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  3. one equal to two ?

one equal to two ?

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  • Kornfeld Eliyahu PeterK Kornfeld Eliyahu Peter

    It has nothing to do with 0 equal to 0 or not (which is a mathematical game only), but the fact that division by zero is not permitted (or more precisely the outcome is undefined)...

    Skipper: We'll fix it. Alex: Fix it? How you gonna fix this? Skipper: Grit, spit and a whole lotta duct tape.

    T Offline
    T Offline
    tayoufabrice
    wrote on last edited by
    #43

    1=1 a=a <=> a/a=1 a²=a² ?? a=b <=> a/b=1 a²=b² ?? then a=b a-c = b-c ? a-a = a-a (assuming a=b and a=c) ?? Where is the division by zero ? I've added c at both sides of =

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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 ?

      R Offline
      R Offline
      R Giskard Reventlov
      wrote on last edited by
      #44

      tayoufabrice wrote:

      Where is the error ?

      In between your ears: obviously the answer is 42.

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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 ?

        P Offline
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        phil o
        wrote on last edited by
        #45

        a - a = 0

        You cannot get anything useful from a multiplication once it has involved a zero term.

        5 x 0 = 12012 x 0

        does not mean that

        5 = 12012

        There are two kinds of people in the world: those who can extrapolate from incomplete data.

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        • T tayoufabrice

          1=1 a=a <=> a/a=1 a²=a² ?? a=b <=> a/b=1 a²=b² ?? then a=b a-c = b-c ? a-a = a-a (assuming a=b and a=c) ?? Where is the division by zero ? I've added c at both sides of =

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          P Offline
          PIEBALDconsult
          wrote on last edited by
          #46

          At the final step, you effectively have a=0, which means that a/a=1 on the second line is division by zero. But why were b and c introduced? It's just nonsense. Additionally, a²=b² certainly does not mean a=b.

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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 ?

            P Offline
            P Offline
            pt1401
            wrote on last edited by
            #47

            The error is simultaneously in line 5, your misunderstanding of basic maths, and your posting of this question.

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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 ?

              Mike HankeyM Offline
              Mike HankeyM Offline
              Mike Hankey
              wrote on last edited by
              #48

              The error is I haven't had coffee yet.

              New version: WinHeist Version 2.1.0 There's a fine line between crazy and free spirited and it's usually a prescription. I'm currently unsupervised, I know it freaks me out too but the possibilities are endless.

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              • P PIEBALDconsult

                At the final step, you effectively have a=0, which means that a/a=1 on the second line is division by zero. But why were b and c introduced? It's just nonsense. Additionally, a²=b² certainly does not mean a=b.

                T Offline
                T Offline
                tayoufabrice
                wrote on last edited by
                #49

                a=0 never means a/a=0 (a can never be 0). a=a <=> a/a=a/a <=> 1=1 I could fix the post as : Given a C ]--;0[ U ]0;++[ (meaning 0 excluded)

                Kornfeld Eliyahu PeterK 1 Reply Last reply
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                • P phil o

                  a - a = 0

                  You cannot get anything useful from a multiplication once it has involved a zero term.

                  5 x 0 = 12012 x 0

                  does not mean that

                  5 = 12012

                  There are two kinds of people in the world: those who can extrapolate from incomplete data.

                  T Offline
                  T Offline
                  tayoufabrice
                  wrote on last edited by
                  #50

                  very very TRUE:thumbsup::thumbsup:

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                  • R R Giskard Reventlov

                    tayoufabrice wrote:

                    Where is the error ?

                    In between your ears: obviously the answer is 42.

                    T Offline
                    T Offline
                    tayoufabrice
                    wrote on last edited by
                    #51

                    the answer is 42 :laugh: :laugh:

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                    • T tayoufabrice

                      a=0 never means a/a=0 (a can never be 0). a=a <=> a/a=a/a <=> 1=1 I could fix the post as : Given a C ]--;0[ U ]0;++[ (meaning 0 excluded)

                      Kornfeld Eliyahu PeterK Offline
                      Kornfeld Eliyahu PeterK Offline
                      Kornfeld Eliyahu Peter
                      wrote on last edited by
                      #52

                      Which means that you have an equation system and not a single equation...It's a different thing to solve...

                      Skipper: We'll fix it. Alex: Fix it? How you gonna fix this? Skipper: Grit, spit and a whole lotta duct tape.

                      "It never ceases to amaze me that a spacecraft launched in 1977 can be fixed remotely from Earth." ― Brian Cox

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                      • P PIEBALDconsult

                        Yes, but the damage is done before that.

                        Kornfeld Eliyahu PeterK Offline
                        Kornfeld Eliyahu PeterK Offline
                        Kornfeld Eliyahu Peter
                        wrote on last edited by
                        #53

                        You mean, when OP went to learn math?

                        Skipper: We'll fix it. Alex: Fix it? How you gonna fix this? Skipper: Grit, spit and a whole lotta duct tape.

                        "It never ceases to amaze me that a spacecraft launched in 1977 can be fixed remotely from Earth." ― Brian Cox

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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 ?

                          R Offline
                          R Offline
                          RedDk
                          wrote on last edited by
                          #54

                          This is best asked here: http://www.codeproject.com/script/Answers/List.aspx?tags=900&alltags=true[^]

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                          • P PIEBALDconsult

                            Abbott and Costello said it better.

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                            M Offline
                            Mark_Wallace
                            wrote on last edited by
                            #55

                            PIEBALDconsult wrote:

                            Abbott and Costello_, who were on first,_ said it better.

                            It was missing something.

                            I wanna be a eunuchs developer! Pass me a bread knife!

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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
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                                        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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