0.0 significant figures

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Dear all,

Me and my friends are arguing how many significant figures does 0.0 and 0.0000 have?
Me and my other friend say it does not have any significant figure and the others say it's 1 significant number and exact, a special case. Please advise for the correct one.
Thanks!
 

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  • #2
Simon Bridge
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You and your friend are incorrect:

If I had a difference of two measurements z=x-y

and if x=1.0m and y=1.0m then z=0.0m

but if x=1.000m and y=1.000m, then z=0.000m

What the sig fig and dp are telling you is how close to being the same the measurements are ... in the first case they are the same to one part in 10 while in the second they are the same to one part in 1000. That's an important difference.

eg. if x=1.091m and y=1.090m then z=0.0 to some sig fig and 0.001 to different sig fig.

Later on you'll learn about uncertainties in measurement and it will become clear.
At the moment, this whole thing about sig figs and so on are a stop-gap rule-of-thumb until your statistics knowledge catches up.
 
  • #3
jbriggs444
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The notation is telling you that the measurements are the same to one tenth of a meter and to one thousandth of a meter respectively. Saying that they are the same to "one part in ten" or to "one part in a thousand" gives an incorrect implication.

The "one tenth of a meter" wording indicates an absolute error bound. "one part in ten" implies a relative error bound. The usual assumption is that a relative error is relative to the measured value.

I would say that 0.0 is accurate to the first decimal place, but that it has no significant figures. And I would say that 0.000 is accurate to the third decimal place but that it has no significant figures.


Significant figures (and relative error) matter when you are multiplying or dividing numbers.

0.0 * 5.0 * 7.1 = 0 and the result has no significance
0.000 * 5.000 * 7.100 = 0 and the result still has no significance


The number of accurate decimal places (and absolute error) matters when you are adding or subtracting.

0.0 + 5.0 + 7.1 = 12.1 the result is accurate to one decimal place and has three significant figures.

0.000 + 5.000 + 7.100 = 12.100 and the result is accurate to three decimal places and has five significant figures.
 
  • #4
Meir Achuz
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Dear all,

Me and my friends are arguing how many significant figures does 0.0 and 0.0000 have?
Me and my other friend say it does not have any significant figure and the others say it's 1 significant number and exact, a special case. Please advise for the correct one.
Thanks!
0.0 has 1 sf just like 0.8 would.
0.0000 has 4 sf, just like 0.2345 would.
 
  • #5
jbriggs444
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0.0 has 1 sf just like 0.8 would.
0.0000 has 4 sf, just like 0.2345 would.
This disagrees with my understanding.

If you multiply 0.0000 by 0.2345, how many sf do you claim that the result has?
 
  • #6
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2 significant figures => relative uncertainty somewhere around 0.5-5% (first for 9.x, last for 1.x)
1 significant figure => relative uncertainty somewhere around 5%-50%
If I extend that:
50%-500% relative uncertainty would correspond to "0 significant figures".
500%-5000% relative uncertainty would correspond to "-1 significant figures".

0.0, 0.0000 or anything similar has no upper limit for the relative uncertainty, so it is a bit like "minus infinity significant figures" - that is not a number, so I would not use the concept of significant figures here.
 
  • #7
bcrowell
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Cute question. Leading zeroes are placeholders, not sig figs, so 0.01 has 1 sig fig. Trailing zeroes after the decimal place are significant, so 1.0 has 2 sig figs. Since your number consists of nothing but zeroes, there's no way to tell which are leading and which are trailing.

I agree with mfb that it should be interpreted as minus infinity sig figs.
 
  • #8
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This disagrees with my understanding.

If you multiply 0.0000 by 0.2345, how many sf do you claim that the result has?
My first thought it that the result has an undefined precision because the relative error of the first term (0.0000) is undefined, so you can't use a standard approach... then again intuition tells me it's simply 0.00000000
 
  • #9
jbriggs444
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My first thought it that the result has an undefined precision because the relative error of the first term (0.0000) is undefined, so you can't use a standard approach... then again intuition tells me it's simply 0.00000000
We agree that the relative error on the 0.0000 term is undefined and that as a result the relative error on the product is also undefined.

