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Sig Fig's Guide to Significant Figures (or 0 is not nothing)

So just how do I know if a digit is significant?

Easy. I know a digit is significant when I read it from a measuring device like a meter stick, stopwatch, or thermometer. Feast your eyes on the scale below. I enlarged a section from a meter stick between the 2 and 3 centimeter marks. Let's take readings at the positions of the A, B, and C pointers. Now whenever I read from a measuring device, I read to a fraction of the smallest division. For a meter stick, the smallest division is 0.001 m (1mm). I can read to a tenth of that, which is 0.0001 m (0.1mm). So my meter stick readings will have 4 digits to the right of the decimal point.

Here are my readings:

A: 0.0218 m

B: 0.0260 m

C 0.0302 m

I have to use my judgement to estimate the last digit. For pointers A and C, someone else might read 0.0217 m and 0.0303 m. That just shows that there's an uncertainty of ±1in the last digit. That's cool. The last significant digit of any measurement is expected to have a bit of uncertainty.

Now I know you're asking..."Why do I have to bother with writing the last 0 in 0.0260 m?" Well, let's see if I can explain this. People only know as much as you tell them. If you write 0.026 m, then you're telling people that the last digit you read was the 6, to the nearest millimeter. So you're reporting a less precise measurement than 0.0260 m. When you write that last 0, you're telling people that you read to the nearest tenth of a millimeter, and that digit just happened to be a 0. Just remember, in measurement, 0 is not nothing. Oops, that was a double negative.

Now what about that 0 between the 3 and 2 in 0.0302 m? Is that 0 significant? Sure is. That tells you that the reading is between 0.030 and 0.031 m. In other words, that 0 is necessary to locate the position on the scale.

How about the zeros in front of the non-zero digits, for example, the 0.0 in 0.0218 m? Well, the 0 to the left of the decimal point isn't significant. That's just there to tell you, "Watch out! There's a decimal coming up." The first 0 to the right of the decimal point isn't significant either. Now, I know what you're thinking, "Of course that 0 is significant. It has to be there or it's a different number. 0.0218 m certainly isn't the same thing as 0.218 m." Well, ok, but you have to remember that I'm using the word significant in a very restricted sense here. Scientists sometimes take everyday words and appropriate them for their own special uses. In the process, they restrict the meanings to something more specific than the general population recognizes. For physics and other sciences, you need to train yourself to think of a significant digit as a digit that is read from a measuring instrument. Now let's look at that second 0 in 0.0218 m more carefully.

Do you see that I could write 0.0218 m as 2.18 cm or 21.8 mm instead? Of course, you do. All I did was convert units. A units conversion doesn't change what was measured, so it can't change the number of significant figures in the measurement. All three versions have the same 3 significant figures. By the way, what if I wanted to convert the measurement to microns? (1 m = 106 μm) What would I write? The answer is 2.18 x 104 μm. Likewise, reading B in microns would be 2.60 x 104 μm. In scientific notation, the number of digits I write are exactly the number that are significant.

Now here's the trickiest area for expressing or identifying significant figures. Suppose I tell you that I have an orchard that measures 620 ft x 480 ft. What about those final 0s? Are they significant? They could be there because the measurements were made to the nearest foot, and the nearest foot just happened to be 0. In that case, they would be significant. Or maybe the measurements were only made to the nearest 10 feet. In that case the 0s wouldn't be significant. Still, they would have to be expressed in order to serve as placeholders. Obviously, it wouldn't make sense to leave off the 0s and write 62 ft x 48 ft. I can think of two ways to make it clear whether those final 0s are significant. Here they are. I hope you're still paying attention.

Put a decimal point after the final 0 to indicate that it's significant. So you would write 620. ft if you wanted to send a message that the 0 was significant.

Write the measurement in scientific notation. Express the 0 only if you want to show that it's significant. In 6.20 x 102 ft, the 0 is significant. If the 0 isn't supposed to be significant, you would write 6.2 x 102 ft.

OK, I think that pretty much covers when a digit is significant and when it isn't.

Now I'll go on to talk about what happens to significant figures when measurements are multiplied, divided, added, and subtracted.

Here's something that simplifies this aspect of the subject.

As far as determining the number of significant figures in the result is concerned, addition and subtraction obey the same rule. Likewise, multiplication and division obey the same rule (but it's a different rule from addition and subtraction.)

