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1200-year-old problem

Topic closed. 16 replies. Last post 10 years ago by JADELottery.

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for99c's avatar - Lottery-058.jpg
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Posted: December 8, 2006, 9:20 pm - IP Logged

Dr James Anderson, from the University of Reading's computer science department, says his new theorem solves an extremely important problem - the problem of nothing.


4/4 = 1 3/3 = 1 2/2 = 1 1/1 = 1 0/0 = 1

    MADDOG10's avatar - smoke
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    Posted: December 8, 2006, 9:40 pm - IP Logged

    That was actually solved 1199 years ago, by Dr.goodfornuttin...!

                                                 

                                                   "  When Injustice Becomes Law, Resistance Becomes Duty "

      for99c's avatar - Lottery-058.jpg
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      Posted: December 8, 2006, 10:22 pm - IP Logged
        JADELottery's avatar - MeAtWork 03.PNG
        The Quantum Master
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        Posted: December 10, 2006, 10:17 pm - IP Logged

        I see two incorrect statements:

        1. 0-1 = (1 / 0) = ¥ , is not correct

        2. ¥-1 = (1 / ¥) = 0 , is not correct

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          JADELottery's avatar - MeAtWork 03.PNG
          The Quantum Master
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          Posted: December 10, 2006, 11:03 pm - IP Logged

          I see two incorrect statements:

          1. 0-1 = (1 / 0) = ¥ , is not correct

          2. ¥-1 = (1 / ¥) = 0 , is not correct

          0 -1 = (1 / 0) = ¥ and ¥ -1 = (1 / ¥) = 0 are incorrect and 0 -1 is a new number I, the Hyper-infinite number.

          1 - Reciprocal Operation with Inequality

                a > b

                taking the reciprocal of both side causes the > to change to <.

                reciprocal of  a > b  is  a-1 < b-1

          a > ba-1 < b-1
          aba-1b-1
          10.251.000004.00000
          10.51.000002.00000
          210.500001.00000
          21.50.500000.66667
          1050.100000.20000
          1090.100000.11111
          10005000.001000.00200
          10009000.001000.00111

          2 - Apply Reciprocal Operation of Inequality to the Inequality of  ¥ -1 > 0

                ¥ -1 > 0 

                the reciprocal of  ¥ -1 > 0  is  ¥ < 0 -1

                if the inequality  ¥ < 0 -1  is true, then 0 -1 is a new number, I, and exists outside the set of Real numbers or Hyper-infinite.

                the number I is analogous to the number i = Ö-1

          3 - Some other additional inequalities and equalities

                0 ¹ ¥ -1

                0 -1 ¹ ¥

                I = 0 -1  Hyper-infinite number

                0 ´ I = 1

                1 ´ I = I

                n ´ I = nI

                0 ´ nI = n

                a is Real and bI is Hyper-infinite, then the sum of a and bI is complex and is  a + bI

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            JADELottery's avatar - MeAtWork 03.PNG
            The Quantum Master
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            Posted: December 11, 2006, 5:15 am - IP Logged

            Something of a proof why ¥ -1 ¹ 0 and ¥ -1 > 0

            1 - If ¥ -1 = 0 , then 0 + ¥ -1 = 0

                  0 + ¥ -1 = 0

                  0 + ( 1 / ¥ ) = 0

                  0 ( ¥ / ¥ ) + ( 1 / ¥ ) = 0

                  ( ( 0 * ¥ ) / ¥ ) + ( 1 / ¥ ) = 0

                  ( 0 * ¥ + 1 ) / ¥ = 0

                  0 * ¥ + 1 = 0 * ¥

                  ( 0 * ¥ ) - ( 0 * ¥ ) + 1 = ( 0 * ¥ ) - ( 0 * ¥ )

                  0 + 1 = 0

                  1 = 0

                  1 = 0 , is false \  ¥ -1 ¹ 0

            2 - If ¥ -1 > 0 , then 0 + ¥ -1 > 0

                  0 + ¥ -1 > 0

                  0 + ( 1 / ¥ ) > 0

                  0 ( ¥ / ¥ ) + ( 1 / ¥ ) > 0

                  ( ( 0 * ¥ ) / ¥ ) + ( 1 / ¥ ) > 0

                  ( 0 * ¥ + 1 ) / ¥ > 0

                  0 * ¥ + 1 > 0 * ¥

                  ( 0 * ¥ ) - ( 0 * ¥ ) + 1 > ( 0 * ¥ ) - ( 0 * ¥ )

