1.4: Significant Figures

Suppose I tell you that one stick is 1.0 meters long, and that it is 4.7 times longer than another stick. How long is the second stick? Writing the words as equations (see previous section), you might write:

\(\ l_{1}=4.7 l_{2}\)

\(\ l_{1}\) is what you know (1.0 meters), and \(\ l_{2}\) is what you’re looking for, so solve the equation for \(\ l_{2}\):

\(\ l_{2}=\frac{l_{1}}{4.7}\)

plug in the numbers and solve for the answer:

\(\ l_{2}=\frac{1.0 \mathrm{~m}}{4.7}=0.212765957447 \mathrm{~m}\)

That answer is wrong! Why? Because it is expressed with too many significant figures.

Think about the original problem. I told you a stick was 1.0 meters long. Notice that I didn’t say 1.00 meters long; only 1.0 meters long. That means that I was only willing to commit to knowing the length of the stick to within a tenth of a meter. It might really be more like 1.04 meters long, or perhaps 0.98 meters long, but I’ve rounded to the nearest tenth of a meter. Since I only know the length of the stick to about ten percent, and since I used that number to calculate the length of the second stick, I can’t know the length of the second stick to the huge precision that I quote above— even though that is the “right” number that my calculator gave me. Given that I only know that the first stick is 1.0 meters long, and it is 4.7 times the length of the second stick, all that I can say I know about the length of the second stick is:

\(\ l_{2}=0.21 \mathrm{~m}\)

By saying this, I’m implicitly saying that I don’t know the length of this second stick to better than the hundredths place. . . and I don’t! Implicitly, I’m saying that I know the length of the second stick to about one part in 21. That’s actually a bit better than I really know it (which is just to one part in 10, or to 10%, as that’s all the better I know the length of the first stick), but this is the best you can do with just significant figures. (To do better, you have to keep track of not just units, but also uncertainties on every number. Doing so is an important part of the analysis of data in physics experiments. However, propagating uncertainties is beyond the scope of this course.)

How well you know a given number you write down is the reasoning behind significant figures. The basic idea is that you shouldn’t report a number to more significant figures than you know are right. The rules can sometimes seem arbitrary, but if you think about them in terms of the basic idea behind them, they can start to make sense. There are four basic rules of significant figures:

1. When multiplying or dividing numbers, the answer has as many significant digits as that member of the product or quotient that has fewer significant digits. So, if I multiply 3.14159 by 2.0, the answer is 6.3; I round the answer to two significant figures, because 2.0 (the member of the product with fewer significant figures) only has two. This rule is an expression of the percent uncertainty in the figures that are going into your result. If you only know a number to within (say) 5%, then you will generally only have two significant figures on that number. You can’t know the result of anything you multiply or divide by that number to better than 5% either, so the result won’t have more significant figures than the number that went into it.

   Sometimes, it makes sense to report your result to one more or one less significant figure than what went into the calculation. This will make sense if you understand the “percent uncertainty” reasoning behind the number. For instance, if I tell you one stick 95 meters long, and another stick is exactly 1/9 as long as the first stick, the significant figure rule would suggest that you should only keep two figures, and report the answer as 11 meters long. However, the two significant figures on the first number means that you know it to about one part in 95. It would be better to report the answer as 10.6 meters long, since a result that is implicitly good to one part in 106 is much closer to your true precision than a result that is implicitly good only to one part in 11.

2. When adding or subtracting numbers, the answer is precise to the decimal place of the least precise member of the sum. If I add 10.02 meters to 2.3 meters, the answer is 12.3 meters. The second number was only good to the first decimal place, so the sum is only good to the first decimal place. Notice that the number of significant figures here is different from either number that went into the sum. When multiplying, it is the number of significant figures that is important; when adding, it is the decimal place that is important.

   Note that if I were to add 10.02 meters to 2.30 meters, the answer would be 12.32 meters; in this case, both members of the sum are significant to the hundreds place. It is possible to gain significant figures doing this. If you add 6.34 meters to 8.21 meters, each significant to three figures, the result is 14.55 meters, now significant to four figures.

   This rule makes sense again if you remember that significant figures represent the precision of a number. To what decimal place do you know all the things that you are adding or subtracting? You can’t know the result to better than that decimal place.

3. A number which is exact should not go into considerations of significant figures. For example, suppose you’re doing a unit factor conversion, and you multiply by the factor (12 in/1 ft). Your answer need not be limited to two significant figures because of this; there are exactly 12 inches in one foot. That’s a definition; there is no uncertainty associated with it. In the first rule above, when I told you that the second stick was exactly 1/9 as long as the first stick, the 9 in 1/9 was a “perfect” number: you were told it was exact. Thus, that there is only one significant figure in the number 9 did not come into consideration for the number of significant figures in the answer.
4. Always keep at least two or three more figures during intermediate calculations than you will report as significant figures in your final answer. This is one of the two most common mistakes I observe in student work. (The other is thoughtlessly reporting your answer to however many digits your calculator gave you.) Otherwise, “round-off” errors will accumulate, and you may get the final answer wrong even though your general method and equations were correct. Consider, for example, summing the numbers 6.1 and 5.3, and multiplying the overall result by 4.1. The sum will be good to the first decimal place, and the final number will only be good to two significant figures because of the two significant figures in 4.1. The result is:

   \(\ 5.3+6.1=11.4\)

   \(\ (11.4)(4.1)=46.74=47 \text { to two sig figs }\)

   If, however, you round too soon, and don’t keep the .4 at the end of the 11.4:

   \(\ (11)(4.1)=45.1=45 \text { to two sig figs }\)

   In fact, you’re now wrong! Even though both 11 and 4.1 are good to two significant figures, your result is incorrect to two significant figures. This is an example of “roundoff” error, where you lose precision by rounding numbers too soon.

You don’t always have to get the number of significant figures exactly right. Significant figures are, after all, just an approximation of correctly taking into account and propagating your uncertainties, which is a topic that those who do more advanced studies in physical science will have to address. Just be reasonable, and make sure you understand the rationale behind why an answer might have a limited number of significant figures. It will often be acceptable to report an answer to one too many significant figures. However, it is technically incorrect to report a number that obviously has too many significant digits; in that case, you’re misrepresenting your knowledge. By the same token, don’t report a number with too few significant figures either, as in that case you’re underselling what you know!