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"It is not transparent to me why this definition is sensible." Because if I have a six digit meter and I measure 5.0000 volts I know that it is 5.0000 volts. I

"It is not transparent to me why this definition is sensible." Because if I have a six digit meter and I measure 5.0000 volts I know that it is 5.0000 volts. I don't see why that is confusing at all. "Here, yyy has correct one and three significant digits relative to xxx, but incorrect 2 significant digits, since the truncation at x2x_2x2 and y2y_2y2 do not agree even to the first significant digit. " This doesn't make sense. Y has the correct one? I think you might be confused about the purpose of significant digits. SD are intended to be paired with a tolerance. If there is no tolerence quote with the number, it is meaningless to talk about significant digits. You arbitrarily defined the tolerance as +/0.5 (half a unit). That is not always true, almost never true. Consider test equipment, not every multimeter is going to be +/0.5mV, some might be +/-.3uV. It varies.

leephillips 6y ago
Yes, I think all we've learned from this article is that the author never learned about significant digits.
PhaseLockk 6y ago
Exactly this. It may be useful for OP to consider what the hypothetical opposite of significant digits would be: insignificant digits, ie. digits that tell you nothing. A leading zero can be omitted from a number without any loss of information. A trailing zero, when correctly used in a position that is not below the tolerance, gives you significant information. On the other hand, if you start combine numbers with different tolerances, but do not truncate the result, you will have a bunch of trailing digits that are meaningless in practice because they are below the combined tolerance.
wsh 6y ago
I agree that it’s better to state the uncertainty [1] together with the measurement, but when this isn’t done, there is a standard convention for significant digits, explained in ISO 80000-1:2009 [2], subclause 7.3.4: When a number is given without any further information, it is generally interpreted so that the last digit is rounded with a rounding range equal to 1 in the last digit (see Annex B). Thus, for example, the number 401 008 is generally assumed to represent a value between 401 007,5 and 401 008,5. In this case, the maximum magnitude of the error in the number 401 008 is 0,5. This is consistent with the author’s “less than half a unit.” [1] JCGM 100:2008, Evaluation of measurement data — Guide to the expression of uncertainty in measurement, https://www.iso.org/sites/JCGM/GUM-JCGM100.htm https://www.iso.org/sites/JCGM/GUM-JCGM100.htm [2] ISO 80000-1:2009, Quantities and units — Part 1: General
- deleted 6y ago
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ticklemyelmo 6y ago
I'm not entirely clear on the author's thinking. But consider the case of watching a real world voltmeter read 1.02v, 1.01v, 1.00v, 0.99v. Did we really lose a digit of precision there? The leading digit transition between 1 and 9 is a boundary condition that messes up that rule.
- taeric 6y ago
  I think the catch is that you aren't keeping the number normalized. Significant digits somewhat assume scientific notation. So that last number is 9.9e-1. Two digits.
- ska 6y ago
  You didn't lose a digit, you wrote it that way. If your instrument has 3 digits, you should have written 0.990v
  - alisonkisk 6y ago
    You can't make your meter add an extra digit (unless you can turn the dial to a different scale). The digit is lost. But of course it isn't a big deal to lose that dgiti because a trailing digit is 0.1 as much error fraction when the leading digit is 9 vs 1.
    - ska 6y ago
      I'm not sure what you mean here, probably because I worded it poorly. Of course your instrument will have different scales, and you should use the appropriate one. But the measurement was done with a number of significant figures, this is independent of how you write it. Nothing is "lost", if measurement had 2 sig figs it is 0.99, if it had 3 it was 0.990. Or 0.991 or whatever. Am I missing something?
      - taeric 6y ago
        You skipped the example. They said if your device measures 1.00, it is three significant digits. If it measures 0.01, it is one significant digit. From the same measuring device. So it is counter intuitive that the measurement just going down below 1 somehow causes it to "lose" significant digits. It didn't lost precision, though. Just the number of digits that are significant.
yudlejoza 6y ago
When it comes to measurement, 1.7320 is not the same as 1.732 The trailing zero was part of the measurement and, hence, is included in the significant digit count. That also means you can't just measure 1.7320, and report 1.73200 instead (or 1.732). You have to report only, and exactly, what you measured. (I'm ignoring rounding in what I said).
taeric 6y ago
I think you missed what they found weird. The subject for that was that 0.0491 only has three significant digits. It is not obvious why, if you don't normalize the number.
- Trombone12 6y ago
  No, what is weird is that the original author seems to think you can look at a number and know how many significant digits it has. This is almost completely wrong: the number of significant digits is something given by metadata, you cannot look at a number and know that it has been properly rounded to the correct number of significant digits. You always need to be told this special circumstance because most numbers you encounter have in fact not been properly rounded.
  - jhallenworld 6y ago
    In the world of general chemistry, the significant figures rule is assumed. It's not a great rule because in this case: .000300 you have 3 significant figures. But in this case: 300000 you could have 1 to 6 significant figures. The convention to deal with this is always to use scientific notation: 3.00x10^5 has three significant figures.
  - alisonkisk 6y ago
    It's assumed that you shouldn't write down insignificant digits.
    - Trombone12 6y ago
      Not sure how you're supposed to do floating point math while following that rule... you'll have better luck with writing C code with all semicolons in place.
  - taeric 6y ago
    You lost me on this one. :( There is the odd case of trailing zeroes in non decimal numbers, but otherwise if a number is written, when is that not assumed to be significant? That is, there are fairly standard rules for conveying significant digits, I thought?
sengstrom 6y ago
I think it was more a matter of attitude. Never a great way to actually explain something.
mkl 6y ago
> "Here, y has correct one and three significant digits relative to x, but incorrect 2 significant digits, since the truncation at x_2 and y_2 do not agree even to the first significant digit. " > This doesn't make sense. Y has the correct one? I think that's a language issue, like "that x̂ agress to x upto p" in the next section. I think what they're trying to say is the following. If you round x = 0.9949 to 1, 2, or 3 significant digits, you get 1, 0.99, or 0.995 respectively. If you round y = 0.9951 to 1, 2, or 3 significant digits, you get 1, 1.0, or 0.995 respectively. So x and y look equal if you round to 1 or 3 significant digits, but not if you round to 2 significant digits.

