Lost significant figures through division by exact numbers

Main Question or Discussion Point

I came across a significant figures problem today that I need information on. The problem is this: "What volume of water can a cylindrical container hold of it is 13.0 cm tall and 12.0 cm in diameter? Show your work and express the answer in scientific notation using significant figures."

Of course the volume formula requires radius. When the radius is calculated, the precision allows estimation of 0.1 cm. Because of this, the radius (as I understand it) would have to be recorded as 6.0 cm.

It is evident that one significant figure is lost from the original measurement although it can be seen that the original 12.0 cm has three significant figures. When the final calculation is done, the answer is limited to two significant figures because significant figures must be determined after each step in the process. (The reduction in significant figures was precipitated by division by an exact number, not a measured value with only two significant figures.)

Is my understanding of this correct, and if so, isn't the precision of the final answer artificially reduced from what in real life could be observed?

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I think this example shows everything I never understood about sig figs.

If you just multiply it out what do you get: $13 * (\pi/4)12^2 = 1470.34$

The 13.0 height could really be anywhere from 12.95 to 13.05; and the 12.0 diameter could be 11.95 to 12.05.

So the minimum volume would be $12.95 * (\pi/4)11.95^2 = 1452.43$ and
the maximum would be $13.05 * (\pi/4)12.05^2 = 1488.25$

I would call that 1470 +/- 18. How do you express that via significant figures? "1500" doesn't get it, to me that means 1450 to 1550 which just isn't right for a value that can't exceed 1488. "1470" doesn't work for me either, as it implies 1465 to 1475, which is too narrow a range.

12.0 is a measurement with 3 sig figs.
To get the radius, we divide by 2. This "2" is not measured, it is perfect, with infinite sig figs.
The result will have the lesser of the two sig fig counts. In this case, that's 3 sig figs.
So the radius is not "6.0" (with 2 sig figs), it is "6.00" (with 3 sig figs).

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I would actually write it out before putting numbers in:

$$\frac{\pi}{4} hd^2$$

I would not convert to radius, round, and then plug in. Round once at the end.

Dale
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How do you express that via significant figures?
Indeed, this is why uncertainties are reported explicitly and not via significant figures in the professional scientific literature.

DaveC426913
Gold Member
12.0 is a measurement with 3 sig figs.
To get the radius, we divide by 2. This "2" is not measured, it is perfect, with infinite sig figs.
The result will have the lesser of the two sig fig counts. In this case, that's 3 sig figs.
So the radius is not "6.0" (with 2 sig figs), it is "6.00" (with 3 sig figs).
This.

If we allow 6.00, that definitely is higher precision than the original instrument. I think that you can't go beyond 6.0 because even dividing by a perfect 2 cannot increase precision regardless of the rules about significant digits. It seems incorrect to say that your original instrument could measure 12.0 cm which would mean that the graduations (calibrations) were in cm which allowed an estimated digit of 0.1 cm, but then dividing by 2 would allow an estimated digit of 0.01 with a graduation (calibration) of 0.1 cm. At least that appears to me to be what 6.00 cm would imply about the measuring instrument which is not correct according to the 12.0 cm that was originally reported by the scientist who made the measurement.

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If we allow 6.00
Round once at the end.

So what your saying is that a "tally" of the significant figures at the end of each step should be kept in the margin (or some other convenient place) and rounding should be done after the last calculation to the number of significant figures that have made it through to the end? (12.0 divided by 2 is exactly 6, so 6.00 is not rounded, it just implies more precision than the instrument actually has which is exactly what significant figures is designed to avoid.)

If we allow 6.00, that definitely is higher precision than the original instrument. I think that you can't go beyond 6.0 because even dividing by a perfect 2 cannot increase precision regardless of the rules about significant digits. It seems incorrect to say that your original instrument could measure 12.0 cm which would mean that the graduations (calibrations) were in cm which allowed an estimated digit of 0.1 cm, but then dividing by 2 would allow an estimated digit of 0.01 with a graduation (calibration) of 0.1 cm. At least that appears to me to be what 6.00 cm would imply about the measuring instrument which is not correct according to the 12.0 cm that was originally reported by the scientist who made the measurement.
As a previous commenter mentions, giving the error bars is much more informative than "significant figures"; significant figures is inherently biased on a base ten number system.

