Expand Messages
• I am wondering if this is positive or saying that it is not strong enough. I also think I may be unclear so I restated it again below. ... From: djbroadhurst
Message 1 of 3 , Aug 3, 2002
• 0 Attachment
I am wondering if this is positive or saying that it is not strong enough. I
also think I may be unclear so I restated it again below.

----- Original Message -----
Sent: Saturday, August 03, 2002 4:05 AM

> Let p and q be successive primes.
> Andrica conjectures that
> q - p < sqrt(q) + sqrt(p)
> whereas Tschebysheff proved merely that
> q - p < p
> which is far weaker.

rewriten as q < 2p
include p itself p < q < 2p

take sqrt, sqrt (p) < sqrt (q) < sqrt(2)*sqrt(p) --- length of side with
square q
take out the factor sqrt (p), 1< sqrt (q) / sqrt(p) < sqrt(2)

subtract the area p from q by
take the sqrt, sqrt (p)
take out the factor sqrt (p), 1 --- length of side with square p
so
1-1 = 0 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1
1 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1 < 1

using the difference of two squares
q - p = (sqrt(q) + sqrt(p)) (sqrt(q) - sqrt(p))
(q - p) / (sqrt(q) + sqrt(p)) = (sqrt(q) - sqrt(p))
and
(sqrt (q) - sqrt(p)) < (sqrt(2) sqrt(p)) - (sqrt(p))
(sqrt (q) - sqrt(p))/sqrt(p) < (sqrt(2) - 1)(sqrt(p))/sqrt(p)
S = sqrt (q) / sqrt(p) -1 < sqrt(2) - 1 < 1
so the first number with p = 2 is < 1 and the limit S as p -> oo = 0.
QED

I'm just a college student in physics so it would help if I could get some
feedback.

Thanks David.

John
• What David is saying is that you have merely re-worked Tschebysheff s proof. Without having put in anything extra into the system, all you have done is find
Message 2 of 3 , Aug 4, 2002
• 0 Attachment
What David is saying is that you have merely re-worked Tschebysheff's proof.

Without having put in anything extra into the system, all you have done is
find new expressions for existing formula.

As a Mathematician, your proof is lacking in several areas:

#rewriten as q < 2p
#include p itself p < q < 2p

This could be written as: there exists a prime q, p<q<2p

#take sqrt, sqrt (p) < sqrt (q) < sqrt(2)*sqrt(p) --- length of side with
#square q
#take out the factor sqrt (p), 1< sqrt (q) / sqrt(p) < sqrt(2)

OK.

#subtract the area p from q by
#take the sqrt, sqrt (p)
#take out the factor sqrt (p), 1 --- length of side with square p
#so
#1-1 = 0 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1
#1 < (sqrt (q) / sqrt(p) - 1) < sqrt(2) - 1 < 1

Line 2 makes no sense, nor does line 3
Line 5 is OK, but Line6 contains obvious errors.

At this point what we have is sqrt(q)/sqrt(p) - 1 < 1, which you managed to
get to in your first proof, but seem to have failed to in this one.

From here, we can get sqrt(q)-sqrt(p)<sqrt(p), but this is far from
Andrica's conjecture.

#using the difference of two squares
#q - p = (sqrt(q) + sqrt(p)) (sqrt(q) - sqrt(p))
#(q - p) / (sqrt(q) + sqrt(p)) = (sqrt(q) - sqrt(p))
#and
#(sqrt (q) - sqrt(p)) < (sqrt(2) sqrt(p)) - (sqrt(p))
#(sqrt (q) - sqrt(p))/sqrt(p) < (sqrt(2) - 1)(sqrt(p))/sqrt(p)
#S = sqrt (q) / sqrt(p) -1 < sqrt(2) - 1 < 1
#so the first number with p = 2 is < 1 and the limit S as p -> oo = 0.
#QED

With such a dodgy start, most Mathematicians would be rolling about on the
floor by now in tears, but persistence is to be encouraged.

Line 1 is good, but you can begin to see why this conjecture is considered
hard.
Line 2 is good.

After here, you need to explain where your subtitutions are coming from.
Also needed is a statement declaring how the algebra will lead to a proof of
the conjecture. This assists any reader in following your work, and takes

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• ... Jon, I don t know if you re trying to be Me Too to David s Big Dog , http://www.winternet.com/~mikelr/flame3.html ? However, you re not really in a
Message 3 of 3 , Aug 4, 2002
• 0 Attachment
--- Jon Perry <perry@...> wrote:
> With such a dodgy start, most Mathematicians would be rolling about
> on the
> floor by now in tears, but persistence is to be encouraged.

Jon,
I don't know if you're trying to be "Me Too" to David's "Big Dog",
http://www.winternet.com/~mikelr/flame3.html ?
However, you're not really in a position to make snide comments about
proofs, particularly in such a patronising way, and especially
considering some analyses of your own proof style (

Phil
(No bonus points for spotting the irony herein!)

=====
--
The good Christian should beware of mathematicians, and all those who make
empty prophecies. The danger already exists that the mathematicians have
made a covenant with the devil to darken the spirit and to confine man in
the bonds of Hell. -- Common mistranslation of St. Augustine (354-430)

__________________________________________________
Do You Yahoo!?
Yahoo! Health - Feel better, live better
http://health.yahoo.com
Your message has been successfully submitted and would be delivered to recipients shortly.