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Re: Coprime Arithmetic

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  • p_for_pa
    Message 1 of 16 , Nov 1, 2009
    • 0 Attachment
      --- In math_club@yahoogroups.com, Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...> wrote:
      >
      > Solve this prob: if a â€" b = 2n where n is
      > any integer then a and b is co-prime.
      >
      >
      >
      > ________________________________
      > From: Tanvir Prince <tanvirp123@...>
      > To: math_club@yahoogroups.com
      > Cc: Prince <tanvirp123@...>
      > Sent: Tue, October 20, 2009 7:52:40 AM
      > Subject: Re: [math_club] Modular arithmatic prob
      >
      >
      > we will use the following theorem which is a generalization of Fermat's theorem:
      >
      > If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)
      > here phi(n) is the euler phi function which count the number of integer from 1 to n-1 which are relatively prime to n.
      >
      > So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.
      >
      > Now since p>3 , p and 6 are relatively prime.
      > So by the above theorem:
      >
      > p^(phi(6))=p^ 2 = 1 (mod 6)
      > add 1 to both side:
      > p^2+1 = 2 (mod 6)
      >
      > So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)
      > This means we need to show 2^p-2 is divisible by 6 or 2(2^(p-1) -1) is divisible by 6 or
      > 2^(p-1) - 1 is divisible by 3 or 2^(p-1) = 1 (mod 3)
      > but 2 = -1 (mod 3). So 2^(p-1) = (-1)^(p-1) = 1 (mod 3) (since p being odd, p-1 must be even)
      > And we are done.
      >
      > --- On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:
      >
      >
      > >From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
      > >Subject: [math_club] Modular arithmatic prob
      > >To: math_club@yahoogrou ps.com
      > >Date: Monday, October 19, 2009, 9:15 AM
      > >
      > >
      > >
      > >Can anyone prove this assertion: 2p ≡ p2 + 1(mod 6), where p is any prime greater than 3.
      > >
      > >
      > >
      > >
      > ________________________________
      > From: avik roy <avik_3.1416@ yahoo.co. in>
      > >To: math_club@yahoogrou ps.com
      > >Sent: Fri, October 16, 2009 12:04:49 AM
      > >Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
      > >
      > >
      > >
      > >I am not using that fact. The fact that |1+ik| = |1-ik| leads ta calculation as follows...
      > >
      > >|1+ik| / |1-ik| =1
      > >
      > >=>|1+ik|^2 / |(1+ik)(1-ik) | =1
      > >=> |1+ik|^2 = 1 + k^2
      > >
      > >--- On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:
      > >
      > >
      > >>From: Tanvir Prince <tanvirp123@yahoo. com>
      > >>Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
      > >>To: math_club@yahoogrou ps.com
      > >>Date: Thursday, 15 October, 2009, 10:42 PM
      > >>
      > >>
      > >>
      > >>Good proof. I have only one concern:
      > >>You are using the fact that zz* = |z|^2 where z* is the conjugate of z and the fact that zz*=|z|^2 depends on pythagorean theorem
      > >>
      > >>--- On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:
      > >>
      > >>
      > >>>From: avik roy <avik_3.1416@ yahoo.co.. in>
      > >>>Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
      > >>>To: "mathclub clubmath" <math_club@yahoogrou ps.com>
      > >>>Date: Thursday, October 15, 2009, 12:36 AM
      > >>>
      > >>>
      > >>>
      > >>>Here, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.
      > >>>
      > >>>let,
      > >>>z1 = x
      > >>>z2 = ikx [k is any arbitrary real constant]
      > >>>
      > >>>|z1|^2 + |z2|^2 = (1+k^2)*x^2
      > >>>[u need not use the Pythagoras here, the concept of length does the job]
      > >>>
      > >>>|z1+z2| = x|1+ ik|
      > >>>
      > >>>A simple observation leads that |z1+z2| = |z1-z2|
      > >>>=>|1+ik|=|1-ik|
      > >>>A simple calculation proves that, |i+ik|^2 = 1+k^2
      > >>>
      > >>>This gives us that |z1|^2 +|z2|^2 = |z1 + z2|^2
      > >>>
      > >>>I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.
      > >>>
      > >>>
      > >>>________________________________
      > >>>From cricket scores to your friends. Try the Yahoo! India Homepage!
      > >>
      > >________________________________
      > >Add whatever you love to the Yahoo! India homepage. Try now!
      > >
      >i cant seem to understand wd u written in da equation there r crazy notations
    • Tanvir Prince
      The assumption is not right. For example, 16 - 8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a
      Message 2 of 16 , Nov 1, 2009
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        The assumption is not right. For example, 16 - 8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:
        Let d divide both a and b. We will show that d = 1 and this will prove our assumption.
        Since d divide both a and b and a-b=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.
        But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.
        So we must have k =1 and thus d = 2^0 = 1 and we are done.