"(12.0 +/- .05) / 2 == (6.00 +/- .025)", which is a lot different than "6.00 +/- .005" suggested by "3 sig figs". But note that the error between the two (.025 vs .005) is within "an order of magnitude".

I think of "sig figs" as more of a guideline for guessing the accuracy approximations than a "hard" result.

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So what your saying is that a "tally" of the significant figures at the end of each step should be kept in the margin
No, I';m saying what I wrote. Round once at the end.

But "round once at the end" doesn't address the fact of the loss of a significant figure when dividing a measured number by a number of infinite precision. This was my original question.

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You have received two excellent answers: one is that significant figures is a shortcut to proper error propagation, and the other is that intermediate rounding - what you asked about - should not be done. And yet you are repeating your question.

If those answers don't satisfy you, what will?

Not trying to be hard to get along with but 1) I don't believe I used the word "rounding" until after you brought it up. "Estimation" is of course something that the scientist is allowed to do to give a best judgment on where the measurement falls between the calibration marks to allow another level of precision beyond the instrument calibrations. Rounding is not part of this process and is undesirable. 2) Also, I asked about the matter of dividing 12.0 cm, a measured number, by an exact infinitely precise number 2. This operation requires no rounding, but results in a loss of precision. That is what my question is about. Any math beyond that was offered by other posters on this thread. When you stated to "round at the end," I thought that using that process would avoid having to lose the precision by "keeping tabs" on precision without having to actually write 6.0 and use two significant figures from then on out. It was an honest question. 3) I understand the concept of "significant figures is a shortcut to proper error propagation." I certainly am willing to accept the fact that precision is unnecessarily (unfortunately) lost in the math, but the hope is to not propagate unmerited precision rather than to preserve due precision. I think my junior high and high school students would understand that explanation as well. It think only one poster, Dale, actually gave an answer that basically credited the problem to imperfections in the system. But he wasn't the first response and others suggested that the 6 should be recorded as 6.00, which I'm not sure is correct since that obviously would imply a greater precision than the original instrument. In summary, a) I'm not sure how rounding applies to a single division that doesn't require rounding in the first place; b) I can accept that there are weaknesses in the system that we have to live with. I'm just wanting to make sure that is the correct answer.

I understand that the correct number of significant figures should be tracked throughout the calculation process (even when rounding is done at the end). Here is the essence of my question. When the number of significant figures required actually adds unmerited precision to the measurement, should I maintain the number of significant figures required by the rules rather than the precision dictated by the measuring instrument?

Where are "the rules" defined? The Wiki entry says
Only measured quantities figure into the determination of the number of significant figures in calculated quantities. Exact mathematical quantities like the π in the formula for the area of a circle with radius r, πr2 has no effect on the number of significant figures in the final calculated area. Similarly the ½ in the formula for the kinetic energy of a mass m with velocity v, ½mv2, has no bearing on the number of significant figures in the final calculated kinetic energy. The constants π and ½ are considered for this purpose to have an infinite number of significant figures.
If your students are facing a standardized test where the "correct" answers are determined by the "rules," then you should (IMO) teach them how to follow the rules. You could/should follow up with a discussion of the pitfalls in the rules.

I'm still thinking about whether the mult/div by a pure number rule makes sense, in light of your question.

Okay. Thanks for the answer. I've seen that rule about exact numbers in a number of places. This problem is not original to me, I'm just trying to decide how to resolve the issue in my own mind of how 12.0 divided by 2 can result in an answer 6.00 which has an unjustified significant figure. I have to decide whether to tell students to disregard this particular example (which states that because of the precision of the measuring instrument, this intermediate calculation would result in 6.0 cm) or to explain the exception. I think that there is validity in the thought that it seems contrary to the purpose of the significant figures rule structure to allow an unmerited level of precision. Even if rounding is held until the end, one must write the number 6 on the paper (or 6.0 or 6.00 or 6.000 or whatever) to do the math. Should the student note in the margin that there are three significant figures in this number even though 6.00 is not justified by the instrument used to make the original measurement and just ignore that fact for the sake of the rules? My main concern at the moment is how to handle confusion caused to students by an example in a book when the book offers a seemingly valid justification for using 6.0 cm (which of course affects the precision of the final answer because the division reduced the number of significant figures from 3 to 2.)

Here's my thought. If you measure a stack of identical objects, say there are 12 of them, and you measure the stack height as 66.6 cm, then each individual is 66.6/12 = 5.55 cm. They have to be, given that they are identical. They each take an identical portion of that 0.6 cm.

