Please correct me if I’m wrong, but I’ve read that the reason why we use significant figures is to avoid making the result of a calculation more accurate than the starting values prior to the calculation. For example, if we were seeking the ratio of two weights, lets call them A and B, using a cheap scale and an expensive scale we could get the following varied results:

Cheap Scale $$A = 34 \space oz$$ $$B = 23 \space oz$$

Expensive Scale $$A = 34.0000 \space oz$$ $$B = 23.0000 \space oz$$

Computing the ratio $\frac{A}{B}$ using a calculator we get the following value: $$\frac{34}{23} = 1.47826$$ Now this is a correct ratio depending on the scale used (ie the cheap scale could not have produced such an accurate measurement given the starting values) This leads to the following question:

Why are the zeros behind a decimal point ignored if they are just place holders?

For example, the value 0.000023 oz only has two significant figures, but isn’t that value something only a very fine tuned scale could measure (keeping in line with the previous weight example). So why are these zeros ignored?


2 Answers 2


This is part of the reason I really don't like the way we teach significant figures. What's really important isn't the number of digits you have, but rather the uncertainty in your measurements. I think significant figures are taught as an approximation to the proper uncertainty analysis, but it really seems to be not worth the trouble.

What's really going on is this: when you say $A = 34$ oz, you're actually saying that $A$ is closer to 34 oz than it is to either 35 oz or to 33 oz. In other words, you mean $A = (34 \pm 1)$ oz. Likewise, when you say $A = 34.0$ oz, you're really saying $A = (34.0 \pm 0.1)$ oz. And in the final example you give, when you say $A = 0.000023$ oz $= 2.3 \times 10^{-5}$ oz, you're actually saying $A = (2.3 \pm 0.1)\times 10^{-5}$ oz.

The more precise way to do things is to use the uncertainty you have for each measurement, and use the rules for propagation of uncertainties to get the uncertainty in your final answer. One advantage of using the uncertainties is that your uncertainty won't always be 1 unit of your least significant digit, but the rules for propagating significant digits essentially imply that. Using sig figs usually gets close to the answer you would get from properly using uncertainties, but there are some cases where the result is really weird.

To answer your main question directly, you ignore those placeholder zeros because they don't affect the fractional uncertainty of the quantity you're looking at. Keeping them would give you an answer very different from the proper uncertainty analysis.

but isn’t that value something only a very fine tuned scale could measure?

Not necessarily. The cost of an instrument is much more closely related to the relative precision of its output, rather than the absolute precision. If all you're trying to do is weigh things with masses between $1 \times 10^{-5}$ and $1 \times 10^{-4}$ oz, then you don't care at all if your scale can measure 1 oz. But if you have objects that are 1 oz and $1 \times 10^{-5}$ oz, and you care about the difference, then you'll need a really precise scale.

  • 3
    $\begingroup$ If A=34, and this implies it is closer to 34 than to 33, or 35, the the error implies is 0.5 and not 1. Or am I wrong here? $\endgroup$ Commented Sep 22, 2016 at 4:58
  • $\begingroup$ @Martin-マーチン I think we should interprete closer as most probable. So we are assuming this human or machine is near the mean, but it could be 33 or 35 $\endgroup$
    – user43021
    Commented Sep 16, 2018 at 6:27

In your example 0.000023 oz does indeed have two significant figures. It is indeed probably something only a very fine tuned scale could measure.

Now consider that you add that mass to another sample that you measured with the much less accurate scale, let's say the other scale measured that other weight as 23 oz. Both have two significant figures, but considering the weight read by the inaccurate scale, you have no legitimate idea whether it is really 23.28125 oz or 22.87284 oz, or some other number for that matter.

What you do know is that it probably falls within some range of values, we usually look at the last significant digit and if the uncertainty is not mentioned we can assume it is +/- 2%. That means the value 1.0 oz might be somewhere between 0.98 oz and 1.02 oz.

So now when we add the two masses together, we do not want decimals since 23 oz has none. The result is 23.000023 oz. Which is 23 oz knowing that we can only go down to the ones place. The reason is that you have no clue whether it is 23.38125 oz + .000023 oz, or 22.87284 oz + .000023 oz, or any other values for that matter. The fact is that these zero's are not ignored, they are giving us the magnitude of the number. But when we do something with them in conjunction with some other uncertain numbers, we cannot be sure of the result to high certainty. One rotten apple spoils the whole barrel.

Now in this scenario described above, even if we said 0.000023 oz has 6 significant figures. The other number 23 oz still has only two. Therefore we must go with 2.

When we divide or multiply we count the sig figs. 23 oz has 2 sig figs, so our result of multiplying should have 2 as well. So how many times greater is the inaccurate weight (23 oz) compared to the accurate weight (0.000023 oz). 23 oz / 0.000023 oz ~= 1000000 times greater. Notice that the number 1000000 has only 1 significant figure. We want two significant figures so we write. $1.0 * 10^6$


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