Given n relative prime integers find the number of ways to make a proper fraction (i.e. numerator < denominator).
The answer is 2^(n-1) as for every integer we have two options either it goes in numerator or in denominator so answer is (total combinations/2).
I cannot understand,why the above statement holds?That is, how can we prove the statement mathematically.
Consider all fractions you could make. There are $2^n$ of these, like you mentioned above. Half of these are proper. This is so since every fraction consisting of relative primes will either be proper or improper (i.e. none of them equals exactly 1) and to each improper fraction $\frac{p}{q}$ we have the corresponding proper fraction $\frac{q}{p}$.