## a prime between x^2 and (x+1)^2, proof?

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• Dr. Broadhurst, ... there must exist at least 1 prime between two consecutive squares. ... proof: assume that there are no primes between y^2= (x+1)^2 and x^2;
Message 1 of 13 , Sep 14, 2011
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>> this is very close to how it should be proven! right?

there must exist at least 1 prime between two consecutive squares.
...
proof: assume that there are no primes between y^2= (x+1)^2 and x^2;
we'd have ln(x^2) =~ ln(y^2 -x^2) {{by something less than one}};
clearly, y^2 -x^2 = 2x +1; still... 2*ln(x) =~ ln(2*x), and e^2*x =~
2*x, and... e^2 =~ 2. [not gonna happen, even if we double or triple
the RHS!]
...
thus, there must exist a prime between two consecutive squares.
*QED

Bill
• simpler than that... if no primes exist between x^2 and (x+1)^2, then ln(x^2)= ln(y^2) implies that 2*ln(x)= 2*ln(x+1) implies that x = x +1, or 0 = 1. can t
Message 2 of 13 , Sep 14, 2011
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simpler than that...
if no primes exist between x^2 and (x+1)^2, then ln(x^2)= ln(y^2)
implies that 2*ln(x)= 2*ln(x+1) implies that x = x +1, or 0 = 1.
can't be... at least 1 prime exists between them.

--- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...> wrote:
>
>
> >> this is very close to how it should be proven! right?
>
> there must exist at least 1 prime between two consecutive squares.
> ...
> proof: assume that there are no primes between y^2= (x+1)^2 and x^2;
> we'd have ln(x^2) =~ ln(y^2 -x^2) {{by something less than one}};
> clearly, y^2 -x^2 = 2x +1; still... 2*ln(x) =~ ln(2*x), and e^2*x =~
> 2*x, and... e^2 =~ 2. [not gonna happen, even if we double or triple
> the RHS!]
> ...
> thus, there must exist a prime between two consecutive squares.
> *QED
>
> Bill
>
• So ln(2)=ln(3)? ... [Non-text portions of this message have been removed]
Message 3 of 13 , Sep 14, 2011
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So ln(2)=ln(3)?

On Thu, Sep 15, 2011 at 7:27 AM, leavemsg1 <leavemsg1@...> wrote:

> **
>
>
> simpler than that...
> if no primes exist between x^2 and (x+1)^2, then ln(x^2)= ln(y^2)
> implies that 2*ln(x)= 2*ln(x+1) implies that x = x +1, or 0 = 1.
> can't be... at least 1 prime exists between them.
>

[Non-text portions of this message have been removed]
• if there are no primes between x^2 and (x+1)^2, then pi(x^2) = pi((x+1)^2) and ln(x^2) = ln((x+1)^2) would be appropriate, assuming... (x^2)/ln(x^2) =
Message 4 of 13 , Sep 15, 2011
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if there are no primes between x^2 and (x+1)^2, then
pi(x^2) = pi((x+1)^2) and ln(x^2) = ln((x+1)^2) would
be appropriate, assuming... (x^2)/ln(x^2) = ((x+1)^2)/ln((x+1)^2)
the bottoms of each pi function would be equal... but can't be.

From: Bernardo Boncompagni <RedGolpe@...>
Sent: Thursday, September 15, 2011 1:54 AM
Subject: Re: [PrimeNumbers] Re: a prime between x^2 and (x+1)^2, proof?

So ln(2)=ln(3)?

On Thu, Sep 15, 2011 at 7:27 AM, leavemsg1 <leavemsg1@...> wrote:

> **
>
>
> simpler than that...
> if no primes exist between x^2 and (x+1)^2, then ln(x^2)= ln(y^2)
> implies that 2*ln(x)= 2*ln(x+1) implies that x = x +1, or 0 = 1.
> can't be... at least 1 prime exists between them.
>

[Non-text portions of this message have been removed]

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[Non-text portions of this message have been removed]
• ultimately,... we d have pi(x^2) = pi((x+1)^2 giving us... 0/0 = 1/1... impossible, so there d be a prime between them, always. From: Bill Bouris
Message 5 of 13 , Sep 15, 2011
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ultimately,... we'd have pi(x^2) = pi((x+1)^2
giving us... 0/0 = 1/1... impossible, so there'd
be a prime between them, always.

From: Bill Bouris <leavemsg1@...>
To: pgroup <primenumbers@yahoogroups.com>; Bernardo Boncompagni <redgolpe@...>
Sent: Thursday, September 15, 2011 3:02 AM
Subject: Re: [PrimeNumbers] Re: a prime between x^2 and (x+1)^2, proof?

if there are no primes between x^2 and (x+1)^2, then
pi(x^2) = pi((x+1)^2) and ln(x^2) = ln((x+1)^2) would
be appropriate, assuming... (x^2)/ln(x^2) = ((x+1)^2)/ln((x+1)^2)
the bottoms of each pi function would be equal... but can't be.

