Determining Onto and One-to-One Mappings: Finding Left and Right Inverses

[Shackleford]

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

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

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110714_194543.jpg?t=1310691331

 

[micromass]

Hi Shackleford!

Looks all good! However, for 2n, you never defined what g was...

 

[Shackleford]

Hi Shackleford!

Looks all good! However, for 2n, you never defined what g was...

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.

 

[micromass]

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.

Can't you pick g(x)=2x-1 for all x??

 

[Shackleford]

Can't you pick g(x)=2x-1 for all x??

I don't think so. The mapping is in cases. If x is even, f(x) = x + 1.

 

[micromass]

Yes, but g(x) is always odd... So you only need one case to calculate f(g(x)).

 

[Shackleford]

Yes, but g(x) is always odd... So you only need one case to calculate f(g(x)).

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.