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Has anyone noticed this?

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  • Stphn828@aol.com
    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
    • Jud McCranie
      ... 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)) |
        +-----------------------------------------------------------+
      • Phil Carmody
        ... 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
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        • Jon Perry
          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@...
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            -----Original Message-----
            From: Stphn828@... [mailto:Stphn828@...]
            Sent: 03 January 2001 22:17
            To: primenumbers@egroups.com
            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

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          • Jud McCranie
            ... 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) |
              +--------------------------------------------------------+
            • Gauthier, Jean
              ... 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
              • Jon Perry
                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
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                  -----Original Message-----
                  From: Gauthier, Jean [mailto:Jean.Gauthier@...]
                  Sent: 04 January 2001 19:01
                  To: 'Jud McCranie'; Jon Perry
                  Cc: primenumbers@egroups.com
                  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
                • Jud McCranie
                  ... 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)) |
                    +-----------------------------------------------------------+
                  • Jon Perry
                    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
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                      -----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:

                      In reply to message
                      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.


                      --------------------------------------------------------------
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                    • Gauthier, Jean
                      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
                        To: Primenumbers@Egroups. Com
                        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
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                        -----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:

                        In reply to message
                        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.


                        --------------------------------------------------------------
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                      • Jon Perry
                        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
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                          -----Original Message-----
                          From: Gauthier, Jean [mailto:Jean.Gauthier@...]
                          Sent: 05 January 2001 18:59
                          To: 'Jon Perry'; Primenumbers@Egroups. Com
                          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
                          To: Primenumbers@Egroups. Com
                          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
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                          http://www.egroups.com/group/Delphiadvanced




                          -----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:

                          In reply to message
                          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.


                          --------------------------------------------------------------
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                        • Jud McCranie
                          ... 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)) |
                            +-----------------------------------------------------------+
                          • Gauthier, Jean
                            ... 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
                              To: Primenumbers@Egroups. Com
                              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
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                              -----Original Message-----
                              From: Gauthier, Jean [mailto:Jean.Gauthier@...]
                              Sent: 05 January 2001 18:59
                              To: 'Jon Perry'; Primenumbers@Egroups. Com
                              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
                              To: Primenumbers@Egroups. Com
                              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
                              Subscribe to Delphiadvanced:
                              http://www.egroups.com/group/Delphiadvanced




                              -----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:

                              In reply to message
                              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.


                              --------------------------------------------------------------
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                            • Jon Perry
                              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
                                sqrt(x), which is a contradiction.

                                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@...
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                                -----Original Message-----
                                From: Jud McCranie [mailto:jud.mccranie@...]
                                Sent: 05 January 2001 20:13
                                To: Jon Perry
                                Cc: Primenumbers@Egroups. Com
                                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)) |
                                +-----------------------------------------------------------+



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                              • Gauthier, Jean
                                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.

                                  Then you add:

                                  " ... , 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
                                • Nathan Russell
                                  ... 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
                                  • Jon Perry
                                    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@...
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                                      -----Original Message-----
                                      From: Gauthier, Jean [mailto:Jean.Gauthier@...]
                                      Sent: 05 January 2001 21:27
                                      To: 'Jon Perry'; Primenumbers@Egroups. Com
                                      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.

                                      Then you add:

                                      " ... , 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
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