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Written with StackEdit. Which of the following can be true for a mapping $$ f : \Bbb Z \to \Bbb Q $$
A. It is bijective and increasing
B. It is onto and decreasing
C. It is bijective and satisfied $f(n) \ge 0$ if $n \le 0$
D. It has uncountable image

Here's my attempt at finding a mapping that satisfies C(Following Method to prove countability of $\Bbb Q$ from Stephen Abbot )

Define $A_n = \{ \frac ab : a,b\ge 0, \gcd(a,b) = 1,a+b=n,b \ne 0 \}$

$B_n = \{ \frac {-a}{b} : a,b > 0, \gcd(a,b) = 1, a+b = n \} $

Mapping the negative integers and 0 to the set $A_n \forall n \in \Bbb N$ and positive integers to the set $B_n \ \forall n \ \in \Bbb N$ in the same manner as Stephen Abbot, the mapping will be surjective because the sets $A_n$ and $B_n$ would certainly cover $\Bbb Q$ and would be injective because no two sets have any elements in common.

Correct Answer - C
Source - Tata Institute of Fundamental Research Graduate Studies 2014

Is my proof correct? In any case, I can't imagine such a method to be required in this problem so is there a more 'easy' mapping? Also, why can't we have a mapping as desired in A and B?

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  • $\begingroup$ Your construction for (c) is fine and seems like the natural one. Answers (a) and (b) are impossible because if true, they would imply that $\mathbb Q$ can be linearly ordered. And (d) is impossible because $\mathbb Z$ is countable. $\endgroup$ – Bungo Dec 2 '16 at 19:50
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A. False. Suppose $f\colon\mathbb{Z}\to\mathbb{Q}$ is bijective and increasing. Then $f(0)<f(1)$. Take $q\in\mathbb{Q}$ with $f(0)<q<f(1)$. Then $q=f(n)$: can you see the contradiction?

B. False. Same argument as before.

D. False. The image is a subset of $\mathbb{Q}$.

C. True. Your argument is good, but it can be better formalized.

Take your favorite bijection $g\colon\mathbb{N}\to\mathbb{Q}_{>0}$ (positive rationals). Now define $$ f(n)=\begin{cases} -g(n-1) & \text{if $n>0$}\\[4px] 0 & \text{if $n=0$}\\[4px] g(-n+1) & \text{if $n<0$} \end{cases} $$ (note: $\mathbb{N}=\{0,1,2,\dotsc\}$).

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