Absolute Error Confusion I don't understand the following concept. It was given by the textbook.
The dimensions of a cylinder are measured to the nearest millimeter using a measuring tape. The circumference is measured to be 22.0 cm and height measured to be 15.0 cm. Using these measurements to (a) estimate the volume of the cylinder, and (b) estimate an upperbound for the percentage error in your answer.
I understand part a). We simply substitute the values given to 'estimate' the volume of the cylinder.
For part b) the working out was,
$$\Delta V \approx \frac{\partial V}{\partial C} \Delta C + \frac{\partial V}{\partial h} \Delta h$$
$$= \frac{Ch}{2\pi}\Delta C + \frac{C^2}{4\pi}\Delta h $$
$$= \frac{165}{\pi}\Delta C + \frac{121}{\pi}\Delta h$$
Taking the absolute value of both sides
$$|\Delta V |\approx |\frac{165}{\pi}\Delta C + \frac{121}{\pi}\Delta h|$$
$$\leq\frac{165}{\pi}|\Delta C| + \frac{121}{\pi}|\Delta h|$$
I understand this because the triangular inequality was applied. But for the next step is what I don't understand.
It says The absolute error in each measurement is at most 0.5 mm, which is 0.05 cm. Then,
$$\Delta C \leq 0.05, \Delta h \leq 0.05$$
Where did the 0.05 cm come from? It seems to me it was pulled from the air, am I missing something? Thanks.
 A: As the measurements are taken to the nearest millimetre, the largest absolute error that can be made in the measurement is $0.5$mm. If there was a larger absolute error, the measurement would not have been rounded to the nearest millimetre.
Explanation: Suppose an object is measured and, to the nearest millimetre, it has length $k$mm where $k \in \mathbb{N}$. If the absolute error is $\varepsilon$, then the true length of the object is $k-\varepsilon$ or $k + \varepsilon$. If $\varepsilon > 0.5$, then $|(k - 1)-(k-\varepsilon)| = |\varepsilon - 1| < 0.5$, so $k - \varepsilon$, to the nearest millimetre, is $k - 1$, not $k$. A similar calculation shows that if $\varepsilon > 0.5$, $k + \varepsilon$ to the nearest millimitre is $k + 1$, not $k$. Therefore, if measurements are made to the nearest millimetre, the absolute error in the measurement (i.e. $\varepsilon$) is at most $0.5$mm.
Note, this has nothing to do with millimetres at all. If a measurement is made to the nearest unit, then the absolute error in that measurement is at most half of a unit.
