Imagine you have $1 \Omega$ resistors. You can build a $\frac{5}{3} \Omega$ or $\frac{3}{5} \Omega$ using only 4 resistors. It’s not a challenging puzzle, but it is stunning how far you can stretch it out.  Compare the solutions for $\frac{5}{3}$ and $\frac{3}{5}$. There is a duality–a transition between them. Suppose you have a chain of resistors where you can solve for the resistance using combination rules–formula for series and parallel resistors. You can break that chain into two sub-chains, where the two sub-chains are either in series or in parallel. Change the relation between the two from series to parallel or vice versa. Then apply the same process to the sub-chains recursively. You start with a chain of resistance $R \Omega$, and the final resistance would be $\frac{1}{R} \Omega$. This is provable by induction. For the base case of induction note about the choice of $1$ in $1 \Omega$ resistor. $1$ is the inverse of itself.

That theorem results in a corollary. You can make any rational resistor–$\frac{p}{q} \Omega$ for any positive integers $p$ and $q$. These numbers, however, are the only numbers you can make. Consider any arbitrary network of resistors with finite number of $1 \Omega$ resistors. You can’t always break the network into two nice components. But you can apply node analysis method to solve for the total resistance of the network. It would give you a linear equation per node. Every factor in these equations are rational. So the total resistance must also be rational. With infinitely many resistors you can create ladders. Even a simple ladder can have a $\phi \Omega$ resistor, where $\phi$ is the golden ratio–a solution to $x = 1 + \frac{1}{x}$. You don’t need that hustle to build an irrational number. Write that number in decimal notation. For every digit, build that specific digit with finite number of resistors. Then put them in a series so they add up.

In that network, however, the start and finish terminals are not well defined. The most left digit in an irrational number is always known, but the most right digit is not defined. To solve this issue, instead of a series of resistors, build each digit in parallel. If you initially wanted to build a $R \Omega$ resistor, now try to look at the digits of $\frac{1}{R}$. You just have to build the inverse of every digit in $\frac{1}{R}$. You already saw that it is possible to build inverse of a rational number–a digit in this case. So that’s how you can build a resistance of $\pi \Omega$, with infinitely many–but fortunately countable–number of $1 \Omega$ resistors. 