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

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  • leavemsg1
    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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      Dr. Broadhurst,

      >> 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
    • leavemsg1
      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:
        >
        > Dr. Broadhurst,
        >
        > >> 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
        >
      • Bernardo Boncompagni
        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]
        • Bill Bouris
          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@...>
            To: primenumbers@yahoogroups.com
            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]
          • Bill Bouris
            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@...>
              To: primenumbers@yahoogroups.com
              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]
            • Bernardo Boncompagni
              ... 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]
              • Christ van Willegen
                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
                • Bernardo Boncompagni
                  ... 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]
                  • Christ van Willegen
                    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
                    • Bernardo Boncompagni
                      ... 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]
                      • Christ van Willegen
                        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
                        • Bill Bouris
                          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@...>
                            Cc: pgroup <primenumbers@yahoogroups.com>
                            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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                          • Mark
                            ... 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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