, but it looks good with words too!

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry
Brainbench 'Most Valuable Professional' for HTML
Brainbench 'Most Valuable Professional' for JavaScript
http://www.brainbench.com
-----Original Message-----
From: Gauthier, Jean [mailto:Jean.Gauthier@...]
Sent: 05 January 2001 21:27
Subject: RE: [PrimeNumbers] Has anyone noticed this?

Hi,

I don't want to do a english lesson here but if we want to use word, let's
use it a good way.

> x is a factor of x, so you will find x as a factor of itself,and
x>sqrt(x).

This one great.

> If x is not prime, then x will have at least two factors, both greater
than
> sqrt(x), which is a contradiction.

In that one, something is missing.

You say:
" If x is not prime, then x will have at least two factors, ... "

That's good until that point.

" ... , both greater than sqrt(x), ..."

Here we must stop, because at that point we don't know where you want to go
with
your idea. So it is confusing. Better will be : "..., and if we evaluate the
possibility that
only two factors are greater than x, ..."

Finally:
" ..., which is a contradiction."

According to your sentence, what is the exact contradiction? The fact that
if x is not prime
it can't have at least two factors? Or it is that both cannot be greater
than sqrt(x)?

Sure we all now know the answer, but it is important to express ourself
correctly if
we want to use words. If not, then we do not have to write correctly because
we all know
the answer of every thing, so we can be as confusing as we want and every
body will always
understand us.

If you let me extrapolate again on the subject, maths symbols are use
exactly to prevent
errors of comprehension. Words may have more signification than maths
symbols, so errors
of comprehension may happens during scientifics discussion.

Ex.:
Maths: 1 + 1 = 2; This will never confuse anybody.

Words: Take a number, which is the unit number of the integer class, then
make a copie of that number also,
execute an operation that consist of adding the two numbers you have, then
the result equals two.

Tells me the one you prefer.

Jean
Your message has been successfully submitted and would be delivered to recipients shortly.