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Mathematical Proofs

  1. Sep 25, 2005 #1
    Hi all,

    If I have these two statements given to me, and I have to determine whether they are true or not.

    a) [tex] \forall x \epsilon R [/tex] [tex]\exists y \epsilon R [/tex] [tex](y^2 = x^2 + 1)[/tex]
    b) [tex]\exists y \epsilon R [/tex] [tex]\forall x \epsilon R [/tex] [tex](y^2 = x^2 + 1) [/tex]

    Now, to me, they both mean exactly the same thing, and both can be shown to be false by setting x = 2, then y is not a real number.

    However, seeing that the question specifically asks to prove just those two statements, I'm wondering if perhaps I am interpreting them wrong and they actually mean two different things.

    Thanks in advance for any advice,
    Robbie
     
  2. jcsd
  3. Sep 25, 2005 #2

    TD

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    Homework Helper

    If x = 2 then [itex]y^2 = 2^2 + 1 \Leftrightarrow y = \pm \sqrt 5 [/itex]. Those are real numbers, no?
     
  4. Sep 25, 2005 #3
    lol... yeah...
    I was thinking I was looking for rationals only... whoops...

    Still, is there any difference in the two statements themselves?
     
  5. Sep 26, 2005 #4

    honestrosewater

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    Gold Member

    Try a simpler one to see how the order of the quantifiers makes a difference:

    [tex]\forall x \in \mathbb{R} \ \exists y \in \mathbb{R} \ (x = y)[/tex]

    This says that for every real x that I choose, I can find at least one real y that is equal to that x. This is obviously true, since x = x.

    [tex]\exists y \in \mathbb{R} \ \forall x \in \mathbb{R} \ (x = y)[/tex]

    This says that I can find at least one real y that is equal to every real x. Well, there's more than one real number, so this is false.
     
    Last edited: Sep 26, 2005
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