        --- On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...> wrote:

        From: Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...>
        Subject: [math_club] Coprime Arithmetic
        To: math_club@yahoogroups.com
        Date: Saturday, October 31, 2009, 6:36 AM

         
          Solve this prob: if a – b = 2n where n is any integer then a and b is co-prime.


        From: Tanvir Prince <tanvirp123@yahoo. com>
        To: math_club@yahoogrou ps.com
        Cc: Prince <tanvirp123@yahoo. com>
        Sent: Tue, October 20, 2009 7:52:40 AM
        Subject: Re: [math_club] Modular arithmatic prob

         
        we will use the following theorem which is a generalization of Fermat's theorem:
         
        If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)
        here phi(n) is the euler phi function which count the number of integer from 1 to n-1 which are relatively prime to n.
         
        So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.
         
        Now since p>3 , p and 6 are relatively prime.
        So by the above theorem:
         
        p^(phi(6))=p^ 2 = 1 (mod 6)
        add 1 to both side:
        p^2+1 = 2 (mod 6)
         
        So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)
        This means we need to show 2^p-2 is divisible by 6 or 2(2^(p-1) -1) is divisible by 6 or
        2^(p-1) - 1 is divisible by 3 or 2^(p-1) = 1 (mod 3)
        but 2 = -1 (mod 3). So 2^(p-1) = (-1)^(p-1) = 1 (mod 3) (since p being odd, p-1 must be even)
        And we are done.

        --- On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

        From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
        Subject: [math_club] Modular arithmatic prob
        To: math_club@yahoogrou ps.com
        Date: Monday, October 19, 2009, 9:15 AM

         

        Can anyone prove this assertion:  2p ≡ p2 + 1(mod 6), where p is any prime greater than 3.




        From: avik roy <avik_3.1416@ yahoo.co. in>
        To: math_club@yahoogrou ps.com
        Sent: Fri, October 16, 2009 12:04:49 AM
        Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]

         

        I am not using that fact. The fact that |1+ik| = |1-ik| leads ta calculation as follows...

        |1+ik| / |1-ik| =1

        =>|1+ik|^2 / |(1+ik)(1-ik) | =1
        =>  |1+ik|^2 = 1 + k^2

        --- On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:

        From: Tanvir Prince <tanvirp123@yahoo. com>
        Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
        To: math_club@yahoogrou ps.com
        Date: Thursday, 15 October, 2009, 10:42 PM

         
        Good proof. I have only one  concern:
        You are using the fact that zz* = |z|^2 where z* is the conjugate of z and the fact that zz*=|z|^2 depends on pythagorean theorem

        --- On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:

        From: avik roy <avik_3.1416@ yahoo.co. in>
        Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
        To: "mathclub clubmath" <math_club@yahoogrou ps.com>
        Date: Thursday, October 15, 2009, 12:36 AM

         
        Here, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.

        let,
        z1 = x
        z2 = ikx [k is any arbitrary real constant]

        |z1|^2 + |z2|^2 = (1+k^2)*x^2
        [u need not use the Pythagoras here, the concept of length does the job]

        |z1+z2| = x|1+ ik|

        A simple observation leads that |z1+z2| = |z1-z2|
        =>|1+ik|=|1-ik|
        A simple calculation proves that, |i+ik|^2 = 1+k^2

        This gives us that |z1|^2 +|z2|^2 = |z1 + z2|^2

        I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.



