TITLE: Bijective map from $\Bbb Z$ to $\Bbb Q$
QUESTION [8 upvotes]: There  exists  a  map $f: \Bbb Z\rightarrow \Bbb Q $  such  that  $f$  is 
A. Bijective  and  increasing
B. Onto  and  decreasing
C. Bijective  and  satisfies  $f(n)\ge 0$  if $n\le 0$
D. Has  uncountable images
Now  option  D.  is  absurd . Option  C.  is  given  to  be  the  correct  answer.I  was  thinking  since  both  sets  are  countable  bijection  is  obvious. Now  why  cannot  be  increasing  ? I  could  map  $0$  to  $0$  and  the  negative  integers  to the  negative  rationals and  positive  integers  to  the  positive  rationals. And  if  increasing  would  be  possible  just  interchanging signs  would  give  the  decreasing  map. So  none  is  possible  but  why?

REPLY [6 votes]: For A, some $f(n)$ lies between $f(0)$ and $f(1)$, which requires the integer $n$ to lie between $0$ and $1$.For B, let $M$ be the least positive $n$ for which $f(n)<f(0)$. Then $f(M)<f(m)=(f(M)+f(0))/2<f(0)$ for some integer $m$, so $M>m>0$. But by the definition of $M$, if any integer $m$ satisfies $M>m>0$ then $f(m)\geq f(0)$.