Same with the radius: if you measure the diameter, you know the two radii are identical. They must split the measured diameter.

You've heard of "reduction to absurdity" I'm sure. Hypothetically, if I had a stack of 123 objects that measured 1.27 mm, each would share equally and be 0.010325203252.... mm. Of course the problem here it is a repeating decimal so we now have an infinite number of significant figures if we accept the thought of identical portions as justification for increasing precision beyond the original instrument's precision. Where does it stop?

Where does it stop?
The "rule" says keep the number of figures in the measured value. In this case, three: 1.27/123 = 0.0103. You have a result to ten-thousanths (0.0001), even though your scale reads hundredths (0.01).
.

Right. So if we keep with the rule, I have to discard the example and tell the students that it is incorrect to say that 12.0 cm divided by 2 is 6.0 cm (in order to match the precision of the measuring instrument). I have to stick with the significant figures from the original measurement. I have no problem doing this to maintain consistency with the rules. But the problem still remains in my mind of how do I get a precision of 6.00 cm by calculation which I would not be justified in measuring if I had decided to measure the radius instead of the diameter. Hypothetically speaking, one could get an inflated precision in a final answer even following the rules of significant figures. Whether that happens to any consequence in real life is debatable. But to avoid confusion, it looks like I will have to tell the students that the book has handled it incorrectly (which I have had to do in the past in other courses.) Also, I guess a proper understanding of the purpose of significant figures would be that the system is designed to minimize misrepresenting the precision of your instruments. It is not able to eliminate it all together. What I gathered from Dale's post is: when higher precision is desired in reporting data, a different system is used.

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I understand that the correct number of significant figures should be tracked throughout the calculation process (even when rounding is done at the end).
Where does that "rule" come from? You can - and probably should - use a calculator for the entire problem, and then round at the end.

Here is quote from this site http://www.astro.yale.edu/astro120/SigFig.pdf

"General guidelines for using calculators

"When using a calculator, if you work the entirety of a long calculation without writing down any intermediate results, you may not be able to tell if a error is made and, even if you realize that one has occurred, you may not be able to tell where the error is.
"In a long calculation involving mixed operations, carry as many digits as possible through the entire set of calculations and then round the final result appropriately. For example,

(5.00 / 1.235) + 3.000 + (6.35 / 4.0)=4.04858... + 3.000 + 1.5875=8.630829...

"The first division should result in 3 significant figures; the last division should result in 2 significant figures; the three numbers added together should result in a number that is rounded off to the last common significant digit occurring furthest to the right (which in this case means the final result should be rounded with 1 digit after the decimal). The correct rounded final result should be 8.6. This final result has been limited by the accuracy in the last division.

"Warning: carrying all digits through to the final result before rounding is critical for many mathematical operations in statistics. Rounding intermediate results when calculating sums of squares can seriously compromise the accuracy of the result." (Bold italics added for emphasis.)

Granted that as many digits as possible should be kept when using a calculator, but paragraph 3 (especially what I have highlighted) shows that the significant figures in various steps must be considered to ensure that the final answer does not exceed the least precise number.

Consider this:

If a person measures the diameter as 12.0 cm and the height as 13.0 cm and uses the volume formula generally taught, maintaining the rules of significant figures, he would have to report his answer as 1.47x10^3 cm^3 because the original measurement has 3 significant figures. If he takes same measuring device and measures the radius, he could only measure it to 6.0 cm precision. In this case he would have to report his calculated volume as 1.5x10^3 cm^3. Does a simple mathematical manipulation give a real life precision increase even though the individual used the same instrument for both measurements? If he decides to measure the diameter to gain the perceived precision, is that fudging the results?

If he decides to measure the diameter to gain the perceived precision, is that fudging the results?
I don't think so. If the measuring device has a fixed precision (like, 0.1 cm) then measuring a larger dimension gives a smaller relative error. If your device is a tape rather than a ruler or calipers, you could measure the circumference and use $\frac{h c^2}{4\pi}$, that would be even better. Or, get a pi-tape .

In this specific example, how do you propose to measure the radius? The centerline is an imaginary line floating in space. Where do you put the edge of the ruler?

In the OP you said, the diameter is 12.0 cm, and
Of course the volume formula requires radius.
Not really. As several posts have noted, you can use $\frac{\pi}{4} hd^2$