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      • Mirza Sabbir Hossain Beg
        Yes Tanvir you are right. I forgot to quote a and b as odds. But the fact of letting d as the divisor of both a and b seems very contradicting to me. In my
        Message 3 of 16 , Nov 2, 2009
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          Yes Tanvir you are right. I forgot to quote a and b as odds. But the fact of letting d as the divisor of both a and b seems very contradicting to me. In my statement a and b is co-prime if so then there cannot be any integer d that divides a and b at the same time and even if d = 1 because (a, 1) is always coprime where a > 1 and any integers because they don't have the greatest common divisors. Don't you think its absurd!!!!??!! I can give you a couter example of your deduction that you used to proof the assumption. You said that d divides both a and b and 2^n but 7 - 3 = 2^2 so do we find any d that divides both 7, 3 and 4????!!!!!!!??????!!!!?? Your proof is completely wrong and contradicting. You proof must need a far rectification. Mail me. Thanks. 


          From: Tanvir Prince <tanvirp123@...>
          To: math_club@yahoogroups.com
          Sent: Mon, November 2, 2009 1:41:47 AM
          Subject: Re: [math_club] Coprime Arithmetic

           

          The assumption is not right.. For example, 16 - 8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:
          Let d divide both a and b. We will show that d = 1 and this will prove our assumption.
          Since d divide both a and b and a-b=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.
          But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.
          So we must have k =1 and thus d = 2^0 = 1 and we are done.


          --- On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

          From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
          Subject: [math_club] Coprime Arithmetic
          To: math_club@yahoogrou ps.com
          Date: Saturday, October 31, 2009, 6:36 AM

           
            Solve this prob: if a – b = 2n where n is any integer then a and b is co-prime.


          From: Tanvir Prince <tanvirp123@yahoo. com>
          To: math_club@yahoogrou ps.com
          Cc: Prince <tanvirp123@yahoo. com>
          Sent: Tue, October 20, 2009 7:52:40 AM
          Subject: Re: [math_club] Modular arithmatic prob

           
          we will use the following theorem which is a generalization of Fermat's theorem:
           
          If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)
          here phi(n) is the euler phi function which count the number of integer from 1 to n-1 which are relatively prime to n.
           
          So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.
           
          Now since p>3 , p and 6 are relatively prime.
          So by the above theorem:
           
          p^(phi(6))=p^ 2 = 1 (mod 6)
          add 1 to both side:
          p^2+1 = 2 (mod 6)
           
          So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)
          This means we need to show 2^p-2 is divisible by 6 or 2(2^(p-1) -1) is divisible by 6 or
          2^(p-1) - 1 is divisible by 3 or 2^(p-1) = 1 (mod 3)
          but 2 = -1 (mod 3). So 2^(p-1) = (-1)^(p-1) = 1 (mod 3) (since p being odd, p-1 must be even)
          And we are done.

          --- On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

          From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
          Subject: [math_club] Modular arithmatic prob
          To: math_club@yahoogrou ps.com
          Date: Monday, October 19, 2009, 9:15 AM

           

          Can anyone prove this assertion:  2p ≡ p2 + 1(mod 6), where p is any prime greater than 3.




          From: avik roy <avik_3.1416@ yahoo.co. in>
          To: math_club@yahoogrou ps.com
          Sent: Fri, October 16, 2009 12:04:49 AM
          Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]

           

          I am not using that fact. The fact that |1+ik| = |1-ik| leads ta calculation as follows...

          |1+ik| / |1-ik| =1

          =>|1+ik|^2 / |(1+ik)(1-ik) | =1
          =>  |1+ik|^2 = 1 + k^2

          --- On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:

          From: Tanvir Prince <tanvirp123@yahoo. com>
          Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
          To: math_club@yahoogrou ps.com
          Date: Thursday, 15 October, 2009, 10:42 PM

           
          Good proof. I have only one  concern:
          You are using the fact that zz* = |z|^2 where z* is the conjugate of z and the fact that zz*=|z|^2 depends on pythagorean theorem

          --- On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:

          From: avik roy <avik_3.1416@ yahoo.co. in>
          Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
          To: "mathclub clubmath" <math_club@yahoogrou ps.com>
          Date: Thursday, October 15, 2009, 12:36 AM

           
          Here, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.

          let,
          z1 = x
          z2 = ikx [k is any arbitrary real constant]

          |z1|^2 + |z2|^2 = (1+k^2)*x^2
          [u need not use the Pythagoras here, the concept of length does the job]

          |z1+z2| = x|1+ ik|

          A simple observation leads that |z1+z2| = |z1-z2|
          =>|1+ik|=|1-ik|
          A simple calculation proves that, |i+ik|^2 = 1+k^2

          This gives us that |z1|^2 +|z2|^2 = |z1 + z2|^2

          I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.



