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Nested Quantifier

  1. Jan 9, 2007 #1
    Here is the problem that I'm having trouble solving - I'm not sure where to begin. I need to determine the truth value but don't know how to do that.

    Ax3y(x^2 < y + 1)
     
  2. jcsd
  3. Jan 9, 2007 #2

    matt grime

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    Perhaps you could write that out in plain english. It might even help you understand how to solve your problem.
     
  4. Jan 9, 2007 #3
    For every real number x there exists y, x to the second power is less than y plus 1.

    So basically I need to find a number that is less than y + 1
     
  5. Jan 9, 2007 #4

    matt grime

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    No. Read it out loud inserting all of the words.

    For all x there is a y such that the condition x^2<y+1 is true.
     
  6. Jan 9, 2007 #5

    HallsofIvy

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    In other words, to prove that true, you must prove that for any given x there exist y such that y+ 1>x2. You need to show that, whatever x is you can find a corresponding y.
     
  7. Jan 9, 2007 #6
    So if I stated x = 1, then x2 would be 1 and then would or could I say y is 1 as well? Making y + 1=2
     
  8. Jan 9, 2007 #7

    matt grime

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    Now how can you do this for any x?
     
  9. Jan 9, 2007 #8
    That's what I don't get - I'm not sure what numbers I'm supposed to be inputing here. Please help as I'm really confused
     
  10. Jan 9, 2007 #9
    After working on this - Is this the answer?

    y = x2 + 2x = (x + 1)2 – 1
    x < 0 then (x – 1)2 > x2 so y can be x2 – 2x -2
    x = 0 then y can be 0
     
  11. Jan 9, 2007 #10

    matt grime

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    You're not supposed to put any numbers in that's the point.

    Let's play a game. I'm thinking of a number x. Can you give me a number y(possibly in terms of x) so that y+1 is definitely larger than x^2?
     
  12. Jan 9, 2007 #11
    How about z?
     
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