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I have been reading definitions of significant figures which vary from source to source.

1-The digits in a number that indicate the accuracy of the number are called significant figures or digits---College Algebra by Raymond A. Barnett and Micheal R Ziegler

2-Significant figures by definition are the reliable digits in a number that are known with certainty---Textbook of Chemistry by Prof Muhammad Omar Mangrio

3-The digits which give us idea of quantity and accuracy are significant digits---An Introduction to Error Analysis by John R. Taylor.

4-The significant figures of a number are those digits that carry meaning contributing to its precision---Wikipedia (somewhat close to definition 1)

A rule for determining significant figures in a number is that, Zeros that locate the decimal point in numbers larger than one are not necessarily significant. For instance the number $1570$ has only $3$ significant digits $1,5$ and $7$

I can't figure out that how this rule is consistent with the definitions above. Or simply my question is why $0$ in the number $50$ or $15700$ is not considered to be significant?

Whenever I asked my teacher, he said that zero is not for the purpose of indicating accuracy here. I didn't get him! Any helpful example related to measurements would be greatly valued.

Sufyan Naeem
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    Maybe start by looking at an extreme case. If somebody told you there were 330,000,000 people in the United States, it is unlikely that they were all counted and it was 329,999,999+1 people counted. – turkeyhundt Apr 22 '15 at 15:13
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    When ambiguity may exist, always use scientific notation. In this case $50$ could either be written $5\times 10^1$ or $5.0\times 10^1$. Then you'd immediately be able to tell how many sig figs the number has. Of course some authors are less considerate of possible ambiguities than I. –  Apr 22 '15 at 15:20
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    You once wrote $5 \times 10^1$ and then wrote $5.0 \times 10^1$ I think in the first notation, according to a rule of significant figure, there is one significant digit and in second notation there are $2$ significant digits. Isn't so? – Sufyan Naeem Apr 22 '15 at 15:23
  • This is a question of notation, rather than fundamentals. If fifty is written has simply 50, the 0 is not a significant figure. If it is written as "50.", or as "$5\overline{0}$", the zero is significant. – Paul Apr 22 '15 at 15:23
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    "Zeroes that ... are not necessarily significant" but that doesn't mean they are never significant. It just means that more information about where the number came from is required to know. – JMoravitz Apr 22 '15 at 15:24
  • @SufyanNaeem Yes. In scientific notation you just read off the number of digits of the number before the $\times 10^n$ part and that'll be your number of significant figures. That was my point. $50$ is ambiguous. $5\times 10^1$ and $5.0\times 10^1$ are not. –  Apr 22 '15 at 15:24
  • Yes, if we were talking about integers, we would never write $50$ nor $15700$ in scientific notation, but simply know from the context that all their figures were significant, so two and five significant figures respectively. – String Apr 22 '15 at 15:28

4 Answers4

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I would say that 50 could be to either 1 or 2 significant figures, depending on the context or calculations (etc) of the author (or measurer) of the number. Similarly, 15700 could be 3, 4 or 5 significant figures. For example, a scientist may report in a paper that they have measured the length of an object to be $50m$, to 1 significant figure (making the lower and upper bounds $45m$ and $55m$ respectively), but they may also report instead that the measurement was to 2 significant figures, making the bounds $49.5m$-$50.5m$. Another way this author might express the same thing is to use "plus-minus"; the first would be something like $L=50\pm5 m$, while the second would be $L=50\pm0.5 m$.

If I say "this object is 50m long, to 2 sig.figs." then that is more precise, but just as valid, as saying "this object is 50m long, to 1 sig. fig."

danimal
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I'm not sure, but if you want it to be clear that you mean 50 more specifically than just "approximately 50", you can write 50. with the period after, then I believe 0 is considered a significant digit

Alex Mathers
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  • But then you run into the problem (as exists in your very own sentence!) of whether the period is a full stop or a decimal point. One could, of course, use a separate decimal point, as exists, but there are places, such as the U.S., where that's simply not very common. – Brian Tung Apr 22 '15 at 15:26
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Suppose I have a meter stick with no markings on it. I round my height to the nearest meter. I say "I am 2 meters tall, to the best accuracy I could find with my equipment." This is not very precise. I could be 151 cm tall. I could be 249 cm tall. You don't know.

Now, a meter is 100 centimeters. So, using my equipment, I could say, "I am 200 centimeters tall, to the best accuracy I could find with my equipment." But just because I changed the units, doesn't mean that my measurement was any more precise. I made the measurement with the same stick and I rounded in the same way. The only difference is the way I expressed it. We see that the 0's at the end of the number are arbitrary; an artifact of what unit we wish to use. The only thing real is the "2" and there is one significant figure.

hunter
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To my understanding,"Significance" for mathematics should mean the number have value.For example $0.305$ as a number.The first "0" is not significant because it have a value of "0",but second "0" have significance because it had a value of 0.30.

It is very unclear of what "significance" means because it's definition is not universal.To me,the 0 in $50$ is significant because it holds a digit of "50".It does have a value,a quantity.

ministic2001
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