          From cricket scores to your friends. Try the Yahoo! India Homepage!



          Add whatever you love to the Yahoo! India homepage. Try now!





        • Mirza Sabbir Hossain Beg
          Tanvir I also observed another fact that: if p is any odd prime and a + b = p then a and b must be co-prime. Mail me the proof.
          Message 4 of 16 , Nov 2, 2009
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            Tanvir I also observed another fact that: if p is any odd prime and a + b = p then a and b must be co-prime. Mail me the proof.


            From: Tanvir Prince <tanvirp123@...>
            To: math_club@yahoogroups.com
            Sent: Mon, November 2, 2009 1:41:47 AM
            Subject: Re: [math_club] Coprime Arithmetic

             

            The assumption is not right. For example, 16 - 8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:
            Let d divide both a and b. We will show that d = 1 and this will prove our assumption.
            Since d divide both a and b and a-b=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.
            But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.
            So we must have k =1 and thus d = 2^0 = 1 and we are done.


            --- On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

            From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
            Subject: [math_club] Coprime Arithmetic
            To: math_club@yahoogrou ps.com
            Date: Saturday, October 31, 2009, 6:36 AM

             
              Solve this prob: if a – b = 2n where n is any integer then a and b is co-prime.


            From: Tanvir Prince <tanvirp123@yahoo. com>
            To: math_club@yahoogrou ps.com
            Cc: Prince <tanvirp123@yahoo. com>
            Sent: Tue, October 20, 2009 7:52:40 AM
            Subject: Re: [math_club] Modular arithmatic prob

             
            we will use the following theorem which is a generalization of Fermat's theorem:
             
            If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)
            here phi(n) is the euler phi function which count the number of integer from 1 to n-1 which are relatively prime to n.
             
            So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.
             
            Now since p>3 , p and 6 are relatively prime.
            So by the above theorem:
             
            p^(phi(6))=p^ 2 = 1 (mod 6)
            add 1 to both side:
            p^2+1 = 2 (mod 6)
             
            So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)
            This means we need to show 2^p-2 is divisible by 6 or 2(2^(p-1) -1) is divisible by 6 or
            2^(p-1) - 1 is divisible by 3 or 2^(p-1) = 1 (mod 3)
            but 2 = -1 (mod 3). So 2^(p-1) = (-1)^(p-1) = 1 (mod 3) (since p being odd, p-1 must be even)
            And we are done.

            --- On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

            From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
            Subject: [math_club] Modular arithmatic prob
            To: math_club@yahoogrou ps.com
            Date: Monday, October 19, 2009, 9:15 AM

             

            Can anyone prove this assertion:  2p ≡ p2 + 1(mod 6), where p is any prime greater than 3.




            From: avik roy <avik_3.1416@ yahoo.co. in>
            To: math_club@yahoogrou ps..com
            Sent: Fri, October 16, 2009 12:04:49 AM
            Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]

             

            I am not using that fact.. The fact that |1+ik| = |1-ik| leads ta calculation as follows...

            |1+ik| / |1-ik| =1

            =>|1+ik|^2 / |(1+ik)(1-ik) | =1
            =>  |1+ik|^2 = 1 + k^2

            --- On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:

            From: Tanvir Prince <tanvirp123@yahoo. com>
            Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
            To: math_club@yahoogrou ps.com
            Date: Thursday, 15 October, 2009, 10:42 PM

             
            Good proof. I have only one  concern:
            You are using the fact that zz* = |z|^2 where z* is the conjugate of z and the fact that zz*=|z|^2 depends on pythagorean theorem

            --- On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:

            From: avik roy <avik_3.1416@ yahoo.co. in>
            Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
            To: "mathclub clubmath" <math_club@yahoogrou ps.com>
            Date: Thursday, October 15, 2009, 12:36 AM

             
            Here, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.

            let,
            z1 = x
            z2 = ikx [k is any arbitrary real constant]

            |z1|^2 + |z2|^2 = (1+k^2)*x^2
            [u need not use the Pythagoras here, the concept of length does the job]

            |z1+z2| = x|1+ ik|

            A simple observation leads that |z1+z2| = |z1-z2|
            =>|1+ik|=|1-ik|
            A simple calculation proves that, |i+ik|^2 = 1+k^2

            This gives us that |z1|^2 +|z2|^2 = |z1 + z2|^2

            I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.



