Last time, we introduced what we mean by significant digits (or figures), and touched on why they are defined as they are. Here we will look at how significant digits and decimal places differ, and how they are affected by operations (primarily addition and multiplication). This is another aspect of why they are defined at all.

## Significant digits vs. decimal places

We’ll start again with a basic question, from 1999:

Rules for Significant Figures and Decimal Places I need to know what significant digits are, and what rules go with them.

The mention of rules implied that Ashley wanted to know not merely what they are (as we talked about last time), but how they are used in conjunction with calculations. So I started with an essential distinction:

When we work with numbers that come from the real world (such as measurements from a ruler), the numbers are not exact, but carry some amount of inaccuracy with them (because, for example, no ruler is absolutely, perfectly straight). There aretwo main ways we can describe the accuracy of a measurement: If it isaccurate to N *decimal places*, this means that there areN digits to the right of the decimal point that you can trust. For example, if I measure a length with a ruler marked off with millimeters, then the measurement will be accurate to the nearest millimeter. (If I write it in meters, to three decimal places: 0.001m.) If I claim to have measured it as 1.1293m, you know I was guessing about the 3 ten-thousandths, and you would round it off to the nearest thousandth: 1.129. If I say it was 1.100 m to three decimal places, you know that the two zeroes are not just guesses, but what I actually read from the ruler.The ruler will always produce the same number of decimal places, since there is a certain minimum size it can measure. If a number isaccurate to N *significant digits*(or figures), this means there areN meaningful digits that you can trust. For example, in my 1.129m, there are four digits I consider dependable, based on how I measured. If I had measured 0.024m with the same ruler, there would be only two significant digits. (The zeroes are there only to show the place value of the other digits, and are not 'significant'.) The ruler does not always produce the same number of significant digits, becauseit is better at measuring larger things. If I tried to measure something smaller than a millimeter, it would be useless. It would not give me any significant digits at all!

When we talk about decimal places, we are talking about **absolute precision**: a specific size below which we can’t be accurate (for example, 1 millimeter). When we use significant digits, we are talking about **relative precision**: how the possible inaccuracy compares to the actual value as a ratio (for example, 0.05% of the value).

As we discussed last time, the hard part is how to handle zeros:

Incidentally, be careful about zeroes in a number. If I told you a road was 12300m long, according to my car's odometer which shows tenths of a kilometer, you would know that only three digits are significant, because I read "12.3." The two zeroes, like the zeroes in 0.024, are there only to give the other digits the right meaning. But if I used a more accurate instrument, I might have read all five digits exactly. You do not know unless I tell you how I measured it or how many digits are significant.

This is the ambiguous case, where the way we write the number hides details about precision.

## Addition and multiplication

There are two basic operations you can do, and each has its own kind of precision:

Now, what happens to the accuracy of a number when I use it in a calculation? Or rather,how does the accuracy of the 'inputs' to a calculation affect the accuracy of the 'output'?When you areaddingnumbers, you want to look at the number ofdecimal places. For example, if I add 1.2 and 3.45, with different numbers of decimal places, I do not know what the hundredths place of 1.2 is, or what the thousandths place of 3.45 is. I can put an X for the unknown digits and see what happens: actual with X's 1.2 1.2XX + 3.45 + 3.45X ------ ------- 4.65 4.6XX You see, since I do not know all the hundredths I am adding, I really have no idea what the hundredths place of the result is (an 'unknown' plus 5 is still 'unknown'). So, to be honest, I have to drop the 5 and call the answer 4.6 (or else round it up to 4.7), showing that my answer is accurate to only one decimal place. (Even the tenths might be wrong because of a carry, but it would not be too far off.) So, when I add numbers,the result is only accurate to the smallest number of decimal placesI am adding. In this case, since 1.2 has only one decimal place, that is all I can keep in my sum.

Here, it is the **absolute precision** that matters: If I know the hundredths place in one number, but not the other, then I can’t be sure what goes there in the sum.

The best practice is to carry out the entire addition as if we knew the unknown places were zero, but then at the end we round the result to ignore the inaccurate digits. In my example, we would write the answer as 4.7 (rounded up from 4.65).

On the other hand, if Imultiply numbers, what counts is the number ofsignificant digits. Suppose I run for 1.45 hours at 6.1 miles per hour. Then I have gone 1.45 x 6.1 miles. How accurate is that? Again, I will put an X for the unknown places and see what happens: actual with X's 1.4 5 1.4 5 X x 6.1 x 6.1 X --------- ----------- 1 4 5 X X X X 8 7 0 1 4 5 X --------- 8 7 0 X 8.8 4 5 ----------- 8.8 X X X X You can see that the number of significant digits in the result (two) isthe smaller of the significant digits for the two multiplicands(three and two respectively), so I have to write my product as 8.8, rounding it to two significant digits and dropping two digits that I worked hard for and would otherwise have thought were good. Since 6.1 has only two significant digits, I cannot keep more than that in my product.

