Consider a small pool that contains 200,000.0 litres of water
Imagine a single drop of rain falling in the pool with a volume of 0.2ml (0.0002L)
If we add these together this gives 200,000.0002L
However according to the rules for addition we express the answer to the same number of decimal places as there are in the data with the least number of decimal places i.e. (200,000.0) So the answer is 200,000.0L
At first glance this result may not seem to make sense because it shows no increase in volume but if we focus on the fact that we are adding two measurements we can explain the validity of this result.
The volume of the pool was measured to within +/-0.05 L (ie 200,000.0 +/- 0.05 L) . So if the pool volume was measured before and after the single raindrop fell the difference in volume would not be detectable with this limited resolution. The result of the calculation is reflecting this.
However it should be noted that in the result of this calculation information has been lost (i.e the volume of the raindrop). When carrying out a calculation that has several stages it is important to carry out the stages in the order that preserves the most information in the final result. To understand this, consider the example below.
Imagine that two million identical drops of rain fall into the pool.
If say we were to add the volume of each drop individually to the volume of the pool ( i.e. 200,000.0 +0.0002 + 0.0002 + ..)then as well as this being ridiculously impractical we would loose the information about the volume of the raindrop at each stage (as shown in the previous example) and so our final result would still be 200,000.0 litres!
However if we adopt the correct procedure we would first multiply the volume of each raindrop by the number of raindrops.
2,000,000 x 0.0002 = 400.0 L (using the rules for the precision of the result of multiplication)
Then we add this to the volume of the pool 200,000.0 + 400.0 = 200,400.0 litres.
So the calculation now predicts the increase in volume (which could be confirmed by measurement of the pool volume to the resolution indicated).
This example has just been used to show that the rules above need to be applied carefully to produce sensible results. Measuring the volume of the pool before and after a single drop of rain has fallen will show no measurable difference at the level of precision stated. However there will be a clear measurable difference in the volume of a lake after a millions of drops of rain have fallen. This needs to be reflected by the sensible application of the rules above when performing any calculations!