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I am writing a report and I'm stuck on determining how to write the uncertainty.

I have this:

$$1.0 \times \left( 0.1000 \pm 0.0005 \right).$$

If I multiply the values, I need to keep two significant figures, but then my uncertainty would be beyond the correct amount of decimal points. What would I do in this case?

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  • $\begingroup$ See here for an explanation why $\pm$ is discouraged by several bodies. $\endgroup$ – mhchem Jun 19 '17 at 11:31
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When a number is given without any further information, it is generally interpreted so that the last digit is rounded.

Therefore, the first given value $a=1.0$ is generally assumed to represent a value between $0.95$ and $1.05$, or $a=\left(1.00\pm0.05\right)$. Thus, the assumed uncertainty of $a$ is $u(a)=0.05$.

Conversely, the second given value $b=\left(0.1000\pm0.0005\right)$, where the assumed uncertainty of $b$ is $u(b)=0.0005$, might simply be written as $b=0.100$.

When multiplying numerical values, according to the simplified rules of significance arithmetic, the result is rounded to the number of significant digits in the factor with the least significant figures. Therefore, when you multiply $a=1.0$ (two significant digits) by $b=0.100$ (three significant digits), the result is $\boldsymbol{y=a\cdot b=0.10}$ (rounded to two significant digits). This result may be estimated to represent a value between $0.095$ and $0.105$, or $y=\left(0.100\pm0.005\right)$; i.e. the assumed uncertainty of $y$ is $u(y)=0.005$.


By way of comparison, a generally better result for the uncertainty $u(y)$ may be obtained by using the rules of propagation of uncertainty (or propagation of error) as follows.

When calculating the product of $a$ and $b$

$$y=a\cdot b$$

and if we assume that the uncertainties of $a$ and $b$ are not correlated, the relative uncertainty of the result $y$ may be estimated as

$$\begin{align} \left(\frac{u(y)}{y}\right)^2 &= \left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2 \\[6pt] \frac{u(y)}{y} &= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2} \end{align}$$

Thus, the absolute uncertainty of $y$ is

$$\begin{align} u(y) &= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2} \cdot y \\[6pt] &= \sqrt{\left(\frac{u(a)}{a}\right)^2 + \left(\frac{u(b)}{b}\right)^2} \cdot a \cdot b\\[6pt] &= \sqrt{\left(\frac{0.05}{1.00}\right)^2 + \left(\frac{0.0005}{0.1000}\right)^2} \times 1.00 \times 0.1000\\[6pt] &= 0.005024938\\[6pt] &\approx 0.005 \end{align}$$

The same result may be obtained using the general formula

$$\begin{align} u(y) &= \sqrt{\left(\frac{\partial y}{\partial a}\right)^2 u^2(a)+\left(\frac{\partial y}{\partial b}\right)^2 u^2(b)} \\[6pt] &= \sqrt{b^2\cdot u^2(a)+ a^2\cdot u^2(b)} \\[6pt] &= \sqrt{(0.1000)^2\times (0.05)^2+ (1.00)^2\times (0.0005)^2} \\[6pt] &= 0.005024938\\[6pt] &\approx 0.005 \end{align}$$

Thus, in this case, the result that is obtained by using the simplified rules of significance arithmetic is a good approximation of the result that is obtained by using the rules of propagation of uncertainty.

(Note that the numerical values of the estimate $y$ and its uncertainty $u(y)$ should not be given with an excessive number of digits, although in some cases it may be necessary to retain additional digits to avoid round-off errors in subsequent calculations.)

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