leephillips 6y ago

Yes, I think all we've learned from this article is that the author never learned about significant digits.

PhaseLockk 6y ago

Exactly this. It may be useful for OP to consider what the hypothetical opposite of significant digits would be: insignificant digits, ie. digits that tell you nothing. A leading zero can be omitted from a number without any loss of information. A trailing zero, when correctly used in a position that is not below the tolerance, gives you significant information. On the other hand, if you start combine numbers with different tolerances, but do not truncate the result, you will have a bunch of trailing digits that are meaningless in practice because they are below the combined tolerance.

wsh 6y ago

I agree that it’s better to state the uncertainty [1] together with the measurement, but when this isn’t done, there is a standard convention for significant digits, explained in ISO 80000-1:2009 [2], subclause 7.3.4: When a number is given without any further information, it is generally interpreted so that the last digit is rounded with a rounding range equal to 1 in the last digit (see Annex B). Thus, for example, the number 401 008 is generally assumed to represent a value between 401 007,5 and 401 008,5. In this case, the maximum magnitude of the error in the number 401 008 is 0,5. This is consistent with the author’s “less than half a unit.” [1] JCGM 100:2008, Evaluation of measurement data — Guide to the expression of uncertainty in measurement, https://www.iso.org/sites/JCGM/GUM-JCGM100.htm https://www.iso.org/sites/JCGM/GUM-JCGM100.htm [2] ISO 80000-1:2009, Quantities and units — Part 1: General

deleted 6y ago
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ticklemyelmo 6y ago

