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2 is good. We can get any odd number in the range 1 to $2^{k+1}$ as a numerator of a fraction with denominator $2^{k+1}$ from an odd number in the range 1 to $2^{k}$ by either taking it in series with the element 1/2 for those numbers less than $2^{k}$ or in parallel for those numbers greater than $2^{k}$. If 2 is good then $2^{m}$ will be good because any fraction with its denominator a power of $2^{m}$ will have a denominator a power of 2. This can be extended if any prime $p$ is good by a similar argument $p^{k}$ is good. We have to take the union of this set with all previously constructed sets to take care of cases where the numerator is divisible by a power of 2.

25 is bad. 31/625 cannot be expressed in this system. to get 31/625 one must get it directly or by a product with 31/125. If we get 31/125 it must involve two parallel elements whose product must be 94 and whose denominators are 25 or less but 47 divides 94 and is a prime greater than 25 so we cannot obtain 31/125. If we get 31/625 from parallel elements we must have two elements whose product is 594 if they both have denominator 25 the maximum product is 576, if one has denominator 5 and the other denominator 125 the maximum product is 496. Both of these are less than 594 so we cannot express 31/625 this way. Since we have exhausted all possibilities 31/625 cannot be expressed in this system and 25 is bad.

2 is good. We can get any odd number in the range 1 to $2^{k+1}$ as a numerator of a fraction with denominator $2^{k+1}$ from an odd number in the range 1 to $2^{k}$ by either taking it in series with the element 1/2 for those numbers less than $2^{k}$ or in parallel for those numbers greater than $2^{k}$. If 2 is good then $2^{m}$ will be good because any fraction with its denominator a power of $2^{m}$ will have a denominator a power of 2. This can be extended if any prime $p$ is good by a similar argument $p^{k}$ is good. We have to take the union of this set with all previously constructed sets to take care of cases where the numerator is divisible by a power of 2.

2 is good. We can get any odd number in the range 1 to $2^{k+1}$ from an odd number in the range 1 to $2^{k}$ by either taking it in series with the element 1/2 for those numbers less than $2^{k}$ or in parallel for those numbers greater than $2^{k}$. If 2 is good then $2^{m}$ will be good because any fraction with its denominator a power of $2^{m}$ will have a denominator a power of 2.

2 is good. We can get any odd number in the range 1 to $2^{k+1}$ as a numerator of a fraction with denominator $2^{k+1}$ from an odd number in the range 1 to $2^{k}$ by either taking it in series with the element 1/2 for those numbers less than $2^{k}$ or in parallel for those numbers greater than $2^{k}$. If 2 is good then $2^{m}$ will be good because any fraction with its denominator a power of $2^{m}$ will have a denominator a power of 2. This can be extended if any prime $p$ is good by a similar argument $p^{k}$ is good. We have to take the union of this set with all previously constructed sets to take care of cases where the numerator is divisible by a power of 2.

3 is good. If we have all fractions whose numerators are in the range 1 to $3^{k}$ and whose denominator is $3^{k}$ then we can construct all fractions whose numerators are in the range 1 to $3^{k+1}$ and whose denominator is $3^{k+1}$ by taking fractions whose numerators are in the range 1 to $3^{k}$ and whose denominator is $3^{k}$ in series with $1/3$ and taking the same values in series with $2/3$ together with all fractions whose numerators are in the range 1 to $3^{k}$ and whose denominator is $3^{k}$ in parallel with $1/3$ and then taking the same values in parallel with $2/3$. The values in series with $1/3$ will give the first third of all fractions whose numerators are in the range 1 to $3^{k+1}$. The values in parallel with $1/3$ will give the last third of all fractions whose numerators are in the range 1 to $3^{k+1}$. The values in series with 2/3 will contain all even values of the middle third of all fractions whose numerators are in the range 1 to $3^{k+1}$. The values in parallel with 2/3 will contain all odd values of the middle third of all fractions whose numerators are in the range 1 to $3^{k+1}$. We have to take the union of this set with all previously constructed sets to take care of cases where the numerator is divisible by a power of 3.

Any product of any power of 3 and any power of two is good. We have the fractions $1/2$ and $2/3$ and $1/2$ in the starting set of fractions for any such set. We then use the fracion $1/3$ and $2/3$ to generate the desired set of fractions for any desired power of three whose denominator is the power of three and whose numerator is any number less than the power of three. Then we take this set in parallel with $1/2$ and take the union of this set in series with $1/2$. We repeat this process and we will eventually get the set of numbers whose numerator is is less than the product of the desired power of two and the desired power of three and whose numerator is the desired product of the desired product of two and the desired product of three. We have to take the union of this set with all previously constructed sets to take care of cases where the numerator is divisible by a power of 2, a power of three or both.

If $p$ is a prime of size 5 or greater then it is not good. We can pick a prime $q$ such that $q$ is greater than $p$ and less than $2p-1$. If $p$ is greater than 24 we can find such a number by using a result from this paper: Jitsuro Nagura (1952). "On the interval containing at least one prime number". Proc. Japan Acad. 28: 177–181. The result is that for any number greater than 24 there is always a prime between $n$ and $(1 + 1 / 5)n$. I found this result here.For 5 we have value 7, for 7, 11, for 11, 17, for 13, 17, for 17, 29 and for 23 29. Thus for $n$ equal to 5 or greater we can always find such a value. $q/p^{2}$ has to be represented by a parallel circuit of size two with two parallel elements of the form $a/p$ and $b/p$ with $a$ and $b$ less than $p$ hence it must be of the form $1-ab/p^{2}$ with $a$ and $b$ less than $p$ but $ab$ must be less than $(p-1)(p-1)$ but then smallest value that can be so expressed is $2p-1/p^{2}$ and we have a contradiction.

If $p$ is a prime of size 5 or greater then it is not good. We can pick a prime $q$ such that $q$ is greater than $p$ and less than $2p-1$. If $p$ is greater than 24 we can find such a number by using a result from this paper: Jitsuro Nagura (1952). "On the interval containing at least one prime number". Proc. Japan Acad. 28: 177–181. The result is that for any number greater than 24 there is always a prime between $n$ and $(1 + 1 / 5)n$. I found this result here.For 5 we have value 7, for 7, 11, for 11, 17, for 13, 17, for 17, 29 and for 23 29. Thus for $n$ equal to 5 or greater we can always find such a value. $q/p^{2}$ has to be represented by a parallel circuit of size two with two parallel elements of the form $a/p$ and $b/p$ with $a$ and $b$ less than $p$ hence it must be of the form $1-ab/p^{2}$ with $a$ and $b$ less than $p$ but $ab$ must be less than $(p-1)(p-1)$ but then smallest value that can be so expressed is $2p-1/p^{2}$ and we have a contradiction.

2 is good. We can get any odd number in the range 1 to $2^{k+1}$ from an odd number in the range 1 to $2^{k}$ by either taking it in series with the element 1/2 for those numbers less than $2^{k}$ or in parallel for those numbers greater than $2^{k}$. If 2 is good then $2^{m}$ will be good because any fraction with its denominator a power of $2^{m}$ will have a denominator a power of 2.