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B Why we take the least # of significant digits when multiplying?

  1. Aug 16, 2016 #1
    This question just flew right into my head when I was in the mid of a shower :), and I cannot resist finding out the answer. Question: Why do we take the least number of significant digits when multiplying/ dividing, and the least number of decimal places when adding/subtracting? Thank you!
     
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  3. Aug 16, 2016 #2

    Orodruin

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    Have you tried figuring the answer out for yourself? If so, what were your thoughts?
     
  4. Aug 16, 2016 #3
    I do have a few ideas on this topic. I think that significant digits are there to determine the amount of precision of the given data. I believe that the reason why we take the least number of decimal places when adding and subtracting is because we cannot have a result that is more precise than the least precise number(s). For example, when adding 5.1 + 5.20, the answer should be 10.3. It is because the precision (+- 0.1) of the number 5.1 is less precise than the precision of the number 5.20 which is +- 0.01. However, I cannot think of a good reason why we apply the rule of rounding to the least number of significant digits instead round the number to the least decimal places. I would like an answer from an expert like you. Thank you!
     
    Last edited: Aug 16, 2016
  5. Aug 16, 2016 #4

    Orodruin

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    What happens to the errors when multiplying?
     
  6. Aug 16, 2016 #5
    That is a very tough question from you. I just did a quick check and realized that when we multiply the errors, in this case, 1x10^-1 and 1x10^-2. The result is 1x10^-3, which makes it even smaller. That means the result is more precise than it can actually be. Is that the reason why?
     
  7. Aug 16, 2016 #6

    Orodruin

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    I suggest you think of it this way in terms of errors. A number is given by ##a \pm \delta a##. For multiplication purposes, it is more convenient to write this as ##a (1 \pm \delta a/a)##, here ##\delta a/a## is the relative error and essentially gives you an estimate on how many significant digits you should use. If ##\delta a/a## is (roughly) of order ##10^{-n}##, you use ##n## significant digits.

    Adding two numbers gives you
    $$
    (a \pm \delta a) + (b \pm \delta b) = (a+b) \pm \delta a \pm \delta b = (a+b)\left( 1 + \frac{\pm \delta a \pm \delta b}{a+b}\right).
    $$
    Clearly, the larger of ##\delta a## and ##\delta b## here dominates the error and you should use the larger error (the number with the smaller number of decimals has the larger error) to determine how many decimals you can reasonably use.

    Multiplying two numbers gives you
    $$
    (a \pm \delta a) (b \pm \delta b) = ab (1\pm\delta a/a)(1\pm\delta b/b) \simeq ab (1 \pm \delta a/a \pm \delta b/b),
    $$
    where we have assumed that the errors are relatively small so that the product of the errors is negligible. Clearly, the number with the larger relative error determines the error in the product.
     
  8. Aug 16, 2016 #7

    David Lewis

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    Actually you round to the least number of significant figures after multiplying/dividing. During the process itself I believe you keep one or two more significant figures (if you have them) than the least precise multiplicand/dividend/divisor contained in your product/quotient.
     
  9. Aug 17, 2016 #8

    Orodruin

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    You never round your answer until all computations are done, regardless of the operation.
     
  10. Aug 17, 2016 #9

    jbriggs444

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    *gets up from his front porch chair and shakes his cane at the young whippersnappers walking by*. In my day...

    If you are computing with longhand arithmetic, you will want to round off. Each multiplication would otherwise double the number of figures you are computing with and each division would likely result in an infinite string. If you are using a log or trig table in your handy CRC, those results are already rounded.

    Edit: And if you are a whippersnapper, using a calculator or computer, you are almost certainly rounding off anyway. Which can bite you if you are unaware of it.
     
  11. Aug 17, 2016 #10

    Orodruin

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    I think "never" is accurate enough for a B-level thread in an age where people are using calculators and computing numbers with at most three significant digits ...
     
  12. Aug 17, 2016 #11
    A number like 5.5 can be interpreted as 5.5 +- 0.05. So, if we add two numbers,
    3.24 + 5.5
    we are really adding
    (3.24 +- 0.005) + (5.5 +- 0.05)
    Now, if we don't have any additional information about the distributions of the errors, then we just use the rule of thumb where we "absorb" the smaller error into the larger one, and we get
    8.74 +- 0.05
    which is just 8.7

    Now, as an exercise, try the same thing with multiplication.

    By the way, if you do have more information about the distribution of errors, then you probably shouldn't use the rules of significant figures and instead use explicit propagation of errors.
     
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