Well why didn't you say so! You were stating the process rather than your goal when you were walking down the wrong path (or walking us down the wrong path, rather).
Yes, you need a resistor. THe LED is a diode with a fixed voltage drop across it which is lower than your supply voltage. As a result you need a resistor to drop the extra supply voltage away so that the diode doesn't end up acting like a short-circuit and burning out. THe resistor you pick must drop the extra voltage of the supply while leaving enough voltage to forward bias (turn on) the diode. It must provide the voltage drop at the current you want running through the diode. Obviously the voltage drop will only be X volts for Y current. So if the current changes the voltage drop will too. But since your diode does have a fixed voltage drop this isn't a problem (it would be if the current through your load varies like a digital circuit).
So you use the formula V=IR:
R = (Excess Voltage to be dropped)/(Desired Current) = (Vsupply-Vdiode)/Idesired
Remember...the diodes have a maximum operating current you should stay below.