Multiplying and Dividing

Reference > Science > Significant Figures

In the school where I teach, we have a set of exterior stairs on the second floor that my Physics students drop objects from, in order to do calculations on acceleration and velocity. In order to do their calculations, they need to determine how far it is from the dropping point to the ground. Because of the nature of the stairs, they have to use some unusual measuring techniques, and add measurements together to get a total height. I never give them advice; I merely give them some measuring tools and let them figure it out for themselves. One technique they use is to measure the height of a brick*, multiply by the number of bricks, and then add the height of the exposed foundation wall. This process involves not just adding values, but also multiplying, and how we handle multiplying with significant figures is different than just adding and subtracting.

* To be clear, the students (if they're thinking this through properly) won't actually measure the height of a brick; they'll measure the distance from the bottom of one brick to the bottom of the next. This way they include the layer of mortar between the bricks.

Let's focus, to begin with, on the first part of the process: measuring a brick's height, and multiplying by the number of bricks. We'll say that the students measured the height as 0.0582. This means that their measuring tool had millimeter markings, and they estimated one decimal beyond that.

Let's suppose the students counted that the height of the wall was 35 bricks. Then they will calculate the height of the wall as follows: 0.0582 x 31 = 1.8042 meters.

However, we need to figure out how many of those digits are significant. We use the following rule:

When multiplying or dividing values, determine which value has the least number of significant figures. That is how many significant figures the result must have.

The significant digits in this number are 5, 8, and 2. The number has three significant digits.

35 is a counted, rather than measured, quantity, so we can think of it as 35.000.... It has infinitely many significant digits.

Clearly the measured value has fewer significant digits. Therefore our answer must have three significant digits. This means we must round our answer from 1.8042 to 1.80 meters.

Notice that we're no longer claiming that we know the digit in the thousandths place or the ten-thousandths place, even though we had those decimal places in the brick measurement! Why is this? Remember that the last decimal place was estimated, so we could have been off by 0.0001 in either direction. By multiplying by 31, we've also multiplied the potential error by 31, meaning that we could potentially be off by as much as 0.0031, so it really doesn't make sense to include those last two digits.

Let's try another example.

A farmer has a rectangular field which is very long and very narrow. He knows that the length of the field is 520 meters. He measures the width to be 20.5 meters. What is the field's area?

Since the area of a rectangle is its length times its width, the area of the field is 10660 square meters.

However, we must apply our multiplication rule to this result. We look at our two values: 520 and 20.5 and ask, which one has the fewest significant digits. The answer is: 520 has two significant figures. Therefore, we must limit our result to two significant digits. So we round our answer to the ten-thousands place: 11000 square meters.

Here's another example. A cone's radius is measured as 1.25 meters, and its height is 0.1 meter. What is the cone's volume?

A cone's volume is given as V =

πr²h. So we plug in our values and get the following: 0.1636246173697. Obviously, that's way too many digits! We do our multiplication rule again. The numbers being multiplied are

, π, 1.25, and 0.1. The first two quantities are non-measured, so we don't need to worry about their significant digits (as long as we didn't round them!). So we need to ask which of our two measured quantities has the least number of significant digits. The height has only one significant digit, which means our answer can only have one significant digit! That means we need to round our answer after just one decimal place. The volume is 0.2 cubic meters.

Incidentally, this problem is an excellent example of why it's so important to keep track of your significant figures; it's likely that the person who measured the height and radius probably used the same measuring device for both dimensions, yet one of them has fewer significant figures. If this was the case, the height measurement should have been written as 0.10 meter, to indicate that it has the same precision as the radius measurement. With that extra digit of precision on the height, we would conclude that the volume is 0.16 cubic meters, with an extra digit of precision.