Cardinality of all rational points in $R^3$ Question: Find the cardinality of the set of all points in $R^3$ all of whose coordinates are rational, and justify the answer. 
Idea: Call the set of all points in $R^3$ all of whose coordinates are rational A. Also, I will make a set called B, that is the set of all coordinates (x,0,0) where x is any rational number. Now, from this it is clear that B has a cardinality of $\aleph_0$. Also It is clear that B is a subset of A. So cardinality of A is greater than or equal to cardinality of B. Therefore, cardinality of A is at least $\aleph_0$. I will then create a function $f$ that is injective from A to the set of all natural numbers. This function will be $f(x)= 2^{a_1}*3^{b_1}*5^{c_1}*7^{a_2}*11^{b_2}*13^{c_3}$ where $a_1$ is the integer in the numerator of a, $a_2$ is the integer in the denominator of a,where $b_1$ is the integer in the numerator of b, $b_2$ is the integer in the denominator of b,where $c_1$ is the integer in the numerator of c, $c_2$ is the integer in the denominator c, and a,b,c are any rational numbers.  Therefore, cardinality of A is less than or equal to $\aleph_0$. Therefore, cardinality of A is $\aleph_0$.
Question: would something like this work? Also, is there a way to do this by strictly using a bijection? Thanks in advance. 

I can't comment for my reputation, I don't know if I should do this. But take care of the possibility of $a_1$ being negative. This would imply that for some x (in particular if x has some negative coordinate), f(x) would not even be integer.