I take the position of mfb and bcrowell that significant figures are (roughly) computed as

-log(relative uncertainty) =
-log(uncertainty/measurement)

When the measurement approaches zero while the uncertainty remains non-zero, this blows up toward negative infinity
 
  • #10
Simon Bridge
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Measurement is a process. If I measured my mass on a balance calibrated in tonnes, I'd find my weight to be 0.0T and just shrug .... but if I got 0.0000T then I'd have a problem. Really we would think of this as keeping dp here rather than the measurement having sig fig ... the only time the sig fig comes up in calculations is where two measurements have to be multiplied ... what if I had to use this measurement to determine my weight. g=9806.65N/T has six sig fig ... so is the result better represented by 0N, or 0.0N for the first case?

That is a situation where the zero measurement comes from the coarseness of the measurement.

If you put a ruler against one side of an object you wanted the area of and found the length to be 0.00m you'd conclude that you needed a more accurate ruler rather than that the surface was truly one-dimensional.

How would the (prev example) multiplication 0.0000x0.2345 come about?

How about as z=(a-b)c
Then the calculation is saying that a and b are very close to the same measurement but you cannot be sure that they are exactly the same.

The standard (or absolute) uncertainty on z is given by:
##\sigma_z=c(\sigma_a + \sigma_b)+(a-b)\sigma_c##
(assuming all dependent measurements for the sake of a simple example)

(eg. you may want to find the weight of your cat but have trouble getting it to sit still on the scale. So you weigh yourself and the cat together for measurement a, then yourself alone for measurement b, and subtract them. c, in this context, would be the acceleration of gravity and z the weight. a-b=0.0000kg would mean that you have a very small cat indeed. IRL we'd find another way to do the measurement - probably involving tweezers.)

A common rule-of-thumb is to take the standard error to be half the lowest resolution measurement (yes, I know there are other rules of thumb with fine arguments in support.). So, in this example, (a-b)=(0.0000±0.00005)units (always put the units on your measurements) and c=(0.2345±0.00005)units.

The ##\sigma_z=0.2345\times 0.0001 = 0.0002\text{units}## ... so the answer to the question would be that 0.0000x0.2345=(0.000±0.0002)units ... has three or four sig fig. depending on the rule you want to use to associate sig fig.

In the rule that you keep the smallest number of sig fig in the multiplication - you would put 0.00x0.234=0.00 and 0.000x0.23=0.00 because you want to indicate how accurate the measurement process was that went into it.

What you are actually noticing is that the significant figures rule is not really all that useful for representing the uncertainty in the result. That is why we don't use it IRL.
 
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  • #11
Hurkyl
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Remember that significant figures are not rigorous error analysis: they are merely a quick and dirty ad-hoc heuristic approximation to proper error analysis. It shouldn't be too distressing to find situations where they don't work very well.
 
  • #12
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I think Simon has a very rational and compelling line of thought.

Consider the distance travelled from the south pole by Scott as measured in hundreds of km = 14 (2sf) and the distance to safety from the south pole as measured in the same units =14

I think the difference between (14-14) = 0.0

and (13.8 - 13.6) = 0.2

made all the difference to Scott.
 
  • #13
Simon Bridge
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me said:
That is why we don't use it IRL.
Studiot said:
I think the difference between (14-14) = 0.0
and (13.8 - 13.6) = 0.2
made all the difference to Scott.
... or, when we do use it IRL, things tend to go badly :/
Hurkyl said:
Remember that significant figures are not rigorous error analysis: they are merely a quick and dirty ad-hoc heuristic approximation to proper error analysis. It shouldn't be too distressing to find situations where they don't work very well.
Bears repeating :D
 
  • #14
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Dear all,

Me and my friends are arguing how many significant figures does 0.0 and 0.0000 have?
Me and my other friend say it does not have any significant figure and the others say it's 1 significant number and exact, a special case. Please advise for the correct one.
Thanks!
I would say two and five. But there aren't any conventions about this, and it depends on the possible range of whatever it is you are measuring. It if could be from -100 to 100 then 0.0 has four significant figures. Nobody ever writes 000.0 (though in a way it would make sense).
 
  • #15
jbriggs444
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I would say two and five. But there aren't any conventions about this, and it depends on the possible range of whatever it is you are measuring. It if could be from -100 to 100 then 0.0 has four significant figures. Nobody ever writes 000.0 (though in a way it would make sense).
That is a somewhat interesting approach that might be understood in terms of information theory.

If an actual value is uniformly distributed in the range from -100 to 100 then a measurement that is accurate to ±0.1 provides 3 decimal digits (~10 bits) of information.
 

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