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Are these statements true or false? Prove...

  1. Sep 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Are these statements true or false? Prove/reason why/why not.
    a.)∃x∈R , ∃y∈R : (x2-2)2+y2=1
    b.)∀x∈R , ∀y∈R : y >x2-1
    c.)∀x∈R , ∃y∈R : |x+y|=1
    d.)∃x∈R , ∀y∈R :|x|>y

    3. The attempt at a solution
    a.) true, for example (x=1 and y=0)
    b.) false, because if I choose (x=1 and y=0) I get 0>(1-1) which cannot be true so the statement cannot be always true
    c.)false? Let. x=-y+8 ⇒|8|≠1 x can be any real number
    d.)?
     
  2. jcsd
  3. Sep 13, 2015 #2

    mfb

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    Staff: Mentor

    You cannot choose an x that depends on y, that would reverse the logic.

    d) Can you find an x such that ∀y∈R :|x|>y?
     
  4. Sep 13, 2015 #3
    no so it's false
     
  5. Sep 13, 2015 #4

    mfb

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    Right.
     
  6. Sep 14, 2015 #5
    Okay. The remaining question is: how do I prove c and d are false?

    d.)∃x∈R , ∀y∈R :|x|>y

    its negation: ∀x∈R , ∃y∈R :|x|≤y is true but how i prove it is true?
     
  7. Sep 14, 2015 #6

    Mark44

    Staff: Mentor

    @mfb, are you sure? If y is any arbitrary real number, surely we can find a number x for which |x| > y.
     
  8. Sep 14, 2015 #7
    You need to re-read the statement d. It says: 'there exists an x∈R such that for all y∈R, |x|>y'.

    It is obvious that there isn't an x of which absolute value is greater than every real number, isn't it?
     
    Last edited: Sep 14, 2015
  9. Sep 14, 2015 #8

    Mark44

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    I think I am misinterpreting d) as if it said ∀y∈R, ∃x∈R, :|x|>y
     
  10. Sep 14, 2015 #9
    Tempting, isn't it?

    I am at a loss too, though. How are we supposed to read d as? "There exists an x and for every y..?
    I think what is meant that there is such an x we can couple with Any y such that the implication holds.
     
  11. Sep 14, 2015 #10

    Mark44

    Staff: Mentor

    I'm now convinced that the correct reading is: "There is a number x such that, for any real y, |x| > y." It's not difficult to show that this is not true.
     
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