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Math Help - Show that the recursively defined function z is neither surjective or injective

  1. #1
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    Show that the recursively defined function z is neither surjective or injective

    My question is:

    The function z from  \mathbb{N} to  \mathbb{N} is defined recursively by:

    z(1)=1 and

    z(n+1) = \begin{cases} 1/2(z(n)+3), & \text{if }z(n)\text{ is  odd} \\ z(n)+5, & \text{if }z(n)\text{ is even} \end{cases}

    for all n\geq 1.

    Show that z is neither an injection or a surjection.

    What I've done so far is find the first couple of values of z(n) up to n=7, where I got z(2)=2, z(3)=7, z(4)=5, z(5)=4, z(6)=9, and z(7)=6. At this point, I'm not sure about what to do, so any help would be greatly appreciated!
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  2. #2
    Member ModusPonens's Avatar
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    Re: Show that the recursively defined function z is neither surjective or injective

    z(9)=z(3), so it's not injective.

    From 9 onwards, the sequence is periodic, because z(3+k+6n)=z(3+k), so it's not surjective.
    Thanks from HallsofIvy and sakuraxkisu
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  3. #3
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    Re: Show that the recursively defined function z is neither surjective or injective

    I'm a bit confused about k, could you explain how you got z(3+k+6n) and z(3+k)? Thank you
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