Proof that the set of  odd positive integers  greater then 3 is countable I found  one  problem which  asks  following:

Show that the set of  odd positive integers  greater then 3 is countable.

At the begining I was thinking that  such numbers  could be represented   by $2k+1$,where $k>1$; but  in the answers paper  there was  written  as $2n+3$  or  general function   is
$$f(n)=2n+3$$
and  when I was thinking how to prove that  such answer  is countable, the answer  paper said  this function is a one-to-one correspondence  from the set of positive  numbers set to the set of positive odd  integers  greater to 3. 
My  question is:  is it  enough  to prove a one-to-one  correspondence  between  two sets, that one of them  is  countable. If yes, then once my lecturer ask me to proof that rational numbers are countable, so in this case if I represent  rational numbers  by following  function from set of positive numbers:
$$f(n)=\frac{n+1}{n}$$
 or  maybe   $f(n)=\frac{n}{n+1}$. They both are one-to-one correspondences from the set of positive numbers to the set of rational numbers (positives sure). Please help me, is my logic correct or not?
 A: If you know that a set $A$ is countable and you demonstrate a bijection $f:A\to B$ then you have also shown that the set $B$ is countable; when $A=\mathbb{Z}^+$ this is the very definition of countable. Both of the functions $2k+1$ and $2n+3$ can be used to show that the set of odds greater than $3$ are countable but the former uses the countable domain of $\{2,3,\dots\}$ instead of $\mathbb{Z}^+=\{1,2,3,\dots\}$.
However, neither the functions $f(n)=(n+1)/n$ or $n/(n+1)$ are bijections. Try and express the positive rational number $1/3$ in either of these forms and you will find there is no integer $n$ that works. In order for a function to be a bijection it must be both injective and surjective; your function here is not surjective (it does not obtain every value in the target in the codomain at least once - some values, like $1/3$, are left untouched).
A: The other answers are focused on explicitly constructing an bijection to the natural numbers. In practice, one  would just note that the set in the question is a subset of a countably infinite set, $\mathbb{N}$, and so it is either finite or countably infinite. It is infinite, so it must be countably infinite. 
This requires some justification, but it is not too difficult. See the first chapter of Topology by Munkres, for example.
A: The sufficient condition for proving countability is establishing the 1-1 correspondence of a given set with $\mathbb{N}$. For proving the countability of rationals, a method called as diagonalization is used. Take a look at the following link. Your logic is correct, only that in place of $f(x)=\frac{n}{n+1}$, we use $f(x)=\frac{m}{n}$ and express $\frac{m}{n}$ in mth row and nth column. The representation covers all possible rational numbers and their ordering makes it a set corresponding 1-1 to $\mathbb{N}$.
http://www.homeschoolmath.net/teaching/rational-numbers-countable.php
