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Spivak Calculus (1994)

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data

    2. What is wrong with the following "proof"? Let x = y. Then
    x2 =xy,
    x2 - y2 =xy - y ,
    (x + y)(x -y) = y(x - y),
    x + y = y,
    2y = y,
    2= 1.


    2. Relevant equations

    It's obvious that x + y =/= y, but I do not know how to "prove" this, i.e. which proof from algebra is applicable here. My hunch is that it is the Distributive Property?

    3. The attempt at a solution

    I honestly am not sure where to start with this. All I can see is that dividing (x-y) out from step 3 to step 4 is what causes the inequality.
     
    Last edited: Apr 4, 2012
  2. jcsd
  3. Apr 4, 2012 #2

    tiny-tim

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    welcome to pf!

    hi jrdunaway welcome to pf! :smile:
    that's right! :smile:

    if x = y, that means that you divided by 0, which isn't allowed

    so (if x = y) that step is not valid
     
  4. Apr 4, 2012 #3

    HallsofIvy

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    You mean xy- y2 on the right.

    But you still can't divide by 0.

    5*0= 0 and 3*0= 0 so 5*0= 3*0. It does not follow that 5= 3!
     
  5. Apr 4, 2012 #4

    Mentallic

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    What instead should have been done was

    [tex](x+y)(x-y)=y(x-y)[/tex]

    [tex](x+y)(x-y)-y(x-y)=0[/tex]

    [tex](x-y)(x+y-y)=0[/tex]
    (factorized out (x-y) from both factors)

    [tex](x-y)x=0[/tex]

    So either x-y=0, thus x=y, or x=0

    This is common practice when solving quadratics and such. If you end up with [tex]x^2+x=0[/tex] for example, you don't divide through by x to obtain [itex]x+1=0[/itex] because then you lose the solution of x=0. What you instead do is factorize into the form [itex]x(x+1)=0[/itex] which allows you to find all the solutions.
     
  6. Apr 4, 2012 #5
    PhysicsForums,

    you guys have exceeded my expectations! Thank you! I'll be back soon I'm sure :)
     
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