# Experts of "significant digits" please enter...

Discussion in 'Physics & Math' started by kingwinner, Oct 13, 2005.

1. ### kingwinnerRegistered Senior Member

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796
Hello Aer,

When you round 9.7 up to 10, are you using the rule of the number of significant digits for addition & subtraction, which states that "When quantities are being added or subtracted, the number of decimal places (not significant digits) in the answer should be the same as the least number of decimal places in any of the numbers being added or subtracted. " This quote is directly from the following site.
http://www.physics.uoguelph.ca/tutorials/sig_fig/SIG_dig.htm

Because for a - b defined in your post, a & b should both have 2 significant digits after doing the divisions, and thus both a & b should be considered to have 0 deciminal places when subtraction is required, therefore the final answer after the subtraction should also have 0 deciminal places...is that what you mean?

3. ### AerRegistered Senior Member

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2,250
Yes.

In the context of addition and subtraction, you should say a & b both have significant digits up to the 0th decimal place (hmm, I don't think this is the correct mathematical terminology but anyway..) so when you round 9.7 up to 10, both digits are still significant since they are before any decimal values.

5. ### kingwinnerRegistered Senior Member

Messages:
796
How about this? What would be the correct number of deciminal places in the final answer? Is it 8, with no deciminal places?

(100/4.1) - (100/6.1)

7. ### espRegistered Senior Member

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908
DaleSpam wrote...
Unless of course, you're a professional mechatronic engineer working with the wavelength of light to obtain laser dot gain on a sheet of pure aluminium, in which case, x10^-4, 5 or 6 is very useful.

You may find that some engineers want to work quickly.
But lots of engineers want to be precise. And very few want to please their professors.

Last edited: Oct 24, 2005
8. ### D HSome other guyValued Senior Member

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2,257
This is wrong. The mistake here was to omit the required decimal places in the intermediate results.

James, there are a lot of errors in this. For example,
You usally can't simply compute (say) 100.5/4.15 and 99.5/4.05 to give 23.976 to 24.815. This biases the result for most realistic distributions. For example, if the distance and velocities all have log normal distributions, the ratios also have log normal distribution, with the mean equal the ratio of the means. I.e.,
(100 +/- 0.5) km/(4.1 +/- 0.05) km/sec =~ (24.4 +/- 0.3) km/sec
This calculation exagerates the error. For example, the error resulting from the adding (or subtracting) two normal distributions is the square root of the sum of the squares of the original errors (root sum square or RSS for short).

Finally, If the answer was indeed
it would be better to report the answer as 10 rather than 9.

This too is wrong. Minor mistake: there are too many decimal places in the intermediate results. The big mistake was rounding up 9.7 to 10.

100/4.1 = 24.4 (intermediate answer must maintain at least the tenths)
100/6.8 = 14.7 (ditto)
difference = 9.7 (almost there)
Finally, round 9.7 to two significant digits, resulting in 9.7.

Significant digits calculations play a bit fast and loose with statistics. A careful engineer would give an answer of 9.7 (two significant digits) to the above problem. An even more careful statistician would be more likely to give an answer of 9.7 +/- 0.34, or just 10, with one significant digit). Subtraction often loses precision.

9. ### AerRegistered Senior Member

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2,250
I agree that 9.7 is a better answer than 10. However, I am unaware of any special rules for subtraction as apposed to the rules for addition. Also, concerning "too many decimal places", that makes no difference in the end result, I just copied and pasted the whole lot from the computer.

Also, I believe you are correct when you say subtraction looses precision, but for the matter at hand, I do not believe 9.7 would be correct (remember that we are dealing with elementary significant figure calculations).

10. ### D HSome other guyValued Senior Member

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2,257
The easiest way to do significant figure calculations is to ignore the constraint until the final answer and apply the constraint at that point:

100/4.1 - 100/6.8 = 6750/697 = 9.68436154949785
= 9.7 to two significant digits, not 10 .​

The pre-calculator technique is to retain the decimal places during the intermediate calculations:

100/4.1 - 100/6.8 = 24.4 - 14.7 = 9.7
= 9.7 to two significant digits, not 10 .​

Rounding 9.7 up to 10 is an error. 10 has but one significant digit.

There are no 'rules' for subtraction versus addition. But consider
100/4.1 - 100/4.2​
for which 0.6 is a 'better' answer than 0.58.

11. ### AerRegistered Senior Member

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2,250
That is the shortcut method. Or rather, you are only applying the rules of multiplication and ignoring the rules for addition and subtraction.

That is the method I always use, however, I do not believe it is the method that is correct for kingwinner's class. At any rate, only kingwinner's teacher/professor has the final say.

12. ### James RJust this guy, you know?Staff Member

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37,193
D H:

I've been thinking about this, and I agree that 10 is a better answer than 9. Whether 9.7 is a better answer than 10 depends on factors such as whether we take 100 as having 1 or 3 significant figures, which is ambiguous.

You are completely correct that using significant figures this way ignores complexities of possible distributions and so on, but no such distributions were specified for the particular problem given. Therefore, we make do as best we can with what we're given.

And yes, my error estimates are most probably over-estimates. Better to overestimate uncertainty than to underestimate it, though, right?

13. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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23,198
Perhaps not on thread but interesting to some. Perhaps on thread and with two meanings of "significant":

A college of mine (last name Newton also) wrote a whole book on Ptolemy’s rounding of data in the Almagesto. No original Greek copies of which exist, so I am not sure Newton is correct in claiming, "Ptolemy was a fraud." (Which is approximately the name of his book.)

Actually, all Newton convincingly shows is that most of the observations reported in the Arabic translations that do survive, were not made by Ptolemy in Greece, but were converted to appear as if they were. (Perhaps in the original Greek Almagesto, Ptolemy gave credit to the astronomers who did make the observations, but this got lost in translations and now only appears as if Ptolemy claims it all.)

Newton was interested in empirical data on the rate the day is growing longer, so he became interested in the old astronomical observations, especially in which cities a total eclipse of the sun was observed. That is why he became an expert on Ptolemy (and Chinese astronomer's data).

Newton’s statistical analysis of the least significant figures of Almagesto data showed that it is NOT original Greek data, in most cases, but has been converted from some other observation site.

Ptolemy again rounded the original data AFTER it had already been rounded in the true original data. This second rounding (and sometime also transformed for use of different units) is in the results of the calculations used to convert to "Athens observations". (Certain least significant digits reported in the Almagesto are rare of absent. - I forget all the details - if interested find Newton's book.)