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Why is a leading figure of 1 so special in uncertainties?

  1. Oct 11, 2015 #1
    "The uncertainty should be rounded off to one or two significant figures. If the leading figure in the uncertainty is a 1, we use two significant figures, otherwise we use one significant figure. Then the answer should be rounded to match."

    "Here’s a rule of thumb you can rely on: round the uncertainty to one significant figure. Then round
    the answer to match the decimal place of the uncertainty. One exception to the rule of thumb: If rounding the uncertainty to one significant figure would cause that figure to be a 1, then you keep the next digit as well."

    Both the quotes are taken from leading universities such as Harvard, regarding the number of significant figures to keep (in uncertainties in Physics), and they both say the same thing.

    So my question is: Why is a leading figure of 1 so special in uncertainties (in physics) that the said uncertainty deserves two significant figures, as opposed to just one sig fig?
     
  2. jcsd
  3. Oct 11, 2015 #2

    Svein

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    Look at it this way: 2 is 100% more than 1, while 3 is only 50% more than 2 (and progressively less). Thus, if your uncertainty estimate is better than 50% precise, you should include one extra digit if it is somewhere between 1 and 2.
     
  4. Oct 11, 2015 #3
    Hmm I kinda get what you mean, but not entirely clear. Where are you getting the 50% from (...better than 50% precise)?

    Say we get a value of 5.64 with uncertainty 0.73, we write our answer as 5.6+-0.7 because the extra 0.03 (only 4.3% of 0.7) is meaningless if we're already unsure about the 0.7? Do correct me if I'm wrong.

    Now we have a value of 5.64 with uncertainty 0.13, we write our answer as 5.64+-0.13, because the 0.03 is actually 30% of 0.1? Is this what you mean?

    But then what about 5.64 with uncertainty 0.98? According to the rules described we should not round the uncertainty to one sig fig (and cause the figure to be 1) and get an answer of 6+-1. Instead we should leave it as 5.64+-0.98. Why?
     
  5. Oct 11, 2015 #4

    Svein

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    Yes.
    Well, no. The rules are a bit sloppily described. It should be "If rounding the uncertainty down to one significant figure would cause that figure to be a 1, then you keep the next digit as well."
     
  6. Oct 11, 2015 #5

    Svein

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    By the way - here is a list of numbers between 1 and 10 with constant relative precision (rounded to one decimal).

    1.0
    1.3
    1.6
    2.0
    2.5
    3.2
    4.0
    5.0
    6.3
    7.9
     
  7. Oct 11, 2015 #6
    Hey you're reading taylor too!

    Thanks for the answer btw I was actually curious about that rule too.
     
  8. Oct 11, 2015 #7
    What do you mean by constant relative position? Could you kindly clarify?
    I still don't get where the 50% comes from. Thanks.
     
  9. Oct 11, 2015 #8
    Also my example of 5.64 with uncertainty 0.98, is the correct answer 6+-1, or 5.6+-1.0?
     
  10. Oct 11, 2015 #9

    Svein

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    No, sorry. I just did some maths.
    Precision, not position. Every number is about 27% greater than the previous number.
    5.6 ± 1.
     
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