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Why does 1101 cm  1091 cm = 10 cm with 2 significant figures?
 You stated "when you have an expression that mixes operations, do the operations one at a time. 1101 cm  1091 cm = 10 cm with 2 significant figures..." Why is this 2 significant figures? According to your tutorial, "trailing zeros that all appear to the LEFT of the decimal point can not be assumed to be significant." This would mean that the answer of 10 cm would only have 1 significant digit making the final answer to the problem 5, not 5.0! Please clarify, I am totally confused!
Sherry W. Sherman

Sherry, if you just had the 10 cm, without any other information, you'd be correct. The safest thing to say about a measurement like 10 cm is "This measurement has at least one significant digit".
But you know that the 10 cm is the difference between 1101 cm and 1091 cm, which are uncertain in the 1's place. That means the difference between these numbers will be uncertain in the 1's place.
"Significant figures" means "all digits up to and including the first uncertain digit". While you're carrying significant figure counts through a calculation with a lot of steps, keep an eye on where that first uncertain digit is. Converting the numbers to scientific notation will help you keep things straight by discarding the nonsignificant, placeholding zeros from the beginning. For example:
12100 kg 10100 kg 2000 kg  is better written  1.21 × 10^{4} kg  1.01 × 10^{4} kg 0.20 × 10^{4} kg 
After you convert the numbers to scientific notation, you see exactly where the uncertain place is (it's lit up in red). You have 2 significant figures in the answer. The last digit just happens to be a zero. (It would be best to write the answer as
2.0 × 10^{3} kg).
Reader Comments
 (I disagree)... 11011091 yields 10, and if this number is written as 10, it has only one significant digit and the 0
is a place keeper. (a reader, 01/16/00

1101 is uncertain in the ones place, and 1091 is uncertain in the ones place,
so 11011091 is uncertain in the ones place. If you are given a 10 without knowing
where it came from, then yes, you say "1 significant figure". But if you have other
information showing that the uncertainty is in the ones place, not the tens place,
you have to say that it has two significant figures.
Following your reasoning, if you take 1102  1091, you end up with 2 significant digits
and if you take 11011091 you end up with only one. Why should such a small difference in the
calculation result in such a large increase in the uncertainty?
 Think about this example. What if you have an
equation that looks like this. 1101.0001  1101.0000 divided by 0.00425
Here we have 8 sig figs in the numerator and 3 sig figs
in the denominator. If we are dealing measurements, I am SURE that you
would agree that we must have been using a very detailed
measuring instruments (say for mass in grams for instance). If we follow your
method and subtract the numbers in the numerator
AND round to correct sig figs at that time, the numerator will have only 1 sig fig,
as opposed to the 8 it had before the operation.
Now your answer is limited to only 1 sig fig and your answer will be 0.02. If you
round AFTER completing the math, your answer
will have 3 sig figs and will be 0.0235 (clearly a major difference in the
answers using different rounding approaches). I would think
that you have to agree that it is not the goal to expressing answers in
sig figs to lose any numbers that are relevant. From your
perspective, looking at the example I have given, how could you possibly
justify having only one sig fig in your answer? (a reader, 01/16/00)

Your error is in assuming that no precision is lost in the subtraction
part of the calculation. In fact a subtracting two numbers that are almost
exactly equal results in a serious loss of significant digits. This loss doesn't
arise from rounding off intermediate results (which would obviously be
wrong). It arises from the fact that the difference and the uncertainty in the
original measurements are nearly the same size.
Look at it from the perspective of uncertainties.
There's an uncertainty in the tenthousandths place in 1101.0001 and in 1101.0000.
So the first number is 1101.0001 +/ 0.0001. It could be as low as 1101.0000, or
as high as 1101.0002, right?
The second number 1101.0000 +/ 0.0001 could be anywhere from 1100.9999 to 1101.0001.
The difference between the numbers can be anywhere from
1101.0002  1100.9999 = 0.0003 to 1101.0000  1101.0001 = 0.0001.
The tenthousandths place is uncertain in the difference. Clearly the difference
has only one significant figure.
Now do the whole calculation this way, to show that the range of results
reflects one significant digit *without any intermediate rounding*:
The highest possible result is
(1101.0002  1100.9999)/ 0.00424 = 0.07(0755)
The lowest possible result is
(1101.0000  1101.0001) / 0.00426 = 0.02(3474)
So the uncertainty in the final result is in the hundredths place,
and there is only one significant figure. The lesson is
high precision for measurements doesn't guarantee high precision for results calculated from them!
Author: Fred Senese senese@antoine.frostburg.edu 