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After going all through web and posts I can't get a complete idea of significant figures. I'll try to explain the problem.

The definition that seems more frequent is:

significant figures: number of figures carrying on precision.

It is easy to see that in the number $1000$ zeros are non-significant figures unless we specify it as $1000.$. It is also clear that $0010.$ is two significant figures.

Here it comes when definition shows unuseful (at least to me), because leading zeros as well as non zero numbers talk about precision. For example $0.0017$ and $0.1217$ are same precision. They indicate the measuring instrument can detect variations in the ten-thousands. If not, please explain how. That's the specific problem. I beg you answer with concrete examples.

The most interesting site I've read is this , and I understand how significant figures work, but the previous problem remains.

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    0.0017 and .1217 don't convey the same precision for same reason 1700 (not 1700.) and 1217 don't. Leading zeroes are indicitive of an order of manitude, not a degree of precision. – mxnoqwerty Aug 18 '18 at 17:27
  • @mrnoqwerty that's not true, if you measure 0.0017 you can measure both... –  Aug 18 '18 at 17:28
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    have you seen https://www.purplemath.com/modules/rounding.htm and https://www.purplemath.com/modules/rounding2.htm and https://www.purplemath.com/modules/rounding3.htm? – cgiovanardi Aug 18 '18 at 20:20
  • @cgiovanardi yes, it doesn't touch what the post is asking... –  Aug 18 '18 at 22:08
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    "if you measure 0.0017 you can measure both." Are you sure about that? – David K Aug 18 '18 at 22:22
  • @DavidK im never completely sure, but it seems possible. If you have a rule and measure $0.1$ cm, the zero is meanigful at least to me...; I know it would imply that if we have a 1 cm unit we can measure 1559 and any number, but I hope you can get the point. I cant express it better. –  Aug 18 '18 at 22:24
  • But "cm" is an arbitrary unit of measurement. Suppose we measure the same distance in Angstroms instead. Then instead of $0.1$ cm you can measure $10{,}000{,}000$ Angstroms. Now clearly the zeros are not significant (because you did not measure them when you were doing cm) and there are no leading zeros, so you have one digit of precision. – David K Aug 18 '18 at 22:27
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    Conversely, $0.1$ cm is $0.000001$ km. So did you suddenly gain another five digits of precision, just by stating your result in km instead of cm (without actually doing any new measurement)? – David K Aug 18 '18 at 22:28
  • @DavidK no, that's exactly what i'm not confused with. I know equivalence in units. Look i've described exactly what you say with the thousand. The problem is with real things.If we measure (in reality) $0.1$ units, the zero is meaningful as it is a in $1.01$ or as it is one in $1.2$ isnt it? –  Aug 18 '18 at 22:30
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    An interesting historical note: A mid -20th-century research team measured the height of Mt. Everest above mean sea level, and reported 29,002 feet. Decades later it was revealed that they got 29,000 feet, to the nearest foot, but after some private discussion they reported 29,002 feet. They felt that if they reported 29,000 then many people would assume they had not been competent to do better than a range of 28,950 to 29,050. – DanielWainfleet Aug 19 '18 at 00:53

2 Answers2

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Let's try an actual example. Suppose you have a pile of sand that weighs $0.1217$ on your scale. The sand is a mixture of white and red grains. Your boss asks you to determine what percentage (by weight) of the sand is red.

About how accurate will your answer be? Will you be able to measure the proportion of red sand in tenths of a percent?

The next sample of sand weighs only $0.0017$ on your scale. Again it has white and red grains and your boss wants to know the percentage of red grains by weight.

Now how accurate will your answer be?

It is sometimes true in practice that it doesn't matter that a particular measurement has few digits of precision, because the only use for that measurement is to add it to a much larger measurement. Then only the absolute precision of both measurements matters. But the reason significant figures are so often used is that we are very often multiplying or dividing numbers.

If you write every number in standard scientific notation, even the ordinary numbers such as $10$ or $0.1$ that do not seem to need this notation, then you will get a better idea of significant digits. In scientific notation, all the digits in the mantissa are significant, because we never write leading zeros and we never write more digits than the accuracy of measurement justifies. For example, $0.1$ is $1 \times 10^{-1}.$ The mantissa then is $1$ -- one significant digit.

David K
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  • Im sorry I can't get it, must be stupid. Information is hide in scientific notation below $10^{-1}$...; it might be that zeros are taken as unmeaning because of a definition, not because they are actually unmeaning. They don't denote a quantity (a magnitude), but I can't understand it. Anyways, thanks...(+1). –  Aug 18 '18 at 22:59
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I've understood it by this way:

Suppose we have sand grains of $0.0005$ grams each one, weighed with a 0.0001 precision balance. Of course, the numbers before $5$ have a meaning: they indicate magnitude. But they are not significant figures. It can be seen in this way:

somebody estimate grains' number in $63566$. If we want to estimate the total weight, we should multiply. But there is 'no difference' between $0.0005$ and $0.0006$. Let's see how it modifies the multiplication:

$ 0.0005\times 63566=31783$

$ 0.0006\times 63566=38396$

So we can't even trust the first number, as if only the five $0.0005$ were meaningful. We can see it easily if we write it as $5\times 10^{-4}$

Once understood, $31783$ should be written as $3\times 10^{4}$

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    It seems you do understand. You have explained the distinction between significance and magnitude. I think it would be OK for you to select this as the accepted answer. – David K Aug 19 '18 at 01:35
  • @DavidK Thank you a lot David! I appreciate very much your comment. –  Aug 19 '18 at 15:23