# Sig figs/certainty in percent error calculation

Hello, all. Quick question about how to apply sig figs in percent error problems. Eg. If the actual/target value is 1.95 g and we measure 1.87 g, then should the percent error of our measurement be reported as 4.10% or 4%? Normally, at least abstracting from the problem, after subtracting 1.95 from 1.87 for a difference of -0.08, the zeros would not be considered significant and we would consider this result to have one significant digit, but really, we have certainty about those zeros as values, correct? (So, I'm leaning towards saying three sig figs for the purpose of the subsequent division problem by 1.95. -0.08/1.95 = -4.10% and not -0.08/1.95 = -4%.) Does this seem like an appropriate interpretation of the situation to any of you, or am I forgetting something?

I know the rule technically says to count "placeholding" zeros as non-significant, but here it seems to me they are reporting data about the measurements used in the initial calculation, not just placeholding. Many thanks in advance for your help.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Normally, at least abstracting from the problem, after subtracting 1.95 from 1.87 for a difference of -0.08, the zeros would not be considered significant and we would consider this result to have one significant digit, but really, we have certainty about those zeros as values, correct?
Incorrect. Leading zeros are never significant. Only trailing zeros may be. Significant digits are not really about how many of the digits you have confidence in, it is about relative precision. If you have three significant digits and change the last by one, then the result should not differ more than 1% from your original number.

It is easier to think about significant digits in scientific notation ##x\cdot 10^y##.

As another example, you can write 143 as 0000000000143. It does not change the fact that there are three significant digits even if you know that the number is not 1000000000.

Please help me understand how this applies to this situation, for both the initial subtraction of 1.95-1.87 and the subsequent division by 1.95.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
1.95-1.87 = 0.08, which is one significant digit.

1.95-1.87 = 0.08, which is one significant digit.
And that precision is relative to? The precision of the original measurement which is to the hundredths place? Is that your meaning? That the precision of the result is of equal degree as the precision of the measurement?

mfb
Mentor
These are absolute numbers. Your initial measurements have an uncertainty of 0.01, the difference has a similar uncertainty.

0.01 is 1/8 of 0.08, so the relative uncertainty of your difference is of the order of 10%.

Derrick Palmiter
gmax137
Look at the max and min for the difference:
$$\begin{array}{|c|c|c|} \hline nom&max&min \\ \hline 1.950&1.955&1.945 \\ \hline 1.870&1.865&1.875\\ \hline 0.080&0.090&0.070 \\ \hline \end{array}$$

WWGD and Derrick Palmiter
Can anyone recommend a good text/reading material that explains this concept in a bit more detail than a standard high school physics/chemistry textbook would? Many thanks to all involved for your assistance.

Orodruin
Staff Emeritus