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Freind of mine showed me this 3= 4?

  1. Mar 27, 2005 #1
    what am i missing here? this seems to make sense...

    a + b = c



    4a - 3a + 4b - 3b = 4c - 3c



    4a + 4b - 4c = 3a + 3b - 3c



    4(a+b-c) = 3(a+b-c)



    4 = 3
     
  2. jcsd
  3. Mar 27, 2005 #2
    You divided by zero.

    You can do it even more simply:

    [tex]a=b[/tex]
    [tex]4a = 3a+b[/tex]
    [tex]4a - 4b = 3a - 3b[/tex]
    [tex]4(a-b) = 3(a-b)[/tex]
    [tex]4=3[/tex]

    it's equivalent to saying

    [tex] 3 \cdot 0 = 4 \cdot 0 \Longrightarrow 3 = 4[/tex]

    which is silly
     
  4. Mar 27, 2005 #3
    4(a+b-c) = 3(a+b-c)

    This statement can be true only if (a+b-c) = 0

    If you try to divide both sides by ( a+b-c) then it becomes 0/0 i.e. indeterminate form of the function. hence 4 is never equal to 3, at least not for a mathematician.
     
  5. Mar 27, 2005 #4
    It's even worse. We KNOW [itex]a+b-c=0[/itex]. Note his assumption, [itex]a+b = c[/itex]. Just like in my example, where I assume [itex]a=b \Longrightarrow a-b = 0[/itex].
     
  6. Mar 27, 2005 #5

    Alkatran

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    4(a+b-c) = 3(a+b-c)
    4*0 = 3*0
    ERROR -> 4*0/0 = 3*0/0
    4 = 3
     
  7. Mar 28, 2005 #6
    Thanks for giving me a more "in depth" look at what is most probably basic mathematics to most of you. I'm still learning. :)
     
  8. Mar 28, 2005 #7
    We all start somewhere.
     
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