                  0 + 1 > 0

                  1 > 0

                  1 > 0 , is true \  ¥ -1 > 0

            When dealing with 0 and ¥ it is better to treat them as if they were variables first and do normal math operations first, then work the final equation or inequality. There are times when 02 ¹ 0 and ¥2 ¹ ¥ , much like X2 does not exactly equal X. 

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              JADELottery's avatar - MeAtWork 03.PNG
              The Quantum Master
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              Posted: December 11, 2006, 5:39 am - IP Logged

              Also,  0 * ¥   is a non-reduceable number much like 2i is a non-reduceable number, it is what it is, a number all by itself.

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                The Quantum Master
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                Posted: December 11, 2006, 5:50 am - IP Logged

                In addition, ¥ -1  is the infinitesimal unit and is exactly and only exactly the first number after 0.

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                  JADELottery's avatar - MeAtWork 03.PNG
                  The Quantum Master
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                  Posted: December 11, 2006, 10:04 am - IP Logged

                  Some more:

                    n ¥ = ¥

                    ( n ¥ ) / ¥ = ¥ / ¥

                    n ( ¥ / ¥ ) = ¥ / ¥

                    n ( 1 ) = 1

                    n = 1, \  n ¥ = ¥  true when  n = 1

                   

                    n ¥ ¹ ¥

                    ( n ¥ ) / ¥ ¹ ¥ / ¥

                    n ( ¥ / ¥ ) ¹ ¥ / ¥

                    n ( 1 ) ¹ 1

                    n ¹ 1, \  n ¥ ¹ ¥  true when  n ¹ 1

                   

                    n ¥ = -¥

                    ( n ¥ ) / ¥ = -¥ / ¥

                    n ( ¥ / ¥ ) = -¥ / ¥

                    n ( 1 ) = -1

                    n = -1, \  n ¥ = ¥  true when  n = -1

                   

                    n ¥ ¹ -¥

                    ( n ¥ ) / ¥ ¹ -¥ / ¥

                    n ( ¥ / ¥ ) ¹ -¥ / ¥

                    n ( 1 ) ¹ -1

                    n ¹ -1, \  n ¥ ¹ -¥  true when  n ¹ -1

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                    The Quantum Master
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                    Posted: December 11, 2006, 10:24 am - IP Logged

                    And just a bit more:

                        0 * ¥  = n

                      ( 0 * ¥ ) / ¥ = n / ¥

                      0 ( ¥ / ¥ ) = n / ¥

                      0 ( 1 ) = n / ¥

                      0 = n / ¥ , if and only if n = 0 * ¥

                     

                      0 * ¥ = n

                      ( 0 * ¥ ) / 0 = n / 0

                      ¥ ( 0 / 0 ) = n / 0

                      ¥ ( 1 ) = n / 0

                      ¥ = n / 0

                      ¥ = n / 0 , if and only if n = 0 * ¥

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                      The Quantum Master
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                      Posted: December 11, 2006, 10:39 am - IP Logged

                      Some more:

                        n ¥ = ¥

                        ( n ¥ ) / ¥ = ¥ / ¥

                        n ( ¥ / ¥ ) = ¥ / ¥

                        n ( 1 ) = 1

                        n = 1, \  n ¥ = ¥  true when  n = 1

                       

                        n ¥ ¹ ¥

                        ( n ¥ ) / ¥ ¹ ¥ / ¥

                        n ( ¥ / ¥ ) ¹ ¥ / ¥

                        n ( 1 ) ¹ 1

                        n ¹ 1, \  n ¥ ¹ ¥  true when  n ¹ 1

                       

                        n ¥ = -¥

                        ( n ¥ ) / ¥ = -¥ / ¥

                        n ( ¥ / ¥ ) = -¥ / ¥

                        n ( 1 ) = -1

                        n = -1, \  n ¥ = ¥  true when  n = -1

                       