I'm not entirely clear on the author's thinking. But consider the case of watching a real world voltmeter read 1.02v, 1.01v, 1.00v, 0.99v. Did we really lose a digit of precision there? The leading digit transition between 1 and 9 is a boundary condition that messes up that rule.

taeric 6y ago
I think the catch is that you aren't keeping the number normalized. Significant digits somewhat assume scientific notation. So that last number is 9.9e-1. Two digits.
ska 6y ago
You didn't lose a digit, you wrote it that way. If your instrument has 3 digits, you should have written 0.990v
- alisonkisk 6y ago
  You can't make your meter add an extra digit (unless you can turn the dial to a different scale). The digit is lost. But of course it isn't a big deal to lose that dgiti because a trailing digit is 0.1 as much error fraction when the leading digit is 9 vs 1.
  - ska 6y ago
    I'm not sure what you mean here, probably because I worded it poorly. Of course your instrument will have different scales, and you should use the appropriate one. But the measurement was done with a number of significant figures, this is independent of how you write it. Nothing is "lost", if measurement had 2 sig figs it is 0.99, if it had 3 it was 0.990. Or 0.991 or whatever. Am I missing something?
    - taeric 6y ago
      You skipped the example. They said if your device measures 1.00, it is three significant digits. If it measures 0.01, it is one significant digit. From the same measuring device. So it is counter intuitive that the measurement just going down below 1 somehow causes it to "lose" significant digits. It didn't lost precision, though. Just the number of digits that are significant.

yudlejoza 6y ago

When it comes to measurement, 1.7320 is not the same as 1.732 The trailing zero was part of the measurement and, hence, is included in the significant digit count. That also means you can't just measure 1.7320, and report 1.73200 instead (or 1.732). You have to report only, and exactly, what you measured. (I'm ignoring rounding in what I said).

taeric 6y ago

I think you missed what they found weird. The subject for that was that 0.0491 only has three significant digits. It is not obvious why, if you don't normalize the number.

Trombone12 6y ago
No, what is weird is that the original author seems to think you can look at a number and know how many significant digits it has. This is almost completely wrong: the number of significant digits is something given by metadata, you cannot look at a number and know that it has been properly rounded to the correct number of significant digits. You always need to be told this special circumstance because most numbers you encounter have in fact not been properly rounded.
- jhallenworld 6y ago
  In the world of general chemistry, the significant figures rule is assumed. It's not a great rule because in this case: .000300 you have 3 significant figures. But in this case: 300000 you could have 1 to 6 significant figures. The convention to deal with this is always to use scientific notation: 3.00x10^5 has three significant figures.
- alisonkisk 6y ago
  It's assumed that you shouldn't write down insignificant digits.
  - Trombone12 6y ago
    Not sure how you're supposed to do floating point math while following that rule... you'll have better luck with writing C code with all semicolons in place.
- taeric 6y ago
  You lost me on this one. :( There is the odd case of trailing zeroes in non decimal numbers, but otherwise if a number is written, when is that not assumed to be significant? That is, there are fairly standard rules for conveying significant digits, I thought?

sengstrom 6y ago

I think it was more a matter of attitude. Never a great way to actually explain something.

mkl 6y ago

> "Here, y has correct one and three significant digits relative to x, but incorrect 2 significant digits, since the truncation at x_2 and y_2 do not agree even to the first significant digit. " > This doesn't make sense. Y has the correct one? I think that's a language issue, like "that x̂ agress to x upto p" in the next section. I think what they're trying to say is the following. If you round x = 0.9949 to 1, 2, or 3 significant digits, you get 1, 0.99, or 0.995 respectively. If you round y = 0.9951 to 1, 2, or 3 significant digits, you get 1, 1.0, or 0.995 respectively. So x and y look equal if you round to 1 or 3 significant digits, but not if you round to 2 significant digits.