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Left and Right Inverses

  1. Jul 14, 2011 #1
    1&2b and 1&2n.

    Basically, if they exist, I need to find the left and right inverses. If the mapping f is onto, then the right inverse exists. If the mapping f is one-to-one, then the left inverse exists.

    For the mapping in cases, is my method of determining onto or one-to-one correct?

    Sorry for the crappy quality of 1&2n.

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110714_194112.jpg?t=1310691220 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110714_194153.jpg?t=1310691321 [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110714_194543.jpg?t=1310691331 [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 14, 2011 #2

    micromass

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    Hi Shackleford! :smile:

    Looks all good!! However, for 2n, you never defined what g was...
     
  4. Jul 14, 2011 #3
    For x is even, I tried g(x) = x - 1. However, that only works when x - 1 is even.

    For x is odd, I tried g(x) = 2x - 1. That works for all x in Z.
     
  5. Jul 14, 2011 #4

    micromass

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    Can't you pick g(x)=2x-1 for all x??
     
  6. Jul 14, 2011 #5
    I don't think so. The mapping is in cases. If x is even, f(x) = x + 1.
     
  7. Jul 14, 2011 #6

    micromass

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    Yes, but g(x) is always odd... So you only need one case to calculate f(g(x)).
     
  8. Jul 14, 2011 #7
    Yes. I know it works for all x because the odd case gives the onto. However, I was only able to find the inverse from the x is odd case. I tried to find the inverse from the x is even case, it didn't compute because the g(x)= x - 1 is always odd, but it must be even in this case.
     
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