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            Add whatever you love to the Yahoo! India homepage. Try now!





          • avik roy
            I think the proof is ok. The problem is for odd a and b, if a-b=2^n, a and b are coprime The proof shows that assuming any common divisor d for a and b, we
            Message 5 of 16 , Nov 2, 2009
            • 0 Attachment
              I think the proof is ok.

              The problem is "for odd a and b, if a-b=2^n, a and b are coprime"

              The proof shows that assuming any common divisor d for a and b, we can show that d=1, thus a and b have to be coprime...

              --- On Mon, 2/11/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...> wrote:

              From: Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...>
              Subject: Re: [math_club] Coprime Arithmetic
              To: math_club@yahoogroups.com
              Date: Monday, 2 November, 2009, 2:08 PM

               

              Yes Tanvir you are right. I forgot to quote a and b as odds. But the fact of letting d as the divisor of both a and b seems very contradicting to me. In my statement a and b is co-prime if so then there cannot be any integer d that divides a and b at the same time and even if d = 1 because (a, 1) is always coprime where a > 1 and any integers because they don't have the greatest common divisors. Don't you think its absurd!!!!?? !! I can give you a couter example of your deduction that you used to proof the assumption. You said that d divides both a and b and 2^n but 7 - 3 = 2^2 so do we find any d that divides both 7, 3 and 4????!!!!!!! ??????!!! !?? Your proof is completely wrong and contradicting. You proof must need a far rectification. Mail me. Thanks. 


              From: Tanvir Prince <tanvirp123@yahoo. com>
              To: math_club@yahoogrou ps.com
              Sent: Mon, November 2, 2009 1:41:47 AM
              Subject: Re: [math_club] Coprime Arithmetic

               

              The assumption is not right.. For example, 16 - 8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:
              Let d divide both a and b. We will show that d = 1 and this will prove our assumption.
              Since d divide both a and b and a-b=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.
              But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.
              So we must have k =1 and thus d = 2^0 = 1 and we are done.


              --- On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

              From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
              Subject: [math_club] Coprime Arithmetic
              To: math_club@yahoogrou ps.com
              Date: Saturday, October 31, 2009, 6:36 AM

               
                Solve this prob: if a – b = 2n where n is any integer then a and b is co-prime.


              From: Tanvir Prince <tanvirp123@yahoo. com>
              To: math_club@yahoogrou ps.com
              Cc: Prince <tanvirp123@yahoo. com>
              Sent: Tue, October 20, 2009 7:52:40 AM
              Subject: Re: [math_club] Modular arithmatic prob

               
              we will use the following theorem which is a generalization of Fermat's theorem:
               
              If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)
              here phi(n) is the euler phi function which count the number of integer from 1 to n-1 which are relatively prime to n.
               
              So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.
               
              Now since p>3 , p and 6 are relatively prime.
              So by the above theorem:
               
              p^(phi(6))=p^ 2 = 1 (mod 6)
              add 1 to both side:
              p^2+1 = 2 (mod 6)
               
              So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)
              This means we need to show 2^p-2 is divisible by 6 or 2(2^(p-1) -1) is divisible by 6 or
              2^(p-1) - 1 is divisible by 3 or 2^(p-1) = 1 (mod 3)
              but 2 = -1 (mod 3). So 2^(p-1) = (-1)^(p-1) = 1 (mod 3) (since p being odd, p-1 must be even)
              And we are done.

              --- On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

              From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
              Subject: [math_club] Modular arithmatic prob
              To: math_club@yahoogrou ps.com
              Date: Monday, October 19, 2009, 9:15 AM

               

              Can anyone prove this assertion:  2p ≡ p2 + 1(mod 6), where p is any prime greater than 3.




