Can I Systematically Determine the Cardinality of a Finite Set? Is there any way you can, in a systematically manner, determine the cardinality of a finite set in which the elements are given by a specific expression?
For example, let $A$ be a set $\ A=\{1,2,3,4,5,6\}$ and $B$ be a set defined by:
$$B=\biggl\{\frac{a-b}{a+b} : a,b\in A\biggl\}$$
My question is thus, can I somehow determine the cardinality of the set without having to calculate and enumerate each and every element?
 A: It's really dependent on how you built the set. The thing that can make things rather tricky is when things in the set are equal only one thing is counted. So usually inequalities are easy but the exact number you'd have to eleminate all equivalent elements.

For example, let $A$ be a set $\ A=\{1,2,3,4,5,6\}$ and $B$ be a set
  defined by: $$B=\biggl\{\frac{a-b}{a+b} : a,b\in A\biggl\}$$

Since $|A|=6$ and you are taking two elements in $A$ you get the easy inequality $|B| \le 6^2=36$.
Since $B$ has at least $1$ element $1 \le |B|$
Enumerateing through the values and evaluating them in the function $\frac{a-b}{a+b}$ i get the following table
$\begin{array}{rrrrrr}
0 & \frac{1}{3} & \frac{1}{2} & \frac{3}{5} &
\frac{2}{3} & \frac{5}{7} \\
-\frac{1}{3} & 0 & \frac{1}{5} & \frac{1}{3} &
\frac{3}{7} & \frac{1}{2} \\
-\frac{1}{2} & -\frac{1}{5} & 0 & \frac{1}{7} &
\frac{1}{4} & \frac{1}{3} \\
-\frac{3}{5} & -\frac{1}{3} & -\frac{1}{7} & 0 &
\frac{1}{9} & \frac{1}{5} \\
-\frac{2}{3} & -\frac{3}{7} & -\frac{1}{4} & -\frac{1}{9}
& 0 & \frac{1}{11} \\
-\frac{5}{7} & -\frac{1}{2} & -\frac{1}{3} & -\frac{1}{5}
& -\frac{1}{11} & 0
\end{array}$
So sorting out all the duplicates:
$$B=\left\{0, \pm\frac{1}{2}, \pm\frac{1}{3}, \pm\frac{2}{3}, \pm\frac{1}{4}, \pm\frac{1}{5}, \pm\frac{3}{5}, \pm\frac{1}{7}, \pm\frac{3}{7}, \pm\frac{5}{7}, \pm\frac{1}{9}, \pm\frac{1}{11}\right\}$$
So it looks like in your example $|B|=23$
A: Let's solve the general problem.Consider $A=\{ 1,2,....,n \}$ and    $B=\{ {a-b\over a+b}:a,b\in A \}$. 
Now, note that ${a\over b}= {c\over d}\Leftrightarrow {a-b\over a+b} ={c-d\over c+d}$ i.e there is a one to one correspondence between $B$ and $S= \{ {a\over b} : a,b\in A \} $.So, we have to find cardinality of $S$.
Now, if the least form of $a\over b$ is $c\over d$ then $gcd(c,d)=1$ and $c,d\in A$ as $1\le c\le a\le n$ and $1\le d\le b\le n$.So, the problem is to find the pair of numbers $c,d\in A$ such that $gcd(c,d)=1$. 
Now, for $c>d$, for a chosen $c$, $d$ can be chosen in $\phi (c)$ ways. So, the number $c\over d$ s.t. $c\over d$ $>1$ and $gcd(c,d)=1$ are $\sum_{r=2}^{n} \phi (r)$ in numbers.Now for the numbers $e\over f$ $<1$ and $gcd(e,f)=1$, we just interchange the position of $c$ and $d$ in $c\over d$ $>1$ and $gcd(c,d)=1$.So, they are also in $\sum_{r=2}^{n} \phi (r)$ in numbers.
Now at last just consider ${1\over 1}=1$, which was not in previous cases.Hence the required result is $2 \sum_{r=2}^{n} \phi (r) +1$=$2 \sum_{r=1}^{n} \phi (r) -1$. 
For particularly your problem, $n=6$ and $2 \sum_{r=1}^{n} \phi (r) -1=23$
