I do understand the basic principle of significant figures which is to indicate how accurate the given result is.

However I can't get it into my head why it should be a good idea to actually use them (Bare with me for a moment).

If considering the following mesaurements: $$ m = 0.25~\text{kg} \hspace{2cm} v = 0.25~\frac{\text{m}}{\text{s}} $$ Each of the measurements has 2 significant figures.

If I know want to calculate the momentum of the object (assuming above measurements are from the same object) I'd do $$ p = 0.25~\text{kg} \cdot 0.25~\frac{\text{m}}{\text{s}} = 0.0625~\frac{\text{kgm}}{\text{s}} $$ As can be seen the calculated result now has 3 significant figures. If respecting the rules of significant figures one would have to to say $p=0.063~\frac{\text{kgm}}{\text{s}}$ (which by the way would forbid the use of the equals sign which is always ignored aas far as I have seen).

So in terms of showing to others that the third decimal place is uncertain one actually makes a not so precice result (due to the measured data) even worse by rounding it.

Also if one would need to perform further calculations that include the calculated momentum wouldn't the very same kind of rounding error propagate through the whole calculation ultimately ruining the end-result completely (assuming that the calculation has many intermediate results that are all rounded to the respective significant figure)?

To sum it up: In my opinion it would only make sense to use all digits one can get for calculate the result. After that one would need to perform a error calculation and then after having calculated the error of the result one can round the result because the result is given with the calculated error (e.g. $p = 0.063 \pm 0.001~\frac{\text{kgm}}{\text{s}}$).

But simply using these "magically" significant figures for reasoning that certain digits have to be discarded doesn't seem logical to me.

Furthermore doesn't a calculated result lose all its meaning when given without its error? So why do we need significant figures so that others don't get confused (taking the result as more accurate than it is)?

Or are significant figures some sort of lazy rule of thumb if one doesn't want to calculate the error of one's result?

  • $\begingroup$ More on significant figures. $\endgroup$
    – Qmechanic
    Commented Dec 19, 2017 at 17:35
  • $\begingroup$ That only seems to support my thesis ^^ If they confuse this many people they seem to do more harm than good... $\endgroup$
    – Raven
    Commented Dec 19, 2017 at 17:54
  • 1
    $\begingroup$ ^^ Related to my answer below. I would submit the problem is people saying "0.0625 is wrong". The appropriate evaluation is "0.0625 is unnecessary, 0.063 is enough". So they are not confusing unless people take them to be literal truths; they are guidelines. $\endgroup$
    – levitopher
    Commented Dec 19, 2017 at 18:02
  • $\begingroup$ For one thing, in actual practice it is correct to keep all digits until the very end of the calculation, then round the end result to the proper number of sig figs. Doing so at intermediate steps will lead to unnecessary errors $\endgroup$
    – RC_23
    Commented Apr 5, 2023 at 23:00

2 Answers 2


First of all: always rounding '5' up will lead to a systematic bias (use a random number generator to convince yourself that this is so). To mitigate this, round '5' to the nearest appropriate even digit.

Assuming your 0.25 was actually some number on (0.245, 0.255), uniformly distributed, then the statistical uncertainty is $0.01 / \sqrt{12}$. (also verifiable with a uniform RNG).

Then, using the chain rule on $p=mv$ and adding the terms in quadrature (independent errors):

$$(\delta p)^2=(m\delta v)^2 + (v\delta m)^2$$

Ignoring units (for now, out of strict laziness):

$$(\delta p)^2=2(\frac{1}{4}\frac{1}{100 \sqrt{12}})^2=\frac{1}{6} \frac{1}{400^2} \approx 10^{-6}$$


$$\delta p \approx 0.001 \ \mathrm{kg\cdot m/s}$$

and you can safely claim:

$$p = 0.0625(10)\ \mathrm{kg \cdot m/s}$$

where the $n$-digits in parenthesis represent the uncertainty in the last $n$ digits.