              From: avik roy <avik_3.1416@ yahoo.co. in>
              To: math_club@yahoogrou ps.com
              Sent: Fri, October 16, 2009 12:04:49 AM
              Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]

               

              I am not using that fact. The fact that |1+ik| = |1-ik| leads ta calculation as follows...

              |1+ik| / |1-ik| =1

              =>|1+ik|^2 / |(1+ik)(1-ik) | =1
              =>  |1+ik|^2 = 1 + k^2

              --- On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:

              From: Tanvir Prince <tanvirp123@yahoo. com>
              Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
              To: math_club@yahoogrou ps.com
              Date: Thursday, 15 October, 2009, 10:42 PM

               
              Good proof. I have only one  concern:
              You are using the fact that zz* = |z|^2 where z* is the conjugate of z and the fact that zz*=|z|^2 depends on pythagorean theorem

              --- On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:

              From: avik roy <avik_3.1416@ yahoo.co. in>
              Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
              To: "mathclub clubmath" <math_club@yahoogrou ps.com>
              Date: Thursday, October 15, 2009, 12:36 AM

               
              Here, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.

              let,
              z1 = x
              z2 = ikx [k is any arbitrary real constant]

              |z1|^2 + |z2|^2 = (1+k^2)*x^2
              [u need not use the Pythagoras here, the concept of length does the job]

              |z1+z2| = x|1+ ik|

              A simple observation leads that |z1+z2| = |z1-z2|
              =>|1+ik|=|1-ik|
              A simple calculation proves that, |i+ik|^2 = 1+k^2

              This gives us that |z1|^2 +|z2|^2 = |z1 + z2|^2

              I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.



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            • avik roy
              The same proof is valid in this case.... for any number d to divide a and b, it has to divide p, hence d=1 or p, but a,b
              Message 6 of 16 , Nov 2, 2009
              • 0 Attachment
                The same proof is valid in this case....

                for any number d to divide a and b, it has to divide p, hence d=1 or p, but a,b<p thus d can't be p and thus d has to be 1.

                Q.E.D. 

                --- On Mon, 2/11/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...> wrote:

                From: Mirza Sabbir Hossain Beg <mirzasabbirhossainbeg@...>
                Subject: [math_club] Another Coprime Arithmetic
                To: math_club@yahoogroups.com
                Date: Monday, 2 November, 2009, 2:18 PM

                 

                Tanvir I also observed another fact that: if p is any odd prime and a + b = p then a and b must be co-prime. Mail me the proof.


                From: Tanvir Prince <tanvirp123@yahoo. com>
                To: math_club@yahoogrou ps.com
                Sent: Mon, November 2, 2009 1:41:47 AM
                Subject: Re: [math_club] Coprime Arithmetic

                 

                The assumption is not right. For example, 16 - 8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:
                Let d divide both a and b. We will show that d = 1 and this will prove our assumption.
                Since d divide both a and b and a-b=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.
                But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.
                So we must have k =1 and thus d = 2^0 = 1 and we are done.


                --- On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

                From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
                Subject: [math_club] Coprime Arithmetic
                To: math_club@yahoogrou ps.com
                Date: Saturday, October 31, 2009, 6:36 AM

                 
                  Solve this prob: if a – b = 2n where n is any integer then a and b is co-prime.


                From: Tanvir Prince <tanvirp123@yahoo. com>
                To: math_club@yahoogrou ps.com
                Cc: Prince <tanvirp123@yahoo. com>
                Sent: Tue, October 20, 2009 7:52:40 AM
                Subject: Re: [math_club] Modular arithmatic prob

                 
                we will use the following theorem which is a generalization of Fermat's theorem:
                 
                If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)
                here phi(n) is the euler phi function which count the number of integer from 1 to n-1 which are relatively prime to n.
                 
                So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.
                 
                Now since p>3 , p and 6 are relatively prime.
                So by the above theorem:
                 
                p^(phi(6))=p^ 2 = 1 (mod 6)
                add 1 to both side:
                p^2+1 = 2 (mod 6)
                 
                So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)
                This means we need to show 2^p-2 is divisible by 6 or 2(2^(p-1) -1) is divisible by 6 or
                2^(p-1) - 1 is divisible by 3 or 2^(p-1) = 1 (mod 3)
                but 2 = -1 (mod 3). So 2^(p-1) = (-1)^(p-1) = 1 (mod 3) (since p being odd, p-1 must be even)
                And we are done.

                --- On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

                From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
                Subject: [math_club] Modular arithmatic prob
                To: math_club@yahoogrou ps.com
                Date: Monday, October 19, 2009, 9:15 AM

                 

                Can anyone prove this assertion:  2p ≡ p2 + 1(mod 6), where p is any prime greater than 3.




