Okay. This is a bad practice.
The resistor limits the amount of current.
You should limit the amount of current, it will make the battery last longer.
Also, by not doing it, you are messing up circuit equations (power/heat is being used that you dont want to be used)
So you have your 3 volt battery. All the voltage will equal each other.
Battery = Forward voltage of the LED + wire drop + resistor drop.
3 V = 2.2 V + .8 + 0
The 2.2 V is assumed, you would need to check the data sheet. This set up means your little wire that is not rated for .8V with who knows how much current (which could blow up your LED).
Better approach.
Battery = Forward voltage of the LED + wire drop + resistor drop.
3V = 2.2 V + 0 V + 0.8 V
Now, here we make the resistor soak up all the power. Lets demand that we want 20 mA.
V = I x R
R = 0.8 / 20 ma = 40 ohms (In reality, you would pick 39 Ohms as its the closest value at 10 percent )
http://www.rfcafe.com/references/ele...tor_values.htm
Power of resistor = I * I * R = 0.02 * 0.02 * 40 = 0.016
so a 1/4 (or under 0.25) watt resistor would be fine.
Now just for kicks, if you could add a duty cycle (turn on and off the battery quickly) your eyes will not notice that it is off and you can make the battery last even longer.
Heres more reading and a calculator if the above makes no sense.
http://www.rc-cam.com/led_info.htm