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  3. one equal to two ?

one equal to two ?

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  • T tayoufabrice

    Let read this : 1=1 a=a a²=a² a²-a²=a²-a² a(a-a)=(a+a)(a-a) a=a+a a(1)=a(1+1) 1=1+1 1=2 Where is the error ?

    L Offline
    L Offline
    Lost User
    wrote on last edited by
    #56

    welcome to the lounge. I know this isn't quite your first post - but nearly. And I would like to apologise for the negativity your post received. For someone who hadn't seen that 'proof' before it may have been interesting - as you can see, not only have the majority here seen it (more than once!) but they like to stuff it down your throat - whether to big-note themselves or simply in an attempt to belittle you we cannot tell. They should be ashamed. Merry Xmas

    PooperPig - Coming Soon

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    • T tayoufabrice

      Let read this : 1=1 a=a a²=a² a²-a²=a²-a² a(a-a)=(a+a)(a-a) a=a+a a(1)=a(1+1) 1=1+1 1=2 Where is the error ?

      M Offline
      M Offline
      Mitchell J
      wrote on last edited by
      #57

      How about this... ;P Start with this: 1/9 = 1/9 Then convert one side to decimal equivalent (which is infinitely recurring) 1/9 = 0.11111111111111111111111111111111111111111111111111111111...(etc etc) Then multiply both sides by nine 1 = 0.9999999999999999999999999999999999999999999999999999999999...(etc etc) Therefore, 1 is equal to 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...(where the 9's are in infinite recursion). And yes, this actually is mathematically correct.

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      • L Lost User

        welcome to the lounge. I know this isn't quite your first post - but nearly. And I would like to apologise for the negativity your post received. For someone who hadn't seen that 'proof' before it may have been interesting - as you can see, not only have the majority here seen it (more than once!) but they like to stuff it down your throat - whether to big-note themselves or simply in an attempt to belittle you we cannot tell. They should be ashamed. Merry Xmas

        PooperPig - Coming Soon

        T Offline
        T Offline
        tayoufabrice
        wrote on last edited by
        #58

        Thank you Max and happy new Xear

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        • M Mitchell J

          How about this... ;P Start with this: 1/9 = 1/9 Then convert one side to decimal equivalent (which is infinitely recurring) 1/9 = 0.11111111111111111111111111111111111111111111111111111111...(etc etc) Then multiply both sides by nine 1 = 0.9999999999999999999999999999999999999999999999999999999999...(etc etc) Therefore, 1 is equal to 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...(where the 9's are in infinite recursion). And yes, this actually is mathematically correct.

          T Offline
          T Offline
          tayoufabrice
          wrote on last edited by
          #59

          I agree but here 1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9 I could write 1/9~= 0.111111111111111111111111...... then 1 ~= 0.9999999999999999999999999...... ??

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          • T tayoufabrice

            I agree but here 1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9 I could write 1/9~= 0.111111111111111111111111...... then 1 ~= 0.9999999999999999999999999...... ??

            M Offline
            M Offline
            Mitchell J
            wrote on last edited by
            #60

            tayoufabrice wrote:

            1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9

            Wish I could agree, but I can't... read all about it[^] :-D Even google 0.999999999999999 = 1[^] if you're still unconvinced.

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            • M Mitchell J

              tayoufabrice wrote:

              1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9

              Wish I could agree, but I can't... read all about it[^] :-D Even google 0.999999999999999 = 1[^] if you're still unconvinced.

              T Offline
              T Offline
              tayoufabrice
              wrote on last edited by
              #61

              Ah là là :laugh: Mathematics !! (French laughing)

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              • T tayoufabrice

                Let read this : 1=1 a=a a²=a² a²-a²=a²-a² a(a-a)=(a+a)(a-a) a=a+a a(1)=a(1+1) 1=1+1 1=2 Where is the error ?

                D Offline
                D Offline
                Daniel Pfeffer
                wrote on last edited by
                #62

                a(a-a) = (a+a)(a-a) // divide by (a-a), i.e. divide by 0 a = a+a Division by zero is a no-no because it can lead to "impossible" results like the above.

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                • D Daniel Pfeffer

                  a(a-a) = (a+a)(a-a) // divide by (a-a), i.e. divide by 0 a = a+a Division by zero is a no-no because it can lead to "impossible" results like the above.

                  T Offline
                  T Offline
                  tayoufabrice
                  wrote on last edited by
                  #63

                  I could fix the post as : Given a C ]--;0[ U ]0;++[ (meaning 0 excluded) Now ??

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                  • T tayoufabrice

                    I could fix the post as : Given a C ]--;0[ U ]0;++[ (meaning 0 excluded) Now ??

                    D Offline
                    D Offline
                    Daniel Pfeffer
                    wrote on last edited by
                    #64

                    The value of a is irrelevant; a - a == 0, and factoring out a - a is division by 0, which is forbidden. I am not a mathematician, so I don't know if it is possible to create a self-consistent arithmetic in which division by 0 does not result in nonsensical results. All I know is that in the arithmetic I learnt in school it is forbidden.

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                    • D Daniel Pfeffer

                      The value of a is irrelevant; a - a == 0, and factoring out a - a is division by 0, which is forbidden. I am not a mathematician, so I don't know if it is possible to create a self-consistent arithmetic in which division by 0 does not result in nonsensical results. All I know is that in the arithmetic I learnt in school it is forbidden.

                      T Offline
                      T Offline
                      tayoufabrice
                      wrote on last edited by
                      #65

                      Sure ! :laugh: number can never be divided by zero 0 ; even 0/0 :confused: It is the real error of my process

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