Here are the rules. Sorry, you just have to learn them. At least, there are only two.

Rule for Addition: When adding (or subtracting) measurements, the number of decimal digits in the result is the same as that of the measurement with the smallest number of decimal digits.

Rule for Multiplication: When multiplying (or dividing) measurements, the number of significant figures in the result is the same as that of the measurement with the smallest number of significant figures.

OK, to apply these rules, suppose I have a sheet of metal with these measurements: length of 45.05 cm, width of 9.20 cm, thickness of 0.0014 cm. (You may wonder why the length and width were measured to different precision. That is, the length and width were measured to the nearest 0.01 cm and the thickness to the nearest 0.0001 cm. Well, different measuring devices were used. A meter stick was used for the length, and a device called a micrometer calipers was used for measuring the much smaller thickness.)

I'll calculate the area of the sheet. A = L x W = (45.05 cm)(9.20 cm )= 414.46 cm2, before rounding. Now I know from the Rule for Multiplication that the result must have 3 signficant figures, because 45.05 cm has 4 and 9.20 has 3. So I round to 414 cm2.

Next I'll calculte the volume of the sheet: V = L x W x H = (45.05 cm)(9.20 cm)(0.0014 cm) = 0.580244 cm3, before rounding. This time the thickness determines the number of significant figures in the result. That will be 2. So I round to 0.58 cm3.

Finally, I'll calculate the perimeter of the sheet: P = 2(L + W) = 2(45.05 cm + 9.30 cm) = 2(54.35 cm) = 108.70 cm, before rounding. Now I use the rule for addition and note that both measurements have 2 decimal digits. So I leave the final result with 2 decimal digits. No rounding is necessary in this case. In case you wonder why the factor of 2 had no effect, that's an exact number. It's not a measurement. So I ignored it when applying the significant figure rule.

Now, just to let you know of a situation that crops up frequently and is almost as frequently handled incorrectly...what if I have both addition and multiplication in the same problem? For example, in calculating experimental error. The formula is:

% Error = 100 ∙ |Measured Value - Accepted Value| ÷  (Accepted Value)

You have to apply the rules in the right order in order to get the right number of significant figures. It's like the order of operations in algebra. You carry out the operations in parentheses first. In the case of the above formula, you first apply the addition rule to the operation of subtraction inside the absolute value marks. Then you apply the multiplication rule to the division. Here's an example.

Suppose the volume of the sheet above is measured by a different method. The method is to submerge it in a container of water and see how much the liquid level rises. Suppose the level rises by 0.7 cm3. I'll calculate the experimental error using 0.58 cm3 as the accepted value:

% Error = 100 ∙ |0.7 cm3 - 0.58 cm3| ÷  (0.58 cm3)

First, I carry out the entire calculation in my calculator without any rounding. This gives 20.6896551%. OK, I went wild there with extra digits. I didn't really need to write all those out, but I got carried away. I'm going to have to round off. How do I decide how many digits to round off to? Well, by the addition rule, the result of the subtraction must be rounded to 1 decimal digit, because 0.7 has only 1 decimal digit. This will leave a 1 significant figure number. So applying the multiplication rule to the quotient of a 1 significant figure number with a 2 significant figure number results in a 1 significant figure number. Whew, I used the phrase significant figure number three times in that last sentence. Anyway, I have to round the experimental error to 20%. The 0 here is not significant. I could also write 2 x 101 %.

We're almost done. I just need to tell you a couple of things about rounding. Here they are:

Whenever possible, carry out multi-step calculations completely in your calculator, and round only the final result. The reason for this is that if you have several steps and you round every step of the way, each rounding can generate a little bit of error. This builds on itself, or propagates as we say, and can have the result of magnifying error. Round only once to minimize rounding error.

The other thing I need to tell you is how to round. It's easy. If the digit to the right of the one you're rounding is less than 5, just drop that digit and all others to the right of it. If the digit to right of the one you're rounding is 5 or greater, round up. So, for example, if you're rounding 3.752 to 2 significant figures, the result is 3.8. If you're rounding 3.746 to 2 significant figures, the result is 3.7. Note that the 6 has no role in deciding how to round. It would be incorrect to round the 4 up to a 5 and then round the 7 up to an 8. Such chain rounding, as I call it, is a no no.

Well, that's all I have for you. Think of ol' Sig whenever you're trying to figure out those significant figures.

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