Why does it make sense to round to the smallest significant figure instead to the largest unit?

If I have one measurement made to the nearest millimetre and another to the nearest picometre then why does it make sense to use number of significant figures in multiplication as a measure of precision? Wouldn't the nearest millimetre be the highest possible precision?


It would be a very odd experiment if you're comparing a $\text{mm}$ reading with a $\text{pm}$ reading...

It very much depends on what you want to do. If you want to multiply the two readings together (for whatever reason), then your answer would have units $\text{mm pm}$, and your precision would be given by the significant figures present - for example:

$$2.5572\text{ mm}\times2\text{ pm} \approx 5.1 \text{ mm pm} \text{ (to 2 sf)}$$

(even then perhaps 2 sf is a little generous).

If you want to say add them, then your units would be either $\text{mm}$ or $\text{pm}$, and yes you're right in that case it would not make sense to write something like

$$1\text{ mm} + 1\text{ pm} = 1.000000001\text{ mm}$$

given the precision of the two numbers.

If somehow you had a $\text{mm}$ reading to 10 significant figures, you may be justified in doing the above, but again, a question like this very much depends on context.


My above answer is a little sloppy, in that it does not incorporate errors, which really to answer this question properly you do need.

The way (I use) to calculate addition and multiplication errors are $$A+B=C\Rightarrow(\Delta C)^2 = (\Delta A)^2 + (\Delta B)^2$$ and $$AB = C \Rightarrow\left(\frac{\Delta C}{C}\right)^2 = \left(\frac{\Delta A}{A}\right)^2 + \left(\frac{\Delta B}{B}\right)^2$$

I rewrite the (corrected) example in the comments here...

Say I have a rectangle with sides measured as $1234\pm1\text{ mm}$ (i.e. measured to 4 sf) and $6\pm1\text{ pm}$ (i.e. 1 sf). The area is then (doing proper error analysis) $$1234\pm1\text{ mm}\times6\pm1\text{ pm}=7404\pm1200\text{ mm pm}$$ In other words, the most significant I can write my answer is limited by the least significant measurement (it doesn't matter that it's $\text{pm}$).

The key difference is when combining errors upon adding, you compare their magnitudes (so mm vs pm is important), whereas when multiplying, you compare the fractional errors, which is essentially equivalent to comparing the significant figures.

  • $\begingroup$ Thanks. The example was just arbitrary. I'm bothered in geberal why we should look into the number of sig figures in multiplication when the numbers have different order of magnitude. I see all the time examples like 2x10^(-3) x 1.5x10^(-7) = 3x10(-10) to 1s.f pointing out that the first term has fewer sig figures. Well, this hardly matters here. $\endgroup$ – Eva Oct 9 '18 at 12:40
  • $\begingroup$ Please see my edited answer. $\endgroup$ – Garf Oct 9 '18 at 13:02
  • $\begingroup$ So is the correct answer to the above example 7fm^2 to 1s.f? I know the rules for propagation of uncertainty but for some reason it just doesn't feel reasonable (i.e. intuitive). How can I have 1fm^2 error if my biggest error was 1mm? These have different dimensions indeed but seems illogical to me. $\endgroup$ – Eva Oct 9 '18 at 13:25
  • $\begingroup$ Converting the final answer to one unit (which I agree you should be doing generally) would be $(7.4\pm1.2)\times10^{-12}\text{ fm}^2$ (because $1\text{ pm}=10^{3}\text{ fm}$ and $1\text{ mm}=10^{12}\text{ fm}$). $\endgroup$ – Garf Oct 9 '18 at 13:41
  • $\begingroup$ I think the problem here is you're trying to compare a $\text{fm}^2$ error with a $\text{mm}^1$, which is not so trivial - when you add/subtract, the units don't change. When you multiply, the units do change. $\endgroup$ – Garf Oct 9 '18 at 13:43

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