Say I need to divide a number with a unit by a number without a unit, e.g.: $\frac{16.4\: \text{meters}}{2}$

Following the rules on significant figures, will the correct answer be $8.2$m (since $2$ has no unit), or $8$ m (since $2$ has only one significant figure)?

  • $\begingroup$ It will be 8 m, since the least number of significant digits here is 1 $\endgroup$
    – Bruce Lee
    Commented Feb 8, 2016 at 4:40
  • $\begingroup$ Well it depends on where 2 came from. If this is not a measured quantity it doesn't change the sig figs and would be 2. If it's measured and can only be measured to 1 sig fig, it's 1. Sig figs correspondent to you degree of accuracy in your measurement. $\endgroup$ Commented Feb 8, 2016 at 4:54
  • $\begingroup$ 2 isn't a measured quantity. Does that mean that the correct answer then is 8.2 ? $\endgroup$
    – BlaBlaBla
    Commented Feb 8, 2016 at 5:04
  • $\begingroup$ If the 16.4 doesn't come with any statistical error measure attached, then 16.4/2 is 8.2. You aren't making any mistake that way but you may be throwing significant precision away by rounding. If somebody wants to tell you what the error of a measurement is, then they will do this by specifying it explicitly. $\endgroup$
    – CuriousOne
    Commented Feb 8, 2016 at 5:19

2 Answers 2


It depends on the type of number. If the dimensionless number arises from pure math or counting, then it essentially has infinite precision and doesn't limit the number of digits you report. This is the case for the $2$ relating radius and diameter, or the $1/2$ in the kinetic energy formula, or the $N!$ you often see in statistical mechanics.

$\pi$ is the same, but if you only plug in finitely many digits into your calculator, your are limited by that. Thus $3.14$ has $3$ significant figures, even if you are approximating the exact value of of $\pi$.

Other dimensionless numbers are indeed measured, and thus have limited precision as well. For example, the fine-structure constant or the g-factor related to the electron gyromagnetic ratio are all measured and have uncertainties.

In general the question of having units vs. being dimensionless is orthogonal to the question of being exact vs. having uncertainty.

  • $\begingroup$ If the '2' is unmeasured, do i take into consideration its significant figures? $\endgroup$
    – BlaBlaBla
    Commented Feb 8, 2016 at 5:09
  • $\begingroup$ As the answer says, it depends on where the $2$ comes from. If it is the result of some theory, e.g. in kinetic energy as $\frac{1}{2} m v^2$ then the $2$ is exact and you could regard it as having as many following $0$s as you wish. Similarly, for $\pi$, use as many decimal places as required so that it does not limit the precision. $\endgroup$
    – badjohn
    Commented Dec 2, 2019 at 14:09

Think of it in terms of percentage error.

If a reading is $100 \pm 2$ metres this is $100 \pm 2\%$.

Now divide the 100 metres by 5 to give 20 metres.

The percentage error does not change so the value is $20 \pm 2\% = 20.0 \pm 0.4$ metres.

If you are not given a percentage error then you would have to judge it from the given value.

110 might have the second 1 as significant so approximately $10\%$? Whereas 111 might indicate a $1\%$ error?


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