We’ll be seeing in a later post how to more formally prove the claims here (to the extent that they can be proved); but you can see in the multiplication why only the leftmost two digits are trustworthy. Again, we typically carry out the full multiplication and then round, rather than trying to do only the part of the work that we’ll keep. Back when we had to do the work by hand, we would suggest rounding the more precise numbers to one more significant digit than the least (as I did in my example) to avoid wasting too much work; but with calculators it is no trouble to do the whole calculation.

So those are the rules: 1. When youadd (or subtract), you keep as many *decimal places* as there are in the least accurate number. 2. When youmultiply (or divide), you keep as many *significant digits* as there are in the least accurate number. I should mention that this is only a 'rule of thumb', and itsometimes underestimates the precisionof an answer. If I had demonstrated multiplication with a larger factor in place of the 1.45, you would have seen an extra significant digit because of a carry. There are more careful rules for measuring the accuracy of a result, when you really need to know just how accurate a number is, but significant digits work well as a general rule. In this age of calculators, when you can get many digits in any calculation with no trouble, it is important not to keep all those digits and thus get afalse sense of the precision of your results. We do not want to pretend we know seven digits when we really only know two or three.

We have often seen students believe the calculator, and write out every digit it shows, as if it matters. This is one reason significant digits are especially important today.

I hope this does not overwhelm you! I wanted to give you not just the basic definition, but some background so you could see why it makes sense.

## Exact numbers

A couple years later, students in a math club wrote to ask about a special case; that question was tacked on to the same page:

We reviewed this answer and we still have trouble understanding the rule for dividing. For example, the exact answer for 366/2 = 183. Since 2 has only one significant digit, does that mean that the most accurate answer is 200?

The rule I stated applies equally to multiplication and to division (though I didn’t show why); if it applies to their example, then division doesn’t seem very useful, as we often divide by two (in averaging two numbers, for example, or in finding the area of a triangle). I replied:

It depends on whether 2 is ameasurement, or aknown exact value, as, for instance, if you are calculating a radius from a diameter, so you know you have to divide by exactly 2. In the latter case, you can think of the 2 as havinginfinitely many significant digitswhen you apply the rule, since EVERY possible digit is known exactly. You have three significant digits in 366, so you can keep three in the answer. But if 2 represents, say, the number of hours it took to go 366 miles, then the answer should be 200 mph, since you have only one digit of precision in the time measurement. We usually talk of significant digits only when there are decimal points present to show the assumed precision of the numbers; in fact, we should only do this if weknowsomething about the actual measurements, rather than justassumingthe significant digits from the way a number was written. Also, it's best to write numbers in scientific notation if we want to be completely clear about accuracy. (Does 23000 have 2 or 5 significant digits? If we write it as 2.30 *10^4, we can see that it has 3.) In such a context, writing "2" with no decimal place would make it clear that it is an exact number.

Numbers without a decimal point keep coming up, don’t they? It all comes down to provenance: You have to be told separately how precise it is meant to be. In this case, some formulas imply that a number used in the formula is exact — and they do so by not using a decimal point.

## Only a rule of thumb

I want to include another answer to a very similar question here, because it shows something I mentioned in passing there, namely that my example made things look a little more clear-cut than they really are.

Significant Figures I am having trouble with significant figures. I do not understand why: a) 62.3 multiplied by 5.7 = 360, but 62.30 multiplied by 5.70 = 355. The question says to express your answer with an appropriate number of significant figures. Please help me, as I do not understand why we get 360 and 355.

I gave an answer much like what I said above for multiplication, first using slightly changed numbers:

The idea here is that if one of the numbers you are multiplying is only accurate to two significant digits,you can only trust two significant digits of the result, so you round to that accuracy. When the numbers being multiplied are given as 62.30 and 5.70, there are 4 and 3 significant digits respectively, so you can keep 3 digits in your answer, 355. But when you are only given 62.3 and 5.7, you should only keep 2 significant digits, so you round it up to 360. Here's one way to see why this is.(I'll use a different example, and explain why below.)The multiplication of 12.30 by 5.70 looks like this: 1 2.3 0 * 5.7 0 --------- 0 0 0 0 8 6 1 0 6 1 5 0 ----------- 7 0.1 1 0 0 If we don't know the last digit of each number, but represent each unknown digit by X, your multiplication looks like this: 1 2.3 X * 5.7 X --------- X X X X 8 6 1 X 6 1 5 X ------------- 7 0.X X X X I've written X wherever I don't know what a digit is, because I'm multiplying or adding an unknown digit. The X's show that I can't trust the last digits, and should round off to 70. If you look closely, you'll see that the significant digits in the answer come from the significant digits of the 5.7, the number with the fewest significant digits. So that's the rule: you keep as many significant digits in the product as there are in the factor with the fewest significant digits.

