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How can I write the following without \pm?

(66 ± 4.4) s

I can achieve the same result with \SI[separate-uncertainty=true]{66 \pm 4.4}{\second}, but I thought the plus-minus symbol could be inserted automatically by using multi-part-units. Here's what I tried:

\SI[separate-uncertainty=true,multi-part-units=brackets]{66(4.4)}{\second}
\SI[separate-uncertainty=true,multi-part-units=brackets]{66(44e-1)}{\second}
\SI[separate-uncertainty=true,multi-part-units=brackets]{66(44)}{\second}
\SI[separate-uncertainty=true,multi-part-units=brackets]{66.0(4.4)}{\second}

The problem seems to be getting the decimal in there. I can't seem to find the relevant information in the siunitx documentation.

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  • Your input implies an additional place of accuracy in the main number: \SI[separate-uncertainty=true]{66.0(44)}{\second}. Would this be an acceptable answer?
    – Joseph Wright ♦
    Mar 12, 2014 at 19:07
  • @JosephWright Indeed, 66.0(44) produces the desired output. I'll have to figure out how precision is handled.
    – Cerran
    Mar 12, 2014 at 19:14
  • The logic in siunitx here is that 66 \pm 4.4 doesn't make sense, as the implied accuracy of the main value can be no better than integer level, whereas 66.0 could be \pm 0.1.
    – Joseph Wright ♦
    Mar 12, 2014 at 19:39
  • @JosephWright Ah, I think I see now. Since the precision of the uncertainty is tied to the precision of the main value, there is no need for the uncertainty to specify the position of the decimal point. Thanks for the solution and the explanation.
    – Cerran
    Mar 12, 2014 at 19:52

1 Answer 1

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The 'short' uncertainty format understood by siunitx is based around the idea that the magnitude of the bracketed value is relative to the number of figures in the main value. Thus \num{66(44)} is equivalent to \num{66 \pm 44}, while \num{66.0(44)} is the same as \num{66.0 \pm 4.4}. Notice the additional significant figure in the second pairing.

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