Assuming a simple voltage divider of resistors R1 and R2, which are equal, the voltage in between will be one half the supply, right?
But then if we have two equal lengths of wire that, say, each have a resistance of 1µ ohms, and they are assembled into a voltage divider with a supply of 5V, will the voltage between them really be 2.5V? I somehow doubt it.
You pick your R values based on your current draw of the circuit and what it takes to drop your voltage. Ohms law did not go away here. Current is still related to Voltage and Resistance. You need to know precisely both the voltage and current required.
IF you figure out what you circuit needs to be like to make your very low R values work... you will see that it will become impractical, I think. FYI: You might find that you need a circuit that wastes huge amounts of current or not... do some math.
Also, voltage dividers are not magic and will to only work well in constant current solutions and they are extremely impacted by any significant load current changes.
Yes, it will, provided that the 5V supply can supply 2.5 million Amps and the 1 µ ohm resistors can handle that current. For any power supply you would encounter, it is not going to supply the current, and the voltage will fall to a few millivolts.
[edit]slipped 3 decimal points[/edit]
Further to what gardner said, 25MA (Mega not milli) would be the required amperage, or a 125MW power supply. If the supply were 5V limited to 3A, current becomes the limiting factor, the 2uOhm divider cited at 3A drops 6uV. But guess what... measure the voltage at the middle of the divider, and it would not surprisingly be 3uV. Vout = R2 / (R1+R2) * Vin. Ohm's Law wins 