                        n ¥ ¹ -¥

                        ( n ¥ ) / ¥ ¹ -¥ / ¥

                        n ( ¥ / ¥ ) ¹ -¥ / ¥

                        n ( 1 ) ¹ -1

                        n ¹ -1, \  n ¥ ¹ -¥  true when  n ¹ -1

                      Some more:

                        n ¥ = ¥

                        ( n ¥ ) / ¥ = ¥ / ¥

                        n ( ¥ / ¥ ) = ¥ / ¥

                        n ( 1 ) = 1

                        n = 1, \  n ¥ = ¥  true when  n = 1

                       

                        n ¥ ¹ ¥

                        ( n ¥ ) / ¥ ¹ ¥ / ¥

                        n ( ¥ / ¥ ) ¹ ¥ / ¥

                        n ( 1 ) ¹ 1

                        n ¹ 1, \  n ¥ ¹ ¥  true when  n ¹ 1

                       

                        n ¥ = -¥

                        ( n ¥ ) / ¥ = -¥ / ¥

                        n ( ¥ / ¥ ) = -¥ / ¥

                        n ( 1 ) = -1

                        n = -1, \  n ¥ = -¥  true when  n = -1  <--- correction from previous post

                       

                        n ¥ ¹ -¥

                        ( n ¥ ) / ¥ ¹ -¥ / ¥

                        n ( ¥ / ¥ ) ¹ -¥ / ¥

                        n ( 1 ) ¹ -1

                        n ¹ -1, \  n ¥ ¹ -¥  true when  n ¹ -1

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                        The Quantum Master
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                        Posted: December 11, 2006, 5:10 pm - IP Logged

                        The mathematical operation of X / X = 1 must be continuous for all Real values of X; including {-¥, 0, ¥}.

                        1 - If X / X = 1, then -¥ / -¥ = 1 

                            -¥ / -¥ = 1

                            -¥ = 1 * -¥

                            -¥ = -¥, true \ -¥ / -¥ = 1

                        2 - If X / X ¹ 1, then -¥ / -¥ ¹ 1

                            -¥ / -¥ ¹ 1

                            -¥ ¹ 1 * -¥

                            -¥ ¹ -¥, false \ -¥ / -¥ = 1

                        3 - If X / X = 1, then 0 / 0 = 1 

                            0 / 0 = 1

                            0 = 1 * 0

                            0 = 0, true \ 0 / 0 = 1

                        4 - If X / X ¹ 1, then 0 / 0 ¹ 1

                            0 / 0 ¹ 1

                            0 ¹ 1 * 0

                            0 ¹ 0, false \ 0 / 0 = 1

                        5 - If X / X = 1, then ¥ / ¥ = 1 

                            ¥ / ¥ = 1

                            ¥ = 1 * ¥

                            ¥ = ¥, true \ ¥ / ¥ = 1

                        6 - If X / X ¹ 1, then ¥ / ¥ ¹ 1

                            ¥ / ¥ ¹ 1

                            ¥ ¹ 1 * ¥

                            ¥ ¹ ¥, false \ ¥ / ¥ = 1

                        7 - If Y = 1 and is continuous for all Real values in the function Y = f(X) and f(X) = X / X, then Y = 1 = X / X and is continuous for all Real values

                          Y = 1 = X / X

                          Y * X = 1 * X = X

                          Y * X = 1 * X = X

                          1 * X = 1 * X = X

                          X = X = X, true if and only if Y = 1

                          -¥ = -¥ = -¥, true if and only if Y = 1

                          0 = 0 = 0, true if and only if Y = 1

                          ¥ = ¥ = ¥, true if and only if Y = 1

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                          Posted: December 16, 2006, 8:45 am - IP Logged

                          doug I agree

                            JAP69's avatar - alas
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                            Posted: December 16, 2006, 8:56 am - IP Logged

                            WHEW  Thud

                            I know how to add, subtract and multiply.
                            I think.
                            ROFL

                            MAGA

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                              Posted: December 16, 2006, 9:22 am - IP Logged

                              Mathematics Forum:
                              The place to discuss the strong ties between lotteries and mathematics.

                              Excuse me, but I think I'm missing the strong tie between lotteries and mathematics in this thread.