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  3. one equal to two ?

one equal to two ?

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  • T tayoufabrice

    :laugh: As I know mathematics doesn't approve 0=0 so, the error is a²-a²=a²-a² (meaning 0=0) exactly it is a teacher who showed it to us.

    Kornfeld Eliyahu PeterK Offline
    Kornfeld Eliyahu PeterK Offline
    Kornfeld Eliyahu Peter
    wrote on last edited by
    #41

    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.

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

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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.

      P Offline
      P Offline
      PIEBALDconsult
      wrote on last edited by
      #42

      Yes, but the damage is done before that.

      Kornfeld Eliyahu PeterK 1 Reply Last reply
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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
            P Offline
            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 =

              P Offline
              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.

                1 Reply Last reply
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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:

                      1 Reply Last reply
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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.

                                M Offline
                                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

                                  T 1 Reply Last reply
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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

                                      1 Reply Last reply
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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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