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1-1, onto, both, or neither

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Is the function one-to-one, onto, both, or neither?
    f: Z→Z has the rule of f(n) = 4n^3 - 1

    2. Relevant equations



    3. The attempt at a solution

    My answer: one-to-one

    Is this correct?
     
  2. jcsd
  3. Nov 19, 2012 #2

    Zondrina

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    Well, how would you show it was one to one? How would you show it was onto?
     
  4. Nov 19, 2012 #3
    I believe you are right that it is injective. It be wise that you show a proof to confirm its 1:1 and a counterexample to show its onto.
     
  5. Nov 19, 2012 #4
    Well, the function is not onto because there is no integer n where 4n^3 - 1 = 1.
     
  6. Nov 19, 2012 #5
    One-to-one proof:

    4u^3 - 1 = 4v^3 - 1
    4u^3 = 4v^3
    u^3 = v ^3
    u = v
     
  7. Nov 19, 2012 #6

    Zondrina

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    Good. Now what about onto?
     
  8. Nov 19, 2012 #7
    Thanks.

    The function is not onto because there is no integer n where 4n^3 - 1 = 1.
     
  9. Nov 19, 2012 #8

    Dick

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    That seems fine to me. Neither of those is really a full scale proof. But I doubt you are expected to provide one. They are both good arguments that what you say is correct.
     
  10. Nov 19, 2012 #9
    Yeah, the book doesn't even ask for a proof, but it's nice to know that the answer is right.
     
  11. Nov 20, 2012 #10

    HallsofIvy

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    I disagree. What he gave are "full scale proofs". He showed that if f(u)= f(v) then u= v which is a perfectly good proof that f is "one to one". And he gave a counter example showing that it is NOT "onto".
     
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