- #1
kingerd
- 14
- 0
the following is a problem in the book.
[tex]
f(x)=\left\{\begin{array}{cc}\frac{1}{2}(x+1),&\mbox{ if x is odd}
\\\frac{1}{2}x, & \mbox{ if x is even}\end{array}\right
[/tex]
the solutions that's given is This function is onto because, for any [tex]y\epsilonZ, g(2y)=y.[/tex] It is not one-to-one because g(1) = g(2).
I understand where the numbers come from stating why it's not one-to-one, but I'm a little confused about the onto part, in that they got the 2
[tex]
f(x)=\left\{\begin{array}{cc}\frac{1}{2}(x+1),&\mbox{ if x is odd}
\\\frac{1}{2}x, & \mbox{ if x is even}\end{array}\right
[/tex]
the solutions that's given is This function is onto because, for any [tex]y\epsilonZ, g(2y)=y.[/tex] It is not one-to-one because g(1) = g(2).
I understand where the numbers come from stating why it's not one-to-one, but I'm a little confused about the onto part, in that they got the 2
Last edited: