Resistors in parallel: How do I make this rigorous? In my revision guide it says "when two resistors are in parallel, the current has two paths it can flow through so there is less opposition to the current. This is why the combined resistance is less than the separate resistances."
This is of course a true statement, because $R_{combined}=\frac{R_1\times R_2}{R_1+R_2}<R_1$ $\Leftrightarrow \frac{R_2}{R_1+R_2}<1$ $\Leftrightarrow R_2<R_1+R_2$ (and similarly for $R_2$). However, it is also vague in the sense that I couldn't imagine someone using it to discover that the separate resistances in parallel are less than the combined resistance, without doing any maths.
Basically, it is what Dirac said about philosophy: It creates good explanations for things that have already been discovered, but cannot be used to discover NEW laws.
But I was wondering if we can make the explanation rigorous enough so that it becomes a correct explanation even without knowing it before-hand.
I don't think it's possible to keep it as simple and intuitive as "current has less opposition so resistance is less" while increasing the rigor.
 A: Resistance is Voltage per Current. $R = V/I$ or $V = RI$ if you like.
So if you put two resistors in series, the voltage is that due to the first, plus that due to the second.
You understand that perfectly.
Now, what is the inverse of Resistance, Current per Voltage?
Give it a name, call it Conductance, perhaps. $C = 1/R = I/V$
Then $I = CV$ if you like.
OK, put two resistors in parallel, and what is the conductance?
It is that due to the first resistor, plus that due to the second.
So convert back to $R$ and you have your answer.
So for example, if you have two resistors $R_1$ and $R_2$, their conductances are $1/R_1$ and $1/R_2$, right?
Now put them in parallel - the conductances add, right?
So the total conductance is $1/R_1 + 1/R_2$.
So what's the combined resistance? One over that, or $1/(1/R_1 + 1/R_2)$.
You can put numbers on it if you like, but why is the combined resistance less than either resistor?
Because the combined conductance is more.
A: It perhaps is not as rigorous as you want, but it is simple and intuitive:
$$
R = \frac{\rho L}{A}
$$
Where $\rho$ is the resistivity, $L$ is the length, $A$ is the cross-section area. When you plug resistances in series, you "are" increasing $L$, and thus $R$ increases. If you put in parallel, you "are" increasing $A$, and thus $R$ decreases.