                From: avik roy <avik_3.1416@ yahoo.co. in>
                To: math_club@yahoogrou ps..com
                Sent: Fri, October 16, 2009 12:04:49 AM
                Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]

                 

                I am not using that fact.. The fact that |1+ik| = |1-ik| leads ta calculation as follows...

                |1+ik| / |1-ik| =1

                =>|1+ik|^2 / |(1+ik)(1-ik) | =1
                =>  |1+ik|^2 = 1 + k^2

                --- On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:

                From: Tanvir Prince <tanvirp123@yahoo. com>
                Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
                To: math_club@yahoogrou ps.com
                Date: Thursday, 15 October, 2009, 10:42 PM

                 
                Good proof. I have only one  concern:
                You are using the fact that zz* = |z|^2 where z* is the conjugate of z and the fact that zz*=|z|^2 depends on pythagorean theorem

                --- On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:

                From: avik roy <avik_3.1416@ yahoo.co. in>
                Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
                To: "mathclub clubmath" <math_club@yahoogrou ps.com>
                Date: Thursday, October 15, 2009, 12:36 AM

                 
                Here, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.

                let,
                z1 = x
                z2 = ikx [k is any arbitrary real constant]

                |z1|^2 + |z2|^2 = (1+k^2)*x^2
                [u need not use the Pythagoras here, the concept of length does the job]

                |z1+z2| = x|1+ ik|

                A simple observation leads that |z1+z2| = |z1-z2|
                =>|1+ik|=|1-ik|
                A simple calculation proves that, |i+ik|^2 = 1+k^2

                This gives us that |z1|^2 +|z2|^2 = |z1 + z2|^2

                I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.



                From cricket scores to your friends. Try the Yahoo! India Homepage!



                Add whatever you love to the Yahoo! India homepage. Try now!







                Try the new Yahoo! India Homepage. Click here.
              • Mirza Sabbir Hossain Beg
                Thanks Avik. Now I understood the proof. ________________________________ From: avik roy To: math_club@yahoogroups.com Sent: Mon,
                Message 7 of 16 , Nov 2, 2009
                • 0 Attachment
                  Thanks Avik. Now I understood the proof.


                  From: avik roy <avik_3.1416@...>
                  To: math_club@yahoogroups.com
                  Sent: Mon, November 2, 2009 5:06:45 PM
                  Subject: Re: [math_club] Another Coprime Arithmetic

                   

                  The same proof is valid in this case....

                  for any number d to divide a and b, it has to divide p, hence d=1 or p, but a,b<p thus d can't be p and thus d has to be 1.

                  Q.E.D. 

                  --- On Mon, 2/11/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

                  From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
                  Subject: [math_club] Another Coprime Arithmetic
                  To: math_club@yahoogrou ps.com
                  Date: Monday, 2 November, 2009, 2:18 PM

                   

                  Tanvir I also observed another fact that: if p is any odd prime and a + b = p then a and b must be co-prime. Mail me the proof.


                  From: Tanvir Prince <tanvirp123@yahoo. com>
                  To: math_club@yahoogrou ps.com
                  Sent: Mon, November 2, 2009 1:41:47 AM
                  Subject: Re: [math_club] Coprime Arithmetic

                   

                  The assumption is not right. For example, 16 - 8 = 8 = 2^3 but 16 and 8 is not coprime. So we need to have more on the assumption. If you furthur assume that a and b both are odd then the assumption is true. The proof follows:
                  Let d divide both a and b. We will show that d = 1 and this will prove our assumption.
                  Since d divide both a and b and a-b=2^n, so d also divide 2^n. But then d must be of the form d=2^k for some k<= n.
                  But if k>1 then d=2^k can not divide a or b since by assumption a and b are odd.
                  So we must have k =1 and thus d = 2^0 = 1 and we are done.


                  --- On Sat, 10/31/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

                  From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
                  Subject: [math_club] Coprime Arithmetic
                  To: math_club@yahoogrou ps.com
                  Date: Saturday, October 31, 2009, 6:36 AM

                   
                    Solve this prob: if a – b = 2n where n is any integer then a and b is co-prime.


