A Little More on Significant Figures
(©2002, François G. Amar, All rights reserved) 
 

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Here is some clarification on how Significant Figures work in calculations and a little more optional discussion for those who are interested.

Uncertainty in measurement
    This is is governed by the nature of the apparatus: a ruler marked in millimeters can really
only be trusted to give you one place to the right of the decimal as in 8.9 cm because there are only
10 divisions per centimeter.

Uncertainty in derived results
    We need to have some way of handling the uncertainty that we propagate through a calculation
on the way to a derived result. We shall see that the method of significant figures keeps us from
overestimating the information we have about a derived quantity. Now suppose you have a scale which
allows you to measure with confidence masses between 0 and 100 grams to a precision of 1/100th of
a gram. You would weigh out your compound using a container. Subtracting the tare weight looks
like this:

                (container + compound)       11.53 g



 
                 - container (empty)          9.31 g
                     difference               2.22 g

We have let the number of significant figures that we report in the answer take care of our uncertainty
limit. In this method we will not report ± uncertainty limits. The link to the left is OPTIONAL and will
tell you more about uncertainty than many of you want to know.

Illustration of the addition rule for reporting the right number of significant digits when adding two
numbers. An extreme example is the "battleship" problem. Can you weigh a sailor by first weighing a
battleship empty and then weighing it with the sailor on board? No. Why not? Let's suppose the battleship was in the 130,000 tonne class (where tonne = metric ton = 1000 kg). This means the battleship weighs
about 130 million kg = 1.3 x 10⁸ kg. If we were to add the mass of the sailor and expect to get a reasonable
estimate of her weight, we would have to be able to measure such a large mass to a precision of at least 9
significant digits. No such measuring apparatus exists. The best we might do is to get about 5 figures which might yield
 

        mass of battleship =   131,530,000    kg     <-- here no decimal point implies                                               that the last 4 zeroes are not significant

The true mass might be anywhere from  131,525,000 to 131,534,990 kg. If the mass were less than
the first number, our apparatus would have "rounded down" to 131,520,000 kg and if the mass were
larger than the second number, our apparatus would have "rounded up" to 131,540,000 kg. The range
of uncertainty in the true mass of the battleship is thus of the order of the difference of these limits or
nearly 10,000 kg. Thus the mass of the sailor is only a small fraction of the uncertainty in the mass of
the battleship.
    The rule for addition is that you keep the significant digits of the number with the least number of digits to the right of the decimal point, thus:

                        1000.0     g <--  least number of digits to right of decimal point                      +    31.5462  g
                        1031.5462    g    -->  must be rounded to 1031.5 g

Multiplication
    The multiplication rule is easier to use but some people seem to think it is less intuitive.
The rule is:
    The answer in a multiplication or division has the same number of significant digits as the number with the least number of significant digits.

             10.0   x  15343.1     = 153000 = 1.53x105
 
    sig figs:   3          6                 3

To understand this rule we need to remember how we learned multiplication in grade school. All multiplication is really addition. Let's suppose that the problem above is rewritten as a grade school multiplication problem:
 

                        15343.1
                    x      10.0?
 

The question mark represents the fact the hundredths place or 4th digit of the smaller number
is not known. Let's now work out this problem as we would in second grade:

                        15343.1     <-- certain to 6 digits
                    x      10.0?    <-- certain to 3 digits
                          ???.???
                         0000.00
                        00000.0
                       153431.
      adding           153???.??? <-- certain only to 3 digits

So the answer is 1.53 x 10⁵ as above. The decimal point (.) has gone to the left of the
last 3 digits because the problem was set up with a total of 3 digits to the right in the
"question" (.1 and .0?). [Note that wherever an uncertain digit  (?) multiplies a known
digit (like 3) we have an unknown result (?)].