From: Bernardo Boncompagni <RedGolpe@...>
Sent: Thursday, September 15, 2011 1:54 AM
Subject: Re: [PrimeNumbers] Re: a prime between x^2 and (x+1)^2, proof?

So ln(2)=ln(3)?

On Thu, Sep 15, 2011 at 7:27 AM, leavemsg1 <leavemsg1@...> wrote:

> **
>
>
> simpler than that...
> if no primes exist between x^2 and (x+1)^2, then ln(x^2)= ln(y^2)
> implies that 2*ln(x)= 2*ln(x+1) implies that x = x +1, or 0 = 1.
> can't be... at least 1 prime exists between them.
>

[Non-text portions of this message have been removed]

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[Non-text portions of this message have been removed]
• ... Why would pi(x^2)=pi((x+1)^2) imply ln(x^2)=ln((x+1)^2? pi(8)=pi(10)=4 but obviously ln(8)=/=ln(10). [Non-text portions of this message have been removed]
Message 6 of 13 , Sep 15, 2011
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> if there are no primes between x^2 and (x+1)^2, then
> pi(x^2) = pi((x+1)^2) and ln(x^2) = ln((x+1)^2) would
> be appropriate, assuming... (x^2)/ln(x^2) = ((x+1)^2)/ln((x+1)^2)
> the bottoms of each pi function would be equal... but can't be.
>

Why would pi(x^2)=pi((x+1)^2) imply ln(x^2)=ln((x+1)^2?
pi(8)=pi(10)=4 but obviously ln(8)=/=ln(10).

[Non-text portions of this message have been removed]
• On Thu, Sep 15, 2011 at 12:17, Bernardo Boncompagni ... For as far as I can follow the argument, Pi(x) ~ ln(x). When you are comparing ln(8) and ln(10), you
Message 7 of 13 , Sep 15, 2011
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On Thu, Sep 15, 2011 at 12:17, Bernardo Boncompagni
<RedGolpe@...> wrote:
>>  if there are no primes between x^2 and (x+1)^2, then
>> pi(x^2) = pi((x+1)^2) and ln(x^2) = ln((x+1)^2) would
>> be appropriate, assuming... (x^2)/ln(x^2) = ((x+1)^2)/ln((x+1)^2)
>> the bottoms of each pi function would be equal... but can't be.
>>
>
> Why would pi(x^2)=pi((x+1)^2) imply ln(x^2)=ln((x+1)^2?
> pi(8)=pi(10)=4 but obviously ln(8)=/=ln(10).

For as far as I can follow the argument, Pi(x) ~ ln(x).

When you are comparing ln(8) and ln(10), you failed to take into
account that Bill meant that those ln()'s should be take from x^2 and
(x+1)^2, resp.

You, In this case: ln(9) =!= ln(16) (which is correct), and also Pi(9)
=!= Pi(16) (which is also correct).

Pi(9) == 4, Pi(16) == 6

Hope this helps the confusion?

Christ van Willegen
--
09 F9 11 02 9D 74 E3 5B D8 41 56 C5 63 56 88 C0
• ... That s exactly my point. He said that pi(x^2)=pi((x+1)^2) implies ln(x^2)=ln((x+1)^2). I THINK he says that because he believes that pi(x)=pi(y) implies
Message 8 of 13 , Sep 15, 2011
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> When you are comparing ln(8) and ln(10), you failed to take into
> account that Bill meant that those ln()'s should be take from x^2 and
> (x+1)^2, resp.
>

That's exactly my point. He said that pi(x^2)=pi((x+1)^2) implies
ln(x^2)=ln((x+1)^2). I THINK he says that because he believes that
pi(x)=pi(y) implies ln(x)=ln(y) which I showed is wrong. Otherwise, i have
no idea why he assumes the same exact statement he's trying to demonstrate,
since obviously ln(x^2) can never be equal to ln((x+1)^2).

[Non-text portions of this message have been removed]
• On Thu, Sep 15, 2011 at 13:53, Bernardo Boncompagni ... I think it s exactly Bill s point as well, that ln(x) can never be equal to ln(y), if x y... that s
Message 9 of 13 , Sep 15, 2011
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On Thu, Sep 15, 2011 at 13:53, Bernardo Boncompagni
<redgolpe@...> wrote:
>
>> When you are comparing ln(8) and ln(10), you failed to take into
>> account that Bill meant that those ln()'s should be take from x^2 and
>> (x+1)^2, resp.
>
> That's exactly my point. He said that pi(x^2)=pi((x+1)^2) implies
> ln(x^2)=ln((x+1)^2). I THINK he says that because he believes that
> pi(x)=pi(y) implies ln(x)=ln(y) which I showed is wrong. Otherwise, i have
> no idea why he assumes the same exact statement he's trying to demonstrate,
> since obviously ln(x^2) can never be equal to ln((x+1)^2).

