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Mixed Quantifers confusion! Descrete Math

  1. Sep 17, 2006 #1
    THe directions say< indicate which fo the following statements are true and which are false, Justify your answers as best you can.

    Here is the question:
    [tex] \exists [/tex] x [tex]\in[/tex] R such that [tex]\forall[/tex] [tex]\in[/tex] R, x = y + 1.

    I wrote the following:
    There exists a real number x such that given any real number y the property x=y+1 will be true. True. y = x-1. Then y is a real number, and y + 1 = (x-1)+1 = x.

    I really don't know if i did this right or not but there was an example but slighty different and the book had the following:
    [tex]\forall[/tex] x [tex]\in[/tex] Z, [tex]\exists[/tex] y [tex]\in[/tex] Z such that x = y + 1.

    There answer was:
    Given any integer, there is an integer such that tthe first inteer is one more than the second integer. this is true. Given any integer x, take y = x-1. Then y is an integer, and y + 1 = (x-1) + 1 = x.

    I'm really confused on how to go about tackling these problems. Any help would be great! thanks!
     
  2. jcsd
  3. Sep 17, 2006 #2

    StatusX

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    For the first one, there needs to be a single x that works for all y. Note how this is different from the second one.
     
  4. Sep 17, 2006 #3

    HallsofIvy

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    Doesn't make sense. Did you mean [tex]\for all y[/itex] ??
    If you meant [tex]\forall y[/tex] then y= x-1 works, doesn't it?

    What is true in Z (set of all integers) is not necessarily true in R (set of all real numbers) but the difference is usually a matter of multiplication or division, not addition.

    [/quote]There answer was:
    Given any integer, there is an integer such that tthe first inteer is one more than the second integer. this is true. Given any integer x, take y = x-1. Then y is an integer, and y + 1 = (x-1) + 1 = x.

    I'm really confused on how to go about tackling these problems. Any help would be great! thanks![/QUOTE]
     
  5. Sep 17, 2006 #4
    I thought it was odd that I could solve them exactly the same way. For the first one, if i had to find a single x for all y, u would think i would have to write it differently than if i was finding for all x there exists a y.
     
  6. Sep 17, 2006 #5

    0rthodontist

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    The order of the quantifiers is switched between the example and your problem. Your problem says, (as you correctly interpreted):
    "There exists a real number x such that given any real number y the property x=y+1 will be true."
    Another way of putting it is:
    "There exists a real number x such that no matter what real number y I choose, x = y+1."
    You should intuitively convince yourself that these are the same.

    So let's try an example--say that x = 2. Is it true that no matter what real number y I choose, x = y + 1? No, because if I choose y = 100, then x does not equal 100 + 1 = 101. What if I chose x = 5. Could you find a y that makes the equation false? Is there ANY x that wouldn't have a y that would make the equation false?
     
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