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Please check my (simple) proof. Skeptical of its simplicity

  1. Sep 5, 2011 #1
    x,y are in R. Suppose x2+y2=0. Prove that x=0 and y=0.

    My proof:

    Suppose x[itex]\neq[/itex]0, y[itex]\neq[/itex]0. Then by the field axioms, both x2 and y2 are strictly positive, and so is their sum. This is a contradiction, since we supposed that their sum = 0.

    Thus, x=0, and y=0.

    This problem and proof seem so simple, I think there may be something wrong with it.
     
  2. jcsd
  3. Sep 5, 2011 #2
    What specific geometric shape does it define in the Cartesian ?
     
  4. Sep 5, 2011 #3

    dynamicsolo

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    Homework Helper

    Yes, you're not quite there: you have shown that x and y can't both be zero. Could either one be zero? (Well, of course, they can't, but you still need to include something about this in your argument.)
     
  5. Sep 5, 2011 #4
    Is it a circle? The hw problem is for an intro to analysis class though, so I wonder if a geometric argument / proof would be accepted.

    To dynamicsolo: Thanks, that's what I was looking for. I felt like I was missing something.
     
  6. Sep 6, 2011 #5
    The structure of your method of proof is incorrect in general. You are attempting a proof via contradiction. The negation of a conjunction is not simply the conjunction of the negations of its conjuncts. Finding and using the correct form of your new premise (assumed as a result of contradiction) will lead you to a more traditional and sound argument.
     
  7. Sep 6, 2011 #6
    Hi Syrus. Do you mind clarifying? I don't think I understand what a traditional argument is. What makes a sound proof by contradiction? So far, I show, by contradiction that:

    1.) x [itex]\neq[/itex] 0 ---> x2+y2 [itex]\neq[/itex] 0
    2.) y [itex]\neq[/itex] 0 ---> x2+y2 [itex]\neq[/itex] 0
    3.) x, y [itex]\neq[/itex] 0 ---> x2+y2 [itex]\neq[/itex] 0

    Thus, x=0, and y=0. How do I improve the argument? Thanks.
     
  8. Sep 6, 2011 #7

    HallsofIvy

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    Actually, that's not simple enough! You don't need line (3).
    Although I would add more words:
    "1) if [itex]x\ne 0[/itex] then [itex]x^2> 0[/itex]. [itex]y^2\ge 0[/tex] so [itex]x^2 +y^2> 0[/itex]. Contradiction"
     
  9. Sep 8, 2011 #8
    Well i meant that the negation of (x = 0 AND y = 0( is (x =/= 0 OR y =/= 0). At least that's the way I would look at it and proceed (via proof by cases).
     
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