I am reading the book "Experimentation : an introduction to measurement theory and experiment design" by D. C. Baird and I have come across a paragraph where it says that the absolute uncertainty in an indirect measurement cannot have more significant figures than that of the original direct measurements. Since there are different ways to achieve an estimation of the final uncertainty (and the author has previously explained a calculus-based approach to do so) I wonder why this additional rule regarding significant figures is needed.
I leave the aforementioned paragraph down below. For further clarification, I feel that there is nothing wrong with the expression $R = 9.06 \, \pm \, 0.59 \, \Omega$ because the original uncertainties in $V$ and $I$ have already been taken into consideration when computing the $0.59 \, \Omega$ uncertainty.
Because computations tend to produce answers consisting of long strings
of numbers, we must be careful to quote the final answer sensibly. If, for
example, we are given the voltage across a resistor as $15.4 ± 0.1 $ V and
the current as $1.7 ± 0.1 $ A, we can calculate a value for the resistance.
The ratio $V/I$ comes out on my calculator as $9.0588235 \, \Omega$. Is this the
answer? Clearly not. A brief calculation shows that the absolute
uncertainty in the resistance is close to $0.59 \, \Omega$. So, if the first two
places of decimals in the value for the resistance are uncertain, the rest
are clearly meaningless. A statement like $R = 9.0588235 ± 0.59 \, \Omega$ is,
therefore, nonsense. We should quote our results in such a way that the
answer and its uncertainty are consistent, perhaps something like
$R = 9.06 ± 0.59 \, \Omega$. But is even this statement really valid? Remember
that the originally quoted uncertainties for $V$ and $I$ had the value $±0.1$,
containing one significant figure. If we do not know these uncertainties
any more precisely than that, we have no right to claim two significant
figures for the uncertainty in $R$. Our final, valid, and self-consistent
statement is, therefore, $R = 9.1 ± 0.6 \, \Omega$. Only if we had a good reason
to believe that our original uncertainty was accurate to two significant
figures, could we lay claim to two significant figures in the final uncer
tainty and a correspondingly more precisely quoted value for $R$.
Best Answer
Short story: I assume you have an idea of propagation of errors, then the relative error on the resistance's value is the sum of both relative errors, hence:
$$\Delta R = ((0.1 / 15.4) + (0.1 / 1.7)) \cdot 9.0588235 \Omega \approx 0.59 \Omega$$
So the final answer $R = 9.06\Omega \pm 0.59 \Omega \:$ is perfectly reasonable. You dont need (more over, you shouldn't) to do some obscure magic tricks to manage to get the final answer rounded off to make the decimals fit with the initial raw data. In fact, the decimal digits aren't anything more than a representation of a number, that doesn't mean you should round them off blindly.
Long story: Uncertainties can be a difficult subject if you don't have enough mathematical knowledge. Often, the matter is treated with a very poor focus based on simple but sometimes non-sense rules. I'd recommend you to read this article to get a better understanding. But if this is some sort of homework, ask to your superiors if what that book says is the reciepe they want you to follow and just do that.