I think it's exactly Bill's point as well, that ln(x) can never be
equal to ln(y), if x <> y... that's what he's using for the argument!

Basically, what Bill says is (am I right, Bill?): ln(x^2) <
ln((x+1)^2). Since Pi(w) ~ ln(w) (according to PNT), Pi(x^2) <
Pi((x+1)^2), so there must be a prime between x^2 and (x+1)^2.

Does this clear up the confusion?

Christ van Willegen
--
09 F9 11 02 9D 74 E3 5B D8 41 56 C5 63 56 88 C0
• ... So again, why there should be a prime between, say, x and x+2, since ln(x) ln(x+2)? [Non-text portions of this message have been removed]
Message 10 of 13 , Sep 15, 2011
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> I think it's exactly Bill's point as well, that ln(x) can never be
> equal to ln(y), if x <> y... that's what he's using for the argument!
>
> Basically, what Bill says is (am I right, Bill?): ln(x^2) <
> ln((x+1)^2). Since Pi(w) ~ ln(w) (according to PNT), Pi(x^2) <
> Pi((x+1)^2), so there must be a prime between x^2 and (x+1)^2.
>

So again, why there should be a prime between, say, x and x+2, since
ln(x)<>ln(x+2)?

[Non-text portions of this message have been removed]
• On Thu, Sep 15, 2011 at 15:07, Bernardo Boncompagni ... Good point, No idea. Christ van Willegen
Message 11 of 13 , Sep 15, 2011
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On Thu, Sep 15, 2011 at 15:07, Bernardo Boncompagni
<RedGolpe@...> wrote:
>> I think it's exactly Bill's point as well, that ln(x) can never be
>> equal to ln(y), if x <> y... that's what he's using for the argument!
>>
>> Basically, what Bill says is (am I right, Bill?): ln(x^2) <
>> ln((x+1)^2). Since Pi(w) ~ ln(w) (according to PNT), Pi(x^2) <
>> Pi((x+1)^2), so there must be a prime between x^2 and (x+1)^2.
>>
>
> So again, why there should be a prime between, say, x and x+2, since
> ln(x)<>ln(x+2)?

Good point, No idea.

Christ van Willegen
• I like this naive approach. I know that pi(x) =~ x/log(x), roughly. and the denominators can t force 0 = 1, since (x^2)/ln(x^2) =~ ((x+1)^2)/ln((x+1)^2). It
Message 12 of 13 , Sep 15, 2011
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I like this naive approach. I know that pi(x) =~ x/log(x), roughly.
and the denominators can't force 0 = 1, since (x^2)/ln(x^2) =~
((x+1)^2)/ln((x+1)^2). It appears adequate, since we know that
there's actually quite a gap between x^2 and (x+1)^2 as the pi's
are measured; not like just saying... between x and x+2.

From: Christ van Willegen <cvwillegen@...>
To: Bernardo Boncompagni <RedGolpe@...>
Sent: Thursday, September 15, 2011 8:19 AM
Subject: Re: [PrimeNumbers] Re: a prime between x^2 and (x+1)^2, proof?

On Thu, Sep 15, 2011 at 15:07, Bernardo Boncompagni
<RedGolpe@...> wrote:
>> I think it's exactly Bill's point as well, that ln(x) can never be
>> equal to ln(y), if x <> y... that's what he's using for the argument!
>>
>> Basically, what Bill says is (am I right, Bill?): ln(x^2) <
>> ln((x+1)^2). Since Pi(w) ~ ln(w) (according to PNT), Pi(x^2) <
>> Pi((x+1)^2), so there must be a prime between x^2 and (x+1)^2.
>>
>
> So again, why there should be a prime between, say, x and x+2, since
> ln(x)<>ln(x+2)?

Good point, No idea.

Christ van Willegen

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[Non-text portions of this message have been removed]
• ... Language like appears adequate of course does not cut it in the world of mathematical proof. Is there always a prime between x^1.5 and (x+1)^1.5 ? It
Message 13 of 13 , Sep 15, 2011
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--- In primenumbers@yahoogroups.com, Bill Bouris <leavemsg1@...> wrote:
>
> I like this naive approach. I know that pi(x) =~ x/log(x), roughly.
> and the denominators can't force 0 = 1, since (x^2)/ln(x^2) =~
> ((x+1)^2)/ln((x+1)^2). It appears adequate, since we know that
> there's actually quite a gap between x^2 and (x+1)^2 as the pi's
> are measured; not like just saying... between x and x+2.
> [snip]

Language like "appears adequate" of course does not cut it in the world of mathematical proof.

Is there always a prime between x^1.5 and (x+1)^1.5 ? It turns out not, when x = 20.

How many primes are there between x^2 and (x+1)^2 ? Roughly the same as the number of primes up to x. Seems like a lot. How many primes are there between x^2 and (x+.5)^2 ? Roughly half of that. Still it seems like a lot, and yet there are no primes between (10.7)^2 and (10.7+.5)^2.
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