A simple question about parallel connecting a short circuit (+the physical interpretation) 1. Introduction
We know that
$$1/R_1 + 1/R_2 = 1/R_t \ (equation \ 1)$$
$$\therefore R_t = \frac{R_1\times R_2}{R_1+R_2} \ (equation \ 2)$$
when $R_1$ and $R_2$ are connected in parallel and $R_t$ is the total resistance. Consider a simple DC circuit with the voltage source providing $V$, a constant.
2. Ideal Model of Circuit Theory
2.1 A short circuit is a case in which, say, precisely $R_1 = 0$ and $R_2 = r$ for some nonzero positive number $r$ [Ohms]. We can substitute these numbers into equation 2 to get
$$R_t = \frac{0\times r}{0+r} = 0$$
and for a given constant voltage $V$, the total current is $I_t = V/R_t = \infty$
2.2 We may manipulate the equations in this way to get the current $I_2$ through the resistor R.
$$I_2 = \frac{V}{R_2} = \frac{1}{R_2}\times (I_t \times R_t) = \frac{1}{R_2}\times (I_t \times \frac{R_1\times R_2}{R_1+R_2}) = I_t \times \frac{R_1}{R_1+R_2}$$
$$\therefore I_2 = I_t \times \frac{0}{0+r} = I_t \times 0 = 0$$
precisely, because however big $I_t$ is, it's simply 0 if it's literally multiplied by 0 [Ohms]. From this it looks obvious that when a short circuit is connected parallel with a resistor, the current does not conduct at all through the resistor.
2.3 However, if we see the formula itself, $I_2 = V/R_2 = V/r$, it is not quite clear that precisely $I_2 = 0$ is demanded. Both $V$ and $r$ are positive numbers, so division between them should not yield a complete zero at all. This contradicts the above conclusion. I thought this is because implicitly we allowed division by zero, from the very beginning throughout the whole calculation, e.g. in the first place in equation 1 it's simply not allowed to think about an equation with $R_1$ in the denominator and assert that $R_1$ is defined to be zero.
Is this reasoning correct? If so, how can we improve on it with some naive maths so that things make sense and therefore we can get some more grasp of the physics out of it?
3. Physical model
In reality the case in which $R_1 = 0$ and $R_2$ is nonzero would be parallel connecting a superconductor (at whatever temperature), with for example copper at room temperature. We know that the following equation holds:
$$|J| = e \mu_e n E$$
where J is the current density, e is the electron charge, n is the electron number density in the material, $mu_e$ is the mobility of electron in that material, and E is the electric field strength. I guess we can say that for superconductors $\mu = \infty$ and for copper it obviously has a reasonable mobility value.
This means that although current flowing through the superconductor is infinity, current anyway flows through the copper because in copper, still, the electron mobility is nonzero and the number density is nonzero (as many electrons should be equally distributed over the copper wire) and the electric field $E = -dV/dx$ is nonzero. Regardless of the infinite current through one, physically, current should not be completely zero for the other. Is this the right interpretation to make?
 A: If two resistors are connected in parallel and one of them is 0 that is a short circuit. All the current will flow through the short circuit $R_1$ (the 0-ohm resistor) and none of it through the other one $R_2$.
However, stting a resistor to $0$ will give you some issues and indeterminate forms like $0/0$ and $0\infty$. So you need to choose some conditions to solve them. The easiest thing to imagine is that, if there is no resistence there is no drop in potential hence $V=V_1=V_2=0$.
For the same reason, as the resistor $r$ is shorted, there is no drop in voltage across it and hence, the $V$ in your equation is $V=0$. In symbols,
$V_2=0$ and hence $I_2=V_2/R_2=0$ or $V_2=R_2I_2=0$. No problem here because $R_2$ is finite and we just "assumed" $V_2=0$ because there is no drop in potential as the parallel has $0$ resistence in total.
Instead, for resistor $R_1$ you get $V_1=V_2=0$ (because the two resistors are in parallel)
$I_1=V_1/R_1=0/0$ and $V_1=R_1I_1=0\infty$ which are indeterminate forms - but you know $V_1$ is 0 because you have shorted the circuit by definition: there is no resistance so there is no drop in voltage.
Notice that you get indeterminate forms... that also holds for the total current $I_t=V/R_t=0/\infty$. In real circuits (except superconductors) the resistence is never 0, even a shorted wire has some (very low) resistence, so there always is a finite a current and a finite (small) drop in voltage so that $R_1=0$ is impossible, and one always has $R_1=r_0$ with $r_0$ possibly small but non-zero.
A: Playing with formulae which have zeros and infinities in them can often lead to contradictory results.
It is therefore best to assume a finite value for $R_1$ in this case, do the sums and then look and see what happens when $R_1 \to 0$, ie $R_1$ becomes smaller, and smaller or $R_1 \to \infty$, ie $R_1$ becomes bigger and bigger, compared with other values in the expression.
A: Further thoughts and clarification by the OP. This can be divided into three cases - (1) an impossible case, (2) a possible case that will eventually lead to the burning of the circuit, (3) a clearly possible case.
1. Impossible case

We have assumed that V is nonzero but we are asserting that v is zero. Therefore this does not satisfy KVL and thus is a wrong circuit diagram.
2. Possible but industrially dangerous case

Here we use the approximation that $r_1 -> 0$ but a very small number. We have $V = i_{tot} \times r_{tot} = \infty \times 0 =$ a certain value. This satisfies KVL and KCL. Evidently $i_1 = \infty$ and $i_{tot} = \infty$. The size of infinities cannot be compared so it's fine. Through $r_2$, however, a seemingly normal current flows. This is quite evident -
$i_2 = \frac{r_1}{r_1+r_2} i_{tot} = \frac{1}{r_1+r_2} \times (r_1 \times i_{tot}) = 0 \times \infty =$ a certian nonzero value.
Another way of explaining, $i_2 = V/r_2 = $nonzero. This circuit, though, happens quite often in real electrical industries due to engineers' mistakes. In this case power through $r_1$ is going to be infinity leading to it burning, and if it was in a factory and for example something exploded it might include safety issues and a huge economic cost.
3. A perfectly possible case

This satisfies KVL and KCL. $r_{equivalent} = 0$ for the parallel resistors, and therefore $i_tot = V / r' =$ a finite well-defined value. When it comes to the parallel resistors we can now use the V=0 assumption for a short circuit without causing any contradiction, along with the fact that across parallel resistors the voltage drop is identical. Hence for both $r_1$ and $r_2$ $V=0$ and $$i_1 = \frac{r_2}{r_1+r_2} \times i_{tot} = i_{tot}$$ $$i_2 = \frac{r_1}{r_1+r_2} \times i_{tot} = 0$$
