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Counterexample help

  1. Mar 1, 2009 #1
    Find a counterexample to the statement "For all real numbers u and v, (u + v)^2 is not equal to u2 + v2."
     
    Last edited: Mar 1, 2009
  2. jcsd
  3. Mar 1, 2009 #2

    Hurkyl

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    What work have you done on the problem? Have you tried anything at all? It's actually harder to find an example than it is a counterexample....

    (I assume you meant to write something like [itex](u + v)^2[/itex] -- the proper way to write that when you can't use a superscript is as (u+v)^2, because what you actually wrote is to multiply by 2, and that's a very different problem. :tongue:)
     
  4. Mar 1, 2009 #3
    Hey Hurkyl,
    I've tried a couple of things, but I know they are not right. Should I post them here anyway :(
     
  5. Mar 1, 2009 #4

    Hurkyl

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    Definitely. It is forum policy that we won't offer much help until you've shown that you've worked on a problem.... (p.s. wee the edits in my previous post)
     
  6. Mar 1, 2009 #5

    symbolipoint

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    You want a counterexample for a basic property of Real Numbers? Yes, I know what set of numbers would give a counterexample. It is the set {}.
     
  7. Mar 1, 2009 #6
    Now I'm really confused
     
  8. Mar 2, 2009 #7

    Gib Z

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    Lol he's just being slight cruel.

    Try substitute some small numbers for u and v, see if your Left hand side is equal to your right hand side.
     
  9. Mar 2, 2009 #8

    HallsofIvy

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    No, he's just being completely wrong.

    To find a counter example to "(x+ y)2 is NOT x2+ y2", you must use x and y so that (x+y)2= x2+ y2.
    Presumably you know that (x+ y)2= x2+ 2xy+ y2. That means you must find x and y so that x2+ 2xy+ y2= x2+ y2.

    Notice that the squares on both sides cancel. What does that leave you with?
     
  10. Mar 2, 2009 #9

    symbolipoint

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    From original: "(u + v)^2 is not equal to u2 + v2."

    Actually, I misread the original relation. Either that or the "2" was not shown as exponentiation; but as shown on the right-hand side, the "2" are not shown as exponents.
     
  11. Mar 2, 2009 #10

    HallsofIvy

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    No, I read it as "Find a counterexample to 'for all real numbers x,y it is NOT true that [itex](u+ v)^2= u^2+ v^2[/itex]'" and there is an easy counterexample as I pointed out.
     
    Last edited: Mar 7, 2009
  12. Mar 2, 2009 #11

    symbolipoint

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    That is not what was written originally. The 2's on the righthand side were not exponentiated.
     
  13. Mar 6, 2009 #12
    The notation isn't the problem. The statement "For all real numbers x,y, (x+y)^2!=2x+2y" is still false. I read it the same as Hurkyl.
     
  14. Mar 7, 2009 #13

    Mentallic

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    Why so? try [itex]y=-x[/itex] and [itex]y=2-x[/itex]

    edit: I misread "for all real numbers" as "for what real numbers"
     
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