John Harris, August 27, 2017

THE PROBLEM:  

The significant figures method attempts to give students an appreciation for errors and error propagation in calculations using very simple approximations. However, this method is so crude that it gives students a false sense of the errors in their calculations. The errors can be either much higher or much lower than the method predicts. These problems occur without relying on pathological examples. This is a fundamental issue and clearly we need a better approximation method to teach students. 

Example 1: Suppose I calculate 0.52 + 0.52. By rules of significant figures for addition, this equals 1.04 as the final answer. Now suppose I calculate 0.52 x 2 where the 2 is a constant with infinite precision. From the rules of significant figures for multiplication, the final answer has to be rounded to 1.0 because there can only be two significant figures in the answer. Explain that one to our kids.  

Example 2: Suppose you are told that edge of a cube is 2.0 cm with two significant figures and you ask your students to calculate the volume of the cube. 

After two multiplications, the rules say the volume will is (2.0 cm)^3 = 8.0 cm^3 with two sig figs. So far so good. 

However the two significant figures means that the actual edge length could really be anything from 1.95 cm to 2.05 cm (see note below)

so   1.95^3 = 7.414875

and 2.05^3 = 8.615125

In this problem, the teachers will tell you that you have two significant figures but you have far less in reality. Perhaps there is one significant figure in the results and we should write the answer as a single digit 8 cm^3 but 8 +- 0.5 doesn’t even include the full error range from 7.4 to 8.6. So perhaps there are no significant figures in this answer and it is not even clear how to write the answer down using significant figures. Remember that this calculation only used two multiplications and the significant figures method would fool students into thinking that this volume calculation is much more accurate than it really is.

Example 3: On the other hand, suppose you are given the volume of a cube as 8.0 cm^3 and asked to computer the edge length. Our well-trained students will calculate the edge length as 2.0 cm with two sig figs. However significant figures implies that the volume could actually be anything from 7.95 to 8.05 cm^3.

so   7.95^1/3 is 2.004158016 cm

and 8.05^1/3 is 1.995824623 cm

The “real” answer is 2.00 with 3 significant figures since the error is within +- 0.005. We have created more significant figures in our answer than we had in any of the numbers we were given for the problem. This answer would be marked wrong by the teacher if a student wrote this on an exam. 

I understand exactly why the significant figure rules screw up in these examples—that is not the question. The question is why we teach this nonsense to our kids and what is a better approximation to teach. 

Note 1: There is some confusion as to the error implied by writing a number 2.0 with two significant figures. Is it uniform error +- 0.05 as I assumed? Or is it uniform +- 0.1? Or is it some other distribution? In any case, the overall conclusions above will still hold.

Note 2: What I used in my calculations in Examples 2 and 3 was the so-called “crank three times” method: http://www.av8n.com/physics/uncertainty.htm#sec-crank3  The problem with this is that it does not scale well as you have more and more variable inputs in your calculation. Plus it is also overly pessimistic particularly when the errors between the variables are correlated. 

THE SOLUTION: 

Here is my proposal of how to teach kids: We make calculators (and/or phone apps) cognizant of uncertainty in numbers. Numbers input by default have infinite precision but you can also input numbers with limited precision where you hit a special button immediately after entering the last sig fig. The calculator uses a Monte Carlo analysis and then the result of each calculation comes back looking like this in the display: 3.45356 where the boldface 5 indicates it is the last sig fig in the final answer. The students write down the whole answer indicating the last sig fig with an underline, no rounding required. There is only one number to represent the answer and people can use this number in their calculation without worrying about propagating round off errors since there is no rounding (except for the rounding due to machine precision). The calculator could also automatically introduce this error awareness for ill-conditioned calculations, still giving a result but indicating that the results is low precision (or no precision). This doesn’t fix all the problems but at least the calculator can be more sophisticated about showing the the approximate error level. And we no longer have the problem of saying that something is the “right” answer but don’t dare use the “right” answer in any further calculations. 

© John G. Harris 2017