In that example it is clear that two digits can be trusted. But it isn’t always so simple:

Now here's youroriginal problem: 6 2.3 X * 5.7 X --------- X X X X 4 3 6 1 X 3 1 1 5 X ------------- 3 5 5.X X X X You'll notice that it looks as if we have more valid digits than the rule says! That's becausethe first digits of both numbers are relatively large, so that you get an extra digit. The rule is an approximation, and is a littleon the conservative side, assuming that it's better to keep too few digits than to trust too many in some cases. We could probably modify the rule slightly to take the extra digit into account, but the simple rule has been found to be good enough.

The rules ignore the values of digits; reality is more fuzzy than rules are. This is one reason I don’t feel bad writing an extra digit when it feels appropriate; these rules are not unbreakable, they are only conventions. This is what we mean by “rule of thumb”.

## More than one operation

In real scientific or engineering problems, you do more than one operation. Two issues come up: First, when you have both addition and multiplication, do you use significant digits or decimal places; second, when do you round?

Here’s a good question about that, from 2002:

Calculations Involving Significant Figures My physics class is having a fair amount of trouble with significant digits. We understand what significant digits are and their purpose. We have laboured over the rules for rounding and for multiplication and addition, and so on. However, our BIG problem lies withlengthy calculations that require several intermediate stepsbefore the final result is obtained. In other words, we don't know (even our teacher!)what to do when we have to add and multiply in the same problem. For example: 3.95 x 1.15 + 2.7503 / 8.49. Do you multiply the first two numbers (3.95 x 1.15) and determine the correct number of significant figures (4.59), which you then add to (0.324)? Or do we not round or use significant figures (is there a difference between these two terms?)until the very end?That is, add the raw data (4.5425 and 0.3239458...) and then use the addition significant figure rules. Likewise, when determining slope m = (11.2cm - 10.2cm) / (200g - 0g) do you say that 1.0 / 200 = .0050 cm/g, or do you look at your initial numbers (least amount of significant figures would be 3 becausemasses are constants and contain an infinite number of significant figures) and say that the answer is 0.00500 cm/g? I hope that you can help me (in fact our whole class). Thanks in advance!

There are some good questions in here! I started with some general principles, in addition to the ideas we’ve already discussed:

It sounds as if you are forgetting the fact that significant figures are only appropriate in multiplication, not in addition. This page discusses that: Decimal Places and Significant Figures http://mathforum.org/library/drmath/view/59014.html So you really have tothink about the precision at each step, since each step has a different effect, some via significant figures, others via decimal places. Butyou don't want to round to the appropriate precision at each step, because then you would be introducing error. Significant figures are only a rule of thumb, and rounding too early can allow errors in the dropped digits to affect the result. In these days of calculators you don't have to round anything until the end; in the old days you would have been advised to keep at least a digit or two extra until the end. (Calculators do that too - they have more precision internally than they show, for exactly this reason.)

I used the word think deliberately: you need to **think** about precision at every operation, but not to **do** anything about it until the end. In practice, you can do all the calculations first, but still take precision into account as you read back through it.

Then I got specific:

So let's take your examples: 3.95 * 1.15 + 2.7503 / 8.49 You willdo all the arithmeticwith as much precision as your calculator allows, giving 4.5425 + 0.3239..., and then add them to get 4.8664... . Now youlook at the precision. The product has three significant figures, giving you hundredths, and the quotient has three as well, this time giving thousandths; the sum is accurate only to the hundredths, so you use 4.87. This happens to have three significant figures as well, but that is not necessarily going to happen. (11.2 cm - 10.2 cm) / (200 g - 0 g) Here you first subtract, giving tenths in the dividend and units in the divisor, so you have 1.0 / 200. Since the dividend has only two significant figures (and I'm going to assume the divisor has three, though it could be taken as one), your answer should have two: 0.0050 cm/g.

The masses raise two issues: the ambiguity of writing 200 with no decimal point, and Cassandra’s comment about constants, which had to be answered:

I'm not sure why you said that masses are constants and have infinitely many significant figures; if they are measured, they have finite precision.Only _defined_ constants(like 2 when you are doubling something) are treated asexact. But the 0 probably is exact. I can't tell without knowing the source of these numbers.

I never found out how much precision the 200 really had.

There is more to be said about issues like exact numbers; next time, we’ll dig deeper.