                  From: Tanvir Prince <tanvirp123@yahoo. com>
                  To: math_club@yahoogrou ps.com
                  Cc: Prince <tanvirp123@yahoo. com>
                  Sent: Tue, October 20, 2009 7:52:40 AM
                  Subject: Re: [math_club] Modular arithmatic prob

                   
                  we will use the following theorem which is a generalization of Fermat's theorem:
                   
                  If a and n are positive integer and a and n are relatively prime then a^(phi(n))=1 (mod n)
                  here phi(n) is the euler phi function which count the number of integer from 1 to n-1 which are relatively prime to n.
                   
                  So for example, if n=6 then the number of integer from 1,2,3,4,5 which are relatively prime to 6 are 1 and 5. so phi(6)=2.
                   
                  Now since p>3 , p and 6 are relatively prime.
                  So by the above theorem:
                   
                  p^(phi(6))=p^ 2 = 1 (mod 6)
                  add 1 to both side:
                  p^2+1 = 2 (mod 6)
                   
                  So to show 2^p = p^2 +1 (mod 6) is equivalent to show 2^p = 2 (mod 6)
                  This means we need to show 2^p-2 is divisible by 6 or 2(2^(p-1) -1) is divisible by 6 or
                  2^(p-1) - 1 is divisible by 3 or 2^(p-1) = 1 (mod 3)
                  but 2 = -1 (mod 3). So 2^(p-1) = (-1)^(p-1) = 1 (mod 3) (since p being odd, p-1 must be even)
                  And we are done.

                  --- On Mon, 10/19/09, Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com> wrote:

                  From: Mirza Sabbir Hossain Beg <mirzasabbirhossainb eg@yahoo. com>
                  Subject: [math_club] Modular arithmatic prob
                  To: math_club@yahoogrou ps.com
                  Date: Monday, October 19, 2009, 9:15 AM

                   

                  Can anyone prove this assertion:  2p ≡ p2 + 1(mod 6), where p is any prime greater than 3.




                  From: avik roy <avik_3.1416@ yahoo.co. in>
                  To: math_club@yahoogrou ps..com
                  Sent: Fri, October 16, 2009 12:04:49 AM
                  Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]

                   

                  I am not using that fact.. The fact that |1+ik| = |1-ik| leads ta calculation as follows...

                  |1+ik| / |1-ik| =1

                  =>|1+ik|^2 / |(1+ik)(1-ik) | =1
                  =>  |1+ik|^2 = 1 + k^2

                  --- On Thu, 15/10/09, Tanvir Prince <tanvirp123@yahoo. com> wrote:

                  From: Tanvir Prince <tanvirp123@yahoo. com>
                  Subject: Re: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
                  To: math_club@yahoogrou ps.com
                  Date: Thursday, 15 October, 2009, 10:42 PM

                   
                  Good proof. I have only one  concern:
                  You are using the fact that zz* = |z|^2 where z* is the conjugate of z and the fact that zz*=|z|^2 depends on pythagorean theorem

                  --- On Thu, 10/15/09, avik roy <avik_3.1416@ yahoo.co. in> wrote:

                  From: avik roy <avik_3.1416@ yahoo.co. in>
                  Subject: [math_club] pythagorean theorem using complex numbers [Moon & Saumitra Vi, check this pls]
                  To: "mathclub clubmath" <math_club@yahoogrou ps.com>
                  Date: Thursday, October 15, 2009, 12:36 AM

                   
                  Here, I've tried to avoid the formula of modulus of a complex numbers, that comes from Pythagorean theorem directly.

                  let,
                  z1 = x
                  z2 = ikx [k is any arbitrary real constant]

                  |z1|^2 + |z2|^2 = (1+k^2)*x^2
                  [u need not use the Pythagoras here, the concept of length does the job]

                  |z1+z2| = x|1+ ik|

                  A simple observation leads that |z1+z2| = |z1-z2|
                  =>|1+ik|=|1-ik|
                  A simple calculation proves that, |i+ik|^2 = 1+k^2

                  This gives us that |z1|^2 +|z2|^2 = |z1 + z2|^2

                  I am confused whether this proof contains any statement that relies on Pythagorean Theorem itself.



                  From cricket scores to your friends. Try the Yahoo! India Homepage!



                  Add whatever you love to the Yahoo! India homepage. Try now!







                  Try the new Yahoo! India Homepage. Click here.


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