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One-to-one and onto

  1. Nov 18, 2012 #1
    To me this problem doesn't seem right. Here it is:

    Is the following function one-to-one, onto, both, or neither?
    f: R→N f(x) = ceiling 2x/3

    My answer: onto

    Although, wouldn't this function be invalid since it produces negative numbers and the set of natural numbers doesn't include negatives? Consider f(-1.5) = -1.

    Am I misunderstanding a concept?
     
  2. jcsd
  3. Nov 18, 2012 #2
    A lot of people would consider the ceiling function to be f:R->Z.
    It would be invalid to say it's f:R->N Unless you restrict R to R+
     
  4. Nov 18, 2012 #3
    Well, that's the way is worded in the book, so it must be a typo. Maybe the writers meant to put Z rather than N.

    Would my answer be correct if were R to Z?

    Thanks for the help.
     
  5. Nov 18, 2012 #4
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