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Reading this document, when measuring a thing we have $$\text{relative error} = \frac{\text{absolute error}}{\text{value of thing measured}}.$$

Let's say we have a measurement of the height of an object as $45.2m$ with error of $\pm 0.05m$.

At that point our relative error would be $\frac{0.05m}{45.2m}$.

However, if a person who measures heights always rounds his measurement to three significant digits would it change the relative error?

I am not sure if this is correct, but in other words would the answer $45.2m + 0.05m = 45.25m$ rounded to $45.3m$ as the approximate value. After that would th e absolute error be $0.1m$, so that the relative error would be $\frac{0.1}{45.2}$?

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Three significant digit cancellation wont have an effect in this question and the answer will be same. The answer proposed in the question is the right one as 0.05/45.2 relative error.

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