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We have these rules:

  1. Can't quote answer to more significant figures than original data.
  2. Error quoted to one significant figure unless first significant digit is a one. In which case quote error to 2 significant figure.

But with original data to 3 significant figure, I get the answer 13.15(6).

i.e. 13.15 +/- 0.06

How do I quote this to 3 significant figure without breaking the rules?

If I round the error to 1, I still have to have a trailing 0:

13.20(10)

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Ignore the first rule in this case and just say $13.15 \pm 0.06$. Were you given a rationale behind the rule? –  NowIGetToLearnWhatAHeadIs Nov 22 '13 at 0:20
    
The rationale behind the rule is that the answer can't be more precise that the original data. As an extreme example: if I say the acceleration due to gravity is 10 m^s^2, and an object has fallen for 2.12456s it's not sensible to calculate the distance fallen as 22.5688m as our value of the acceleration is not that precise. –  Cam Nov 27 '13 at 22:08
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1 Answer

Simple, your second rule:

  • Error quoted to one significant figure unless first significant digit is a one. In which case quote error to 2 significant figure.

Mmm.. That just isn't true. If you look at papers that were accepted by the American Journal of Physics (for example), you would find many counterexamples to your second rule.

You must use common sense, that's all.


  • CounterExample:

    counterexample

    I hope you believe me. =)

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The article is paywalled –  lionelbrits Nov 22 '13 at 0:11
    
Oops! Sorry! I didn't remember I have a subscription. Now I edit that part. –  Wikitys Nov 22 '13 at 0:17
    
Thanks, I believe you. I'm not saying this is a general rule though, just that it's a rule explicitly stated in the task we've been given. –  Cam Nov 27 '13 at 21:58
    
Ok! I understand. Then, if you can't brake the second rule, you are in trouble. XD –  Wikitys Nov 27 '13 at 22:26
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