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If $a,\ b$ and $c$ are real numbers and you are required to find $a + b + c$ to $3$ significant figures, to how many significant figures could $a,\ b$ and $c$ be rounded up to to give the result?

kimtahe6
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Jojoba
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2 Answers2

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Each number $a$,... represents a range of possible true values $[a-\mu,a+\mu]$. In the sum of three terms, the width of the range changes thus to $3μ$. Now you want that $3μ<5\cdot 10^{-k-1}$ where $k$ is the number of valid digits...

Disclaimer: Per the other answer and comments in the duplicate question (https://math.stackexchange.com/questions/687358/rounding-up-to-significant-figures), above is only valid if a,b,c have about the same magnitude.

Community
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Lutz Lehmann
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In general, if you want $\mathrm{round}_3(a+b+c) = \mathrm{round}_{3+k}(a)+\mathrm{round}_{3+k}(b)+\mathrm{round}_{3+k}(c)$ you should make $k$ as large as you possibly can.

A simple example is $a=499001,b=499.001,c=0.499001$.

  • If you don't do any rounding before addition you get $\mathrm{round}_3(a+b+c) = \mathrm{round}_3(499001 + 499.001 + 0.499001) = \mathrm{round}_3(499500.500001) = 500000$.

  • If you round $a,b,c$ before addition with $k<3$ you get $\mathrm{round}_3(a+b+c) = \mathrm{round}_{3+k}(499000)+\mathrm{round}_{3+k}(499)+\mathrm{round}_{3+k}(0.499) = \mathrm{round}_3(499499.499)=499000$

This non-intuitive behaviour results from the fact that rounding functions "jump", i.e. they are unstable at the point where they round up or down.

TooTone
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