  • $\begingroup$ This is exactly the kind of logic which leads to thinking that significant figures actually suggest something about confidence intervals. If I make a measurement of a quantity with some underlying uncertainty $\sigma$, I can choose to report fewer or more digits depending on my measurement device. These quantities will not generally be the same, and conflating them is dangerous. $\endgroup$
    – levitopher
    Commented Dec 20, 2017 at 0:58
  • $\begingroup$ @levitopher The logic follows from the pedagogical "Assuming" starting the 2nd paragraph. The only people whom I've ever heard misuse sig figs are news media, when, for instance they're told the fire is 10 square km, and they report it at 3.8681 square miles, or worse: 6.2137 square miles. $\endgroup$
    – JEB
    Commented Dec 20, 2017 at 2:09
  • $\begingroup$ Here is a list of 10 normally-distributed numbers: 2.66, 2.38, 2.08, 1.05, 1.70, 1.33, 2.05, 2.62, 1.94, 1.79, which could represent measurements I made using a ruler. You have no control over what digits I report, of course, but I have reported the sig figs to the best of my ability. The underlying standard deviation is 0.5, the average is 1.96 and the error on the population average is 0.165. Reporting 1.96 +/- 0.165 is pretty nonsensical, 1.96 +/- 0.16 is better, 2.0+/- 0.2 is probably best (I can include citations if you like, but I think this attitude is reasonably agreeable). $\endgroup$
    – levitopher
    Commented Dec 20, 2017 at 4:34
  • $\begingroup$ (Not that I disagree with you about people generally don't screw this stuff up. I'm arguing that sig figs do not tell us anything about the uncertainty in the measurement, or if they do it's very superficial. This is illustrated by my example; the sig figs on 1.96 have no bearing on the actual uncertainty on the measurement) $\endgroup$
    – levitopher
    Commented Dec 20, 2017 at 4:35

I agree with nearly everything you said - and double emphasize that writing $p=0.063$ kg m/s is a violation of the equal sign. When I finish a calculation, I always use $\simeq$ or something similar, and I believe that's the best practice to follow. However, significant figures give us a set of rules to use when dealing with numbers that allow us to set numbers equal to each other in ways which cannot set variables equal to each other.

But I would like to point one thing that you kinda danced around - errors are not the same as significant figures, and we should not expect them to do the same job. If you measure something, you should always give the error on the measurement, and not trust the significant figures to do that for you. When you do a calculation using measured values, you should always propagate the errors through the calculation (or use some other method if you have a population, for instance).

Significant figures do not tell you how accurate a particular number is. When you say "$m=0.25$ kg", you are telling me that's an absolutely perfect number. If I multiplied this by $0.999999999$ m/s (as a perfect number), is the answer really $0.2499999998$ kg m/s? Well maybe, but given that I started with 0.25 kg, the context is such that 0.25 kg m/s is clearly the answer (even though I should probably write $\simeq 0.25$ kg m/s).

Significant figures tell you which digits are significant in a particular situation, which allows you to use the equal sign in a way which we all understand. They are guidelines (so that we don't have to have this conversation every time we write a number), not estimates of error or uncertainty.

  • 1
    $\begingroup$ This guideline-part is actually the part that is left out when speaking about significant figures. But as guidelines this suddenly makes sense! Also one more question: you said that there is a way to set numbers equal to each other in a way variables can't. Could you please give an example for that? I don't think I get what you try to say with that. $\endgroup$
    – Raven
    Commented Dec 19, 2017 at 18:12
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    $\begingroup$ If $m=0.25$, $v=0.999999999$ , and $p=mv$, then $p\neq 0.25$ even though $0.2499999998 =0.25$ (again, I should probably be using $\simeq$ to fix all this). Basically, equal signs with variables should be absolute, but with numbers we are implicitly only including the ones significant in the present context. $\endgroup$
    – levitopher
    Commented Dec 19, 2017 at 18:23
  • $\begingroup$ Another quick comment: I agree, we should not be presenting these as any kind of mathematical rules. Yes, there are good mathematical reasons for doing it and thinking about it, but point of the guidelines are to allow us to be contextual with numbers. $\endgroup$
    – levitopher
    Commented Dec 19, 2017 at 18:25
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    $\begingroup$ No number in physics is ever exactly equal to another number, so no, it doesn't make sense to use the $\approx$ symbol just because we used correct sig figs, nor does it make sense to think of the $=$ symbol as exact just because we used incorrect sig figs. $\endgroup$
    – user4552
    Commented Dec 19, 2017 at 20:33
  • $\begingroup$ @Ben Crowell: I mean, you should probably be careful with saying "No number in physics..." because we have quantization. The energy of two electrons in the same orbits in two different identical atoms satisfies $E_1=E_2$, because all approximate numbers will cancel out leaving just integers. But of course, I generally agree with you. $\endgroup$
    – levitopher
    Commented Dec 20, 2017 at 4:44

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